CHOICE AND CHANCE.
BY THE SAME AUTHOR.
Lately published, Svo., 542 pages, price 16s.,
TKILINEAR COOKDINATES,
AND OTHER METHODS OF MODEEN ANALYTICAL GEOMETRY
OF TWO DIMENSIONS.
" A sound work on a high form of algebraic geometry Trilinear
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CAMBRIDGE: DEIGHTON, BELL AND CO.
LONDON : BELL AND DALDY.
CHOICE AND CHANCE
BY THE REV.
WILLIAM ALLEN WHITWORTH, M. A.
I 1
FELLOW OF ST. JOHN'S COLLEGE, CAMBRIDGE.
iiicm, (Snlargeb.
CAMBBIDGE :
DEIGHTON, BELL AND CO
LONDON : BELL AND DALDY.
1870.
[All rights reserved.]
•\
. >*
rficJZ
LIVERPOOL :
PRINTED EY D. MARPLES,
LORD STREET.
PREFACE.
IN this second edition I have enlarged the appendices
so as to meet the wants of advanced students. I have
also added a collection of upwards of one hundred
miscellaneous examples, which I think will add very
much to the utility of the book.
It should be observed that the two chapters headed
respectively Choice and Chance are simply arithmetic,
and ought not to be beyond the comprehension of the
ordinary reader who has never seen an algebraical sym
bol. But while expressly written for unscientific readers,
they have been found very helpful to the young mathe
matician, when he was about to read in his algebra the
hitherto difficult and embarrassing chapters on permu
tations and combinations, or on probability.
The appendices are addressed entirely to algebraical
students. In the first appendix the usual theorems
respecting permutations and combinations are esta
blished by new proofs, the same reasoning which
was pursued with as little technicality as possible
in the body of the work, being here expressed in
algebraical language.
In the second and third appendices, which are newly
added in this edition, a series of propositions are given
which are not usually found in text books of algebra.
VI PREFACE.
But I can see no reason why examples of such
simple propositions as the xiiith and xxvth should
be excluded from elementary treatises in which more
complex but essentially less important theorems
generally find place.
The classification of a variety of propositions under
the titles of Distribution and Derangement will con
tribute (it is hoped) to disentangle the confusion in
which all questions involving selection or arrange
ment are commonly massed together, and will facilitate
in some degree that precision of language and clear
ness of expression which ought always to be aimed at
in mathematics.
In the fourth appendix I have exhibited the seeming
paradox that a wager which is mathematically fair is
mathematically disadvantageous to both contracting
parties. And I have endeavoured to cast into a simple
and intelligible form the principles upon which the
difficulties of the celebrated Petersburg problem are
explained.
W. ALLEN WHITWOETH.
ST. JOHN'S COLLEGE,
1st January, 1870.
PREFACE TO THE FIRST EDITION.
THE following pages are a reproduction of lectures on
Arithmetic, given in Queen's College, Liverpool, in the
Michaelmas Term, 1866. Many of the students to
whom the lectures were addressed were just entering
upon the study of algebra, and it seemed well, while
the greater part of their time was devoted to the some
what mechanical solution of examples necessary to
give them a practical facility in algebraical work, that
their logical faculties should be meanwhile exercised in
the thoughtful applications of the arithmetical art with
which they were already familiar.
I had already discovered, that the usual method of
treating questions of selection and arrangement was
capable of modification and so great simplification, that
the subject might be placed on a purely arithmetical
basis ; and I deemed that nothing would better serve
to furnish the exercise which I desired for my classes,
and to elicit and encourage a habit of exact reasoning,
than to set before them, and establish as an application
of arithmetic, the principles upon which such questions
of "choice and chance" might be solved.
The success of my experiment has induced me to
publish the present work, in the hope that the expo
sitions already accepted by a limited audience may
Vlll PREFACE TO THE FIRST EDITION.
prove of service in a wider sphere, in conducing to a
more thoughtful study of arithmetic than is common
at present ; extending the perception and recognition
of the important truth, that arithmetic, or the art of
counting, demands no more science than good and
exact common sense.
In the first chapter I have set down and established
as arithmetical rules all the principles usually required
in estimating the choice which is open to us in making
a selection or arrangement out of a number of given
articles under given conditions. In the second chapter
I have explained how different degress of probability
are expressed arithmetically, and how the principles of
the preceding chapter are applied to the calculation of
chances. These two chapters will prove intelligible to
any one who understands the first principles of arith
metic, provided he will consider each step as he goes
on ; not content with the mere statement of any rule,
but careful to follow the explanations given and to
recognise the reason of each successive principle.
For the sake of mathematical students I have added,
as an appendix, a new treatment of permutations and
combinations with algebraical symbols. In my expe
rience as a teacher I have found the proofs here set
forth more intelligible to younger students than those
given in the text books in common use.
LIVERPOOL, 1st February, 1867.
CONTENTS.
PAGE
Explanations xi.
CHAPTER I. CHOICE 1
Examples on Choice 02
CHAPTER II. CHANCE 00
Examples on Chance ... 138
APPENDIX I. Permutations and Combinations treated
Algebraically 142
APPENDIX II. Distributions 163
APPENDIX III. Derangements 185
APPENDIX IV. On the Disadvantage of Gambling ... 199
Miscellaneous Examples 231
Answers to the Examples -251
ERRATA.
Page 22, line 6. For [24, read 24.
Page 89, the 8th and following lines. Head "there are ten which give
a result greater than 8. Hence the required chance is ^ or ~- •
36 18
Page 103, line 8. For X read -+- .
At the head of pages 114, 118, 126. For CHOICE, read CHANCK.
EXPLANATIONS.
The sign = is used to signify that the two expressions
between which it is placed are equal to one another, or represent
the same arithmetical value.
The sum of any given numbers is that which is obtained by
adding the given numbers together. The sign + (plus) is used
to denote that the number which follows it is added to that
which precedes. Thus —
5 + 3 = 8,
or the sum of five and three is eight.
The difference of two given numbers is that which is obtained
by subtracting the smaller of them from the greater. The sign
— (minus) is used to denote that the number which follows it
is subtracted from that which precedes. Thus —
5-3=2,
or the difference of five and three is two.
The product of two given numbers is that which is obtained
by multiplying the given numbers together. The continued
product of three or more numbers is obtained by multiplying
any two of them together, and then multiplying the result by a
third , and so on until all the given numbers have been used
Xll EXPLANATIONS.
The sign x is used to denote that the number which precedes
it is multiplied by that which follows. Thus —
5x3 = 15; 5x3x2 = 30;
or the product of five and three is fifteen: the continued product
of five, three, and two, is thirty.
A full point ( . ) is often used instead of the sign x of multi
plication. Thus —
5.4.3.2.1 = 120.
The quotient of two given numbers is that which is obtained
by dividing the former of them by the latter. The sign --• is
used to denote that the number which precedes it is divided by
that which follows. Thus —
15 -f- 5 = 3,
or the quotient of fifteen and five is three.
Instead of using the sign -f- , the quotient of the two numbers'
is often expressed by writing the former above the latter in
the form of a fraction. Thus —
15 _ IK . K — Q
= 1O -7- O — O .
5
The few other signs which are used in the body of the work
are explained wherever they are first introduced. But the
appendices, being addressed to mathematical readers, involve a
more technical notation, for the explanation of which the student
must be referred to treatises on algebra.
CHOICE AND CHANCE.
CHAPTEE I.
CHOICE.
WE have continually to make our choice among
different courses of action open to us, and upon the
discretion with which we make it, in little matters
and in great, depends our prosperity and our happiness.
Of this discretion a higher philosophy treats, and it is
not to be supposed that Arithmetic has anything to do
with it; but it is the province of Arithmetic, under
given circumstances, to measure the choice which we
have to exercise, or to determine precisely the number
of courses open to us.
Suppose, for instance, that a member is to be
returned to parliament for a certain borough, and
that four candidates present themselves. Arithmetic
has nothing to do with the manner in which we shall
exercise our privilege as a voter, which depends on our
B
SIS CHOTC3.
I ',. - ^
discretion in judging the qualifications of the different
candidates; but it belongs- to Arithmetic, as the science
of counting and calculation, to tell us that the number
of ways in which (if we vote at all) we can exercise our
choice, is four.
The operation is, indeed, in this case so simple
that we scarcely recognise its arithmetical character at
all; but if we pass on to a more complicated case,
we shall observe that some thought or calculation is
required to determine the number of courses open to
us : and thought about numbers is Arithmetic.
Suppose, then, that the borough has to return two
members instead of one. And still suppose that we
have the same four candidates, whom we will distin
guish by names, as A, B, C, D. If we try to note
down all the ways in which it is possible for us to
vote, we shall find them to be six in number 5 thus we
may vote for any of the following : —
A and B, A and C, A and D,
C and D, B and D, B and C.
But we can hardly make this experiment without
perceiving that the resulting number, six, must in
some way depend arithmetically upon the number of
candidates and the number of members to be returned,
or without suspecting that on some of the principles
of arithmetic we ought to be able to arrive at that
result without the labour of noting all the possible
courses open to us, and then counting them up ; a
DIFFERENT SELECTIONS. 3
labour which we may observe would be very great
if eight or ten candidates offered themselves, instead
of four.
In the present chapter we shall establish and explain
the principles upon which such calculations are made
arithmetically. It will be found that they are very
simple in nature as well as few in number. In the
next chapter we shall apply the same principles to the
solution of problems in Probability, a subject of very
great interest, and some practical importance.
We found, by experiment or trial, that there were six
ways of voting for two out of four candidates. So we
may say that, out of any four given articles, six selec
tions of two articles may be made. But we call special
attention to the sense in which we use the words "six
selections." We do not mean that a man can select
two articles, and having taken them can select two
more, and then two more, and so on till he has
made six selections altogether; for it is obvious that
the four articles would be exhausted by the second
selection, but when we speak of six selections being
possible, we mean that there are six different ways
of making one selection, just as among four candidates
there are six ways of selecting two to vote for.
This language may appear at first to be arbitrary and
unnecessary, but as we proceed with the subject we
shall find that it simplifies the expression of many
of our results.
In making the selection of two candidates out of
4 CHOICE.
four, in the case just considered, it was immaterial
which of the two selected ones we took first; the
selection of A first, and then B, was to every intent
and purpose the same thing as the selection of B first,
and then A.
But if we alter the question a little, and ask in how
many ways a society can select a president and vice-
president out of four candidates for office, the order
of selection "becomes of importance. To elect A and
B as president and vice-president respectively, is not
the same thing as to elect B and A for those two
offices respectively. Hence there are twice as many
ways as before of making the election, viz, —
A and B, A and C, A andZ>,
C and D, B and D, B and C,
B and A, C and A, D and A,
D and C, D and B, C and B.
So if four articles of any kind are given us, there will
be twelve ways of choosing two of them in a particular
order; or, as we may more briefly express it, out of
four given articles, twelve arrangements of two articles
can be made. But it must be observed that the
same remarks apply here, which we made on the use of
the phrase " six selections" on page 3. * We do not
mean that twelve arrangements or six selections can
be successively made ; but that if one arrangement
or one selection of two articles have to be made out
of the four given articles, we have the choice of twelve
SELECTIONS AND ARRANGEMENTS. 5
ways of making the arrangement, and of six ways of
making the selection.
We may give the following formal definitions of the
words selection and arrangement, in the sense in which
we have used them : —
DEF. I. — A selection (or combination) of any number
of articles, means a group of that number of articles
classed together, but not regarded as having any
particular order among themselves.
DEF. II. — An arrangement (or permutation) of any
number of articles, means a group of that number
of articles, not only classed together, but regarded as
having a particular order among themselves.
Thus the six groups, —
A B C, B C A, CAB,
AC B, BAG, C B A,
are all the same selection (or combination) of three
letters, but they are all different arrangements (or
permutations) of three letters.
So, out of the four letters A, B, C, D, we can make
four selections of three letters, viz. —
BCD,
CD A,
D A B,
A B C;
but out of the same four letters we can make twenty-
four arrangements of three letters, viz.—
6
CHOICE.
BCD, BDC, CDB,
CD A, CAD, DAC,
DAB, DBA, ABD,
ABC, ACB, BCA,
CBD, DBG, DCB,
DC A, A CD, ADC,
ADB, BDA, BAD,
BAG, CAB, CBA.
Having thus explained the language we shall have
to employ, we may now proceed to establish the
principles on which all calculations of choice must he
founded.
The great principle upon which we shall hase all
our reasoning throughout our work, may he stated as
follows : —
If one thing can be clone in a given number of
different ways, and then another thing in another
given number of different ways, the number of different
ways in which both things can be done is obtained bi/
multiplying together the two given numbers.
We shall first illustrate this principle, and then
proceed to prove it.
Suppose we have a box containing five capital letters,
A, B, C, D, E, and three small letters, x, y, z.
ABODE
x y z
The number of ways in which we can select a capital
letter out of the box is five; the number of ways in
CHOICE IN SUCCESSIVE OPEEATIONS. 7
which we can select a small letter is three ; therefore,
by the principle we have just stated, the numher of
different ways in which we can select a capital letter
and a small one is fifteen, which we find on trial
to be correct, all the possible selections being as
follows : —
Ax, Bx, Cx, Dx, Ex,
Ay, By, Cy, Dij, Ey,
Az, Bz, Cz, Dz, Ez.
Again, suppose there are four paths to the top of
a mountain, the principle asserts that we have the
choice of sixteen ways of ascending and descending.
For there are
4 ways up,
4 ways down,
and 4 x 4 = 16.
We can verify this : for if P, Q, E, 8 be the names
of the four paths, we can make our choice among the
following sixteen plans, the first-mentioned path being
the way up, and the second the way down : —
PandP, PandQ, Pand#, Pand£,
QandP, <2and#, QandjR, Q and S,
EandP, EandQ, jRandE, #and£,
8 and P, 8 and Q, 8 and E, 8 and S.
Or, if we had desired to ascertain what choice we
had of going up and down by different paths, we might
still have applied the principle, reasoning thus :
8 CHOICE.
There are four ways of going up, and when we are
at the top we have the choice of three ways of descend
ing (since we are not to come down by the same path
that brought us up). Hence the number of ways of
ascending and descending is 4 x 3, or twelve.
These twelve ways will be obtained from the sixteen
described in the former case, by omitting the four
ineligible ways,
P and P, Q and Q, E and R, S and 8.
The foregoing examples will suffice to illustrate the
meaning and application of our fundamental propo
sition. We will now give a formal proof of it. We
shall henceforth refer to it as Kule I.
RULE I.
If one thing can be done in a given number of
different ways, and ivhen it is done in any way another
thing can be done in another given number of different
ivays, then the number of different ways in which the
two things can be done is the product of the two given
numl
For let A, B, C, D, E, &c. represent the different
ways in which the first thing can be done (taking as
many letters as may be necessary to represent all the
different ways), and similarly let a, fc, c, d, &c. repre
sent the different ways of doing the second thing.
Then, if we form a table as below, having the letters
PRINCIPLE OF MULTIPLICATION OF CHOICE. 9
A, B, C, D, E &c. at the head of the several columns,
and the letters a, b, c, d, &c. at the end of the several
horizontal rows, we may regard each square in the
table as representing the case in which the first thing
is done, in the way marked at the head of the column
in which the square is taken ; and the second thing in
the way marked at the end of the row.
WAYS OF DOING THE FIEST THING.
A
B
C
D
E
F
a
dsc.
WAYS OF DOING THE SECOND THING.
a
b
*
c
t
(
d
?
dc.
Thus the square marked with the asterisk (*) will
denote the case in which the first thing is done in the
way which we called 0, and the second thing in the
way which we called b ; and the square marked with
the dagger (t) will denote the case in which the first
10 • CHOICE.
thing is done in the way E, and the second in the way
c, and so on.
Now it will be readily seen that all the squares repre
sent different cases, and that every case is represented
by some square or other. Hence the number of possi
ble cases is the same as the number of squares. But
there are as many columns as there are ways of doing
the first thing, and each column contains as many
squares as there are ways of doing the second thing.
Therefore the number of squares is the product of the
number of ways of doing the two several things, and
therefore, this product expresses also the whole number
of possible cases, or the whole number of ways in which
the two events can be done.
This proves the rule.
Question. — If a halfpenny and a penny be tossed,
in how many ways can they fall ?
Answer. — The halfpenny can fall in two ways, and
the penny in two ways, and 2x2=4, therefore they can
fall in four ways.
The four ways, of course, are as follows : —
(1) both heads.
(2) both tails.
(3) halfpenny head and penny tail.
(4) halfpenny tail and penny head.
Question. — If two dice be thrown together, in how
many ways can they fall ?
Answer. — The first can fall in six ways, and the
EXAMPLES OF MULTIPLICATION OF CHOICE. 11
second in six ways, and 6 x 6 = 36; therefore there are
thirty-six ways in which the two dice can fall.
The thirty- six ways may be represented as follows : —
1 and 1,
1 and 2,
1 and 3,
1 and 4,
1 and 5,
1 and 6,
2 and 1,
2 and 2,
2 and 3,
2 and 4,
2 and 5,
2 and 6,
3 and 1,
3 and 2,
3 and 3,
3 and 4,
3 and 5,
3 and 6,
4 and 1,
4 and 2,
4 and 3,
4 and 4,
4 and 5,
4 and 6,
5 and 1,
5 and 2,
5 and 3,
5 and 4,
5 and 5,
5 and 6,
6 and 1,
6 and 2,
6 and 3,
6 and 4,
6 and 5,
6 and 6.
i. — In how many ways can two prizes he
given to a class of ten hoys, without giving both to the
same boy?
Answer. — The first prize can be given in ten ways,
and when it is given the second can be given in nine
ways, and 10 x 9 = 90 ; therefore we have the choice of
ninety ways of giving the two prizes.
Question. — In how many ways can two prizes be
given to a class of ten boys, it being permitted to give
both to the same boy ?
Answer. — The first prize can be given in ten ways,
and when it is given the second can be given in ten
ways ; therefore both can be given in 10 x 10, or 100
ways.
Question. - — Two persons get into a railway carriage
where there are six vacant seats. In how many differ
ent ways can they seat themselves.
Answer. — The first person can take any of the vacant
seats; therefore he can seat himself in six different
12 CHOICE.
ways. Then there are five seats left, and therefore the
other person has the choice of five different ways of
seating himself. Hence there are 6 x 5, or 30 different
ways in which they can take their seats.
Question. — In how many ways can we make a two-
lettered word out of an alphabet of twenty-six letters,
the two letters in the word being different ?
Answer. — We can choose our first letter in twenty-
six ways, and when it is chosen we can choose the
second in twenty-five ways. Therefore we have the
choice of 26 x 25, or 650 ways.
Question. — In how many ways can we select a
consonant and a vowel out of an alphabet of twenty
consonants and six vowels?
Answer. — We can choose the consonant in twenty
ways, the vowel in six ways, both in one hundred and
twenty ways.
Question. — In how many ways can we make a two-
lettered word, consisting of one consonant and one
vowel ? '
Answer. — By the last answer, we can choose our
two letters in one hundred and twenty ways, and
when we have chosen them we can arrange them in
two ways. Hence we can make the word in 120 x 2,
or 240 different ways\
Question.— There are twelve ladies and ten gentle-
CONTINUED MULTIPLICATION. 13
men, of whom three ladies and two gentlemen are
sisters and brothers, the rest being unrelated : in how
many ways might a marriage be effected ?
Answer. — If all were unrelated we might make the
match in 12 x 10, or 120 ways ; but this will include
the 3x2, or 6 ways in which the selected lady and
gentlemen are sister and brother. Therefore the
number of eligible ways is 120 — 6, or 114.
EULE II.
If a series of things can be done successively in
given numbers of ways, the number of ways in which
all the things can be done is the continued product of
all the given numbers.
This rule is only an extension of the former one, and
needs not a separate proof. Its correctness will be
sufficiently evident from considering an example.
a "*
Suppose the first thing can be done in four ways, and
the second in three, then the first and second together
form an event or operation, which can happen (by Kule
I.) in 4 x 3, or 12 ways. Now suppose the third thing
can be done in five ways. Then, since the first and
second together can happen in twelve ways, and the
third in five ways, it follows from Rule I. that the first
and second and the third, can be done in 12 x 5, or 60
ways ; that is, all three can be done in 4 x 3 x 5 ways.
So if a fourth thing can be done in seven ways,
14 CHOICE.
then, since the first three can be done in sixty ways,
and the fourth in seven ways, the first three and the
fourth can be done (by Kule I.) in 60 x 7, or 420 ways;
that is, all four can be done in 4x3x5x7 ways.
And so on, however many things there may be.
The applications of this proposition are very numer
ous and very important ; the solution of almost every
question concerning permutations or combinations
depending upon it, as will presently be seen.
As an example, suppose I have six letters to be
delivered in different parts of the town, and two boys
offer their services to deliver them. To determine in
how many different ways I have the choice of sending
the letters we may reason as follows. The first letter
may be sent in either of two ways ; so may the second ;
so may the third, and so on. Hence the whole number
of ways is, by the rule, 2 x 2 x 2 x 2 x 2 x 2 or 64.
So, if there were three boys, the choice would lie
among 3x3x3x3x3x3 or 729 ways.
The question, in how many ways can six things be
divided between two boys, will be seen to be almost
identical with the question of the six notes sent by the
two boys. The only difference is, that among the 64
ways of sending the notes were included the two ways
in which either boy carried them all. Now six things
cannot be said to be divided among two boys if they
all are given to one. Hence these two ways must be
METHODS OF DIVISION INTO TWO PARCELS. 15
rejected, and there will only be 62 ways of dividing six
things between two boys.
But, again, suppose we are asked in how many ways
can six things be divided into two parcels, the question
seems at first to be identical with the last. But, on
consideration, we observe that if a, b, c, d, e, f repre
sent the six things, one of the ways of dividing them
' between the two boys would be to give
a, b, to the first boy,
c, d, e, /, to the second ;
and another different way would be to give
a, b, to the second boy,
c, d, e, f, to the first ;
but if the question be merely of dividing the six things
into two parcels, with no distinction between them,
corresponding to the two ways noted for the previous
question, we have now only the one way, viz., to put
a, b, into one parcel,
c, d, e, /, into the other.
Hence for every two ways of dividing the things between
two different boys, there is only one way of dividing
them into two indifferent parcels ; and, therefore, we
have the choice in this last case of only thirty- one ways.
The correctness of this result may be more clearly
understood by the following consideration. Suppose we
have six articles to divide between two boys. We may
resolve the operation into the two operations of (1)
dividing the articles into two parcels, and (2) when
16 ' CHOICE.
these parcels are made, giving them to the two boys.
Now we can form our two parcels in thirty-one ways,
and when the two parcels are made, we can give them,
one to each boy, in two ways; hence by Rule L, we
can make the parcels and dispense them in 31 X 2 or
62 ways. ,
Question. — Twenty competitors run a race for three
prizes, in how many different ways is it possible that
the prizes may be given ?
Answer. — The first prize can be given in twenty
ways ; when it is given, the second may be given in
nineteen ; then the third can be given in eighteen
ways. Hence the whole number of ways of giving the
three prizes is 20 x 19 x 18, or 6840.
Question. — In how many ways can four letters be
put into four envelopes, one into each ?
Answer. — For the first envelope we have the choice
of all the letters, or there are four ways of filling the
first envelope ; then there are three letters left, and
therefore three ways of filling the second envelope;
then there are two letters left, or two ways of filling
the third envelope ; so there is only one way of filling
the^last. Hence there are 4 x 3 x 2 x 1, or 24 ways
of doing the whole.
Question. — How many different sums may be formed
with a sovereign, a half-sovereign, a crown, a half-
EXAMPLES. 17
crown, a shilling, a sixpence, a penny, and a half
penny ?
Answer. — Each coin may be either taken or left,
that is, it may be disposed of in two ways, and there
are eight coins. Hence (by Rule II.) all may be
disposed of in
2x2x2x2x2x2x2x2, or 256
ways. One of these ways would, however, consist in the
rejection of all the coins, which would not be a way of
taking any sum. Therefore the number of different
sums that can be made is 255.
Question. — There are twenty candidates for an office,
and seven electors. In how many ways can the votes
be given ?
Answer. — Each man can vote in twenty ways, and
there are seven men to vote. Therefore all the votes
can be given (by Rule II.) in
20 x 20 x 20 x 20 x 20 x 20 x 20 or 1280000000
different ways.
Question. — In how many ways can the following
letters be divided between two persons : —
a, a, a, a, b, b, b, c, c, d ?
Answer. — Of the a, a, a, a, the first person can take
either none, or one, or two, or three, or four. That is,
the a, a, a, a can be divided in five different ways ; so
also the b, b, b can be divided in four ways ; the
c, c in three ways; and the d can be disposed of in
18 CHOICE.
two ways. Hence (by Rule II.) the whole division can
be made in
5 x 4 x 3 x 2, or 120
different ways, including, however, the ways in which
either person gets none and the other gets all. Exclud
ing these two ways, the number of eligible ways is 118.
Question. — In the ordinary system of notation, how
many numbers are there which consist of five digits ?
Ansiver . — The first digit may be any of the ten
except 0. We have, therefore, the choice of nine ways
of determining this digit. Each of the other four digits
may be any whatever, and therefore there are ten ways
of determining each of them. Hence, altogether (by
Rule II. ) the number can be formed in
9 x 10 x 10 x 10 x 10, or 90000
different ways.
Of course these are all the numbers from 10000 to
99999 inclusive.
Question. — The cylinder of a letter-lock contains
four rings, each marked with twenty-six different
letters ; how many different attempts to open the lock
may be made by a person ignorant of the key-word ?
Answer. — The first ring can be placed in twenty-
six different positions ; so may the second ; so may the
third; so may the fourth. Hence (by Rule II.) there
are
26 x 26 x 26 x 26, or 456976
DIFFERENT ORDERS OF ARRANGEMENT. 19
different positions possible, and one of these is the
right one. Hence it is possible to make 456975
unsuccessful trials.
RULE III.
The number of ways in which a given number of
things can be arranged is the continued product of the
given number, and all whole numbers less than it.
Thus, three things can be arranged in 3 x 2 x 1, or
6 ways; four things in 4 x 3 x 2 x 1, or 24 ways;
five things in 5 x 4 x 3 x 2 x 1, or 120 ways.
It will be sufficient to shew the reason of this rule
in a particular case. The reasoning will be of a suffi
ciently general character to apply to any other case.
Take for example the case of five things. We have
then a choice of five ways of filling the first place in
order. When that place is filled there remain four
things, and therefore we have a choice of four ways
of filling the second place. Then there are three
. things left, and we can fill the third place in three
ways. So we can fill the fourth place in two ways,
and the last place in only one way, since we must give
to it the one thing that is now left. Hence (by Rule
II.) all the places can be filled in 5x4x3x2x1
ways, or the whole set of five things can be arranged in
5x4x3x2x1 ways, which shews that Rule III.
is true in this case.
20 CHOICE.
By exactly similar reasoning, we can shew that the
rule is true in any other case. Hence we may accept
it universally.
It is usual to put the mark | round a number to
denote the continued product of that number and all
lesser numbers. Thus —
|2 denotes 2 x 1, or 2 ;
|3 denotes 3 x 2 x 1, or 6;
|4 denotes 4 x 3 x 2 x 1, or 24 ;
|5 denotes 5x4x3x2x1, or 120 ;
|6 denotes 6 x 5 x 4 x 3 x 2 x 1, or 720 ;
and so on.
A great number of questions will be seen, on a little
consideration, to be particular applications of Kule III.
Suppose, for instance, that we have to place six
statues in six niches, it seems, at first sight, that as
the statues and the niches can each of them separately
be taken in any order, we should have to consider the
order of both to determine what choice of arrangement
we have.
But, on consideration, it will be seen that even
though we take the niches in any stated order, yet any
possible result whatsoever may be attained by varying
the order of the statues. We may, in fact, regard the
niches as forming a row in fixed order, and we have
only to consider in how many different orders the six
statues may be taken so as to fill the six niches in
order. Consequently, the number of ways in which it
EXAMPLES. 21
is possible to arrange the six statues in the six niches
is the same as the number of orders in which the
six statues can themselves be taken, which by the
rule is |6, or 720.
This explanation will be the better understood by
comparing the next two questions.
Question. — In how many ways can twelve ladies
and twelve gentlemen form themselves into couples
for a dance?
Answer. — 112. For the first gentleman can choose
a partner in twelve ways ; then the second has choice
of eleven ; the third has choice of ten, and so on.
Therefore they can take partners altogether in
12.11.10.9.8.7.6.5.4.8.2.1, or |12
ways.
Question. — There are twelve ladies and twelve
gentlemen in a ball-room ; in how many ways can they
take their places for a contre-danse?
Answer. — The couples can be formed in |12 ways,
(last question) and when formed, the couples can be
arranged in |12 different orders (Kule III.) There
fore the twelve ladies and twelve gentlemen can arrange
themselves in |12 X |12 different ways.
Or we may reason thus:
The ladies can take their places in |12 different
ways, ( by Rule III.) and so the gentlemen can take
theirs in |12 different ways. Therefore (by Rule I.)
22 CHOICE.
the ladies and gentlemen can arrange themselves in
|12 x 1 1-2 different ways, as before.
Question. — In how many different orders can the
letters a, b, c, d, e, /he arranged so as to begin with
ab?
Answer )ffi^. — For our only choice lies in the
arrangement of the remaining four letters, which can
be put in |4 or 24 different orders (by Kule III.)
Question. — A shelf contains five volumes of Latin,
six of Greek, and eight of English. In how many
ways can the nineteen books be arranged, keeping all
the Latin together, all the Greek together, and all the
English together?
Answer. — The .volumes of Latin can be arranged
among themselves (by Kule III.) in [5 ways, the
volumes of Greek among themselves in |6 ways, and
the volumes of English among themselves in [8 ways.
Also, when each set is thus prepared, the three sets
can be placed on the shelf in |3 different orders.
Therefore, by Rule II., the number of ways in which
the whole can be done is
|5 X [6 x |8 X 8, or 20901888000.
Question. — In how many ways could the same books
be arranged indiscriminately on the shelf ?
Answer. — |19, or 121645100408832000 ways.
DIFFERENT WAYS OF FORMING A RING.
It often requires considerable thought to determine
what is meant by "different ways" of forming a ring.
The next three questions suggest three meanings
which the words in several circumstances will bear.
It will be well to consider them, and compare them
carefully, that the distinctions among them may be
thoroughly recognised.
Question. — A table being laid for six persons, in
how many ways can they take their places ?
Answer. — By Eule III., the number of ways is
(6 or 720.
Question. — In how many ways can six children form
themselves into a ring, to dance round a may-pole.
Answer. — In this case we have not to assign the
six children to particular places absolutely, but only
to arrange them relatively to one another. We may,
in fact, make all possible arrangements, by placing the
first child, A, in any fixed position, and disposing the
others, B, C, D, E, F, in different ways with respect
to him. Thus there is no essential difference between
the three arrangements —
D B E
EC C A F D
* * *
F B D F AC
A E B
24 CHOICE.
but two different arrangements would be —
D D
EC C E
* *
F B B F
A A
And any other essentially different arrangement migbt
be obtained without disturbing A? since absolute posi
tion is not taken into account. Now the five children
B, C, D, E, F can be arranged, by Kule III., in |5
or 120 ways. This, therefore, is the whole number of
ways in which such a ring can be formed.
Question. — In how many ways can six stones be
strung on an elastic band to form a bracelet ?
Answer. — This question is not equivalent to the
preceding one, for if we examine the last two arrange
ments, which we marked down as examples of the
different ways in which the ring could be made, we
shall observe that though they would count as different
arrangements of children round a may-pole, they would
count as the same arrangement of stones in a bracelet,
presenting only opposite views of the same bracelet;
each being, in fact, the arrangement that would be
presented by turning the other completely over. So
the 120 arrangements which we could make according
to the last question, might be disposed into 60 pairs,
each pair presenting only opposite views of the same
ring, and not representing more than one essentially
SUCCESSIVE NUMBERS. 25
different arrangement. Hence the answer is in this
case only 60.
DEFINITION. — Numbers are called successive when
they proceed in order, each one differing from the
preceding one by unity. The numbers are said to be
descending when they commence with the greatest and
continually decrease; they are said to be ascending
when they commence with the least and continually
increase; and such a series of numbers is said to
ascend or descend ( as the case may be ) from the first
number of the series.
Thus 17, 18, 19 are successive numbers ascending
from 17.
So, 17, 16, 15, 14 are successive numbers descending
from 17.
Again, if we speak of a series of five successive
numbers descending from 100, we shall mean the
numbers 100, 99, 98, 97, 96.
So if we speak of seven successive numbers ascend
ing from 3, the numbers
3, 4, 5, 6, 7, 8, 9.
will be meant.
Thus, 1 5 might be described as the continued
product of five successive numbers, ascending from
unity (or descending from 5) ; |7, as the continued
product of seven successive numbers ascending from
unity (or descending from 7), and so on.
•
26 CHOICE.
KULE IV.
Out of a given number of things,
the number of ways in zvhich an arrangement of two
things can be made is the product of the given number
and the next lesser number ;
the number of ways in which an arrangement of three
things can be made is the continued product of three
successive numbers descending from the given number;
the number of ways in which an arrangement of four
things can be made is the continued product of four
successive numbers descending from the given number ;
and so on.
The reason of this rule will he seen at once. For
suppose we have seventeen given things ; then, if we
wish to make an arrangement of two things, we have
the choice of seventeen things to place first, and then
there are sixteen things left, out of which we have to
choose one to place second, and complete our arrange
ment. Hence, hy Rule L, the numher of ways in which
we can make an arrangement of two things, is 17.16.
So if we wish to make an arrangement of three
things, we can place the first two in 17.16 ways,
and we then have fifteen things left, out of which to
choose one to come third, and complete our arrange
ment; therefore, by Rule L, the number of ways in
which we can make an arrangement of three things is
the product of 17.16 and 15, or 17.16.15 : and so on.
DIFFERENT ARRANGEMENTS.
27
Many questions which might be considered under
Kule II. may be answered more directly by this rule.
Thus —
Question. — How many three -lettered words could be
made out of an alphabet of twenty- six letters, not using
any letter more than once ?
Answer.— 26.25.24 = 15600.
Question. — How many four-lettered words ?
Answer.-^ 26.25.24.23 = 358800.
Question. — How many eight-lettered words ?
Answer.— 26.25.24.23.22.21.20.19 = 62990928000.
Question. — Four flags are to be hoisted on one mast,
and there are twenty different flags to choose from :
what choice have we ?
Ansiver. — By Kule IV. we have the choice of
20.19.18.17, or 116280
different ways.
The answer would evidently be the same if the flags
were to be hoisted on different masts, for so long as
there are four different positions to be occupied, the
operation consists in the arrangement in these posi
tions of four out of the twenty flags.
Question. — An eight-oared boat has to be manned
out of a club consisting of fifty rowing members. In
how many ways can the crew be arranged ?
ZO CHOICE.
Answer. — We have simply to arrange eight men in
order out of fifty men. Therefore Rule IV. applies, and
the number of ways is
50.49.48.47.46.45.44.43, or 21646947168000.
RULE V.
The number of ivays in which twenty things can be
divided into two classes of twelve and eight respectively,
is
[20
(M.-B
and similarly for any other numbers.
Suppose that twenty persons have to take their
places in twelve front seats and eight back seats.
By Rule III. they can be arranged altogether in [20
ways. But the operation of arranging them may be
resolved into the following three operations : —
(1) The operation of dividing the twenty into
two classes of twelve and eight.
(2) The operation of arranging the class of
twelve in the twelve front seats.
(3) The operation of arranging the class of
eight in the eight back seats.
Hence, by Rule II., |20 is the product of the
number of ways in which these three several operations
can be performed. But by Rule III. the second can be
performed in |12 ways, and the third in |8 ways;
DIVISION INTO TWO CLASSES. 29
therefore it follows that the first can he performed in
120
J12.J8
ways. This, therefore, expresses the number of ways
in which twenty things can- be divided into two classes,
of which the first shall contain twelve things, and the
second shall contain eight.
And it will be observed that our reasoning through
out is perfectly general, and would equally apply if,
instead of the number twenty, divided into the parts
twelve and eight, we had any other number, divided
into any two assigned parts whatever.
Hence we can write down on the same plan the
number of ways in which any given number of things
can be divided into two classes, with a given number
in each.
Question. — Eight men are to take their places in an
eight-oared boat ; but two of them can only row on
stroke side, and one of them only on bow side ; the
others can row on either side. In how many ways
can the men be arranged?
Ansiver. — The operation of arranging the men may
be resolved into the following three simple and succes
sive operations, viz.,
(1) To divide the five men who can row on
either side into two parties of two and three,
to complete stroke side and bow side re
spectively.
30 CHOICE.
(2) To arrange stroke side when it is thus
completed ; and
(3) To arrange bow side.
The five men who can row on either side can he
divided into two parties of two and three respectively,
in
ways, by Rule V. And when this is done, stroke side,
consisting of four men, can be arranged in _£ or
twenty-four different ways (Rule III.) ; and likewise bow
side in twenty-four ways. Hence the whole arrange
ment can be made in 10 x 24 x 24, or 5760 ways.
RULE VI.
The number of ways in which twenty things can be
divided into three classes of five, seven, and eight,
respectively, is
120
and similarly for any other numbers.
For, by Rule V., the twenty things can be divided
into two classes of twelve and eight in
20
different ways, and, when this is done, the class of
DIVISION INTO CLASSES. 31
twelve can be divided into two classes of five and
seven in
|5.|7
ways. Hence, by Rule L, both these can be done in
|20 |12 |20
x — or
J12.|8 |5.|7 5.|7.|8
different ways.
That is, twenty things can be divided into three
classes of five, seven and eight severally, in
go
|5.|7.|8
different ways; and since our reasoning is perfectly
general, a similar result may be written down when
the numbers are any other.
And it is easily seen that the reasoning may be
extended, in the same manner, to the case of more
than three classes.
Question. — In how many ways can three boys divide
twelve oranges, each taking four?
Answer. — By Rule VI. the number of different ways
in which twelve things can be divided into three classes
of four each, is
112
— ]A or 34650.
4 . 1 4 . \4
32 CHOICE.
Question. — In how many ways can they divide them,
so that the eldest gets five, the next four, and the
youngest three ?
Answer. — By Kule VI. the numher of different
ways is
12
or 27720.
|8.J4.(6
Question. — If there be fifteen apples all alike, twenty
pears all alike, and twenty-five oranges all alike, in
how many ways can sixty boys take one each ?
Answer. — The boys have, in fact, to form themselves
into a party of fifteen for the apples, a party of twenty
for the pears, and a party of twenty-five for the oranges.
They can therefore do it by Kule VI. in
[60
|15.[20.J25
different ways.
Question. — In how many ways can two sixes, three
fives, and an ace be thrown with six dice ?
Answer. — The six dice have to be divided into three
sets, containing 2, 3, 1 severally, of which the first set
are to be placed with six upwards ; the second set with
Jive upwards; and the third set ' with ace upwards.
By Kule VI. it can be done in
|6
• "jsTjaTii or 60
different ways.
APPLICATION OF BULES V. AND VI. 33
Question.— In how many ways may fifty-two cards
be divided amongst four players, so that each may have
thirteen ?
Answer. — By Rule VI.,
A possible error must be guarded against in the
application of Rules V. and VI.
Suppose we are given eight things, say the letters
A, B, C, D, E, F, a, H,
and are asked in how many ways it is possible to
divide them into two parcels of four each. If the
parcels are numbered No. 1 and No. 2, and are de
signed for different purposes, we may apply Rule V.,
and answer that the number of possible ways is
In this case, to put
('A, B, C, D, into the first parcel,
E, F, a, H, into the second;
and to put
(A, B, C) D, into the second parcel,
E, F, G, H, into the first,
will be counted as different ways of disposing of the
eight things. But if the parcels be perfectly indifferent
34 CHOICE.
— if the eight things have simply to he disposed in two
equal heaps, with no distinction between the heaps —
then the two ways just indicated of disposing of the
eight things will hecome identical ; each being merged
into the one way of putting
( A, B, C, D, into one parcel,
\ E, F, G-, H, into another.
In such a case as this, therefore, the Rule V. cannot
be applied without some modification ; we should, in
fact, have to divide by 2 the result given by this rule.
So if there are twelve things to be divided into three
different parcels — as, for instance, twelve oranges to
be divided among three different boys — the Eule VI.
may be applied. But if the parcels are indifferent, and
we are simply asked in how many ways twelve things
can be divided into three equal parts, the rule would
want modification ; we should, in fact, have to divide
our result by |3, the number of different orders in
which the three parcels can be arranged.
When different numbers of things have to be put
into the different parcels — as in the case when twenty
things are to be divided into parcels of five, seven, and
eight — no difficulty or doubt can arise, for the differ
ences of number are sufficient to distinguish the
different parcels, and to give an individuality to each ;
so that in such a case the Rule V. or VI. is always
applicable.
DIFFERENT CLASSES. 35
It must be observed that there is some ambiguity
in the manner in which the words sort and class are
sometimes used, especially when we describe collections
of articles as of different sorts or of the same sort, or of
different classes or of the same class.
Thus, if letters have been spoken of as consonants
and vowels, we may describe the alphabet as containing
twenty letters of one sort, and six letters of the other
sort ; yet if we regard the individual character of each
letter, we shall speak of a printer's fount as containing
twenty- six different sorts of letters. Plainly, there are
either two classes or twenty-six classes, according to the
character adopted as the criterion of class.
For instance, we may describe the letters
a, a, a, x, x,
as three of one sort and two of another sort. But the
letters
a, e, i, x, z,
regarded as vowels and consonants, might also be des
cribed as three of one sort and two of another sort.
Suppose now we are asked in how many different
orders we can write down five different letters, of which
three are of one sort and two of another sort, the an
swer will depend entirely on the sense in which "sort"
is understood. If we suppose the letters to be such as
a} a, a, x9 xt
where those of the same sort are absolutely identical
36 CHOICE.
with one another, having no personal individuality (so to
speak), the answer will be
JL.
H-12
(by Rule V.), since our only choice lies in dividing the
five places into two sets of three and two, for the
a, a, a, and x, x. But if the given letters be such as
a, e, i, x, z>
where the three, a, e, i, are of one sort as vowels, but
each has an individual character of its own, and the
two, x, z, are of one sort as consonants, but these also
like the vowels distinct in their identity, then the
answer becomes 5, by Eule in., since the five letters
are for the purposes of arrangement all different.
We shall avoid this ambiguity as much as possible,
by speaking of things as of one sort, when there is no
individual distinction amongst them, and of one class
when they are united by a common characteristic, but
capable, at the same time, of distinction one from
another.
KULE VII.
The number of orders in which twenty letters can be
arranged, of which four are of one sort (a, a, a, a,
suppose J, Jive of another sort (b, b, b, b} b, suppose J,
ARRANGEMENT OF THINGS NOT ALL DIFFERENT. 37
two of another sort (c> c, suppose}, and the remaining
nine all different, is
|20
|4.|6.|2
and similarly for any other numbers.
For the operation of arranging the letters in order
may be resolved into the following : —
(1) To divide the twenty places into four sets,
of four, five, two, nine, respectively.
(2) To place the a, a, a, a, in the set of four
places.
(3) To place the b, b, b, b, b, in the set of five
places.
(4) To place the c, c in the set of two places.
(5) To arrange the nine remaining letters in the
set of nine places.
Now by Rule VI., the operation (1) can be done in
|20
I4.J6.J2.I9
different ways.
The operation (2) can be done in only one way, since
the letters are all alike.
So the operations (3) (4) can be done in only one
way each.
And the operation (5) can be performed in |9 ways
by Rule III.
38 CHOICE.
Therefore by Rule II. the whole complex operation
can be performed in
120 120
- L_ - x i9 or - L___
ll-IMM?. I1-I6.-P
different ways.
And in the same way we can reason about any other
case. Hence in any case, to find the number of orders
in which a series of letters can be arranged which are
not all alike, we have only to write down the fraction,
having in the numerator the total number of the letters,
and in the denominator the number of letters of the
several sorts ; each number being enclosed in the
mark .
Question. — In how many orders can we arrange the
letters of the word indivisibility ?
114
Answer. — —=- =14.13.12.11.10.9.8.7 = 121080960.
Question. — In how many orders can we arrange the
letters of the word parallelepiped?
Answer.- " =201801600.
Question. — In how many orders can we arrange the
letters oT the word Hang oil en ?
Answer. — 75600.
SELECTIONS OR COMBINATIONS. 39
BULE VIII.
Out of twenty things, a selection of twelve things can
be made in the same number of ways as a selection of
eight things ( 'where 12 + 8 = 20^ ; and the number of
ways is
120
and similarly for other numbers of things.
For the selection of twelve (or eight) things out of
twenty, consists of the operation of dividing the twenty
things into two sets of twelve and eight, and rejecting
one of the sets. Therefore (hy the last rule), which
ever set be rejected, the operation can be performed in
|20
112.18
different ways.
Question. — Out of one hundred things, in how many
ways can three things be selected ?
Answer. — By Rule VIII.,
[100
|97.|3;
40 CHOICE.
or striking out from the numerator and the denomi
nator all the successive factors from 1 to 97,
100 . 99 . 98
We observe that the numerator 100.99.98 expresses
(Kule V.) the number of ways in which an arrange
ment of three things might be made out of one
hundred things.
This suggests the following rule for the number of
ways of selecting any number of things out of a larger
number, which will often be found more convenient
than Kule VIII., although both of course lead to the
same result.
KULE IX.
Out of any given number of things,
the number of selections of two things may be
obtained from the number of arrangements of two
things, by dividing by |2 ;
the number of selections of three things may be
obtained from the number of arrangements of three
things, by dividing by |3 ;
the number of selections of four things may be
obtained from the number of arrangements of four
things, by dividing by |4;
and so on.
SELECTIONS OB COMBINATIONS. 41
It will be sufficient to shew the reason of this rule
in any particular case.
Suppose we have to make a selection of three things
out of a given number of things ; what is our choice in
this case, compared with our choice in making an
arrangement of three things.
The operation of making an arrangement of three
things may be resolved into the two operations following,
viz. : —
(1) To make a selection of three things out of
the given things.
(2) To arrange in order the three selected
things.
Therefore, by Kule I., the number of ways of making
an arrangement of three things is equal to the number
of ways of making a selection of three things, multi
plied by the number of ways of arranging the three
selected things.
But by Eule III., three things can be arranged
in |3 different ways.
Hence, the number of arrangements of three things,
out of a greater number, is equal to the number of
selections multiplied by |3.
Or the number of selections of three things is equal
to the number of arrangements divided by |3.
And the same reasoning would apply if the number
of things to be selected were any other instead of 3.
Therefore the rule is true always.
42 CHOICE.
The student being in possession of the two rules
(VIII. and IX.) for writing down the number of ways
in which any number of things can be selected out of
a larger number, will, in any particular case, use the
rule which may seem the more convenient. It will be
observed that Rule VIII. gives the result in the more
concise form when the number of things to be selected
is a high number ; but the fraction thus written down,
though more concisely expressed, is not in such low
terms as that which would be written down by Rule
IX. Consequently, when the actual numerical value
of the result is required, Rule IX. leaves the less
work to be done, in cancelling out common factors
from the numerator and the denominator. In many
cases, it is simplest to take advantage of the principle
of Rule VIII., that out of twenty things (suppose) the
number of ways in which seventeen things can be
selected is the same as the number of ways in which
20 — 17 or three things can be selected, and then
to apply Rule IX. For, comparing the different forms
of the result in this case, we observe that Rule VIII.
gives
|20
while Rule IX. gives
20.19.18.17.16.15.14.13.12.11.10.9.8.7.6.5.4
1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.16.17
TWO RULES FOE SELECTIONS. 43
which might be simplified by dividing the numerator
and denominator by the factors
4.5.6.7.8.9.10.11.12.13.14.15.16.17
But if we recognise the teaching of Kule VIII. , that
the number of ways of selecting seventeen things is
the same as the number of ways of selecting three
things, and then apply Rule IX. to find the number of
ways of selecting three things, we can at once write
down the result in the simple form
20.19.18
Question. — Out of a basket of twenty pears at three
a penny, how many ways are there of selecting six
pennyworth ?
Answer. — By Rule VIII., we can select eighteen
out of twenty in as many ways as we can select two ;
and, by Rule IX., this can be done in
20.19
- or 190
1.2
ways.
Question. — In how many ways can the same choice
be exercised so as to include the largest pear ?
Answer. — Taking the largest pear first, our only
choice now lies in selecting seventeen out of the
44 CHOICE.
remaining nineteeen, which can be done (Rules VIII.
and IX.) in
ways.
Question. — In how many ways can the same choice
be exercised without taking the smallest pear ?
Answer. — We have now to select eighteen pears out
of nineteen. Therefore (Rules VIII. and IX.) our
choice can be exercised in nineteen ways.
Question. — In how many ways can the same choice
be exercised so as to include the largest, and not to
include the smallest pear ?
Answer. — Taking the largest pear first, we have then
to choose seventeen more out of eighteen, which can be
done (Rules VIII. and IX.) in eighteen ways.
Question. — Out of forty- two liberals and fifty con
servatives, what choice is there in selecting a committee
consisting of four liberals and four conservatives ?
Answer. — The liberal committee-men can be chosen
(by Rule VIII.) in
42.41.40.39
1.2.3.4
different ways, and the conservative committee-men in
50.49.48.47
1.2.3.4
or 230300
SELECTIONS UNDER CERTAIN RESTRICTIONS. 45
different ways. Hence (by Rule I.) the whole choice
can be exercised in
111930 x 230300, or 25777479000
different ways.
Question. — A company of volunteers consists of a
captain, a lieutenant, an ensign, and eighty rank and
file. In how many ways can ten men be selected so as
to include the captain.
Answer. — Since the captain is to be one of the ten,
the only choice lies in the selection of nine men out of
the remaining eighty-two, which can be done (Rule
VIII. or IX.) in
[82
J9.J73
82.81.80.79.78.77.76.75.74
T"
different ways.
Question. — In how many ways can ten men be
selected so as to include at least one officer ?
Answer. — By Rule VIII. ten men can be selected
out of the whole company in
J88
[10.J73
ways altogether. But the number of different ways that
will include no officer will be the number of ways in
46
CHOICE.
which ten can be selected out of the eighty rank and
file, that is, (by Kule VIII.)
|80
110.^70
These must be subtracted from the whole number of
ways in which ten men might be selected, and the
remainder
|83 |80
[10 . 1 78 1 10 . [70
will be the number of ways in which they may be
selected so as to include at least one officer.
Question. — In how many ways can ten men be
selected so as to include exactly one officer?
Answer. — The nine rank and file can be selected in
|80
IMZi
ways, and the one officer in three ways. Therefore the
ten can be selected in
3 x |80
different ways.
Question. — There are fifteen candidates for admission
into a society which has two vacancies. There are
SELECTIONS OUT OF DIFFERENT CLASSES. 47
seven electors, and each can either vote for two candi
dates, or plump for one. In how many ways can the
votes be given ?
Answer. — Each voter can plump in fifteen ways,
and can vote for two candidates in
15.14
or 105
L.
ways (Eule IX). Therefore each elector can vote
altogether in 120 ways. And there are seven electors ;
therefore all the votes can be given (by Kule II.) in
120 x 120 x 120 x 120 x 120 x 120 x 120
or 358318080000000
different ways.
It will be well to notice particularly the points
which distinguish the next three examples.
In all of them we suppose twenty things of one
class and six things of another class set before us, the
individuals of each class being distinct ; and in all of
them a selection has to be made of three things out of
each class. But while the first is a case of simple
selection, in the second each set of three things has
separately to be arranged in order, and in the third
the whole six selected things have to be together
arranged in order.
Question. — Out of twenty men and six women, what
choice have we in selecting three men and three women ?
48 CHOICE.
Answer. — The men can be selected in
20.19.18
T^ or mo
different ways (Kule IX.), and the women in
6.5.4
is§ or20
different ways. Therefore, we have the choice of
1140 X 20 or 22800
different ways of making our selection.
Question. — Out of twenty men and six women, what
choice have we in filling up six different offices, three of
which must be filled by men, and the other three by
women ?
Answer. — We can allot the first three offices to three
men in 20.19.18 or 6840 different ways (Kule IV.) ;
and we can allot the other three offices to three women
in 6.5.4 or 120 different ways. Therefore, we have
the choice of
6840 x 120 or 820800
different ways of making our arrangement.
Question. — Out of twenty consonants and six vowels,
in how many ways can we make a word, consisting of
three different consonants, and three different vowels ?
SELECTION OUT OF DIFFERENT CLASSES. 49
Answer. — "We can select three consonants in
20.19.18
different ways, and three vowels in
different ways. Therefore (by Rule L), the six letters
can be selected in
1140 x 20, or 22800
different ways, and when they are so selected, they can
be arranged (by Rule III.), in [6 or 720 different
orders. Hence (by Rule I.), there are 22800 x 720 or
16416000 different ways of making the word.
Question. — Out of the twenty-six letters of the
alphabet, in how many ways can we make a word
consisting of four different letters, one of which must
be always a ?
Answer. — Since we are always to use a, we must
choose three letters out of the remaining twenty-five.
This can be done in
25.24.23
ways. Then the whole set of four letters can be
arranged in [4 or 24 different orders. Hence we have
the choice of
2300 x 24, or 55200
different ways of making the word,
E
50 CHOICE.
The last answer might have been arrived at in
another way, as follows : —
Without the limitation, we could make a word of
four different letters in 26.25.24.23 or 358800 different
ways. The question is how many of these will contain
a. Now if all the 358800 words were written down on
paper, since each is made of four letters, our paper
would contain 358800 x 4, or 1435200 letters. And
since no letter of the alphabet has been used with more
favour than any other, it follows that each would occur
1435200 -v- 26, or 55200 times. Therefore a must
occur 55200 times ; and since no word contains a more
than once, 55200 words must contain a. That is, the
number of words formed of four different letters of
which a is one is 55200, as before.
Question. — Out of the twenty-six letters of the
alphabet, in how many ways can we make a word
consisting of four different letters, two of which must
be a and b ?
Answer. — We can choose the other two letters out
of the remaining twenty-four in
24.23
-TT or 276
ways, and then we can arrange the whole set of four
letters in |4 or 24 different orders. Hence we have
the choice of
276 x 24, or 6624
different ways of making the word.
SELECTION UNDER CERTAIN RESTRICTIONS. 51
Question. — Out of twenty consonants and six vowels,
in how many ways can we make a word consisting of
three different vowels and two different consonants, one
of the vowels being always a ?
Answer. — We can choose the other two vowels in
ways, and the two consonants in
ways ; hence our letters can be selected in 1900 ways,
and when they are selected the set of five can be
arranged in |5 or 120 ways. Hence the whole number
of ways of making the word is
1900 x 120, or 228000.
Question. — There are ten different situations vacant,
of which four must be held by men, and three by
women ; the remaining three may be held by either
men or women. If twenty male and six female candi
dates present themselves, in how many ways can we
fill up the situations ?
Answer. — The men's situations can be filled up in
20.19.18.17 or 116280 different ways, and the women's
in 6.5.4 or 120 different ways. When this is done,
there are nineteen persons left, all of whom are
eligible for the other three situations. Hence these
three can be filled up in 19.18.17 or 5814 different
'52 CHOICE.
ways. Therefore, the whole election can be made in
116280 X 120 x 5814, or 81126230400 different ways.
ARRANGEMENTS OUT OF A NUMBER OF THINGS NQT
ALL DIFFERENT.
We considered under Kule VII. the modifications
of the case of Rule III., when the things out of which
the arrangement is to be made are not all different.
The corresponding modifications of Rule IV. are too
intricate to be treated by a general rule in an ele
mentary treatise on Arithmetic; but each case, as it
.arises, may be resolved into cases to which the
preceding rules will apply. The manner of proceeding
will be sufficiently illustrated by the following questions.
Question. — In how many ways can an arrangement
of four letters be made out of the letters of the words
choice and chance?
Answer. — There are fifteen letters altogether, of
eight different sorts, viz., c, c, c, c ; li, h ; a, a; n, n ;
.e, e; o ; i; d. The different ways of selecting the
ifour letters may, therefore, be classified as follows :
(1) all four alike,
(2) three alike and one different,
(3) two alike and two others alike,
(4) two alike and the other two different,
(5) all four different.
Now, the selection (1) can be made in only one way
ARRANGEMENTS OUT OF THINGS NOT ALL DIFFERENT. 58
(viz. by selecting c, c, c, c), and when this selection of
letters is made, they can be arranged in only one order ;
therefore (1) gives rise to only one arrangement.
The selection (2) can be made in seven ways, for
three letters alike can be selected in only one way (viz.
c, c, c), and one different one in seven ways (a, e, i, o,
h, n, d). And when this selection of letters is made,
they can be arranged in four ways (Kule VII.) ;
therefore (2) gives rise to 7 x 4, or 28 arrangements.
5.4
The selection (3) can be made in — — or 10 ways
(Eule IX.), since we have to select two out of the five
pairs, cc, hh, aa, nn, ee. And when this selection of
14
letters is made, they can be arranged in — - or 6 ways
|2.|2
(Rule VII.) ;
therefore (3) gives rise to 10 x 6, or 60 arrangements.
The selection (4) can be made in 5 x 21, or 105
ways, for we can select one of the five pairs, cc, hh, aa,
nn, ee in five ways, and two out of the seven different
rj n
sorts of letters that will then be left, in — — or 21
1 . 2t
ways, (Rule IX.) ; and when this selection of letters is
I*
made, they can be arranged in — or 12 ways (Rule
|2
VII.) ;
therefore (4) gives rise to 105 x 12, or 1260 arrange
ments*
54 CHOICE.
The selection (5) of four different letters must be
made out of the eight, c, h, a, n, e, o, i, d, therefore
the number of arrangements which will come from such
a selection must be 8 . 7 . 6 . 5 or 1680.
Hence, the whole number of arrangements of four
letters out of the fifteen given letters is
1 + 28 + 60 + 1260 + 1680, or 3029.
Question. — In how many ways can an arrangement
of three things be made out of fifteen things, of which
five are of one sort, four of another sort, three of
another sort, and the remaining three of another sort?
Answer. — The three selected things may be either —
(1) all three alike,
or (2) two alike and one different,
or (3) all different.
Now, the selection of all three alike can be made in
4 ways, since we can take one of the four different
sorts. And when this selection is made, the selected
things can be arranged in only one order;
therefore (1) gives rise to only four arrangements.
The selection of two alike and one different can be
made in 4 x 3, or 12 ways (Kule I.) ; for the two
alike can be of any of the four sorts, and the one
different of any one of the remaining three sorts. And
when this selection is made, the things selected can be
13
arranged in ^-, or three ways (Kule VII.) ;
therefore (2) gives rise to 12 x 3, or 36 arrangements.
ARRANGEMENTS OUT OF THINGS NOT ALL DIFFERENT. 55
And if all three selected things are to be different
we shall have 4.3.2, or 24 arrangements (Rule IV.).
Hence, the whole number of arrangements of three
things out of the fifteen given things is
4 + 36 + 24, or 64.
tion. — In how many ways can an arrangement
of five things be made out of the fifteen things given in
the last question ?
Answer. — The different ways of selecting five things
may be classified as follows : —
(1) all five alike,
(2) four alike and one different,
(3) three alike and two others alike,
(4) three alike and two different,
(5) two alike, two others alike, and one different,
(6) two alike and three different.
Now by the application of Rules II., VII., IX., as
in the preceding questions, it will be easily seen that
the selection (1) can be made in one way, and leads
to one arrangement :
the selection (2) can be made in six ways, and leads
to 6 x 5 or 30 arrangements :
the selection (3) can be made in twelve ways, and
leads to 12 x 10 or 120 arrangements :
the selection (4) can be made in twelve ways, and
leads to 12 x 20 or 240 arrangements :
the selection (5) can be made in twelve ways, and
leads to 12 x 30 or 360 arrangements :
56 CHOICE.
the selection (6) can be made in four ways, and leads
to 4 x 60 or 240 arrangements.
Hence the whole number of different arrangements
is
1 + 30 + 120 + 240 + 360 + 240, or 991.
RULE X.
The whole number of ways in which a person can
select some or all (as many as he pleases} of a given
number of things, is one less than the continued product
of 2 repeated the given number of times.
For since he is at liberty to take none or all or as
many as he pleases of the different things, he can
dispose of each thing in two ways, for he can either
take it or leave it. Now suppose there are five things,
then he can act altogether in
2x2x2x2x2
different ways. But if he is not to reject all the
things, the number of courses open to him will be
one less than this, or
2x2x2x2x2-1.
And the same reasoning would apply if the number
of things were any other number instead of five.
Hence, the rule will be true always.
TOTAL NUMBER OF SELECTIONS. 57
Question.— One of the stalls in a bazaar contains
twenty- seven articles exposed for sale. What choice
has a purchaser?
Answer. — He may buy either one thing or more,
and there are twenty- seven things : therefore (by
Rule X.), the number of courses open to him is one
less than the continued product of twenty-seven tivos,
or 134217727.
•
Question. — What is the greatest number of different
amounts that can be made up by selection from five
given weights ?
Answer. — By Eule X., 2 x 2 x 2 x 2 x 2 - 1 or 31.
The different selections will not always produce
different sums. Hence, we cannot always make
thirty- one different sums. But under favourable cir
cumstances, as, for instance, when the weights are
lib., 21bs., 41bs., 81bs., 161bs., all the different
selections will produce different sums, and then the
number of different sums is thirty-one. Hence,
thirty- one is the greatest number of different weights
that can be made by a selection from five given
weights.
In the case of the five weights, lib., 21bs., 41bs.,
81bs., 161bs., the thirty-one different amounts that
can be weighed consist of every integral number of
pounds from one to thirty-one.
Thus, with single weights, we can weigh the
58 CHOICE.
following numbers of pounds, viz., 1, 2, 4, 8, 16;
and then we have
3-1+ 2, 7 = 1 + 2+ 4, 15 = 1 + 2 + 4+ 8,
5-1+ 4, 11 = 1 + 2+ 8, 23-1 + 2 + 4 + 16,
6-2+ 4, 13-1 + 4+ 8, 27 = 1 + 2 + 8 + 16,
9-1+ 8, 14=2 + 4+ 8, 29 = 1+4 + 8+16,
10=2+ 8, 19 = 1 + 2 + 16, 30 = 2 + 4 + 8 + 16,
12-4+ 8, 21-1 + 4 + 16,
17=1 + 16, 22-2 + 4 + 16, 31 = 1 + 2 + 4 + 8 + 16.
18 = 2 + 16, 25 = 1 + 8 + 16,
20-4 + 16, 26=2 + 8 + 16,
24=8 + 16, 28=4 + 8 + 16,
It may be observed that, if we had a 321bs. weight,
by adding it to each of the sets already obtained, we
should get all the numbers from 33 to 63 inclusive;
hence all the weights
lib. 21bs. 41bs. 81bs. 161bs. 321bs.
would enable us to weigh any number of pounds from
1 to 63.
Then the addition of a 641bs. weight, would enable
us to weigh any number up to 127, and so on.
Question. — What is the greatest number of different
amounts that can be weighed with five weights, when
each weight may be put into either scale ?
Answer. — This is not a direct example of our rule,
but it may be solved on a like principle to that by
which the rule itself was established.
COMBINATIONS OF GIVEN WEIGHTS. 59
Each weight can be disposed of in three ways, that
is, it can be placed either in the weight-pan, or in the
pan with the substance to be weighed, or it can be left
out altogether. Hence, all the weights can be disposed
in 3 x 3 x 3 x 3 x 3, or 243 ways (Kule II). But
one of these ways would consist in rejecting all the
weights ; this must be cast out, and then there remain
242 ways. But in the most favourable case, half of
these ways would consist in placing a less weight in
the weight-pan than in the other, and these must be
cast out. Hence there remain 121 different amounts
that can be weighed under the most favourable circum
stances with five weights, when it is permitted to place
weights in the pan with the substance to be weighed.
The weights lib., 31bs., 91bs., 271bs., Sllbs., will
afford an instance of the most favourable case. In this
instance, the 121 amounts that can be weighed consist of
every integral number of pounds from 1 to 121. Thus —
2=3-1, 15 = 27 - 9 - 3,
4=3 + 1, 16 = 27 - 9 - 3 + 1,
5-9-3-1, 17 = 27 - 9 - 1,
6=9-3, 18 = 27 - 9,
7=9-3 + 1, 19 = 27 - 9 + 1,
8=9-1, 20 = 27 - 9 + 3 - 1,
10 = 9 + 1, 21 = 27 - 9 + 3,
11 = 9 + 3 - 1, 22 = 27 - 9 + 3 + 1,
12 = 9 + 3, 23 = 27 - 3 - 1,
13 = 9 + 3 + 1, 24 = 27 - 3,
14 = 27 - 9 - 3 - 1, &c.
60 CHOICE.
RULE XI.
The whole number of ways in which a person can
select some or all fas many as he pleases ) out of a num
ber of things ivhich are not all different, is one less than
the continued product of the series of members formed
by increasing by unity the several numbers of things of
the several sorts.
Thus, suppose we have the letters —
a, a, a, a, a,
b, b, b,
d,
viz., five of one sort, three of another, four of a third
sort, one of a fourth sort, and one of a fifth.
The numbers of letters in the several classes are
5, 3, 4, 1, 1, and these, severally increased hy unity,
give the new series of numbers 6, 4, 5, 2, 2. The rule
states that the whole number of ways in which a person
may take some or all (as many as he pleases) of the
given letters, is
6x4x5x2x2-1, or 479.
The reason of the rule will be seen from the following
considerations. Suppose the person were at liberty to
take none, or all, or as many as he pleased of the
SELECTIONS OF THINGS NOT ALL DIFFRENT. 61
letters. He could then dispose of the five a, a, a, a, a
in six different ways, for he might take 5 or 4 or 3 or 2
or 1 or none of them. So he could dispose of the three
b, by b, in four ways, for he might take 3 or 2 or 1 or
none of them. Similarly he could dispose of the
c, c, c, c in five ways, and of the d in two ways, and of
the e in two ways. Hence he might act altogether, in
6x4x5x2x2
different ways (Rule II). But if he is not to reject all
the things, the numher of courses open to him will he
one less than this, or
6x4x5x2x2-1.
And the same reasoning would apply to any other
case. Hence we may accept the rule as true always.
Question. — In how many ways can two booksellers
divide between them 200 copies of one book, 250 of
another, 150 of a third, and 100 of a fourth?
Ansiver. — Either man can take any number of
books, but not either none or all. Therefore, the
number of ways is one less than that given by the
rule : i. e.} the division can be made in
201 x 251 x 151 x 101 - 2
different ways; or in
769428201-2, or 769428199
different ways.
62 CHOICE.
EXAMPLES ON CHOICE.
1. — Having four seals and five sorts of sealing wax,
in how many ways can we seal a letter ?
2. — There are five first-class carriages, eight second-
class, seven third-class and three luggage -vans. In
how many ways can a train be made consisting of one
of each ?
3. — How many changes can be rung upon eight
bells? And in how many of these will an assigned
bell be rung last ?
4. — Out of a class of twelve boys, in how many ways
can three boys be called up to say lessons ?
5. — In how many ways can a set of twelve black
and twelve white draught-men be placed on the black
squares of a draught-board?
6. — In how many ways can a set of chess-men be
placed on a chess-board?
7. — In how many ways can we arrange the letters
of the word possessions ?
8. — In how many ways can we arrange the letters
of the words choice and chance ?
EXAMPLES. 63
9. — In how many ways can a triangle be formed,
having its angular points at three of the angular
points of a given hexagon?
10. — There are three teetotums, having respectively
6, 8, 10 sides. In how many ways can they fall? and
in how many of these will two aces he turned up ?
11. — A company of soldiers consists of three officers,
four sergeants and sixty privates. In how many ways
can a detachment be made consisting of an officer, two
sergeants and twenty privates? In how many of
these ways will the captain and the senior sergeant
appear?
12. — In how many ways can four persons sit at a
round table, so that all shall not have the same
neighbours in any two arrangements?
13.— In how many ways can seven persons sit as in
the last question ? And in how many of these will two
assigned persons be neighbours? And in how many
will an assigned person have the same two neighbours?
14. — Out of a party of twelve ladies and fifteen
gentlemen, in how many ways can four gentlemen and
four ladies be selected for a dance?
15. — Out of twenty consonants and six vowels, in
how many ways can a word be made, consisting of three
64 CHOICE.
different consonants and two different vowels, without
placing all the consonants together?
16. — Out of twenty consonants and ten vowels, in
how many ways can a word he formed consisting of
three different vowels and three different consonants,
the vowels and consonants being placed alternately ?
17.— In how many ways can the foregoing questions
he arranged, so that no question of combination shall
come before any question of permutation ?
18. — A plaything consists of eighteen cubical blocks ;
on each side of five of them a head is painted, on each
side of seven a body, and on each side of six a pair of
legs. How many different figures can be made by
piecing them together ?
19. — Having five pairs of gloves, in how many ways
can a person select a right-hand and a left-hand glove
which are not pairs ?
20. — How many numbers less than 10,000 have a
five in their arithmetical expression, and how many of
them are divisible by five without remainder ?
21. — In how many ways can a school of ninety boys
divide themselves, so that twenty-four play football,
twenty-two play cricket, thirty drill, four play racquets,
and ten take a walk ?
EXAMPLES. 65
22. — From five apples, six pears, and three oranges,
in how many ways can a person take fruit ?
23. — A man has* ten shares in the Great Western
Kailway Company, twelve in the North Western, seven
in the Great Northern, two in the Great Eastern, five
in the South Western. In how many ways can he sell
shares ?
24.— How many different signals can be made with
a set of ten flags, using four at a time, (1) on a
single mast, and (2) on a three-masted ship ?
CHAPTEE II.
CHANCE.
"There is very little chance of fine weather."
"Is there much chance of his recovery?"
"There is no chance of finding it."
" There is a great probability of war."
"This is a more probable result than the other."
"That is more likely to be mine than yours."
"There is less chance of her coming than of his."
— These are expressions in common use amongst us ;
the very commonness of their use shows that people
in general have some idea of chance, and some
conception of different degrees of probability in the
occurrence of doubtful events. All understand what is
meant by much chance and little chance; they dis
tinguish events as very probable, probable, improbable,
or very improbable ; but no attempt is made in common
conversation to measure with any accuracy the amount
of probability attaching to any given event/| If a
Doctor is asked what chance there is of a patient's
recovery, he may answer that there is much chance or
little chance, but he cannot express with any precision
the exact magnitude of his hope or of his fear. Yet his
DEGREES OF CHANCE. 67
expectation of the event has a certain magnitude. He
has a greater expectation of this patient's recovery than
he has of the recovery of another, whose symptoms are
more aggravated, and less expectation than in another
case where the constitution is stronger. His expecta
tion has a definite value, and if he were a sporting man,
he would be prepared to offer or take certain definite
odds on the event. But in common language, this
definite amount of expectation or probability cannot be
precisely expressed, because we have no recognised
standard with which to compare it, no recognised
amount of expectation or probability by which to
measure it.
In fact, in describing the magnitude of any expecta
tion which we entertain, we are in the same position
as if we had to describe the length of a room, or the
height of a tower, to a man who was not acquainted
with a foot or a yard, or any of our standards of
length. We could speak of the room as very long or
very short, we could speak of the tower as very high
or very low, but without some standard length recog
nised alike by ourselves and those whom we addressed,
we could not give an accurate answer to either of
the questions, How long is the room ? or How high is
the tower ?
So when we are asked what chance we think there
is of a fine afternoon, we may say that there is much
chance or little chance, or we may even go further,
and establish in our own minds a scale of expressions,
68 CHANCE.
distinguishing the different degrees of probability in
some such way as follows : —
It is certain not to rain.
It is very unlikely to rain.
It is unlikely to rain.
It is as likely to rain as not.
It is likely to rain.
It is very likely to rain.
It is certain to rain.
but these expressions except the first, fourth, and last,
are vague and indefinite, nor can we ever be sure that
those with whom we are conversing attach exactly the
same idea to each expression that we do.
This vagueness is of little consequence in common
life, because in most cases it is impossible to make an
accurate estimate of a chance, and the expressions
are, perhaps, as accurate as the estimates themselves
which we wish to express. But there are other classes
of events concerning which it is possible to form
accurate estimates of their degree of probability or
likelihood of happening, and in these cases it is well
to have some more precise method of expressing
different degrees, than is afforded by the common
expressions which we have quoted.
We must observe at the outset, that we use the
words chance and probability as strictly synonymous.
In common language, it is usual to prefer the former
word when the expectation is small, and the latter
CHANCE IMPLIES IGNORANCE. 69
when it is large. Thus we generally hear of " little
chance," or of " great probability," but not so often of
"great chance," or "little probability." This distinc
tion, however, is not universal, and we shall entirely
disregard it, using the two words chance and pro
bability in the same sense.
will be seen that probability always implies some
ignorance on the part of the person entertaining the
expectation, and the amount of probability attaching
to any event will depend upon the degree of this
ignorance. With omniscience, degrees of probability
are incompatible; for omniscience implies certainty, and
certainty precludes doubt, and degrees of probability
are the measures of doubt.
Hence, there is no such thing as the absolute pro
bability of an event, all probability being conditional on
our ignorance, and varying when that condition varies.
Thus the same event will be unequally probable td
different persons, whose knowledge of the circum
stances relating to the event is different. And to the
same person, the expectation of any event will be
affected by any accession of knowledge concerning the
event.
For instance, suppose we see a friend set out with
five other passengers in a ship whose crew number
thirty men : and suppose we presently hear that a
man fell overboard on the passage and was lost. So
long as our knowledge is confined to the fact that one
70 CHANCE.
individual only has been lost out of the thirty- six on
board, the probability that it is our friend is very
small. The odds against it would be said to be thirty-
five to one. But suppose our knowledge is augmented
by the news that the man who has been lost is a pas
senger ; though we still feel that it is equally likely to
be any of the other five passengers, yet our appre
hension that it is our friend becomes much greater
than it was before. The odds against it are now
described as five to one. Thus the probability that
our friend is lost is seen to be entirely conditional on
the respective degrees of our knowledge and ignorance ;
and so soon as our ignorance vanishes — so soon as
we know all about the event, and become as far as
that event is concerned omniscient, — then there no
longer remains a question of probability ; the probability
is replaced by certainty.
This example will also illustrate the meaning of the
•ratio of probabilities. Since each of the passengers
was equally likely to have been lost, it was evidently
always six times as likely that the man lost was some
passenger, as that it was our friend. So it was five
times as likely that it was a passenger, but not our
friend, as that it was our friend. Therefore, also, the
probability that it was a passenger, but not our friend,
was to the probability that it was a passenger in the
ratio of 5 to 6.
Let us suppose another case. A number of articles
are placed in a bag, and amongst them are three balls,
RATIO OF PROBABILITIES. 71
alike in all respects, except that two of them are
coloured white and the third black : all the other articles
we will suppose to be coins, or anything distinguishable
without difficulty from balls.
We present this bag to a stranger, and we give him
leave to put in his hand in the dark, and to take out
any one article he likes. But before he does this, we
may consider what chance there is of his taking out a
ball, or what chance there is of his taking out the black
ball. Obviously we cannot form any accurate estimate
of this chance, because it must depend- upon the wants
or the taste of the stranger influencing his will, whether
he will prefer to take a ball or a coin, and being igno
rant of his will in the matter, we cannot say whether it
is likely or unlikely that he will select a ball.
But it is axiomatic, that if he draws a ball at all, it
is twice as likely to be a white ball as to be a black one,
or the respective chances of his drawing white or black
are in the ratio of 2 to 1, and these chances are respec
tively two-thirds and one-third of the chance that he
draws a ball at all.
We now proceed to show how the magnitude of a
chance may be definitely expressed. We have already
pointed out that the expressions used in common
language are wanting in definiteness and precision,
and we compared the expedients by which degrees of
probability are usually indicated to the attempts which
we should make to give an idea of the length of a room
72 CHANCE.
to a person unacquainted with the measures of a foot
and a yard.
Now we observe, that the difficulty in this latter case
ceases, so soon as the person with whom we are
speaking agrees with us in his conception of any
definite length whatever. If he can once recognise
what we mean by the length of a hand, for instance,
we can express to him with perfect accuracy the length
of the room as so many hands; or, if he have an
idea of what a mile is, we can precisely express the
length of the room as some certain fraction of a mile.
So, also, as soon as we have fixed upon any standard
amount of probability that can be recognised and
appreciated by all with whom we have to do, we shall
be able to express any other amount of probability
numerically by reference to that standard. The
numbers 2, 3 would express probabilities twice or
three times as great as the standard probability; and
the fractions |-, J, § would express probabilities half,
one-third, or two-thirds of the standard.
Now, it matters not how great or how small the
standard be, provided it be a probability which all
can recognise, and which all will alike appreciate.
This is, indeed, the one essential which it has to fulfil;
it must be such that all persons will make the same
estimate of it. And that which best satisfies this
condition, and, therefore, the most convenient standard
with which to compare other probabilities, is that
CERTAINTY A MEASURE OF PROBABILITY. 73
supreme amount of probability which attaches to an
event which we know to be certain to happen. All
understand what certainty is : it is a standard which
all estimate alike. Certainty, therefore, shall be our
unit of probability; and other degrees of probability
shall be expressed as fractions of certainty.
But it may be asked, Is certainty a degree of
probability at all, or can smaller degrees of probability
be said to have any ratio to certainty? Yes. For
if we refer to the instance already cited of the six
passengers in the ship, we observe that the chance
of the lost man being a passenger is six times as
great as the chance of his being our friend. This
is the case however great our ignorance of the cir
cumstances of the event; and it will evidently remain
true until we attain to some knowledge which affects
our friend differently from his fellow-passengers. But
the news that the lost man was a passenger does not
affect one passenger more than another. Therefore,
after receiving this news, it will still hold good that the
chance of the lost man being a passenger is six times
as great as the chance of its being our 'friend. But
it is now certain that the lost man was a passenger;
therefore the probability that it was our friend is one-
sixth of certainty. Again in the instance of the balls
and coins in the bag, we have already noticed that the
chances of drawing white or black are respectively two-
thirds and one-third of the chance of drawing a ball
at all. And this is the case whatever this last chance
74 CHANCE.
may be. But suppose the man tells us that he is
drawing a ball, not a coin, then this last chance
becomes certainty ; and therefore the chances of draw
ing white or black, become respectively two-thirds and
one -third of certainty. Thus it is seen that certainty,
while it is the supreme degree, is some degree of
probability, or is such that another degree of probability
can be compared to it and expressed as a fraction of it.
Of course, when we use unity to express certainty,
the probability of the lost passenger being our friend
will be expressed by the fraction -g , and the chances of
the ball drawn being white or black, will be expressed
by the fractions ^ and -3 .
After the explanations which we have already given,
the reader will have no difficulty in accepting the
following axiom.
AXIOM.
If an event can happen in a number of different
ways (of which only one can occur J, the probability of
its happening at all is the sum of the several proba
bilities of its happening in the several ways.
For instance, let the event be the falling of a coin.
It can fall either head or tail, and only one of these
ways can occur. The probability that it falls at all
must be made up by addition of the probability that it
falls head and the probability that it falls tail.
ADDITION OF PROBABILITIES. 75
Again, let the event be that either A, B, or C should
win a race in which there are any number of com
petitors. The event can happen in three ways, viz., by
A winning, by B winning, or by C winning ; and only
one of these ways can occur. The probability that one
of the three should win is equal to the sum of the pro
babilities that A should win, that B should win, and
that C should win.
This is only saying that if a man would give £2 for
A's chance of the prize, £3 for B's chance, and £4 for
C's chance, he would give £2 + £3 + £4, or £9 for
the promise that he should have the prize if any one of
the three should win.
. 1
Again, if -^ be the chance of a shot aimed at a
target hitting the bull's eye, | the chance of its hitting
the first ring, and ^ the chance of its hitting the
outer ring, the chance that it hits one of these, i. e, the
chance of its hitting the target at all, is -JQ- + g + 4,
31
RULE I.
The probability of an event not happening is obtained
by Subtracting from unity the probability that it will
happen.
For it is certain that it will either happen or not
happen, or the probability that it will either happen or
76 CHANCE.
not happen is unity; and only one of these two (the
happening and the not happening) can occur. There
fore, by the axiom, unity is the sum of the probabilities
of the event happening and not happening ; or the
probability of its not happening is obtained by sub
tracting from unity the probability of its happening.
EXAMPLES. — If the chance of an event happening
2 23
is 1= , the chance of its not happening is 1 — -g , or g .
ri
If the chance of a plan succeeding is -^, the chance
of its failing is 1 — -^-, or -^-.
q-l
If the chance of a shot hitting a target be ~6^-, the
9Q
chance of its missing is -^.
If the chance of A winning a race be g, and the
chance of B winning it -g, the chance that neither
should win is ^T. For, by the axiom, the chance that
one of them should win is g + -g , or -^ ; and therefore,
by Rule I.,, the chance that this should not happen is
i JL 17
i — 24 , or 24 .
DEFINITION I. — Two probabilities which together
make up unity, are called complementary probabilities.
DEFINITION II. — When it is said that the odds are
three to two against an event, it is meant that the
RELATION OF CHANCE AND ODDS. 77
chance of the event failing is to the chance of its hap
pening as three to two ; and when it is said that the
odds are three to two in favour of an event, it is meant
that the chance of its happening is to the chance of its
failing as three to two ; and so for any other numbers.
RULE II.
If the odds be three to two against an event, the
chance of the event not happening is
_8_
3 + 2'
and the chance of its happening is
2 .
3 + 2'
and so for any other numbers ; the numerators of the
two fractions being the two given numbers, and their
common denominator the sum of the numbers.
For the two fractions satisfy the condition required
by Rule I., viz., that their sum should be unity, and
that required by the definition, viz., that their ratio
should be the same as the ratio expressing the given
odds. Similarly, —
If the odds be three to two in favour of an event,
the chance of the event happening is
3
8 +~2'
78 CHANCE.
and the chance of its not happening is
2
3 + 2'
and so for any other numbers ; the numerators of the
two fractions being the two given numbers, and their
common denominator the sum of the numbers.
EXAMPLES. — If the odds be ten to one against an
event, the chance of its happening is -jj, and the
chance of its failing is ^-.
If the odds be five to two in favour of the success
5
7>
of an experiment, the probability of success is -=
and the probability of failure is |.
RULE III.
If an event can happen in jive ways, and fail in seven
ways, and if these twelve ways are all equally probable,
and only one of them can occur, the odds against the
event are seven to five, and the chances of its happening
and failing are respectively
5 7
and similarly for any other numbers.
For since the event must either happen or fail, one
of the twelve ways must occur ; therefore the sum of
FUNDAMENTAL KULE. 79
their several probabilities is unity. But all the twelve
ways are equally probable. Therefore the chance of the
occurrence of any particular one is -j^-, and the chance
of the occurrence of one of the five which cause the
event to happen is five times this, or -^. So the
chance of the occurrence of one of the seven which
17
cause the event to fail is -^ •
Suppose, for example, that a die has twelve faces, of
which five are coloured white and seven black. A
person throws the die, and is to receive a prize if it
fall white.
The odds are seven to five against his winning the
prize. The chance that he wins is -^ , and the chance
17
that he loses is jg.
For all the twelve faces are equally likely to turn up,
and one must turn up. Therefore the chance of any
particular face turning up is -jg, and the chance of a
white face turning up is five times this, or -^ .
Or we might put it thus : — Since there are five white
and seven black faces, it is axiomatic that the chance of
a white face is to the chance of a black face as five to
seven. Now as soon as it is certain that the die is to
be thrown, it is certain that either a white or a black
face must turn up. The two chances must therefore
80 CHANCE.
now make up unity. But they still retain the ratio of
five to seven, therefore they become respectively
5 7
777 and IT?'
And in the same way we might reason if the numbers
were any other.
Question. — A party of twenty-three persons take
their seats at a round table ; shew that the odds are ten
to one against two specified persons sitting together.
Answer. — Call the two specified persons A and B.
Then besides A's place (wherever it may be) there
are twenty- two places, of which two are adjacent to
J.'s place and the other twenty not adjacent. And B
is equally likely to be in any of these twenty-two
places. Therefore (Kule III.), the odds are twenty to
two, or ten to one, against his taking a place next to A.
The last rule may be expressed in a somewhat
different form as follows : —
RULE IV.
If there be a number of events of which one must
happen and all are equally likely, and if any one of
a ( 'smaller ) number of these events will produce a
certain result which cannot otherwise happen, the
probability of this result is expressed by the ratio of
of this smaller number to the whole number of events.
FUNDAMENTAL RULE. 81
For instance ; if a man has purchased five tickets in
a lottery, in which there are twelve tickets altogether
and only one prize, his chance of the prize would he
expressed by the ratio 5 : 12, or by the fraction >T62-.
For convenience of reference we have given distinct
numbers to the two Rules III. and IV., although they
are only different statements of one and the same
principle. This will be immediately seen, by con
sidering the case of the lottery just instanced. We
might at once have said that there were twelve ways
of drawing a ticket, and five of these would cause the
man to win, while the other seven would cause him to
lose. Rule III. is therefore immediately applicable.
Question. — The four letters a, e, m, n are placed in a
row at random : what is the chance of their standing in
such order as to form an English word ?
Answer. — The four letters can stand in |4 or
twenty-four different orders (Choice, Rule III.) : all are
equally likely and one must occur. And four of these
will produce an English word —
mane, mean, name, amen.
Hence by the rule, the required chance is -^ or -6 .
Question. — What is the chance of a year, which is
not leap year, having fifty-three Sundays ?
Answer. — Such a year consists of fifty-two complete
weeks, and one day over. This odd day may be any of
G
82 CHANCE.
the seven days of the week, and there is nothing to
render one more likely than another. Only one of
them will produce the result that the year should
have fifty -three Sundays. Hence (Kule IV.), the
chance of the year having fifty-three Sundays is ^ .
Question. — What is the chance that a leap year,
selected at random, will contain fifty-three Sundays?
Ausiver. — Such a year consists of fifty-two complete
weeks, and two days over. These two days may be
Sunday and Monday,
Monday and Tuesday,
Tuesday and Wednesday,
Wednesday and Thursday,
Thursday and Friday,
Friday and Saturday,
Saturday and Sunday,
and all these seven are equally likely. Two of them
(the first and last) will produce the required result.
Hence (Rule IV.) the chance is 7.
Question. — What is the chance that a year which is
known not to be the last year in a century should be
leap year?
Answer. — The year may be any of the remaining
ninety-nine of any century, and all these are equally
likely ; but twenty-four of them are leap years. There
fore (Rule III.) the chance that the year in question is
, . 24 8
a leap year is or ^3- .
EXAMPLES OF RULE IV. 83
Question. — Three balls are to be drawn from an urn
which contains five black, three red, and two white
balls. What is the chance of drawing two black balls
and one red?
Ansiver. — Since there are ten balls altogether, three
balls can be drawn in T~Z~, or 120 different ways,
1.2.3
all equally likely. Now, two black balls can be selected
in -— , or ten ways, and one red in three ways, Hence,
1.2
two black balls and one red can be drawn in 10 x 3,
or 30 different ways. Thus we have 120 different
ways of drawing three balls, whereof 30 ways will
give two black and one red. Hence, when three balls
are drawn the chance that they should be two black and
one red is (by Kule IV.)
30 1
120 °r ?
Question. — If from a lottery of thirty tickets, marked
1, 2, 3, &c., four tickets be drawn, what is the chance
that those marked 1 and 2 are among them ?
Ansiver. — Four tickets can be drawn out of thirty
30.29.28.27
m — ways. Four tickets can be drawn, so
1.2.3.4
28 27
as to include those marked 1 and 2, in — '- — ways.
84 CHANCE.
Hence, when four are drawn, the chance that these two
are included is
28.27 30.29.28.27 3.4 _2_
1.2 1.2.3.4 ~ 29.30 ~ 145*
The odds are, therefore, 143 to 2 against the event.
Question. — A has three shares in a lottery where
there are three prizes and six blanks. B has one
share in another, where there is but one prize and two
blanks. Shew that A has a better chance of winning
a prize than B, in the ratio of 16 to 7.
Answer. — A will get a prize unless his three tickets
all prove blank. Now, three tickets can be selected
9.8.7
in , or 84 ways; and they can be selected so as
A..ZI.O
£%. K A
to be all blank in r1-^, or 20 ways. Hence the
1.2.3
20 5
chance that they should be all blank is — or — ; and,
84 21
therefore, the chance that this should not be so, or
5 16
that A gets at least one prize, is 1 - — , or — . But
21 21
it is evident that the chance that B gets a prize is
1 7
(Rule IV.) ~- or — . Therefore, A has a better chance
O 21
than B in the ratio of 16 to 7.
Question. — If four cards be drawn from a pack,
what is the chance that there will be one of each suit ?
EXAMPLES. 85
- Answer. — Four cards can be selected from the pack
52 51 50 49
in — •' - or 270725 ways (Choice, Rule IX.); but
1.2.3.4
four cards can be selected so as to be one of each suit
in only 13 x 13 x 13 x 13 or 28561 ways (Choice,
Rule II.). Hence the chance is
28561 1
270726
Question. — If four cards be drawn from a pack,
what is the chance that they will be marked one, two,
three, four?
Answer. — There are 4x4x4x4, or 256 ways
of drawing four cards thus marked, and 270725 ways
of drawing four cards altogether. Hence, the chance is
256
270725'
or the odds are more than 1000 to 1 against it.
Question. — In a bag there are five white and four
black balls. If they are drawn out one by one, what
is the chance that the first will be white, the second
black, and so on alternately?
Answer. — There are nine balls, five of one sort, four
of another : they can, therefore, be arranged in
iff " ""
different orders (Choice, Rule VII.). The balls are
86 CHANCE.
equally likely to be drawn in any of these orders ;>
therefore, the chance that they should be drawn in
the particular order, white — black — white — c£c., is r~~m
That this order of colour corresponds to only one of
the 126 arrangements is a direct consequence of our
having disregarded all individuality among balls of the
same colour when we calculated that number. (See
Choice, page 33.).
Question. — In a bag are five red balls, seven white
balls, four green balls, and three black balls. If they
be drawn one by one, what is the chance that all the
red balls should be drawn first, then all the white ones,
then all the green ones, and then all the black ones ?
Answer, — The nineteen balls can be arranged in
119
different orders (Choice, Rule VII.) • All these are
equally likely, and therefore the chance of any par
ticular order is
This will be the chance required, for all individuality
among balls of the same colour has been disregarded ;
only one of the different arrangements will give the
order of colours prescribed in the question.
Question. — Out of a bag containing 12 balls, 5 are
drawn and replaced, and afterwards 6 are drawn. Find
THROWS WITH TWO COMMON DICE. 87
the chance that exactly 3 balls were common to the
two drawings.
Answer. — The second drawing could be made alto
gether in
112
!6>OT924
ways. But it could be made so as to include exactly
3 of the balls contained in the first drawing, in
|5 J7
or 350
|3.|2
ways ; for it must consist of a selection of 3 balls out
of the first 5, and a selection of 3 balls out of the
remaining 7 (Choice, Rules VIII. and II.). Hence, the
chance that the second drawing should contain exactly
350 25
3 balls common to the first, is -^ or -^.
As the respective probabilities of various throws,
with two common dice, are of practical interest, in their
bearing upon such games as Backgammon, it may be
well to discuss this case with some completeness.
It will be observed that as each die can fall in six
ways, the whole number of ways in which the two dice
can fall is 6 x 6 or 36. But these 36 different ways
are not practically different throws, since, for example,
it makes no difference in practice whether the first die
falls six and the second five, or the first five and the
second BIX. The number of practically different throws
88 CHANCE.
is, in fact, only 21, the 36 different ways of the dice
falling being made up of six unique ways —
landl, 2 and 2, 3 and 3, 4 and 4, 5 and 5, 6 and 6,
and 30 other ways, consisting of 15 essentially different
throws, each repeated twice : thus —
1 and 2, 1 and 3, 1 and 4, 1 and 5, 1 and 6,
2 and 1, 3 and 1, 4 and 1, 5 and 1, 6 and 1,
2 and 3, 2 and 4, 2 and 5, 2 and 6,
3 and 2, 4 and 2, 5 and 2, 6 and 2,
3 and 4, 3 and 5, 3 and 6,
4 and 3, 5 and 3, 6 and 3,
4 and 5, 4 and 6,
5 and 4, 6 and 4,
5 and 6.
6 and 5.
Since each die is equally likely to fall in all different
ways, the 36 different ways of the two dice falling are
all equally likely; and, therefore, when the dice are
thrown the probability of any particular way is -^>
But it cannot be said that all throws are equally pro
bable, because six-Jive results practically in two ways
out of the 36 ways of the dice falling, whereas
six-six results in only one way. The correct state
ment is, that the probability of any assigned throw
TWO COMMON DICE. 89
is g6 if that assigned throw be doublets; but it is twice
as much or -^ if the assigned throw be not doublets.
Thus the chance of throwing six-three is ^, but the
chance of throwing three-three is -^r.
Question. — When two dice are thrown, what is the
chance that the throw will be greater than 8 ?
Ansiver. — Out of the 36 ways in which the dice can
fall, there are six which give a result greater than 8,
viz. : —
5 and 4, 5 and 6, 5 and 5, A
4 and 5, 6 and 5, 6 and 6.
Hence the required chance is-g^~or ~6.
Question. — What is the chance of throwing at least
one ace ?
Answer. — Of the thirty-six ways in which the dice
can fall, eleven give an ace. Hence, the chance is ^Q.
Question. — What is the chance of making a throw
which shall contain neither an ace nor a six?
Answer. — Of the thirty- six ways, there are sixteen
which involve neither one nor six. Hence, the chance
. 16 4
IS -gg 01' g.
This question, as well as the preceding one, may be
more conveniently solved by Rule VI.
Question. — What are the odds against throwing
doublets ?
90 CHANCE.
Answer. — Of the thirty-six ways in which the dice
can fall, six give doublets. Therefore, the chance for
doublets is -^ or g, and the chance against doublets
Q (Kule III.). Therefore, the odds are five to one
against doublets.
Or we might reason thus: — However the first die
fall, the second die can fall in six ways, of which only
one way will give the same number as on the first die.
Hence, the odds are five to one against the second die
falling the same way as the first, or the odds are five to
one against doublets.
Question. — In one throw with a pair of dice, what
is the chance that there is neither an ace nor doublets ?
Answer. — The dice can fall in thirty-six ways, but in
order that there may be neither an ace nor doublets, the
first die must fall in one of five ways (viz. 2, 3, 4, 5, 6),
and the second, since it may be neither an ace nor the
same as the first, may fall in four ways. Hence, the
number of ways which will produce the required result,
is 5 x 4 or 20. And, therefore, the chance of this
,, . 20 5
result is ijg- or ^.
Question. — What is the chance of throwing exactly
eleven ?
Answer. — Out of the thirty-six ways, there are two
ways which produce eleven ; therefore, the chance
. 2 i
18 0r'
TWO COMMON DICE. 91
On the principle of the last answer, the reader will
have no difficulty in verifying the following statements :
In a single throw with two dice, the odds are —
35 to 1 against throwing 2,
17 to 1 „ „ 3,
11 to 1 „ „ 4,
8 to 1 „' „ 5,
6i- to 1 „ „ 6,
5 to 1 „ „ 7,
6i to 1 „ „ 8,
8 to 1 „ „ 9,
11 to 1 „ „ 10,
17 to 1 „ „ 11,
35 to 1 „ „ 12.
Thus the most frequent throw will be seven.
In some cases the purpose of a throw is equally
answered, whether an assigned number appear on one
of the dice, or whether it be the numbers of the two
dice together make it. Let us consider, for example,
the chance of throwing five in this way.
The chance of making a throw so that one die shall
turn up Jive is -^, and the chance of making a throw
which shall amount to Jive is -^r. Therefore the chance
of throwing five in one of these ways is -^ + -gg- or -gg.
On this principle the following statements may be
easily verified.
92 CHANCE.
In a single throw with two dice, ivhen the player
is at liberty to count either the sum of the numbers
on the tivo dice, or the number on either die alone,
the odds are —
25 to 11, against throwing 1,
24 to 12, or 2 to 1 „ „ 2,
23 to 13, „ „ 3,
22 to 14, or 11 to 7 „ „ 4,
21 to 15, or 7 to 5 „ „ 5,
20 to 16, or 5 to 4 „ „ 6,
5 to 1 „ „ 7,
61 to 1 „ „ 8,
8 to 1 „ „ 9,
11 to 1 „ „ 10,
17 to 1 „ „ 11,
35 to 1 „ „ 12.
Thus the number which there is the greatest chance
of making is six.
DEFINITION. — If a person is to receive a prize on
condition of some event happening, the sum of money
for which his chance might equitably be sold before
hand is called his expectation from the event.
RULE V.
The expectation from any event is obtained by
multiplying the sum to be realised on the event hap
pening, by the chance that the event ivill happen.
EXPECTATION. 93
This rule may be illustrated as follows: Suppose a
person holds five tickets in a lottery, where the whole
numher of tickets is twelve ; and suppose there be only
one prize, and let its value be one shilling.
The person in question gains the prize, if it happen
that one of his tickets be drawn. The chance of this
event is -^ ; therefore, according to the rule, the per
son's expectation is -^ of a shilling, or five-pence. And
the correctness of this result may be immediately seen ;
for we observe, that if the person had bought all the
twelve tickets he would have been certain of winning
a shilling, and, therefore, he might, equitably, have
given a shilling for the twelve tickets ; but all the
tickets are of equal value, and are equally valuable
whether the same man hold one or more. Hence, each
of them is worth a penny, and, therefore, the five in
question are worth five-pence (as long as it is unknown
which is drawn). Five-pence, therefore, is the sum
that might equitably have been given for the assigned
person's chance, and, therefore, by the definition this
is his expectation.
Question. — A bag contains a £5 note, a ^610 note,
and six pieces of blank paper. What is the expectation
of a man who is allowed to draw out one piece of
paper ?
94 CHANCE.
Answer. — Since there are eight pieces of paper the
probability of his drawing the £5 note is g ; therefore,
his expectation from the chance of drawing this note
is g- of £5} or ^ of a pound. Similarly, his ex
pectation from the chance of drawing the ^610 note
is g of <£10, or ^ of a pound. Therefore, his whole
expectation is g- of a pound, or £1 17s. 6d.
Question. — What is the expectation of drawing a
coin from a bag which contains one sovereign and seven
shillings ?
Anstver. — The expectation from the chance of
drawing the sovereign is g of a sovereign, and the
expectation from the chance of drawing a shilling is g
of a shilling. Hence, the whole expectation is 3s. 4Jd.
Question. — A person is allowed to draw two coins
from a purse containing four sovereigns and four
shillings. What is the value of his expectation ?
n rj
Ansiver. — Two coins can be drawn in ~- or 28
1.2
ways : of these --- or 6 ways will give two sovereigns,
1.2
4 x 4 or 16 ways will give a sovereign and a shilling,
and the other 6 ways will give two shillings.
CONCURRENT EVENTS, 95
Therefore—
f\
Chance of drawing 40 shillings = gg-,
Chance of drawing 21 shillings = ||-,
Chance of drawing 2 shillings = -^.
Therefore the expectation is —
6 60
from the first chance, — x 40, or — shillings ;
28 i
"i £?
from the second chance, — x 21, or 12 shillings ;
28
fi o
from the third chance, — x 2, or - shillings.
2o i
nr\ o
Hence the whole expectation is — + 12 + -, or 21
shillings ; or one-fourth of the whole sum in the hag.
This result might have been inferred at once from
the consideration that, if all the eight coins had heen
drawn two and two, no drawing could be more likely to
exceed in sovereigns than in shillings : (the number of
sovereigns and shillings being the same). Hence the
expectation from each of the four drawings must be the
same ; and therefore each must be one fourth of the
whole sum to be drawn.
EULE VI.
The chance of two independent events both hap
pening, is the product of the chances of their happening
severally.
90 CHANCE.
That is, if the chance of one event happening be 6>
and the chance of another independent event happening
be g-, the chance that both events should happen is
5 7 35
6^ •< 8 Or 48"-
This may be proved as follows : —
The chance of the first event is the same as the
chance of drawing white from a bag containing six
balls, of which five are white (Kule IV.)
The chance of the second event is the same as the
chance of drawing white from a bag containing eight
balls, of which seven are white.
Therefore the chance that both events should happen
is the same as the chance that both balls drawn should
be white.
But the first ball can be drawn in six ways, and the
second in eight ways. Therefore (Choice, Kule I.),
both can be drawn in 6 x 8, or 48 ways.
So the first can be white in five ways, and the second
can be white in seven ways. Therefore both can be
white in 5 x 7, or 35 ways.
That is, the two balls can be drawn in forty- eight
ways (all equally likely), and thirty-five of these ways
will give double-white. Hence (Eule IT.) the chance of
double-white is -^r, and therefore the chance of the two
or
given events both happening is -^.
And the same reasoning would apply if the numbers
were any others. Hence the rule is true always.
CONCURRENT EVENTS. 97
EXAMPLE. — Suppose it is estimated that the chance
that A can solve a certain problem is g , and the chance
that B can solve it is -^ ; let us consider what is the
chance of the problem being solved when they both try.
The problem will be solved, unless they both fail.
Now the chance that A fails is 3 : and the chance
that B fails is -^ .
Therefore the chance that both fail is
177
v f\v
3 12' 36*
The chance that this should not be so, is
1 7 >r 29
86' a86-
This is, therefore, the chance that the problem gets
solved.
In the case just considered, four results were possible,
viz. : —
(1) That A and B should both succeed :
(2) ,, A should succeed and B fail :
(3) ,, A should fail and B succeed :
(4) „ A and B should botfh fail.
We may calculate the chance of these four events
separately. Thus we have
2 1
Chance of A's success = ~ , of A'& failure = 5 ;
o o
5 7
,, I>'s success = r^, of Z>'s failure = ^ •
98
CHANCE.
Therefore, by the rule
(1) Chance that A and B both succeed
= 2 x 5 ._ 10.
3 12 36 '
(2) Chance that A succeeds and B fails
= 2 x 7 ._ 14.
3 12 36 '
(3) Chance that A fails and B succeeds
= l > A - A .
3 12 36 *
(4) Chance that A and B both fail
1 x 7 = -7
3 12 36*
We observe that
10 + 14 + * + 7 = 36 =
36 36 36 36 36 "
or the sum of the four probabilities is unity, as it ought
to be, since it is certain that one of the four results
must happen.
Further, we notice that the problem will be solved if
any of the first thr^e events out of (1), (2), (3) and (4)
occur. Hence the chance of the problem being solved,
might have been obtained by adding together the sep
arate probabilities of these three events. Thus —
10 14 5 = 29
86 86 36 36'
or the probability is -^r , as before.
EXPECTATION OF I^IFE, 99
It may be said that on an average
Ten persons will die in the next ten years :
out of every 62 whose present age is 30,
„ » 45 „ „ 40,
„ 35 „ „ 50,
„ 25 „ „ 60.
We may apply such results as these to the solution
of questions affecting Insurances and Life Annuities.
Question. — What are the odds against a person aged
thirty living till he is sixty ?
Answer. — The chance that he dies between thirty
and forty is -^ ; that he lives to forty and dies between
forty and fifty is -^ x — ; that he lives to fifty and dies
between fifty and sixty is || x || x ™ . Therefore the
chance that he dies between thirty and sixty is
10 52 10 52 35 10 149
62 + 62 * 45 + 62 ' 45 ' 35' C L' 279 '
Hence the odds are 149 to 130, or about 8 to 7
against his living to be sixty.
Question. — What are the odds against a person at
the age of forty living for thirty years?
Answer. — Proceeding as in the last question, we find
the chance of his dying within thirty years to be
10 35 10 35 25 10 Qr _2_
45 45 ' 35 45* 35 ' 25' 3 *
100 CHANCE.
Therefore the odds are two to one against his living
for thirty years.
Question. — What is the probability that two persons,
A and B, aged respectively thirty and forty, will be
alive ten years hence ?
Answer. — The chance of A dying in the next ten
years is -^ , and the chance of his living -— . So the
chance of B dying within ten years is ^ , and the
chance of his living is -jg .
Therefore the chance that A and B will be both
alive is
52 85 182
62 45 ' r 279 '
Question. — If it be eight to seven against a person
who is now thirty years old living till he is sixty, and
two to one against a person who is now forty living till
he is seventy ; find the probability that one at least of
these persons will be alive thirty years hence.
Answer. — One at least will be living unless both be
dead. The chance that the first be dead is y6, and the
chance that the second be dead is 3 : therefore the
chance that both be dead is ^ x -_ , or .,- , and the
chance that this should not be so, or that one at least
16
4y
, ,. . -, 16 29
be alive is 1 - , or
TWO CONCURRENT EVENTS NOT INDEPENDENT. 101
RULE VII.
If there be two events which are not independent, the
chance that they should both happen is the product of
the chance that the first should happen, and the chance
that when the first has happened the second should
happen also.
For instance, suppose we are asked what is the
probability of drawing first a consonant and then a
vowel, when two letters are drawn at random out of
an alphabet of twenty consonants and six vowels.
The second event is dependent on the first ; for if a
consonant be drawn the first time, there are twenty-five
letters left, of which six are vowels, and the chance that
the second letter should be a vowel is -^; but if a
vowel be drawn the first time, there are twenty-five
letters left, of which five are vowels, and the chance
that the second letter should be a vowel is -^ .
According to the rule, however, we have to multiply
the chance of the first event, which is -g, by the
chance of the second event happening when the first
has already happened, which is therefore -~ , and thus
we obtain the result —
20 6 12
26 25' Cl> 65*
The truth of this result may be seen in another way.
It is possible to select two letters in order, out of the
alphabet, in 26 x 25 ways (Choice, Rule IV.), and all '
102 CHANCE.
these are equally likely. But we can select two letters
so that the first is a consonant and the second a vowel
in only 20 x 6 ways (Choice, Kule I). Hence when
two letters are drawn in order, the chance that the first
is a consonant and the second a vowel is (as before)
20 x 6 12
/yw
26 x 25' 65*
Indeed it will appear that this rule follows directly
from the preceding one ; for, since we have only to find
the chance that both events should happen, we have not
to do with the second event at all, except in the case
when the first has happened. The probability of the
double event must therefore be the same as if the
chance of the second were always what it is when the
first has happened (since we are not concerned with the
case when the first has not happened). But if the
chance of the second event can be treated as if it were
always the same, without reference to the first event, it
is to all intents and purposes independent of the first,
and Eule VI. is therefore applicable.
Question. — One purse contains five sovereigns and
four shillings ; another contains five sovereigns and
three shillings. One purse is taken at random and a
coin drawn out. What is the chance that it be a
sovereign ?
Answer. — The chance that the first purse be selected
is ) and if it be selected, the chance that the coin be
EXAMPLES OF CONCURRENT EVENTS. 103
a sovereign is 9 : hence the chance that the coin drawn
out be one of the sovereigns out of the first purse is
Similarly the chance that it be one of the sovereigns
out of the second purse is
1 5 5_
2 ' 8 ' 16 '
Hence the whole chance of drawing a sovereign is
5 £ 5 or 85
18 ' 16 ' 144 '
Question. — What is the expectation from the drawing
of the coin in the last question ?
Qt
Answer. — The chance that it is a sovereign is :TTT,
and therefore the expectation from the chance of draw-
. . 85 f n 1700 ,.,,.
ing a sovereign is ^ of a pound, or -^ shillings.
If the coin drawn be not a sovereign, it must be a
shilling, therefore the chance of drawing a shilling
must be 1 - ^ , or ^ (Kule . I.) Hence the expecta
tion from the chance of drawing a shilling is ^ of a
shilling. Therefore the whole expectation from the
drawing is
1700 , j>9 1759
144 r 144' C 144
shillings, or 12s. 2T\d.
104 CHANCE.
Question. — What would have been the chance of
drawing a sovereign if all the coins in the last case
had been in one bag, and what would have been the
expectation ?
Ansiver. — There would have been ten sovereigns
and seven shillings in the bag; therefore, the chance
of drawing a sovereign would have been ^-, and the
chance of drawing a shilling -~ (Rule I.) The expec-
pectation would therefore have been
200 7 207
TT + 17 or-ir
shillings, or 12s. 2T2Td.
The chance of drawing a sovereign is therefore in
this case a little less, and the whole expectation very
slightly less than in the former case.
Question. — There are three parcels of books in
another room, and a particular book is in one of them.
The odds that it is in one particular parcel are three to
two ; but if not in that parcel, it is equally likely to be
in either of the others. If I send for this parcel, giving
a description of it, and the odds that I get the one I
describe are two to one, what is my chance of getting
the book ?
Answer. — The chance of getting the parcel described
is |, and the chance that the book is in it is 5 ;
EXAMPLES Otf CONCUKRENT EVENTS. 105
therefore, the chance of getting the book in the
described parcel is g x ^ or -^ .
The chance of getting a parcel not described is
£-, and the chance that the book is in it is ^ ;
therefore, the chance of getting the book in a parcel
not described is ^ x - , or -^ ...
Therefore, the, whole chance of getting the book at
all is -jg- + -jg , or -^ ; or the odds are eight to seven
against getting it.
Question. — In a purse are ten coins, of which nine
are shillings and one is a sovereign; in another are
ten coins, all of which are shillings. Nine coins are
taken out of the former purse and put into the latter,
and then nine coins are taken from the latter and
put into the former. A person may now take which
ever purse he pleases ; which should he select ?
Answer. — Since each purse contains the same num
ber of coins, he ought to choose that which is the more
likely to contain the sovereign. Now the sovereign can
only be in the second bag, provided both the following
events have taken place, viz. —
(1) That the sovereign was among the nine coins
taken out of the first bag and put into the
second.
(2) That it was not among the nine coins taken
out of the second bag and put into the first.
106 CHANCE.
Now the chance of (1) is j^, and when (1) has hap
pened the chance of (2) is ||j-; therefore, the chance of
both happening is -^ x -^ , or -^ . This, therefore
is the chance that the sovereign is in the second bag,
and therefore (Eule I.) the chance that it is in the
first is 1 - — or -J-. Hence, the first bag ought to
be chosen in preference to the other.
EULE VIII.
The chance that a series of events should all happen
is the continued product of the chance that the first
should happen, the chance that (when it has happened}
then the second should happen, the chance that then the
third should happen, and so on.
This is a simple extension of the last rule. For
suppose there be four events, and let ^ be the chance
that the first should happen, and when the first has
o
happened, let -g be the chance that the second should
happen, and when these have happened, let g be the
chance that the third should happen, and when these
have happened, let -g be the chance that the fourth
should happen ; by Rule VII., the chance that the first
-JO o
and second should both happen is 2 x 4 , or g . We
SERIES OF CONCURRENT EVENTS. 107
iay now treat this as a single event, and then, again
applying the same rule, we get | x |, or ||- as the
chance that the first, second, and third should all
happen. Treating this compound event as one event,
15
64
we can again apply the same rule, and obtain -«j x -g.
or 25(3 as the chance that all the four events should
happen. Thus the chance of all the events is
1351
2 X 4 X 8 X V
the continued product of all the given chances.
Question. — There are three independent events whose
231
several chances are 3 , g , ^ . What is the chance
that one of them at least will happen ?
Answer. — One at least will happen, unless all fail.
121 1
The chance of all failing is ^ X ^ x ~, or -^ .
d O A J.O
1 14
Hence the required chance is 1 - — , or -— .
15 15
Question. — There are three independent events
whose several chances are 3, ^, ^. What is the
chance that exactly one of them should happen ?
Answer. — The chance that the first should happen
and the others fail is
__ __ __ __
"3" x y x y or so
108 CHANCE.
So the chance that the second should happen and the
others fail is
A x 1 x 1, or A.
5 3 2 30
And the chance that the third should happen and the
others fail is
1 x 1 x 1, or A.
2 3 5 30
Hence, the chance that one of these should occur —
that is, that exactly one of the three events should
happen — is
432 9 3
- + - + — , or — , or — .
30 30 30 30 10
Question. — When six coins are tossed, what is the
chance that one, and only one, will turn up head ?
Answer. — The chance that the first should turn up
head is g, and the chance that the others should turn
up tail is 2 for each of them. Therefore, the chance
that the first should turn up head and the rest tail is
JLX !x !x Ixlxl, orX
2 2 2 2 2 2 64
And there will be a similar chance that the second
should alone turn up head, or that the third should
alone turn up head, and so on.
EXAMPLES. 109
Hence, the whole chance of some one, and only one,
turning up head is
1_ + JL JL JL + JL + JL or~
64 64 64 64 64 64* 64*
Question. — When six coins are tossed, what is the
chance that at least one will turn up head ?
Answer. — The chance that all should turn up tail is
— x — x — x — x — x JL , or — .
2 2 2 2 2 2 64
The chance that this should not he so, or that at least
one head should turn up, is (Kule II.)
1 63
Question. — A person throws three dice, what are the
respective chances that they should fall all alike, only
two alike, or all different ?
Ansiver. — The chance that the second should fall
the same as the first is g, and the chance that the
third should also fall the same is g. Hence, the
chance that all three fall alike is
1 x A, or 1.
6: 6 36
The chance that the second should fall as the first,
and that the third should fall different, is
155
x or — ;
6 6 36
110 CHANCE.
and there is the same chance that the second and third
should be alike, and the first different ; or that the first
and third should he alike, and the second different.
Hence, the chance that some two should he alike, and
the others different, is
A + A + A, or I5.
36 36 36 36
The chance that the second should be different from
the first is g, and the chance that the third should be
different from either is g. Hence, the chance that all
three are different is
, .
6 6 86
Therefore, the three chances required are -^, -^r, -^
respectively, their sum being unity, since the dice must
certainly fall in some one of the three ways.
Question. — A person throws three dice, and is to
receive six shillings if they all turn up alike, four
shillings if two only turn up alike, and three shillings
if all turn up different, what is his expectation ?
Answer. — Referring to the last question, the chance
of all turning upjjjgffgsnt is -^ ; his expectation from
this event is therefore -^ of six shillings, or two pence.
The chance of two only turning up alike is gg- or -^
and his expectation from this event is therefore -jg-
NOTATION. Ill
of four shillings, or twenty pence. The chance of all
turning up different is -^ or <,, and his expectation
from this event is therefore 9 of three shillings, or
twenty pence. Therefore his whole expectation is
2 + 20 + 20, or 42 pence, or three shillings and
sixpence.
We shall find the following notation very con
venient : —
The symbol 32 means 3 x 3, or 9 :
52 means 5 x 5, or 25 :
53 means 5 x 5 x 5, or 125 :
24 means 2 x 2 x 2 x 2, or 16 :
25 means 2 x 2 x 2 x 2 x 2, or 32 :
and so on, whatever he the numbers ; the small figure
above the line denoting the number of times the other
number is to be repeated, and the sign of multiplication
being understood before every repetition.
2228
'
a
So
and so on.
|4_8X888_81.
- 4 X 4 X 4 4 " 256'
Question. — A person goes on throwing a single die
until it turns up ace. What is the chance (1) that he
will have to make at least ten throws ; (2) that he will
have to make exactly ten throws ?
112 CHANCE.
Answer. — (1) The chance that he fails at any
5
particular trial to throw an ace is -~- . The chance
that he should fail the first nine times (by Rule VIII.)
/5\e
18 ( . This, therefore, is the probability that he
will have to throw at least ten times.
/5\9
(2) Since ~ is the chance that he fails the
first nine times, and -^ the chance that he succeeds
D
/5V 1
the next time, therefore by Rule VII., Igj x ^g"
is the chance that he will have to throw exactly ten
times.
Question. — A die is to be thrown once by each of
four persons, A, B, C, D, in order, and the first of
them who throws an ace is to receive a prize. Find
their respective chances, and the chance that the prize
will not be won at all.
Answer. — Since A has the first throw, he wins if he
throws an ace ; his chance is therefore g.
So B wins provided A fails and he succeeds. The
chance of A failing is ^, and of B succeeding is 6.
Therefore J5's chance of winning is
5 l A
6 X 6' °r 36 '
EXAMPLES. 113
So C wins provided A and B both fail, and he
succeeds. The chance of A and B both failing is
6 x g , or -gg- ; and then the chance of C succeeding
is g . Therefore O's chance of winning is
25 1 j^
36 X 6 ' CT 216 '
So D wins provided A, B, and C all fail, and he
succeeds. The chance of A, B, and C all failing is
g x x ^ , or -g^g- j a*id then the chance of D
succeeding is 6. Therefore Z)'s chance of winning is
125 1 125
216 X 6' >r 1296*
The prize is not won at all, provided all four fail to
throw an ace. The chance that this should be the
case is
5555 625
6666' ' 1296'
Question. — Two persons, A and B, throw alternately
with a single die, and he who first throws an ace is to
receive a prize of £1. What are their respective
expectations ?
Answer. — The chance that the prize should be won
at the first throw, is g :
at the second throw, is x :
114 CHOICE.
at the third throw, is « x
/5V 1
at the fourth throw, is I « 1 x » :
\6/ 6
/5V 1
at the fifth throw, is -5 x ~ :
\6/ 6
/5V 1
at the sixth throw, is [ 5 I x 5 :
\o/ b
and so on.
But the first, third, and fifth, &c., throws belong to
A, and the second, fourth, sixth, &c., belong to B.
Hence A1 a chance of winning is
1 /5V 1 /5V 1 &c .
6 \6j • 6 \6) ' 6
and jB's chance is
5 1 /5V 1 /5V 1 o
6-6 + VBJ '6 + lej '6 + &C';
that is, B's chance is equal to ^.'s multiplied by |.
Hence -B's expectation is g of A'&, or B's is to A' a
in the ratio of five to six. But their expectations must
f*
together amount to £1. Hence A's expectation is ^y
of a pound, and JB's -^- of a pound.
Question. — What is the chance that a person with
two dice will throw aces exactly four times in six trials ?
EXAMPLES. 115
Answer. — The chance of throwing aces at any par
ticular trial is -^ , and the chance of failing is ^-'
Hence the chance of succeeding at four assigned trials,
/ 1 Y /35\2
and failing at the other' two is ^ X ^ . But
\db/ \ow
aces will be thrown exactly four times if they be thrown
at any set of four trials which might be asssigned out
of the six trials, and if they fail at the remaining two.
And (Choice, Kule IX.) it is possible to assign four out
of six in ' ' , or fifteen ways. Hence the chance
1.2.3.4
required is fifteen times the chance of succeeding in
four assigned trials, and failing at the other two.
Therefore it is
1 V x ( 35 V x 15 or 6125
x x Lb or
36 J V 36 J 725594112
therefore the odds are more than 100,000 to 1 against
the event.
Question. — If on an average nine ships out of ten
return safe to port, what is the chance that out of five
ships expected, at least three will arrive ?
Answer. — The chance that any particular ship
Q
returns is ^. The chance that any particular set
/ 9 \3
of three ships should all arrive is 1 .57? ) "j and the
116 CHANCE.
(1 ^ 2
—
Therefore the chance that a particular set of three
/ 9 \3 / 1 \2 729
should alone arrive is , or ---.
And out of five ships a set of three can be selected in
1~2~S or ^ ways. Hence the chance that some one
of these sets of three should alone arrive is
79,Q 79Q
I -^J * * 1,1 I -*-J » '
___ y* TO f)1* B
100000 10000'
This is therefore the chance that exactly three ships
should arrive.
Similarly the chance that any particular set of
/9V 1 6561
four should alone arrive is 1 ^ x ^ , or -
and the chance that some one of the five possible sets
, ,, . 6561 , K 32805
of four should alone arrive is ^^^ x 5 , or --
100000 100000 '
This is therefore the chance that exactly four ships
should arrive.
And the chance that all the five should - arrive is
9_\* 59049
16; ' r 100000 '
But the chance that at least three should arrive is
the chance that either three exactly, or four exactly,
or five exactly should arrive : and is therefore the
SKILL IN GAMES. 117
sum of the several chances of these exact numbers
arriving : that is, the required chance is
7290 32805 59049 99144 12393
100000 100000 100000 ' °r 100000 ' °" "12500 *
Question. — A and B play at a game which cannot be
drawn, and on an average A wins three games out of
five. What is the chance that A should win at least
three games out of the first five ?
Answer. — The chance that A wins three assigned
/3\3 /2\2 108
games, and B the other two, is I -= } • {-„} or ^77?.
\O/ w/ dl^O
But the three may be assigned in j^g > or 1° wavs
(Choice Rule IX.). Hence the chance that A should
win some three games and B the other two is
108 ,0 1080
3125 3125'
Similarly the chance that A should win some four
games and B the other one is
162 x 5 or 81°
3125 3125 '
And the chance that A should win all five games is
243
5J ' °r 3125
118 CHOICE.
Therefore the chance that A wins either three, or four,
or all out of the first five games is
1080 810 243 ._ 2133
3125 " 3125 3125 " 3l25 '
or the odds are rather more than two to one in J.'s
favour.
EULE IX.
If a doubtful event can happen in a number of differ
ent ways, any accession of knowledge concerning the
event which changes the probability of its happening
will change, in the same ratio, the probability of any
particular way of its happening.
It follows from the axiom that the probability of the
event happening at all must be equal to the sum of the
probabilities of its happening in the several ways.
First, suppose for simplicity that all the ways are
equally likely. Let there be seven ways, and let the
chance of each one severally occurring be ^ : then the
chance of the event happening at all is seven times
this, or ~.
But suppose that our knowledge is increased by the
information that the event happens nine times out of
ten, or by such other information as brings our estimate
of its probability up to ^ instead of -^, thus increasing
the probability in the ratio of seven to nine.
ACCESSION OF KNOWLEDGE. 119
It is still true that there are only seven ways of the
event happening, all of which are equally likely : hence
the probability of the event happening in any particular
one of these ways is ^ of -^, or -^ , with our new in
formation. Hence our information concerning the
event, has increased the chance of its happening in an
179
assigned way from -^ or -^ to -^-, that is, it has in
creased it in the ratio of seven to nine, the same ratio
in which the probability of the event itself was increased.
And the same argument would hold if the numbers
were any others, and therefore the rule is true, provided
all the ways of the event happening are equally probable.
Secondly, suppose the ways are not equally probable.
We may in this case regard them as groups of subsidiary
ways, which would be equally probable. Then, as we
have shewn, the chance of each one of these subsidiary
ways would be increased (or decreased) in the same
ratio as the chance of the event itself, and therefore the
sum of the chances of any group of these subsidiary
ways would be changed in the same ratio.
For instance, if the event could happen in any one
of three ways, whose respective chances were g-, g, ^»
or -jj, -jg-, -jg , we might divide the first of these ways
into four subsidiary ways, the next into two ways, and
the other into three ways, and the chance of each of
these subsidiary ways would be ^ . If» therefore, by
an accession of knowledge, the chance of the whole
event were diminished in the ratio of three to two, each
120 CHANCE.
subsidiary way of the event's happening would have a
diminished probability of g of ^, or -yg-, and the prob
abilities of the three given ways would become respec-
423 211
tively, -jj, -jg, -jg- , or ^, g, g : that is, they would be
diminished in the same ratio as the chance of the event
itself.
Thus we see that the rule is true always.
Question. — A bag contains five balls, which are
known to be either all black or all white — and both
these are equally probable. A white ball is dropped
into the bag, and then a ball is drawn out at random
and found to be white. What is now the chance that
the original balls were all white ?
Answer. — The probabilities are here affected by the
observed event that a ball drawn out at random proved
to be white.
We will first calculate the probabilities before this
event was observed (which we will call a priori proba
bilities), and then consider how they are affected by
the accession of knowledge produced by the observation
of the event. (Probabilities modified by this knowledge
may be distinguished as a posteriori probabilities.)
The event might happen in two ways ; either by the
balls having been all white, and any one of them being
drawn, or by the five original balls having been black,
the new one alone white, and this one drawn.
The a priori probability that all are white is | , and
then the chance of drawing a white ball is 1 (or
EXAMPLE. 121
certainty). Hence the chance of the event happening
in this way is 2 x 1 , or ^ •
So the a priori probability that the first five were
black is ^ > and then the chance of drawing a white
ball is | . Hence the chance of the event happening
in this way is ^ x g > or "^ •
Therefore the whole a priori chance of the event
happening is \ + ^-, or ^-.
But when the ball is drawn and observed to be white,
this knowledge immediately increases the chance from
^ to 1 (or certainty) : that is, it increases the chance
in the ratio of 7 to 12. Therefore, by Rule IX., the
chances of the event happening in the several ways
are increased in the same ratio.
Hence the a posteriori chance of the event having
happened in the first way is ^ x ~Y > or 7 » an<^ ^e
a posteriori chance of its having happened in the
second way is -^ x -y- , or ^ . Or the chance of the
original balls having been all white is now -^ , and the
chance of their having been all black is y.
Question. — A penny is tossed ten times in suc
cession, and always falls head. Supposing that one
penny in every million that are coined has two
heads, what is the chance that the penny in question
has two heads?
122 CHANCE.
Answer. — The penny has either two heads or a head
and a tail : and the respective chances of these two
. . I -, 999999 T ,, n
cases a priori are ^^ and 1000000 . In the first
case the chance that head should fall ten times in
succession is unity; in the second case it is (--) , or
•jo^-. Therefore we have a priori, —
(1) the chance that there should be two heads,
and head should fall ten times is -100QOOO :
(2) the chance that there should be head and
tail, and head should fall ten times is
and the chance that from one or
1024000000 *
other the same result should happen is
1 999999 1001023
1000000 1024000000' 1024000000
But after our knowledge is augmented by the obser
vation of the fact that the penny falls head ten times
in succession, this latter chance becomes unity, that
., T ii- T j i 1024000000 TT /-n ,
is, it becomes multiplied by 1001093 - • Hence (Rule
IX.) the chances (1) and (2) become multiplied in the
same ratio. Therefore we have a posteriori, —
(1) the chance that there should be two heads,
and head should fall ten times is
1024000000 1024
lAniAOO 9
1000000 1001023 1001023'
EXAMPLES OF AUGMENTED KNOWLEDGE. 123
(2) The chance that there should be head and
tail, and head should fall ten times is
999999 1024000000 999999
1024000000 1001023 1001023*
The required chance (after the observed event) is
consequently
1024 1
, or rather more than --7^ .
1001023' " 1000
Question. — A purse contains ten coins, each of
which is either a sovereign or a shilling : a coin is
drawn and found to be a sovereign, what is the chance
that this is the only sovereign ?
Answer. — A priori, the coin drawn was equally
likely to be a sovereign or a shilling, therefore the
chance of its being a sovereign was -^ .
A posteriori, the chance of its being a sovereign is
unity: or the chance is doubled by the observation of
the event. Therefore (Rule IX.) the chance of any
particular way in which a sovereign might be drawn
is also doubled.
Now the chance that there was only one sovereign
was a priori
10
or
1024 '
and in this case the chance of drawing a sovereign
would be - .
124 CHANCE.
Hence the chance that there should be only one
sovereign, and that it should be drawn was a priori
jo_ x A J_
1024 10' 1024*
And the a posteriori chance that a sovereign should
be drawn in this way is the double of this : i. e.,
or 5l2 » w^cn is therefore the required chance.
Question. — A purse contains ten coins, which are
either sovereigns or shillings, and all possible numbers
of each are equally likely : a coin is drawn and found to
be a sovereign, what is the chance that this is the only
sovereign ?
Answer. — A priori, the coin drawn was equally
likely to be a sovereign or a shilling, therefore the
chance of its being a sovereign was ^ •
A posteriorly the chance of its being a sovereign is
unity : or the chance is doubled by the observation of
the event.
Therefore (Rule IX.) the chance of any particular
way in which a sovereign might be drawn is also
doubled.
Now, a priori, eleven cases were equally probable,
viz., that there should be
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 sovereigns.
10, 9, 8, 7/6, 5, 4, 3, 2, 1, 0 shillings.
Therefore the chance of there being exactly one
EXAMPLES. 125
sovereign was ^-, and in this case the chance of drawing
i
a sovereign was -^ .
Hence the chance that there should be only one
sovereign, and that it should be drawn was, a priori,
ii x 16' or no *
And the a posteriori chance that a sovereign should
be drawn in this way is the double of this, that is,
^ or -gg-, which is, therefore, the chance required.
Question. — Keference is made to a year which con
tained fifty- three Sundays, and was not the last year of
a century. What are the odds against its being a leap-
year?
Answer. — Of the ninety-nine years, excluding the
last in any century, twenty-four are leap-years. Hence,
before we consider the fact that the year in question
contained fifty-three Sundays, the a priori chance that
o
it was a leap-year is -^ , and that it was not a leap-
25
year -^ .
The chance that a leap-year has fifty-three Sundays
is 7, and the chance that another year has fifty-three
Sundays is ~ .
126 CHOICE.
Hence the chance that the year in question should he
a leap-year, and have fifty-three Sundays, is
8 2 16
33 X 7 ' r 231 '
And -the chance that it should not he a leap-year, and
yet have fifty-three Sundays, is
25 x * or 25
33 7 ' 231 '
Hence the whole a priori probability that the year
should have fifty-three Sundays, is
!§. 2£ 41_
231 231 ' 231
But a posteriori this chance becomes certainty, or the
231
probability gets multiplied by -^ . Hence also the
probability that the fifty-three Sundays resulted from a
leap-year is multiplied in the same ratio, and becomes
JL6^ 231 16
231 41 41 '
and the chance that it is not a leap-year becomes
25 x ?31 or 25
231 41 ' 41 '
Thus the odds are now 25 to 16 against the year
in question being a leap-year.
Question. — A, B, C were entered for a race, and
their respective chances of winning were estimated at
IT' IT' Tf * ^u^ circumstances come to our knowledge
EXAMPLES. 127
in favour of A, which raise his chance to ^; what are
now the chances in favour of B and C respectively ?
Answer. — A could lose in two ways, viz., either by
B winning or by C winning, and the respective chances
of his losing in these ways were a priori -^j and -yj-i
and the chance of his losing at all was -yp But after
our accession of knowledge the chance of his losing
at all becomes ^ *na^ ig> it becomes diminished in
the ratio of 18 : 11. Hence the chance of either way
in which he might lose is diminished in the same
ratio. Therefore the chance of B winning is now
11
18
and of C winning
5 11 5
n x is' or is-
These are therefore the required chances.
Question. — One of a pack of fifty-two cards has
been removed; from the remainder of the pack two
cards are drawn and are found to be spades ; find
the chance that the missing card is a spade.
Answer. — A priori, the chance of the missing card
being a spade is ^, and the chance that then two cards
drawn at random should be both spades is rr^, or
128 CHANCE.
Hence the chance that the missing card should be a
spade, and two spades be drawn is
! x _132_ 1!
4 2550' 850*
The chance of the missing card being not a spade is
o
£, and the chance that then two spades should be
drawn is or <. Hence the chance that the
missing card should be not a spade, and two spades
be drawn, is
3 x 156 or -??-
4 2550' 850'
Therefore the chance that in one way or the other
two spades should be drawn is
11 39 50 1
850 " 850' 850' 17*
But after the observation of the event this chance
becomes certainty, or becomes multiplied by 17.
Therefore the chance of either way from which the
result might occur is increased in the same ratio.
So the chance that the given card was a spade
becomes a posteriori,
* »• - -
Questions as to the credibility of the testimony of
witnesses will depend for their solution upon the last
rule, and may be answered in a manner similar to that
CBEDIBILITY OF TESTIMONY. 129
of the questions just considered. In most questions of
this class, the testimony given, or the assertions made,
constitute a phenomenon which might have occurred
whether the event reported occurred or not, or in what
soever manner it occurred. We may first investigate
the a priori probabilities of such testimony being given,
on the several hypotheses possible with respect to the
occurrence of the event, and by summing them we may
deduce the a priori probability of the testimony being
given at all. If we then take into consideration the
fact that the testimony has been given, this accession
of knowledge raises the last probability into certainty,
and therefore increases it in a definite ratio, which can
be calculated. In the same ratio (by Rule IX.) must
the probabilities be increased of the several ways in
which the testimony may have been generated, or in
which the event in question may have happened. Thus
we obtain the a posteriori and final probability of any
assigned manner in which the event could possibly have
occurred. A few examples will fully illustrate this.
Question. — A speaks truth three times out of four,
B four times out of five ; they agree in asserting that
from a bag containing nine balls, all of different colours,
a white ball has been drawn ; shew that the probability
96
that this is true is -^ .
Answer. — We will consider those chances as a priori,
which are independent of the knowledge that A and B
130 CHANCE.
make the report in question, and those as a posteriori
which are subsequent to this knowledge.
The a priori chance that a white ball should be
drawn is ^ , and in this case the chance .that A and B
should both assert it, is ^ x - ; hence the chance that
A and B should both truly assert a white ball to be
drawn, is a priori,
134 1
9 : 4 : 5'°rl5'
The a priori chance that a white ball should not be
Q
drawn is g, the chance that A should make a false
report is ^ , and that he should select the white ball
out of the eight which might be falsely asserted to have
come up is g ; hence in this case the chance that he
asserts that the white is drawn is ~8 x ^ , and the
chance that B should make the same assertion is g x g I
therefore the chance that A and B should both falselj
assert a white ball to be drawn is
Ixixixlxi, !
98485' 1440'
Consequently the a priori chance that they both assen
either truly or falsely that a white ball should b<
drawn is
JL + 1 . 97
15 1440 ' C " 1440 '
EXAMPLES OF TESTIMONY. 131
But a posteriori this chance becomes certainty, or it
becomes multiplied by -^ . Hence the chance of each
way in which they may make the assertion, is multi
plied in the same ratio. Therefore, a posteriori, the
chance that they make the assertion truly is
_!_ 1440 96 .
15 97 97 '
and the chance that they make it falsely is
1 x 144? or 1
1440 ' 97 97'
Question. — A gives a true report four times out of
five, B three times out of five, and C five times out of
seven. If B and C agree in reporting that an experi
ment failed which A reports to have succeeded, what is
the chance that the experiment succeeded ?
Ansiver. — The chance that the given reports should
be made upon the experiment having succeeded is
tl 4 2 2 16
v — v v or
2 " 5 " 5 '* 1' 350'
The chance that the given reports should be made on
the experiment having failed is
1 1 8 5 . 16_
2 5 5 7' * 350*
The a priori chance that in one way or other the
given reports should be made is
I6 11 or *1
350 350 ' 350 '
132 CHANCE.
But, a posteriori, this is certain, or the chance is
QCA
multiplied by -^ . Hence, also, the chance of each
way in which the reports could be made is multiplied
, 350
bv si-
Therefore, a posteriori, the chance that the experi
ment succeeded is
16^ 350 16
350 X 31 ' °r 31 '
We will conclude this chapter with some illustrations
of the principles of probability, drawn from the game
of whist.
This game is played with a pack of fifty-two cards,
consisting of four suits of thirteen cards, marked differ
ently. The cards are all dealt out to four players,
of whom two and two are partners, so that each has
thirteen cards. One of the dealer's cards is turned
up, and the suit to which this card belongs is called
trumps. Four particular cards in this suit — the ace,
king, queen, and knave— are called honours.
It follows, from Rule IV., that the chance that the
turned up card is an honour is -j^r, and that it is not
an honour is ^-.
Io
Question. — What is the chance that each party in
the game should have two honours?
Answer. — Besides the turned up card, there are
fifty-one cards, of which twenty-five belong to the
GAME OF WHIST. 133
dealer and his partner, and twenty-six to their ad
versaries.
First, — Suppose the turned up card is an honour.
The chance of this is -^. Then the chance that one
other honour should be among the twenty-five, and the
remaining two among the twenty-six, is
3 25 26 25.
' 51 * 50 ' 49 '
therefore the chance that the turned up card should be
an honour, and the honours equally divided, is
o 25 26 25 4_ 100
'51 "50 ' 49 ' 13' r 833*
Secondly, — Suppose the turned up card is not an
Q
honour. The chance of this is -^ • Then the chance
that two of the honours should be among the twenty-
five, and the remaining two among the twenty- six, is
4.3 25 24 26 25 .
1 . 2 ' 51 ' 50 ' 49 ' 48 '
therefore the chance that the turned up card should not
be an honour, and the honours be equally divided, is
6 25 24 26 25 _9^ 225
' ' 51 ' 50 ' 49 ' 48 ' 13 ' r 833 '
Hence the whole chance that the honours should be
equally divided is
100 225 325
833 " 833' 833 '
184 CHANCE.
In the same manner we may write down almost at
sight the chances of the occurrence of other arrange
ments of the cards. We give a few examples : —
1. — If an honour turns up, the respective chances
that the dealer and his partner have between them
exactly one, two, three, or four honours, are respectively
26.25.24 3 25.26.25 3 25.24.26 25.24.23 .
. 51.50.49 ' ' 51.50.49 ' ' 51.50.49 ' 51.50.49 '
312 975 936 276
or
2499' 2499' 2499' 2499'
the sum of these fractions being unity, for one of the
four cases must certainly occur.
2. — If an honour does not turn up, the respective
chances that the dealer and his partner have none, or
exactly one, tivo, three, or four honours, are
26.25.24.23 , 25.26.25.24 6 25.24.26.25
51.50.49.48 ' : " 51.50.49.48' ' ' 51.50.49.48'
25.24.23.26 25.24.23.22
4.
51.50.49.48' 51.50.49.48'
299 1300 1950 1196 253 .
4998 ' 4998 ' 4998 ' 4998 ' 4998 '
the sum of these fractions being unity, for one of the
five cases must certainly occur.
GAME OF WHIST. 135
3. — Before it is known whether an honour will turn
up, the respective chances that the dealer and his
partner have between them none, or exactly one, two,
three, or four honours, are
207 1092 1950 1404 _845_ .
4998' 4998' 4998' 4998' 4998'
the sum of the fractions being unity.
Hence, speaking approximately, we may expect that
on the average, for every one hundred times the cards
are dealt, the dealer and his partner will have four
honours seven times, and the other players four times.
The dealer and his partner will have three honours
twenty-eight times, and the other players twenty-two
times. And each party will have two honours the
remaining thirty-nine times.
4. — The chance that each of the four players should
have one honour is
.., 4.12.133 |3.133
either _!= or L. ,
51.50.49.48 51.50.49
(which happen to be equal) according as the turned up
card is not, or is, an honour. Before the turned up
card be seen, the chance is
— M2.1B3 + A L8-138 or 9633
13 ' 51.50.49.48 13 " 51.50.49' 249900*
136 CHANCE.
5. — If an honour turns up, the respective chances
that the dealer should hold exactly one, two, three, or
four honours are
39.38.37 3 12.39.38 3 12.11.39 12.11.10.
51.50.49' 1X 51.50.49 ' 1 X 51.50.49' 51.50.49'
54834 53352 15444 1320
or
124950' 124950' 124950' 124950*
The sum of these fractions is unity, it being certain
that the dealer has at least one honour.
6. — If an honour does not turn up, the respective
chances that the dealer should hold none, or exactly
one, two, three, or four honours, are
39.38.37.36 , 12.39.38.37 6 12.11.39.38
" ''
51.50.49.48 ' " 51.50.49.48' '51.50.49.48'
4
'
12.11.10.39 12.11.10.9*
or
51.50.49.48' 51.50.49.48'
82251 109668 48906 8580 465
249900' 249900' 249900' 249900' 249900*
The sum of these fractions is unity, since one of the
five cases must certainly occur.
7. — Before it is known whether an honour will turn
up, the respective chances that the dealer should hold
GAME OF WHIST. 137
none, or exactly one, or two, or three, or four honours
are
56943 109668 66690 15444 1155
249900' 249900' 249900' 249900' 249900'
the sum of these fractions being unity.
8. — If an honour turns up, the respective chances
that a player, who is not the dealer, should hold no
honour, or exactly one, or two, or three honours are
38.37.36 3 13.38.37 8 13.12.28 13.12.11.
51.50.49' '5L5O49' '5L5O49' 51.50.49'
50616 54834 17784 1716
or
124950' 124950' 124950' 124950'
the sum of these fractions being unity ; since the player
must certainly hold none, or one, or two, or three
honours.
9. — If an honour does not turn up, the respective
chances that the player, who is not the dealer, should
hold none, or one, or two, or three, or four honours are
38.37.36.35 , 13.38.37.36 g 13.12.38.37
51.50.49.48 ' * 51.50.49.48 ' ' ' 5T.50.49.48 '
4 13.12.11.38 13.12.11.10.
'51.50.49.48' 51.50.49.48'
73815 109668 54834 10868 715
or
249900' 249900' 249900' 249900' 249900
138 CHANCE.
10. — Before it is known whether an honour will
turn up, the respective chances that a player, who is
not the dealer, should hold none, or one, or two, or
three, or four honours are
82251 109668 48906 _8_580_ _495_
249900' 249900' 249900' 249<TOO' 249900'
the sum of these fractions being unity.
EXAMPLES ON CHANCE.
1. — If ten persons form a ring, what is the chance
that two assigned persons will be together ?
2. — If ten persons stand in a line, what is the chance
that two assigned persons will stand together ?
3 — If two letters are selected at random out of the
alphabet, what is the chance that both are vowels ?
4. — Compare the chances of throwing four with one
die, eight with two dice, and twelve with three dice,
having two trials in each case.
5. — A bag contains four red balls and two white.
Three times in succession a ball is drawn and replaced.
Find the chance that a red ball is drawn each time.
6. — What is the probability of throwing not more
than eight in a single throw, with three dice ?
EXAMPLES. 139
7. — A bag contains six black balls and one red. A
person is to draw them out in succession, and is to
receive a shilling for every ball he draws until he draws
the red one. What is his expectation ?
8. — There are ten tickets, five of which are num
bered 1, 2, 3, 4, 5, and the other five are blank. What
is the probability of drawing a total of ten in three
trials, one ticket being drawn out and replaced at each
trial ?
9. — What is the probability in the preceding question
if the tickets are not replaced ?
10. — A person has ten coins, which he throws down
in succession. He is to receive one shilling if the first
falls head, two shillings more if the second also falls
head, four shillings more if the third also falls head,
and so on, the amount doubling each time ; but as soon
as a coin falls tail, he ceases to receive anything. What
is the value of his expectation ?
11. — A and B play at chess, and A wins on an
average two games out of three. Find the chance of A
winning four games out of the first six that are not
drawn.
12. — A and B play at chess, and A wins on an
average five games out of nine. Find ^4's chance of
140 CHANCE.
winning a majority (1) out of three games, (2) out of
nine games, (3) out of four games, drawn games not
being counted.
13. — If the odds on every game between two players
are two to one in favour of the winner of the preceding
game, what is the chance that he who wins the first
game, shall win at least two out of the next three ?
14. — A) B, C play at a game in which each has a
separate score, and the game is won by the player who
first scores two points. If the chances are respectively
2, 3, g, that any point is scored by A, B, C, find the
respective chances of the three players winning the
game.
15. — Assuming the results stated on page 99, what
are the odds against a person aged thirty living to be
seventy ?
16. — What is the chance that three persons, aged
respectively 30, 40, and 50, will be all alive twenty
years hence, and what is the chance that at least two
of them will be alive ?
17. — Four flies come into a room in which there are
four lumps of sugar, of different degrees of attractive
ness, proportional to the numbers 8, 9, 10, 12 ; what is
the chance that the flies will all select different lumps ?
EXAMPLES. 141
18. — Shew that the odds are eleven to three against
a month selected at random, containing portions of six
different weeks.
19. — Reference is made to a month which contains
portions of six different weeks ; what is the chance that
it contains thirty-one days ?
20. — A living man is known to be between thirty and
fifty years old, and the odds are estimated at three to
two that he is over forty. If he now die, what do these
odds become ? (See page 99.)
APPENDIX.
PERMUTATIONS AND COMBINATIONS TREATED
ALGEBRAICALLY.
DEFINITION I. A collection of r things in a par
ticular order is called a permutation or arrangement of
r things.
DEFINITION II. A collection of r things without
regard to order is called a combination or selection
of r things.
PEOPOSITION I.
If one operation can be performed in m ways, and
('when it lias been performed in any way) a second
operation can then be performed in n ways, there will
be mn ways of performing the two operations.
For if we confine our attention to the case in which
the former operation is performed in its first way, we
can associate with this way any of the n ways of per
forming the latter operation : and thus we shall have
n ways of performing the two operations, without re-
PERMUTATIONS AND COMBINATIONS. 143
cognising more than the first way of performing the
former one.
Then, if we consider the second way of performing
the former operation, we can associate with this way
any of the n ways of performing the latter operation :
and thus we shall have n ways of performing the two
operations, using only the second way of performing the
former one.
And so, corresponding to each of the m ways of
performing the former operation, we shall have n ways
of performing the two operations.
Hence, altogether we shall have m times n, or mn
ways of performing the two operations. Q. E. D.
PROPOSITION II.
If one operation can be performed in m ways, and
then a second can be performed in n ways, and then a
third in r ways, and then a fourth in s ways (and so
on), the, number of ways of performing all the opera
tions will be m x n x r x s x &c.
For by Prop. L, the first and second can be per
formed in mn ways.
Then if we treat these two as forming one complete
operation, and associate with it the third operation
(which can be performed in r ways), it follows again
from Prop. I., that both these can be performed in
mn x r different ways. That is, the first, second, and
144 APPENDIX.
third of the original operations can be performed in
mnr ways.
Again if we treat these three as forming one complete
operation, and associate with it the fourth operation
(which can he performed in s ways), it follows again
from Prop. I. that both these can be performed in
mnr x s different ways. That is, the first, second,
third, and fourth operations can be performed in mnrs
ways, and so on. Q. E. D.
COROLLARY. — If there be x operations which can be
performed successively in m ways each, then all can be
performed in mx ways.
This follows from the proposition, by considering the
particular case in which m, n, r, s, &c., are all equal.
EXAMPLES. — If there are p candidates for the office
of president, s candidates for that of secretary, and t
candidates for that of treasurer, the election of the
three officers can be made in pst different ways.
If a telegraph has m arms, and each arm is capable
of n different positions, including the position of rest,
the number of signals that can be made is nm-l.
If there be x things to be given to n persons, nx will
represent the whole number of different ways in which
they may be given.
PERMUTATIONS AND COMBINATIONS. 145
PKOPOSITION III.
The number of different orders in which n different
things can be arranged is
n (n-l) (w-2) 3.2.1.
For having to arrange the n things, we may arrive
at any possible arrangement, by taking them one by
one, and placing them in the n places in order.
The first place may be filled up by any of the n
things : that is, it may be filled up in n different ways.
Then the second place may be filled up by any of
the n-l things that are left: that is, it may be filled
up in n-l different ways.
Then the third place may be filled up by any of the
w-2 things that are now left : that is, it may be filled
up in n- 2 different ways.
Similarly the fourth place may be filled up in n- 3
ways, the fifth in n—4 ways, and so on ; and ultimately
the last place may be filled up in only one way.
Hence (Prop. II.) the whole number of ways of
filling up all the places, or making the whole arrange
ment, is the continued product of all these numbers, or
n (n-l) (n-2) 3.2.1.
NOTE. — The continued product of all integers from
1 to n is generally denoted by the symbol \n.
146 APPENDIX.
COROLLARY. — If n given things have to be devoted to
n given objects, one to each, the distribution can be
made in n ways.
EXAMPLES. — The number of ways in which n persons
can stand in a row is |w. The number of ways in
which they can form a ring is \n ~ 1. (See page 23.)
The number of ways in which m ladies and m gentle
men can form a ring, no two ladies being together,
is m, \m - 1.
PKOPOSITION IV.
Out of n different things, the number of ways in
which an arrangement of r things can be made is
n (n-l) (w-2) . . . to r factors,
or n (n-l) (n-%) . . . (n-r + 1).
For we have to fill up r different places in order with
some of the n given things. As in the last proposition,
the first place can be filled up in n ways, the second
in n-l ways, the third in ft -2 ways ; and so on for all
the r places.
Hence the whole number of ways of filling up all
the r places, or making the required arrangement, is
n (n-l) (?i-2) ... to r factors.
PEKMUTATIONS AND COMBINATIONS. 147
Or, observing that the
1st factor is n,
2nd „ n-1,
3rd „ n-2,
&c. &c.
rth ,, n-(r-l) or n-r + 1,
we may write the result, —
n (w-1) (w-2) . . . (n-r + 1).
EXAMPLE. — The number of times a company of mn
men can form a rectangular column, having m men in
front, so as to present a different front each time, is
mn (nut— 1) (mn — 2) .... (mn — m -\- 1).
PROPOSITION V.
Out of n different things, when each may be repeated
as often as ive please, the number of ways in which an
arrangement of r things can be made is nr.
For the first place can be filled up (as before) in n
ways, and when it is filled up the second place can also
be filled up in n ways (since we are not now precluded
from repeating the selection already made) ; and so the
third can be filled up in n ways, and so on, for all the
r places.
Hence (Prop. II., Cor.) all the r places can be
filled up, or the whole arrangement can be made, in
nr different ways.
148 APPENDIX.
EXAMPLE. — In the ordinary scale of notation 10r- 1
different numbers can be made, each consisting of not
more than r figures.
PKOPOSITION VI.
The number of ways in ivhich x + y things can be
divided into two classes, so that one may contain x and
the other y thing s9 is
\x + y
x \y
For suppose N represents the number of ways in
which the division could be made ; then the things in
the first class can be arranged in x different orders
(Prop. III.), and the things in the second class in \y_
different orders, and therefore the whole set of x + y
things can be arranged in x places of one class, and y
places of another class, in N . x . \y different ways
(Prop. II.). But this must be the same as the number
of ways in which the whole set of x + y things can be
arranged into any x + y different places, which, by
Prop. III., is \x + y. Hence we have the equation
or N
= \z_+_y,
\x + y
PERMUTATIONS AND COMBINATIONS. 149
That is, the number of ways in which the required
division can be made is
y
which was to be proved.
PKOPOSITION VII.
The number of ivays in which x + y + z things can
be divided into three classes, so that they may contain
x, y, and z things severally, is
\x + y + z
\x_ \y I*
For, by Prop. VI., the x + y + z things can be
divided into two classes, containing x and y + z
things in
ways ; and then the class of y + z things can be sub
divided into two classes, containing y and z things in
\y + *
Tif"
Therefore the three classes of x, y, z things can be
made in
\x + y + z \y_±_^ \x + y + z
or
x y + z \y z ' \x \y \z_
ways (Prop. I.), which was to be proved.
150 APPENDIX.
COROLLARY. — We might similarly extend the reason
ing if there were any more classes. Thus, the number
of ways in ivliich v + w + x + y + z things can be divided
into jive classes, containing respectively v, w, x, y, z
things, is
\v + w + x + y + z
\v w x \y \z
EXAMPLES.— The number of different ways in which
2w boys can divide themselves into two equal parties, to
play a game, is
The number of ways in which mn things can be
divided into m parcels, of n things each, is
mn
PEOPOSITION VIII.
The number of different orders in which n things can
be arranged, whereof p are all alike (of one sort}, q
all alike (of another sort), r all alike (of another
sortj, and the rest all different is
IP \q
PERMUTATIONS AND COMBINATIONS. 151
For the operation of making this arrangement may
be resolved into the several operations following : —
(1) to divide the n places which have to be
filled up into sets of p places, q places, r places, and
n-p-q-r places respectively:
(2) to place the p things all alike in the set of
p places, the q things all alike in the set of q places,
the r things all alike in the set of r places :
(3) to arrange the remaining n-p-q-r things
which are all different in the remaining set of n-p-q-r
places.
Now the operation (1) can be performed, by Prop.
VII., in
\n
\P \q r \n-p-q-r
different ways : the operation (2) can be performed in
only one way: the operation (3), by Prop. III., in
\n-p-q-r ways.
Hence, (Prop. II.) the whole operation can be per
formed in
n -p-q- r , or 7 [..
\p \q \r \n—p—q—r \p \q \r
different ways. Q. E. D.
COROLLARY. — The same argument would apply if
the number of sets of things alike were any other than
152 APPENDIX.
three. Thus, for instance, the number of orders in
which n things can be arranged) whereof p are alike,
q others alike, r others alike, s others alike, and t others
alike, is
\P |« \r |s |*
EXAMPLE. — If there be m copies of each of n
different volumes, the number of different orders in
which they can be arranged on one shelf is
PKOPOSITION IX.
Out of n different things, the number of different
ivays in which a selection of r tilings can be made, is
the same as the number of different ways in which a
selection of n-r things can be made, and is
r n-r
For either operation simply requires the n things to be
divided into two sets of r and n—r things respectively,
whereof one set is to be taken and the other left.
Therefore (by the last proposition), whichever set be
rejected, the operation can be performed in
n
r n-r
different ways.
PERMUTATIONS AND COMBINATIONS. 153
The expression
\r n-r
may be written
n (n-1) (n-2) . ^ 3.2.1
r. (n-r) (n-r-1) . . . 3.2.1
or, dividing the numerator and denominator of the
fraction by all the successive integers from 1 to n-r,
n (n-1) (n-2) (n-r + 1)
This result might have been obtained quite inde
pendently, as follows : —
Let x represent the number of ways of making a
selection of r things out of n things. The r things
thus selected might be arranged (Prop. III.) in \r
different orders. Therefore (Prop. I.) x x r is the
number of ways in which r things can be selected out
of n things, and arranged in order. But by Prop. IV.
this can be done in n (n-1) (n-2) (n-r + 1)
different ways. Therefore we have the equation
x x |r = n(n-l) (n-2) (n-r+1),
which gives us
n (n-1) (n-2) (n-r+1)
x = - ——
\r
the required expression.
154 APPENDIX.
EXAMPLES. — The number of ways in which a com
mittee of p liberals and q conservatives can be selected
out of m liberals and n conservatives, is
m
\P \m-p \q \n-q '
If there be n-l sets containing 2&, 3&, 4a, . . (n-l)a
things respectively, the number of ways in which a
selection can be made, consisting of a things out of
each set, is
|8a \4a
x ---=— x _~" x &c. . . . x
la la
a
\na
or
PROPOSITION X.
The ivhole number of ways in which a selection can
be made out of n different things is 2n — 1.
For each thing can be either taken or left ; that is, it
can disposed of in two ways. Therefore (Prop. II.,
Cor.) all the things can be disposed of in 2n ways.
This, however, includes the case in which all the
things are rejected, which is inadmissible ; therefore
the whole number of admissible ways is 2W- 1. Q. E. p.
PERMUTATIONS AND COMBINATIONS. 155
PROPOSITION XI.
The whole number of ways in which a selection
can be made out of p + q + r + &c., things, whereof
p are all alike (of one sortj, q all alike (of another
sortj, r all alike (of another sort), dc., is
(p + l)(q + 1) (r + 1) dc. - 1.
For, of the set of p things all alike, we may take
either 0 or 1 or 2 or 3 or &c. or p, and reject all
the rest; that is, the p things can be disposed of in
p + 1 ways. Similarly, the q things can be disposed of
in q + 1 ways, the r things in r -f- 1 ways, and so on.
Hence (Kule II.) all the things can be disposed of in
(p + I) (q + 1) (r + 1) ways. This, how
ever, includes the case in which all the things are
rejected, which is inadmissible ; therefore the whole
number of admissible ways is
(P + 1) (q + 1) (r + 1) . . . . dc. - 1.
Q. E. D.
EXAMPLE. — If there be m sorts of things and n
things of each sort, the number of ways in which a
selection can be made from them is (n + l)m — 1.
If there be m sorts of things, and one thing of the
first sort, two of the second, three of the third, and so
on, the number of ways in which a selection can be
made from them is w + 1 - 1.
156 APPENDIX.
PEOPOSITION XII.
A NEW PROOF OF THE BINOMIAL THEOREM.
The Binomial Theorem was first published by Sir
Isaac Newton, who was Lucasian Professor of Mathe
matics at Cambridge from 1669 to 1702. It furnishes
a ready method of raising any given binomial ex
pression to any required power.
We proceed to consider a question of combinations,
and, from our results, to deduce a proof of the binomial
theorem, applicable to all cases in which the exponent
is a positive integer.
i. — A painter has x + y colours, of which x
are dark and y are light colours. He has to paint n
croquet-balls (all of different sizes), each ball being one
colour, but as many balls as he pleases the same colour.
In how many ways can he paint the balls ?
Answer I. — Since each ball can be painted with any
one of the x + y colours, and there are n balls, the
whole number of different ways in which the work can
be done is (by Prop. II., Cor.)
(x + y)\
Answer II. — If he paint all the balls dark, each can
be painted in x different ways ; therefore the work can
be done in xn different ways.
If he paint one light and the rest dark, the selection
of the one to be light can be made in n ways; then the
PERMUTATIONS AND COMBINATIONS. 157
n— 1 can be painted dark in xn~l ways, and the one
light in y ways; therefore the work can be done in
n xn~ly, or, as we will write it for the sake of
symmetry,
different ways.
If he paint two light and the rest dark, the selection
of the two to be light can be made in
n(n-~L)
1.2
ways (Prop. IX.), then the ?i-2 can be painted dark in
xn~2 ways, and the two light in y* ways : therefore the
work can be done in
n_2 o
ways.
If he paint three light and the rest dark, the selec
tion of the three to be light can be made in
n(»-J)(n-2)
1.2.3
ways (Prop. IX.), then the n-3 can be painted dark in
xn~* ways, and the three light in y3 ways : therefore the
work can be done in
n- ,-..
1.2.3
ways.
158 APPENDIX.
And so on; until finally we consider the case in which
all are light, in which case the work can be done in yn
ways.
Hence the whole number of ways in which the work
can be done is the sum of the series
which will have n + 1 terms altogether.
The Binomial Theorem. — The two answers to the
question just investigated must give the same result
numerically, or the two algebraical results must be
equal : therefore we have
/ \n n , n n-l , n (n~^-} n-2 2
(x + y)n = xn + - xn l y + V12 xn 2 if
We are thus furnished with a formula by which we
can write down any power of a binomial expression,
as a series of terms, consisting of powers of the two
original terms. The statement of this formula is
called the Binomial Theorem.
EXAMPLES. —
= a? -f Sofy + Sxy' + y3.
PERMUTATIONS AND COMBINATIONS. 159
So, (a-Zb?
16 Z>4.
The theorem will hold equally if we have any two
fractions, - and T- suppose, instead of the numbers x
and 11. For
r\n (ps + qr\n (ps + qr)n
9j \ qs ) (qs)n '
and since ps and qr are integers, we may apply the
theorem, and write
(ps + qr)n = (ps)n + 1
+ &c. . . . + (qr)";
and therefore dividing by (qs)n we have
P TY _ (P^" - nfv\" ~r n (n~]
_
— 1 ~T~ i \ _. /
ql^l\ql s^ 1.2
which shews that the expansion follows the same law
when any fractions are substituted for the terms x
and y.
EXAMPLES. —
1 \«
x
160 APPENDIX.
_flY_ -,_4M 4.3/aV 4.3.2/W 4.3.2.1/aV
b) ~ l\b) + 1.2\6y 1.2.3W 1A8.4\6/
* 4. ^Y-K+3^y^u8-2r
^~To/ — \ ct ~r-i\n I ttl T « ct\ n
3/ 1-22; 1.2.3V3
It may here be observed that if x be a small fraction,
the powers #2, xz, &c., will be smaller still, and will
rapidly become inconsiderable when the index is
increased. Hence, a few terms of the expansion
- n (n-1) (rc-2) ,
l 1.2.3
will give an approximately true value for (1 + x)n when
x is small compared with unity.
It is proved in treatises on algebra that the formula
of expansion
still holds when n is a fractional or negative index.
But in this case the series is interminable, and is only
of practical use when x is a proper fraction, when a
finite number of terms will give an approximate value.
PEEMTJTATIONS AND COMBINATIONS. 161
EXAMPLES. —
To find the square root of \-x in ascending powers
of x.
We have
&c
- __ _ __ -*
2 1.2 V2y " 1.2.8 W " 1.2.8.4 " c*
To find the square root of 2.
We have
98 100 _98 100/ 1 \
"49" 49*100 49 V 50/
Therefore,
50
But, as in the last example,
/1~T~:L_1 _ JL -1 fJLY- — f — Y_
V 50" 100 L2UOOJ 1.2.8\10oJ "
- 1 - -01 - -00005 - -0000005 - -00000000625- &c.
= •9899494936
and therefore, multiplying by y
-v/2 = 1-414213562...
This is the readiest method of extracting the square
root of 2, when a very high degree of accuracy is
required. Very little labour would extend the result to
20 or 30 places of decimals. The binomial theorem
162 APPENDIX.
may be similarly applied to find the square root of any
other number, or to find cube roots or any other roots.
Indeed when the fifth, seventh, or any higher root of
a number is required, this method is the only prac
ticable one, unless tables of logarithms are employed ;
and it has the advantage over the method by loga
rithms of bringing the result to any degree of accuracy
required.
APPENDIX II.
DISTKIBUTIONS.
Most of the questions of Permutations and Combina
tions which we have considered have involved the
division of a given series of things into two parts, one
part to he chosen, and the other rejected. The theorem
expressed arithmetically in Rule VI. (page 30), and
algebraically in Proposition VII. (page 149), is the
only one in which we have contemplated distribution
into more than two classes. But as the number of
things to be given to each class was in the terms of that
theorem assigned, the problem was reduced to a case of
successive selection, and was therefore classed with
other questions of combinations. But when the number
of elements to be distributed to each several class is
unassigned, and left to the exercise of a further choice,
the character of the problem is very much altered, and
the problem ranks among a large variety which we class
together as problems of Distribution.
Distribution is the separation of a series of elements
into a series of classes. The great variety that exists
among problems of distribution may be mostly traced
to five principal elements of distinction, which it will
164 APPENDIX II.
be well to consider in detail before enunciating the
propositions on which the solution of the problems will
depend.
I. — The things to be distributed may be different or
indifferent. The number of ways of distributing five
gifts among three recipients, will greatly depend upon
whether the gifts are all alike or various. If they are
all alike (or, though unlike, yet indifferent as far as the
purposes of the problem are concerned), the only ques
tions will be (i.) whether we shall divide them into sets
of 2, 2, 1 or 3, 1, 1, and (ii.) how we shall assign the
three sets to the three individuals. If on the contrary
the five gifts are essentially different, as a, b, c, d, e, then
they may be divided into sets of 2, 2, 1 in 15 ways, and
into sets of 3, 1, 1 in 10 ways, and then we shall have
to assign the three sets which are in this case all
essentially different (because their component elements
are so), to the three individuals. In the first case, the
sets could be formed in 2 ways, and when formed in
either way they could be assigned in 3 ways, thus
giving a complete choice of 6 distributions. In the
second case, the sets could be formed in 25 ways, and
when formed in any way they could be assigned in 6
ways, thus giving a complete choice of 150 dis
tributions.
II. — The classes into which the things are to be dis
tributed may be themselves different or indifferent. We
DISTRIBUTIONS. 165
here use the adjectives different or indifferent to qualify
the abstract classes regarded as ends or objects to which
the articles are to be devoted, without any reference to
a posteriori differences existing merely in differences of
distribution into the classes.
Where five gifts were to be distributed to three
recipients, the distinct personality of the three recipients
made the classes characteristically different, quite apart
from the consideration of the differences of the elements
which composed them. But if we had only to wrap up
five books in three different parcels, and no difference
of destination were assigned to the parcels, we should
speak of the parcels as indifferent. The problem would
be simply to divide the five things into three sets, with
out assigning to the sets any particular order. The
distribution could be made in 2 ways if the things
themselves were indifferent, and in 25 ways if they
were different.
III. — The order of the things in the classes may be
different or indifferent, that is, the classes may contain
permutations or combinations. Of course this distinc
tion can only arise when the things themselves are
different, for we cannot recognise any order among
indifferent elements. We shall avoid confusion by dis
tinguishing arranged and unarranged classes respectively
as groups and parcels. If three men are to divide a
set of books amongst them, it is a case of division into
parcels, for it does not matter in what order or arrange-
166 APPENDIX II.
ment any particular man gets his books. But
if a series of flags are to be exhibited as a signal
on three masts, it is a case of division into groups,
for every different arrangement of the same flags
on any particular mast would constitute a different
signal.
IV. — It may or may not be permissible to leave some
of the possible classes empty. It will entirely depend
upon the circumstances out of which the problem arises,
whether it shall be necessary to place at least one
element in every class, or whether some of them may
be left vacant ; in fact, whether the number of classes
named in the problem is named as a limit not to be
transgressed, or as a condition to be exactly fulfilled.
If we are to distribute five gifts to three recipients, it
will probably be expected, and unless otherwise ex
pressly stated it will be implied, that no one goes away
empty. But if it be asked how many signals can be
displayed by the aid of five flags on a three-masted
ship, it will be necessary to include the signals which
could be given by placing all the flags on one mast, or
on two masts.
V. — It may or may not be permissible to leave some
of the distributable things undistributed. This will be
illustrated by a comparison of the Propositions XXV.
and XXVI. below.
DISTRIBUTIONS. 167
Propositions XIII., XIV., XV., and XVI. apply to
the distribution of indifferent things.
The XVII. and following propositions embrace the
different cases which arise in the distribution of
different things.
The case in which the parcels are indifferent as well
as the things to be distributed into them, is reserved to
the last, as presenting peculiar difficulty. It will be
found treated of in Prop. XXVIII.
PROPOSITION XIII.
The number of ways in ivhich n indifferent things
can be distributed into r different parcels (blank lots
being inadmissible) is the number of combinations of
n — 1 things taken r—1 at a time.
For we may perform the operation by placing the n
things in a row, then placing r— 1 points of partition
amongst them, and assigning the r parts thus created,
in order, to the r parcels in order.
Hence the number of ways is the number of ways
of placing r— 1 points of partition in a selection
out of n—~L intervals. Therefore it is the same as the
number of combinations of n — 1 things taken r— 1 at
a time. Q. E. D.
PROPOSITION XIV.
The number of ways in which n indifferent things
168 APPENDIX II.
can be distributed into r different parcels (blank lots
being admissible) is the number of combinations of
n-\-r— 1 things taken r—~L at a time.
For the distribution of n things, when blank lots are
admissible, is the same as the distribution of n + r
things when they are not admissible, since in the
latter case we have to place one thing in each of the r
parcels, and then to distribute the remainder as if
blank lots were admissible. Hence, writing n + r for
n in the result of Proposition XIII. , we obtain the
number required.
EXAMPLES. — Twenty shots are to be fired; the work
. 19 . 18. 17
can be distributed among four guns m-^ — ^ — =- or
1 . -a . O
969 ways, without leaving any gun unemployed. Or,
neglecting this restriction, the work can be done in
23.22.21
1.2.3
or 1771 ways.
Again, five partners in a game require to score 36
to win. The number of ways in which they may share
this score (not all necessarily contributing), is
40.39.38.37
1.2.3.4
or 91390 different ways.
Again, in how many ways can five oranges be
distributed amongst seven boys? Evidently two or
DISTRIBUTIONS. 169
more of them will get none. The answer is 462, viz.,
11.10.9.8.7.6
1.2.3.4.5.6'
PBOPOSITION XV.
The number of ways in which n indifferent things
can be distributed into r different parcels, no parcel to
contain less than q things, is the number of combina
tions of n — l — r(q — l) things taken r at a time.
For if we first place q things in each of the r parcels
we shall have n — qr things left, and it will only remain
to distribute them among the same r parcels according
to Proposition XIV., which shows that the number of
ways of making the distribution is the number of com
binations of n — qr + r— 1 things taken r at a time.
Q. E. D.
PKOPOSITION XVI.
The number of ways in which n indifferent things
can be distributed into r different parcels, no parcel to
contain less than q things, nor more than q+z — 1
things is the coefficient of xn~qr in the expansion of
'I- of
For if we multiply together r factors, each repre
sented by
170 APPENDIX II.
we shall have in our result a term xn for every way in
which we can make up n by the addition of one index
q or q + l or g + 2 or &c., or q+z — 1 from each of the
r factors. Hence we shall have xn as many times as
there are ways of distributing n into r parts, no part
less than q nor greater than q+z—I. Therefore the
number of ways of so distributing n is the coefficient
of xn in the expansion of
or of xqr(l+x+x*+...xz-1Y
which is the coefficient of xn~qr in the expansion of
-xK
Q-E.D.
EXAMPLE. — The number of ways in which four persons,
each throwing a single die once, can score 17 amongst
them is the coefficient of #17'4 in the expansion of
\l-x
Now
'-&c.
And coefficient of x13 in the product
= !J14.15.16-4.8.9.10+6.2.3.4J
= 104.
DISTRIBUTIONS. 171
PROPOSITION XVII.
The number of ways in which n different things can
be distributed into r different parcels is rn, when blank
lots are admissible.
For each of the n different things can be assigned to
any one of the r parcels without thought of how the
others are disposed of. Hence the n things can be
(severally and) successively disposed of in r ways each,
and therefore (Choice, Rule II.) all can be disposed of in
r7* different ways. Q. E. D.
PROPOSITION XVIII.
When blank lots are not admissible) the number of
ways in which n different things can be distributed into
r different parcels is [n_ times the coefficient of xn in the
expansion of (ex-l)r,
Let Nr denote the number of ways in which n things
can be distributed into r different parcels, blank lots
being inadmissible.
Then rNr-i will be the number of ways in which
the distribution might be made if one parcel were left
blank.
r (r — 1)
So -V~?| -ZVr-a will be the number of ways in which
the distribution might be made if two parcels were left
blank.
And so on,
172 APPENDIX II.
But we know that if any number of blank lots were
admissible, the distribution could be made in rn different
ways. Therefore
rn = Nr + | AUi +T-^~- Nr., + &c. + rNlf
or, if we establish the convention that Np is always to
be replaced by Np) we may write
Similarly, (r-l)n =
and so on.
Now multiply these equations in order by the coeffi
cients in the expansion of (1 — x)r, and add (having
regard to algebraical sign) ; then the first member of the
resulting equation will be
rn-^(r-l)ra+r(7,'~1)(r-2)n-&c. (tillit stops)
1 I. la
and the second member, since the sum of the coefficients
in the expansion of (1 - x)r is zero, will be { (N+ 1) — l} r
or Nr or Nr. Therefore
Nr = r-_^(r-l)n+r^^(r-2r-&c. till it stops.
that is, (Todhunter's Algebra, Art. 549.)
Nr = \n times the coefficient of xn in the expansion of
(ex-l)r. Q. E. D.
EXAMPLES. — The number of ways in which five
different commissions can be executed by three mes
sengers is 35 or 243. But if no one of the messengers
DISTRIBUTIONS. 173
is to be unemployed, the number of ways will be [5
times the coefficient of x5 in the expansion of (ex— I)3.
But
=^+F+?+-
Hence the number of ways will be _ x [5 or 150.
PKOPOSITION XIX.
The number of r-partitions of n different things, i. e.
the number of ways in which n different things can be
distributed into r indifferent parcels, ivith no blank lots,
is \n times the coefficient of xn in the expansion of
For every way of distributing the things into r indif
ferent parcels, must give rise to \r ways of distributing
them when the parcels are different. Hence, if $r
denote the number of partitions, we have, by compa
rison with Prop. XVIII.,
= \n times the coefficient of xn in the expansion of
(«•- iy
\r_ Q. E. D.
174 APPENDIX II.
EXAMPLE. — To divide the letters a, b, c, d, e into
three parcels. The numher of ways will be [5 times
the coefficient of x* in the expansion of (ex — I)3 -f- [3 ;
£
that is (as in the last example), 5.4 x ^ = 25.
The twenty-five divisions are easily seen to be ten
such as abc, d, e, and fifteen such as ab, cd, e.
PKOPOSITION XX.
The total number of ways in which n different things
can be distributed into 1, 2, 3 ... or n indifferent parcels
/
is \n times the coefficient of xn in the expansion of — .
t/
For with the notation, of preceding theorems, we have
$1 = I w times the coefficient of xn in e* ~~1>
Li
f,= («*-!)'
12
»,= («•-!)'.
L5
and so on.
Therefore by addition
& + &+$.+ - (Ml it stops, i. e. ... JJ
is equal to \n times the coefficient of xn in the expan
sion of
e>-l (<?-!)* (e'-iy
' ~~ ~~ + '
DISTRIBUTIONS. 175
this last series being carried to infinity if we please,
since the terms beyond the nttl do not involve xn} and
therefore the inclusion of them will not affect the
coefficient of xn.
But this series is the expansion of ee~— 1, and the
coefficient of xn therein is the same as in the expansion
of eeX ~* or ee* -f- e. Hence
&+§,+».+ «.+!.
is equal to \n times the coefficient of xn in the expan
sion of J°
e Q. E, D.
PROPOSITION XXI.
The number of ways in which n different things can
be arranged in r different groups (with no blank lots)
is
n I w — 1
[»— r [r— 1
For they can be arranged in one row in \n ways, and
then the r — l points of partition can be placed in a
\n- 1
selection of the n — 1 intervals in L — =-- ways.
\n-r\r-\
But the number of ways in which n things can be
arranged in r different groups must be the product of
these two numbers (Choice, Rule L), or
\n \n—I
\n-r [r-1* Q. E. D.
N
176 APPENDIX II.
PROPOSITION XXII.
The number of ways in which n different things can
be arranged in r indifferent groups with no blank lots
is
\n\n-l
[r \n—r [r— 1
For it is plain that for any one arrangement in this
case we must have [r arrangements when the groups
are not indifferent. Hence the result is p of that in
\r
Proposition XXI., or
lnln-1
\r \n — r \r — \
Q. E. D.
PROPOSITION XXIII.
The total number of ivays in ivhich r different
arranged groups can be made out of m things all
differ entf is the coefficient of xm~T in the expansion of
\m ex
d-*)''
blank groups being inadmissible.
For if we use n of the things at a time the groups
can be made (by Theorem XI.) in
?n n — l
\m—n \n—r \r—\
DISTRIBUTIONS. 177
ways. And n may have any value from r to 7??
inclusive. Hence the required number is
to m — r + 1 terms k
which is equal to the coefficient of xm~r in the pro
duct of the two series
r r
| ?7i xm^ Imx™-*-1 \m ,
and ...+ ~ 1.±= 1 — L_ _|_ &c<j
\m— r [m — r—l [m — r— 2
which are respectively the expansions of (\—x)~r and
;m ex. Hence the required number is the coefficient
of xm~T in the expansion of
(l-x)r Q. E. D.
PROPOSITION XXIV.
The total number of ways in which r indiiferent
arranged groups can be made out of m tilings all
different is the coefficient of xm~r in the expansion of
[rn^e*
[r_(l-xY'
blank lots being inadmissible.
This follows from the previous theorem, as Proposi
tion XXII. from Proposition XXI.
178 APPENDIX II.
PKOPOSITION XXV.
The number of ivays in which n different things can
be arranged in r different groups (blank groups being
admissible ) is
[n+r-1
[r^T~
For they can be arranged in one row in [n ways, and
then the r — 1 points of partition can be placed in the
n+1 intervals (including the ends of the row) in
ways by Proposition XIV.
The number of ways required is the product of these
two numbers (Choice, Rule L), or
\n+r-l
Q. E. D.
PROPOSITION XXVI.
The total number of signals that can be made by
displaying arrangements out of m flags on a set of r
masts, where each mast will hold any number of flags,
is one less than the coefficient of xm in the expansion of
[m ex
tt-*r'
If we use all the r masts, the number of signals is
DISTRIBUTIONS. 179
(by Prop. XXIII.) the coefficient of xm~r in the
expansion of
[m ex
(!-«)'
\m exxr
K— the coefficient of xm in expansion of L? __
a-*)
So, if we use r — 1 given masts, the number of signals
I jft X r—1
is the coefficient of xm in the expansion of L =
and the r—1 masts can be selected in r ways, therefore
the number of signals
r I m ex x*~l
— the coefficient of xm in expansion of ^-. b=. _ —r)
and so on for (r-2), (r — 3), &c. masts. Hence, the
total number of signals, including the case when no
flag is hoisted is the coefficient of xm in the expan
sion of
or
or
to r+1 terms }-,
Hence, excluding the case when no flag is hoisted,
the number of signals is one less than the coefficient
of xm in the expansion of
[m ex
(l~x)r ' Q. E. D.
180 APPENDIX II.
PKOPOSITION XXVII.
To find the number of ways in which n indifferent
things can be distributed into r indifferent parcels (no
blank lotsj.
OB
To find the number of different r-partitions of n.
Let Pn,r denote the number of r-partitions of n, or
the number of ways of distributing n indifferent ele
ments into r indifferent parcels.
Suppose that in any distribution, x is the smallest
number found in any parcel. Then setting aside a
parcel which contains x, all the other parcels contain
not less than x, and therefore more than .7;-!. If we
place x — 1 in each of these r—l parcels, the distribution
can then be completed by distributing the remaining
n— x— (r — !)(#— l)or n — 1 — r(x— 1) things among the
same r — l parcels, and this can be done in Pn-i-r(X-\\ r-i
ways. In this way we shall obtain all the distributions,
by giving x successively all its possible values. But
since x is the smallest number found in any parcel, x
cannot be greater than the greatest integer in -.
Denote this integer by '— .. Then x must have all values
from 1 to this integer, and therefore
Pn,r = -P»-l r-l + Pr>-l-r, r-l + P*-i-*rt r-1 + • - • to '- terms.
DISTRIBUTIONS. 181
Now it is plain that Pn>l — 1 for all values of n.
n
Hence Pn,z=Pn-i,i+Pn-3,i+Pn-5,i+ ...to ~ terms
Again Pn3=Pn_
\n-l lw-4
... to 57 terms
dl
n — 7 In .
-— + .- to L terms.
2 | 2
The summation will depend upon the form of n; thus,
2
If n= 6
12
n2-4
12
Therefore Pn>8 is always the integer nearest to =-5
whether in excess or defect.
This integer is conveniently denoted by the symbol
12
Again, Pn)4=Pn-i,3+^-5.3+^-9>3+ ... to Lj terms
+ ... to -r, terms.
12
12
12
182 APPENDIX II.
The summation will depend upon the form of n : thus,
n3_|_3^2
If n = 12q then Pn>i= ~^f~
w=12^ + l PM= jg-
^ = 12^ + 2 PM~r"
144
144
P _^3+3rc
^ 144
144
P^3-^2-*'
144
144
144
144
144
Therefore PM is always the integer nearest to
when w is even, and the integer nearest to -
when n is odd : or, with the notation introduced above,
*«M =
DISTRIBUTIONS.
n*+3n2
183
144
when w is even :
144
when n is odd.
By a like process we may deduce successively Pn)5,
Pn>6, &c., and thus we may find Pn>r for any values of n
and r, although we cannot write down a general expres
sion for Pn>r in any simple terms.
EXAMPLES. — There are twelve 3-partitions of 12, viz.
11 10 147 237 336
129 156 246 345
138 228 255 444
There are fifteen 4-partitions of 12, viz.
1119
1227
2226
1128
1236
2235
1137
1 2 4 0
2244
1146
1335
2334
1155
1344
3333
The number of 4-partitions of 13 is the integer
nearest to (rc3+3^2-9w)-^144 when n=lB. Therefore
there are 18 partitions, viz.
1
1
1
10
1
2
3
7
2
2
2
7
1
1
2
9
1
2
4
6
2
2
3
6
1
1
3
8
1
2
5
5
2
2
4
5
1
1
4
7
1
3
3
6
2
3
3
5
1
1
5
6
1
3
4
5
2
3
4
4
1
2
2
8
1
4
4
4
3
3
3
4
184 APPENDIX II.
To find the number of ^-partitions of 13.
We have P1W=P1M+PW
= 15 + 3
-18.
These eighteen partitions may be exhibited as follows
11119
11236
12244
11128
11245
12334
11137
11335
13833
11146
11344
22225
11155
12226
22234
11227
12235
22333
APPENDIX m,
DERANGEMENTS
WHEN we place a series of elements in a particular
order we are said to arrange them. But if they have
been already arranged, or if they have a proper order
of their own, and we place them in other order, we are
said to derange them. Thus derangement implies a
previous arrangement in which each element had its
own proper place, either naturally belonging to it or
arbitrarily assigned to it.
The following Notation is useful : —
It is proved, in treatises on algebra, that if e be the
base of Napierian logarithms then, whatever be the value
of x positive or negative,
ex=l + x + f^- + *~+&c.
Lf L?
This series is continued to infinity, but by suffixing
an integer to en we obtain a symbol which conveniently
expresses the sum of the same series continued only as
far as the term in which that integer is the index of x.
Thus—
186 APPENDIX III.
Let J denote the operation of changing any factorial
into the next inferior factorial, as \n into \n — 1, or
\n — 1 into [w — 2. And let J J or J2 indicate that the
operation is to be performed a second time upon the
result of the first, so that J2 operating on \n produces
\n — 2, and so on.
Then J [n = \n - 1
J2 [n = [n - 2
J3 [n = |n-3
Jr [n = [n — r
Thus we may write —
[n.ex= [n + [n-La; + n ~1} [n
= (1+
where it is understood that (l-\-xJ)n is to be expanded
by the law of the binomial theorem, and every term is
then to operate upon \n.
As a particular case we may write
1.^ = (l-J)\[n.
DERANGEMENTS. 187
PKOPOSITION XXVIII.
The number of derangements of a set of elements is
one less than the number of permutations of the elements.
For of all the permutations, one must give the proper
order of the elements, and all the rest must be derange
ments.
PKOPOSITION XXIX.
The number of icays in which a row. of n elements
may be so deranged that no element shall be in its
proper place is [n.e~l.
Let a, /3, y, .... x denote the n elements, and let N
represent the number of ways in which they can be
permuted when unrestricted by any condition.
Also let (A) express the condition that a is in its
proper place and (a) the condition that a is out of its
proper place. Let (B) and (b) denote the same conditions
with respect to |8 ; and so on.
[With this notation N (ACdk) will stand for the words,
" The number of permutations of the n things subject
to the conditions that a and y are in their proper places
and 8 and K not in their proper places"}
Then we have
N = [n
and N(A) = \n-l
But since every permutation must satisfy one and
188 APPENDIX III.
only one of the conditions expressed by (^4) and (a) it
follows that
N(A) + N(a) = N
therefore N(a) = [n — \n — 1
Now if we introduce the condition (B) our choice
will be the same as if /3 did not exist, and therefore the
same as if we had n — 1 elements to deal with instead of
n elements. Hence writing n — I for n we obtain from
the last equation
N(aB) = [n-l - [n-2
And by subtraction — remembering that the conditions
(B) and (b) are complementary —
N(ab) = [n - 2 [n-l + [ro-2
Repeating our former operation, writing n — 1 for n
on the introduction of the condition (C) we have
N(abC) ~ \n- 1 - 2[w-2 + |w-3
and by subtraction
A"(a6c). = [w - 8 [ro-1 + 3 [n-2 - [w-8
Similarly
N (afccd) = [w - 4 |n-l + 6 [ii-2 - 4 [ro-3 + \n-4
and so on, the coefficients following the same law as in
the Binomial theorem.
Hence finally,
N(abcd...k) = [n -n-
n(n-l)(n-2)r
-^ 2 3 - [n — 6 +
DERANGEMENTS. 189
= M1~i:T+lo~ [Q + &C. to n + 1 terms.}
( \i Lg L£ J
= [n.e~1. Q. E. D.
The student who is familiar with the use of symbols
of operation will arrange the foregoing proof briefly as
follows :
N=[n
N(A) = J[n
subtracting
N(a) = (l-J)[n
then N(aB)=J(l-J)[n
and subtracting
N(ab) = (l-
And so every introduction of a condition such as (A)
produces a factor of operation J, and every introduction
of a condition such as (a) produces a factor of operation
(1-J)
Hence N(abc...k) = (l-J)n[n = [n.e~\
ANOTHER PROOF OF PROPOSITION XXIX.
Let /„ denote the number of ways of deranging a set
of n elements so that no element may be in its proper
place.
Then the number of derangements so that exactly
r elements may be in their proper places will be
190 APPENDIX III.
since such derangements are obtained by first selecting
r elements to retain their proper places and then
deranging the remaining n — r.
But all the \n permutations of the n elements must
be made up of those in which 0, 1, 2, &c., have their
proper places.
Hence
[» =/. + " /„ + -nlr /- + •••• + n/, + 1
And, if we establish the convention that/r is always to
be replaced by fr we may write
[n= (/+!)"
Similarly \n - 1 = (/+ 1)" '
|n-a= (/+!)-
and so on.
Now multiply these equations in order by the coeffi
cients in the expansion of (1 — x)n and add (having
regard to algebraical signs) and we obtain an equation
of which the first member is
n r n (n — 1)
. r
[n + j \n - I +
1.2
n — 2 — &c.
or [n \l - . j + ,x ~" IQ + &c- to w + 1 terms. I
or [w-e^"1
while the second member is , (/ + 1) — 1 [ n or fn which
by our convention represents jn
Therefore /„ = [n.e^1 Q. E. D.
DERANGEMENTS. 191
COROLLARY I. — The number of ways in which n
elements can be deranged so that not any one of r
assigned elements may be in its proper place (the rest
being unrestricted) is
\n — - [w — 1 + • \n — 2 — . . . ± [»i — r
+ &c.
[2 ii (»-•!) |8»(fi-l)(ft--.2)
to w+1 terms, r
This is established passim in the first proof of the
Proposition.
COROLLARY II. — The number of derangements of
m-\-n elements so that m are displaced and n not
displaced is
COROLLARY III. — If an arrangement of n elements
be re-arranged at random, the chance that no element
will be in its original position is e~*.
COROLLARY IV. — If an arrangement of an infinite
number of elements be re-arranged at random the
chance that no element will be in its original position
is e -1 or -.
EXAMPLE.-— Suppose we have the four elements
o
192 APPENDIX III.
abed; the number of derangements, so that all may
be displaced, is by the proposition
1 2 6 ' 24,
These nine derangements are as follows :
b d a c c a d b d c a b
bade c d b a d c b a
b c d a c d a b d a b c
If it be required to derange the same terms so that
two may remain in situ and two be displaced, the
number of derangements is, by Corollary II.
12J1 -1+|}= 6.
These six derangements are as follows :
a b d c a d c b a c b d
b a c d c b a d d b c a
PKOPOSITION XXX.
The number of ways of deranging a series of n terms
so that no term may be followed by the term which
originally followed it is [n en + \n — 1 e~n-i
Let a, |8, y, . . .x represent the n terms. Then amongst
the \n arrangements of which the terms are capable
there will be \n — 1 in which any assigned sequence
a/3 occurs : (for the arrangements will be obtained by
regarding a/3 as one term and then arranging it with
the remaining w-2 terms.) Similarly any two
DERANGEMENTS. 193
sequences which can consistently occur (as a/3, /3y, Or
a/3, yS)* will be found in [rc-2 different arrangements.
Any three consistent sequences will be found in
\n — 8 different arrangements : and so on.
Hence there are [n arrangements altogether, among
which we should find
a/3 in \n — 1 of them
/3y in \n-~L — \n - 2 more of them
yS in [n — 1 — 2 [n — 2 + [w - 3 more
3 2
$e in [w — 1 — 3 [w — 2 + -—. \n — 3 — | w — 4 more
- JL • ^ ~~ — -•' "
and so on for all the n — 1 sequences. Therefore the
number of arrangements free from any of these
sequences is
. n(n — l)r ~ w(ft — l)(ft — 2).
-
&c.
= [n . e71-1
or, adding and subtracting unity,
= \n . e~l + \n-l . e~1^
Q. E. D.
The foregoing result may be written in very con
venient form by the use of the notation explained on
page 186. Thus
\n e~l + [n-1 e^ = (l-J)n\n + (l-J^n-l
* Of course such sequences as a/3 ay could not consistently occur,
as a could not at the same time be followed by ft and y.
194 APPENDIX III.
or, since the operation (1— J)n is equivalent to the
operations (l-J)n~l (1-J)
- n-l - (l-J)n~l [n-l
Or we may establish the result in this form inde
pendently as follows.
PKOPOSITION XXXI (otherwise.)
The number of ways of deranging a series of n terms
so that no term may be followed by the term which
originally followed it, is (1— J")""1 \n
For there are \n arrangements of the n terms a, 0, y . . . *
and we should find
a/3 in J [n of them
/3y in (J-JZ) [n or (1-J) J [n more of them
y& in (J-2J2+J3) [n or (l-J)*J[n more:
and so on.
Hence the whole number of arrangements containing
at least one of the n — 1 sequences a/3,/3y,y§, ... IK is
and therefore the number of admissible arrangements is
(!-</)"->
Q. E. D.
DERANGEMENTS. 195
COROLLARY. — The number of derangements of a series
of n terms, free from any of r assigned sequences
(which might occur simultaneously in one arrangement)
is(l-J)r[n
EXAMPLE. —Let us derange the series of four elements
abed so as to exclude the sequences ab be cd.
By the proposition the number of derangements is
[4 - 3 [3 + 3 [2 - [1, or 11.
And they are found on trial to be
acbd bdca cadb dbac
adcb lade chad dcba
Mac cbda dacb
PROPOSITION XXXII.
The number of ways of deranging the series of n
terms a, /3, y, . . . . *, *, so that none of the n sequences
«/3, /3y, ... ix, xa may occur is
[n .
For, as before, there are [n arrangements of the n
terms, and
a/3 occurs in [n - 1 of them,
|8y in [w — 1 — \n— 2 more of them,
y& in [n — l - 2[w^2_ + fo — 3 more,
and so on for all the n sequences, except that in the
case of the last one the final term [0 must be rejected
196 APPENDIX III.
since there cannot be any arrangements containing all
the n sequences.
Therefore the whole number of admissible arrange
ments is
[n — n \n-l -{ -j— o~ [ft -2 - &c. to n terms.
a result which may also be written
n(l-J)*~l\n-l
EXAMPLE. — Let us derange the series of four elements
abed so as to exclude the sequences ab be cd da.
By the proposition the number of derangements is
f 1 1 _ 1)
L- j II 12 [3 I or ^ : an(^ *key are f°und ^0 be
acbd bdca cadb dbac
adcb bade cbad dcba
PROPOSITION XXXIII.
If n terms be arranged in circular procession the
number of ways in which they can be deranged so that
no term may be followed by the term which originally
followed it, is
(1 1 , 1 1 1)
-\n ~ n-l + [2 (w-2) ~ [3 (w-3) "* t [»j
For the whole number of arrangements of n things
in circular procession is \n — 1 ; and the sequence
DERANGEMENTS. 197
a/3 occurs in |w-2 of them,
/3y in |n — 2 - [w— 8 more,
y § in [w — 2 — \n — 3 + |n— 4 more
and so on for all the n sequences, provided *we replace
| -1 in the last term by unity. Hence the number of
arrangements free from any of these sequences is
|n-l -n [w-JJ +^7rV-3 - &c. to (w+1) terms.
Q. E. D.
If we establish the convention that Jn \n-~L = [— 1 is
to be replaced by unity, the above series is seen to be
the algebraical expansion of (1- J)n \n — l, in which
form the result is easily remembered.
EXAMPLES. —
If n = 3, the number of derangements may be written
[2 - 8[1 + 3[0-1 = 1
or the only available derangement is the one in which
the order of the terms is reversed.
If n — 5 we have
|4 - 5 [8 +10 [2 - 10 [1 + 5 [0 - 1 =8
If abode represent the original order the eight derange
ments may be exhibited as follows
acbed aebdc acedb aecbd adceb
adbec acebd aedcb
198 APPENDIX III.
If n = 6 the number of derangements is 36 : if n = 7
it is 229 : and if n = 11 it is 1214673. And always if
n be a prime number the number of derangements, with
2 added, is. a multiple of n.
In the foregoing propositions we have investigated
the number of ways of deranging groups of elements
subject to various laws. But as there can scarcely be a
limit to the variety of laws which might be proposed to
regulate the distribution in different cases, it would be
an endless task to undertake a strictly complete discus
sion of the subject, or to make our treatise exhaustive.
The cases which we have considered are those which
most obviously arise, and the methods which we have
applied to them will be easily adapted to a variety of
other cases, or will suggest other methods of still wider
applicability.
APPENDIX IV.
ON THE DISADVANTAGE OF GAMBLING.
"If (says Professor Rogers,) we are to understand
the very elements of political economy, we must get
rid of the impression, that if the contract be voluntary
and the service be mutual, one man's gain is another's
loss The real truth is exactly the reverse ; for
one man's gain in all acts of free exchange is another
man's gain.*" A fair bargain is a mutual benefit to the
persons between whom it is made. If this were not so
all commerce would be immoral, for no man could seek
his own commercial profit without compassing the
injury of his neighbour and so violating the law of
civilised humanity.
But as a fair bargain is an advantage to both the
contracting parties, so, speaking generally, a fair wager
is a disadvantage to each party who enters into it. By
a fair wager, we mean one in which each party's stake
is equal to his mathematical expectation, calculated
according to the principles laid down in the chapter on
Chance : for instance, the wager is fair, if two men stake
* Manual of Political Economy, Oxford, 1868, p. 4.
200 APPENDIX IV.
equal sums, and either is to take the whole stakes
according as a coin falls "head" or "tail." It is a
common impression that such a wager as this, being
obviously of the same import to each contracting party,
can be neither an advantage nor a disadvantage to
either, and that consequently, any the slightest odds
must make it expedient for the party favoured by the
odds to enter into the wager. For instance, according
to this view, it must be decidedly expedient for any man
to stake a pound against a pound and a penny, (if he
can find any one foolish enough to give the odds of 241
to 240) that a coin will fall on an assigned side. Of
course a wager may be made on such unequal terms
that it may be decidedly expedient (from a selfish point
of view) for one party to enter into it, but it must then
be still more decidedly inexpedient for the other party.
We do not deny that a man who does not scruple to
take advantage of the ignorance or folly of another,
or to exert against his neighbour the intellectual
violence of superior knowledge or cunning,* may profit
ably enter into gambling speculations ; but we combat
the notion that there is a neutral advantage or dis
advantage in a fair wager, and that the contract to
play a game of pure chance for equal stakes, though it
be not expedient, cannot be branded as inexpedient,
and the delusion that, though no good be done, at least
* " Thou wouldst not take, by force or stealth,
What is not lav/fully thy right ;
But in the race for power and wealth
No wrong is done by mental might ! " — Monsell.
THE DISADVANTAGE OF GAMBLING, 201
no harm is done. It is this notion which palliates
gambling. If it were only recognised that a fair wager
were disadvantageous to each contracting party, it would
be regarded as disreputable for one man to cast this
disadvantage on another, even though he were accepting
the like himself. But at present the evil motives
which may lead men to gamble are covered by a
reputable cloak, in the charitable hope that each
party may be entering into a contract not disadvan
tageous to the other.
Every prospect of receiving anything of value, how
ever doubtful the prospect be, we must regard as having
itself some value. If a man have a chance, however
small, of receiving £100, he will not relinquish his
title to it without receiving something in return. He
may take £75, or £50, or it may be £2 or £1 for his
chance, according to his estimate of the probability of
his getting the prize, but if he has any chance at all,
his prospect must be worth something. When a man
makes a wager, he is buying such a prospect as this.
He pays down a certain sum of money and receives in
return a doubtful prospect of a larger sum. Whether
his bargain be advantageous or disadvantageous will
depend upon whether the sum that he pays down is
worth to him less or more than the prospect which he
buys, and we can only decide this by considering
whether he would gain or lose in the long run, if he
repeated his operation on the same terms for a very
great number of times. It is our object to prove
202 APPENDIX IV.
that if the sum which he pay be the mathematical
expectation of his prospect (and this is plainly the only
sum which he can fairly pay in justice to the other
contracting party), the bargain is disadvantageous to
him : in the long run he will lose by the repetition of
it.
For instance : if there be twenty tickets in a lottery
for a prize which is worth £1, the value of the expecta
tion of a man who holds one ticket is one shilling;
and it is plain that the organiser of the lottery cannot
without loss sell the tickets for less than one shilling
each : the fair price of a ticket is one shilling. But
according to our principle, it is inexpedient for a man
to give so much as one shilling for a ticket ; and though
the twenty tickets together are undoubtedly worth
twenty shillings, yet a single ticket is no more worth a
shilling than a single glove is, by itself, worth half the
price of a pair of gloves.
The twenty tickets in the hands of the original
holder were worth a pound, but when distributed to
twenty men (we say) they are not worth twenty shillings.
The establishment of the lottery to effect the distribu
tion was therefore inexpedient; the distribution itself
was on the whole disadvantageous.
But if the distribution of the tickets be disadvan
tageous, their collection must be in the same degree
advantageous. If therefore there exist in the nature of
things such chances as those represented by the tickets
of which we have spoken, it will be a beneficial and
THE DISADVANTAGE OF GAMBLING. 203
profitable act to collect those chances together. And
this is precisely what an Insurance Company does, when
it issues a policy undertaking to indemnify an owner
against accidents which may befal his property. The
man who insures his house against fire exchanges an
uncertain position for a certain one. The man who
buys a lottery ticket exchanges a certain position for an
uncertain one. Insurance is the reverse of gambling,
and is only wise in that gambling is foolish. The
consent of the civilised world to the proposition that
insurance is expedient is a tacit acknowledgment of
the truth of the cognate proposition that gambling is
inexpedient.
This will be seen more clearly by the consideration
of an example.
Suppose that out of every twenty ships which make a
particular voyage one is lost, and the remaining nine
teen come safely to port. And suppose there is one
ship making this voyage which with its cargo is worth
£20,000. The value of the owner's expectation, ac
cording to our chapter on Chance, is just £19, 000.
But, according to the hypothesis which we are illustrat
ing, it would be expedient for the owner to take a less
sum than this for his expectation, say £19,000 — x.
He may in consequence prudently pay £1,000+2 to an
Insurance Company in return for a guarantee that his
£20,000 shall be secured him in full. And the Insu
rance Company, collecting together a great number of
such risks, may profitably accept the bargain, their
204 APPENDIX IV.
profit being entirely dependent on the fact that the
shipowner is ready to accept for his contingent pros
pect an uncontingent sum, which is less than his
mathematical expectation. For if the Insurance Com
pany were to insure all the ships, securing to each
owner his mathematical expectation, their own mathe
matical expectation of profit would be zero. They could
only hope that the premiums received would in the
long run balance the claims upon them, without
leaving any profit to remunerate them for their trouble.
Thus, the continued existence of Insurance Companies
commercially successful is a standing witness to the
fact, that a prudent man will commute a contingent
prospect of value for less than the sum measured by
his mathematical expectation.
But some one will object that mathematics must be
utterly at fault, if an expectation is always worth less
than the value which mathematics would assign to it.
Not so. But if the mathematical result seems to con
tradict the conclusions of experience, or to violate the
plain dicta of prudence, it is not that the mathematics
have failed, but that two different problems have been
confused. The mathematician solves one problem : the
speculator seizes the result, and expects it to answer to
another. It is not a true statement of the principle we
have been enforcing to say that an expectation is
always worth less than the value which mathematics
would assign to it. The price which a» speculator may
THE DISADVANTAGE OF GAMBLING. 205
prudently give for a contingent prospect of value
depends always upon the amount of money he has to
speculate with. But that which is commonly called
the mathematical expectation of the contingency is the
price which a man of infinite means might prudently
pay for it. Therefore, the true statement of the case is
rather that an expectation is worth less to a man of
limited means, than the value which mathematics
assign to the like expectation to a man whose means
are unlimited.
The value of an expectation to any particular man, is
the stake which he may prudently lay down for the
sake of the expectation. That this depends upon the
man's means is evident, as soon as we consider an
extreme case. However foolish it may be for a man
who is possessed of thousands of pounds to make a bet
of £100, we are sensible that if the same bet were made
by a man who possessed nothing in the world but the
£100 which he risked the folly would be very much
greater. Or, however advantageous it might be for the
rich man to stake £100 in a speculative venture, it
would scarcely be prudent for the poor man to do the
same, and to run a risk of absolute ruin.
But the reader will expect that we appeal not to his
sense of what is prudent, but that we rather shew him,
by close mathematical reasoning, how the value of an
expectation depends upon the means of the speculator.
The question whether a particular speculation is
206 APPENDIX IV.
advantageous or disadvantageous to a particular man,
can only be tested by considering whether, if he repeat
the operation continually, he will in the end gain or
lose. But what is to be understood by repeating the
operation? If a boy, who has only a shilling, tosses
for sixpence, he is staking half his property. Sup
pose he wins, and so becomes the possessor of eighteen-
pence. If we now speak of his repeating the operation,
we may mean either of two things (1), that he is again
to stake sixpence ; or (2), that he is again to stake half
his property, which will now be ninepence. So also, if
he loses, there will be the like ambiguity as to whether
he is to repeat the operation by staking his remaining
sixpence, or only by staking threepence.
And so always, as soon as we take into consideration
the funds at the disposal of a speculator, the stake
which he lays down in any venture may be considered
either as an absolute amount, or as a certain fraction
of his entire fund. And by the repetition of his ven
ture, we may either understand that he stakes the same
sum again and again, or that he always stakes the same
fraction of the fund which he holds at the time of
making the venture.
But, whichever of these views we take, we are led
mathematically to the same conclusion, that a fair
wager is disadvantageous to a man whose means are
limited, because, if it be repeated indefinitely, he will
in the long run be the loser.
THE DISADVANTAGE OF GAMBLING. 207
First, — Suppose that the same sum is repeatedly
staked, and for the sake of simplicity suppose the sum
to be gained the same as the sum staked, as in the case
of an even bet. The speculator repeatedly stakes (say)
a pound against a pound, so that every venture issues
in making him either a pound richer or a pound poorer,
each of these results being equally likely. If he con
tinue his operation a great number of times, the balance
of profit and loss will be continually varying. At one
time he may have gained a considerable number of
pounds, at another time he may find himself much
poorer than when he began. And if his means were
unlimited, he might go on gambling for ever; and
there would be no reason to expect him to leave off
a loser rather than a gainer. But his means being
limited, this equilibrium of chances is disturbed. He
has only (say) n pounds to begin with. The balance of
profit and loss may oscillate : now he may have gained
x pounds, and now he may have lost y pounds, and the
balance may again recoil ; but as soon as ever a loss of
n pounds is reached, there is no more hope of restitu
tion, for he has nothing more to venture. If his funds
had been unlimited, there would have been an equal
prospect of his leaving off richer or poorer, but the
limitation of his resources, being apt to put a sudden
termination on his operations at a time when the balance
is most against him, interferes with the equality of
chances, and occasions a presumption that he will leave
off the loser. If, indeed, he were gambling with another
p
208 APPENDIX IV.
speculator whose means were also limited, we should
have to set against the prospect of the game abruptly
terminating against him the counter prospect of its
abruptly terminating in his favour by the exhaustion of
the other man's funds; but, in fact, no one is restricted
to gambling with one single opponent ; the speculator
deals with the public at large, with a world whose
resources are practically unlimited. There is a prospect
that his operations may terminate to his own disadvan
tage, through his having nothing more to stake ; but
there is no prospect that it will terminate to his
advantage through the exhaustion of the resources of
the world. Every one who gambles is carrying on an
unequal warfare : he is ranged with a restricted capital
against an adversary whose means are infinite.
We have said there is a chance of the man being
ruined who with limited means continues to stake a
pound against a pound in a fair wager. But if he
prolong his play indefinitely, the chance of his being
ruined is not distinguishable from certainty : it ap
proaches nearer to certainty than by any assignable
difference ; and the chance that he should escape ruin
is less than any assignable chance. Thus : —
PKOPOSITION XXXIII.
If a man with limited funds repeatedly stake a pound
against a pound, in a fair wager > the chance of his
not being ultimately ruined is less than any assignable
chance.
THE DISADVANTAGE OF GAMBLING. 209
Suppose liis original funds are n pounds: and let
RT represent the chance of his being ultimately ruined
when he has x pounds in hand. At the same time
l—Rx will be the chance that he will escape ruin.
If at any time he have only one pound, the next
venture must either ruin him or double his fund : the
chance of either of these issues is one-half, and in the
former case his chance of ruin is RQ or unity, in the
latter case it is R2. Hence,
or " 2E, = 1 + -Ra
Similarly if at any time he have x pounds a single
venture must leave him with either x—1 or #4-1 and
so we have
This being true for all values of x it follows that
l,Rl}R2tR3.... are in arithmetical progression, and
if we write
B, = i-p
then Rn = 1—np
and if we take the case of any one else whose funds are
z pounds his chance of ruin is Rz — 1—zp, where p is
a constant quantity whatever be the value of z. But
however great a man's funds may be, his chance of ruin
can never be negative. Hence, however great a value
we assign to zy Rs can never be less than zero, and there
fore zp (which is 1 — Rt) can never be greater than unity.
210 APPENDIX IV.
Therefore p is not greater than * , and np not greater
than - however great z may be. But by choosing z
large enough while n is finite we may make £ less than
any assignable quantity, and therefore np which is
invariable and not greater than - must be less than any
assignable quantity. Therefore 1 - Rn is less than any
assignable quantity, and therefore the chance that the
man is ultimately ruined differs from certainty by less
than any assignable chance, and the chance that he
escapes ruin is less than any assignable chance.
Q. E. D.
The next proposition is not necessary to our present
argument, but is added here as illustrating this part of
the subject. It reduces to the case of the foregoing
proposition when /* = 1.
PROPOSITION XXXIV.
If a man with limited funds (£n) repeatedly stake a
pound against a pound in a wager in which the odds
on every venture are p : 1 in his favour, the chance that
he is ultimately ruined is -^.
Arguing as in the last proposition, and observing
that the chances of his losing or winning any venture are
1 I*
respectively and -, we have
+
1+jM,
THE DISADVANTAGE OF GAMBLING. 211
or (!+/*) ^ = 1+1^
whence JR. - 722 = - (1 — -ft,) »
jLfc
So £2 - R3 = * (1^-iy
and generally R^-R* = (R^-R^
Therefore by continued multiplication
•frc-i - Rx = -*=i (1-^)
Now give x successively all the values from 1 to n and
add, and we get
But Rn can be zero only when n is infinite, therefore
,.,.3
and consequently
i
-| T"> -| *
n~ ' ~ ' [S
or
Q. E. D.
212 APPENDIX IV.
To resume our argument :— Proposition XXXIII.
shews that, on the strictest mathematical principles,
a man wfto continues to stake a constant sum in a
fair wager must expect to be ruined in the end.
Such a course of gambling — unregulated by any con
sideration of the gambler's disposable fund, until the
absolute exhaustion of the fund places a peremptory
limit on his play — is therefore manifestly inexpedient.
And if it be inexpedient in the long run, it cannot be
expedient even when carried to any limited extent.
For any limited extent of play which may be thought
to be expedient, must leave the gambler either poorer
or richer than he was at first. If it leave him poorer,
it cannot be expedient. If it leave him richer, the
same course of gambling which was expedient at first,
must be, a fortiori, expedient now ; and, therefore, it
cannot be expedient for him to stop. Hence, no play,
consisting of simple repetitions of the same venture,
which is proved to be inexpedient in the long run, can
be expedient when confined to any assignable extent.
But it may be said that a wise speculator will always
regulate his ventures by his means. If, when he
posesses a shilling, he stakes sixpence, the repetition
of the act is not to be sought for in a constant
staking of sixpence, but in a constant staking of half
his fund. And if he always keeps as much as he
stakes, it is plain that he can never be absolutely
ruined. The reasoning of Proposition XXXIII. will
not now apply. We must seek some other proof, if
THE DISADVANTAGE OF GAMBLING. 213
we would show that a course of gambling thus regulated
is in the end disadvantageous.
The proof is easily found. Let us suppose that the
man having n pounds begins as before by staking one
pound against one pound in an even wager. He stakes
one nth part of his fund, and he continues the operation
by staking every time not one pound but always one
nth part of the fund which he then possesses. Every
venture which issues in his favour increases his fund
by one nth part, or multiplies it by (l + H. Every
venture which turns out against him decreases his fund
by one nth part, or multiplies it by f 1 - -J. A
favourable and an unfavourable issue will therefore
together multiply his fund by
that is, the two operations will decrease his fund by
I — )th part: and this will be the case whether the
\n*J *
gain precede or follow the loss. Thus a gain and a loss
do not balance one another, but they leave a net loss.
And in any number of ventures in which there are the
same number of profitable and unprofitable issues,
there will be a resultant loss, greater as the number
of ventures is greater. Now the man cannot expect
in the long run to win oftener than he loses, and
therefore if he repeat his operations indefinitely he
must expect to lose in the end.
214 APPENDIX IV.
For example, if he begin with £100 and always stake
one tenth of his fund, a single gain will raise his
fund to £110 ; he will then stake £11, and if he lose
he will only have £99 left. Or if he had lost the first
venture he would have had £90 left : then he would
have staked £9, and if he had won he would then have
had £99 as before. In either order the gain and the
loss reduce £100 to £99, and ten gains and ten losses
in any order would reduce it to £90 8s. lljd.
PKOPOSITION XXXY.
There are m tickets in a lottery for one prize of value
A. To determine ivhat price may be paid for a ticket
by a man ivhose available fund is nA, so that by
repeating his operation an average number of gains may
balance an average number of losses.
" Kepeating his operation" will here mean that when
the first venture is decided the man will purchase a
ticket in another lottery in which the prize is the nth
part of the amount of the fund which he then holds.
xA
Suppose - - the sum which he may pay for a ticket.
m
Then every unsuccessful venture multiplies his fund by
IT
1 • and every successful venture multiplies it by
mn
1 x
H — . But in the long run he will have m - 1
n mn
unsuccessful issues for one successful one. Therefore
THE DISADVANTAGE OF GAMBLING. 215
the average multiplier for m ventures will be
1+---— 1- —
n mn } \ mn
This must be unity since the gains, on the average,
balance the losses. Therefore
1 T I X \-(m~^
1+--— =(!---)
n mn \ mnj
an equation to determine x.
If n be a large number (as is usually the case) we may
obtain an approximate solution. Thus
n mn mn
f\ m-l
or 0 = - 1 + # + ~— -
Zmn
~v, 1
whence x — 1 —
Consequently it is inexpedient for one to give more than
/ _m—\\A
for the chance ( — ] of the prize (A).
\mj
COROLLABY. — If he buy two tickets in the same
lottery his chance of a prize is — . Hence writing ~
for m in the result of the proposition we obtain the
price he may pay for two tickets, viz.
/ m-2\2.4
\ Zmn ) m
216 APPENDIX IV.
or he may pay for each of them
^m-avi
2mn Jm
So if he purchase r tickets, he may pay for each of them
'm
or if he purchase all the m tickets, he may pay for each
its full representative value — .
PKOPOSITION XXXVI.
There are a + b + c+ ...=m tickets in a lottery, and
there are a blanks, b prizes each worth /3, c prizes each
worth y, and so on. To determine ivhat price a man
whose available fund is n may prudently pay for a
ticket.
Let a) = b (B + cy + ... = the sum of the prizes,
and let £l = bfi*+ cy*+ ...
Then the absolute value of a ticket is — .
m
Let be the price which the man in question ought
to pay for the ticket,
Then if we proceed as in the last question we find
the average multiplier for m ventures, viz.
mn n mn \ n mn
This must be equal to unity. Therefore
THE DISADVANTAGE OF GAMBLING. 217
mn n mn
+ ... -0.
from which equation x is to be found. If, as is usually
the case, the amount at stake is small compared with
the speculator's whole funds, n must be a very large
number ; therefore we shall obtain an approximate
value of x, by neglecting the terms involving high
negative powers of n, in our equation.
Thus, expanding the logarithms, we have
2x-x2 «>2
&c.
2 —
* m %n*
whence x==1
Consequently, the price which the man may prudently
pay for a ticket is
(
— co. co
COROLLARY. — The price which a man whose available
fund is n pounds may prudently pay for a share in a
speculation in which p^ will be his chance of winning
PI, pz his chance of winning P2, and so on (where
be
This is derived immediately from the result of the
proposition, by writing
and - = S
m m
218 APPENDIX IV.
EXAMPLES. — Having n shillings, what may one pru
dently pay to be entitled to a number of shillings equal
to the number turned up at a single throw of a die ?
Here
=4(1+4 + 9 + 16+25+36) = ^
D D
And the required number of shillings is approximately
7,
Again. If the throw be made with two dice, we
shall have (see page 91)
2(pP) = 7 2 (pP*) = 2023-^-36
And the required number of shillings will be approxi
mately
72 n '
It must be remembered that the results of the two
foregoing propositions and the corollary are only obtained
on the hypothesis that n is very large compared with
2(pP) and with S^P2) -5-2(pP). This requires that
the man's original fund should be very large compared
not only with the amount which he stakes, but also
with the amount which he has a chance (though it may
be a very small chance) of winning.
THE DISADVANTAGE OF GAMBLING. 219
In all practical cases the former conditions will be
fulfilled, as no one would think of staking, in any single
venture, his whole property, or a sum bearing any
considerable ratio to his whole property. But cases will
be likely to arise in which the latter condition is not
satisfied, as when a man may purchase, for a sum com
paratively small compared with his means, a ticket in
a lottery in which there is a prize very many times
larger than his whole property. In this case, the
approximate results obtained above cannot be applied,
and we must have recourse to the original equation in
the form in which it was presented before we introduced
the approximations which are now inadmissible. We
may, however, still approximate, in virtue of the con
dition that the stake must be small compared with the
speculator's whole funds. Thus : —
PEOPOSITION XXXVII.
To find an approximate formula for the sum which
a speculator may pay for any defined expectation,
without assuming that his funds are necessarily large
compared ivith the value of the prizes.
Let X (small compared with n, though not necessa
rily small compared with P) be the sum which a man
whose available fund is n may prudently pay for the
chances pv pz, p3... of receiving prizes worth Plf P^
P8, ... respectively, as in the last corollary.
220 APPENDIX IV.
Then the rigorous equation to determine X is
n n j \ n n \ n n
which may be written
p
1+
n \ n n
w> /.
J t
V1 (-I--*— Yp° fi x VP8
ij \ w+pj v n+pj •••
or, since X is small compared with n,
jTp~+ n+P + "j
P\P* f PVS
+ =-*} 1+^-5 1...-1
w / v » /
whence X =-—
This formula is equally applicable when there is a
possibility of not receiving any prize, as the failure to
receive a prize may be treated as the receiving of a prize
of zero value : i. e. one of the quantities P19 P2, P3 ...
will be in this case zero.
COROLLARY. — In the case when there is a single
prize P, and the chance of gaining it is p, the formula
becomes
X
p +1-P
n + P n
n(B*P) I/, , -PV
n + (l-p)P\\ n I
THE DISADVANTAGE OF GAMBLING. 221
EXAMPLES. — A man possessing a pound is offered a
ticket in a lottery, in which there are 99 blanks, and
one prize worth 100 pounds. What may he prudently
pay for the ticket ?
Here n=l P=100
x *0472 ='0519
Hence he can only afford to pay about a shilling for
the ticket.
Again. How much may a man with 10 pounds pay
for the same ticket ?
i
Here n = 10 P = 100 p = -
1100 /100 /IT
x " 109 ( V J
1100
Hence he can afford to pay about 4s. lOJd. for the
ticket.
Again. How much may a man with 100 pounds pay
for the same ticket ?
Here rc=100 P=100
_ 20000 /100 /y
A " 199 ( V *
= 20020 x -0069 = -693.
199
Hence he may pay nearly 14 shillings for the ticket.
222 APPENDIX IV.
Again. Suppose the man has 1000 pounds. This
sum is large compared with the value of the prize.
Hence we may apply the simpler formula given in
Proposition XXXVI. (Cor.), which leads to the result
Hence he may pay 19 shillings for the ticket.
If we had applied to this last case the formula
of the present corollary we should have obtained the
slightly more correct result '9507, the difference between
the two results being nearly one-fifth of a farthing.
The formula of Proposition XXXVII. is applicable
in the case of the Petersburg problem, a problem of
some intrinsic interest, but chiefly of importance on
account of its having been repeatedly made the ground
of objections to the mathematical theory of probability.
This celebrated problem may be stated as follows :
THE PETERSBURG PROBLEM.
A coin is tossed again and again until a tail is
turned up. If the first throw give a head one is to
receive a florin, if the second also give a head one is to
receive two florins more, if in addition the third throw
give a head one is to receive four florins more, and so
on, doubling the sum received every time ; but as soon
as a tail is turned up the play stops and one receives
nothing further. The question is, what ought one to
give for the expectation ?
THE DISADVANTAGE OF GAMBLING. 223
The absolute value of the expectation is easily seen
to be infinite. Thus —
The chance of winning the florin at the first throw
is j : the value of the expectation is therefore one
shilling.
The chance of winning the two florins at the second
throw is J : the value of the expectation is therefore
J of two florins, or one shilling.
The chance of winning the four florins at the third
throw is £ : the value of the expectation is therefore
J of four florins, or one shilling, and so on.
Hence the value of the expectation attaching to each
throw to which there is a possibility of the play
extending, is one shilling. But the play may possibly
extend to a number of throws larger than any assignable
number. Therefore the whole expectation is worth
a number of shillings larger than any assignable
number : that is, it is infinitely great.
Or we may analyse the expectation by considering
the various total sums which it is possible to receive,
and the chance of each being received. Thus—
If a tail turn up the first time, nothing is received :
and the chance of this is j-.
If a tail turn up the second time and not before, one
florin is received : and the chance of this is 5.
If a tail turn up the third time and not before, 1 + 2
florins are received : and the chance of this is £.
And so on. Hence
Q
224 APPENDIX IV.
J = chance of receiving 0
jp-= 1 = 2 - 1 florins.
|r = 1 4- 2 = 22-l florins.
gr* 14-2 + 4 = 23-l florins.
and so on, ad infinitum.
Therefore the value of the whole expectation is (in
florins)
!_ J^
2 ~" W
Now the sum of this series to r terms* is
r-l fl\r+1
~o — ^ 1 o ) ' a ^° an infinite number of terms
it is infinity.
Hence the value of the mathematical expectation is
infinite, as we showed before.
But here a difficulty is raised. The mathematical
expectation has been found to be of infinite value, and
yet (it is objected) no one in his senses would give even
such a moderate sum as ^£50 for the prospect defined in
the problem.
The fallacy of this objection has been pointed out
already. We have shown that the absolute value of . a
* It is to be observed that the sum of this series to r terms does
not exactly correspond with the expectation when the play is limited
to r throws, because when the play is thus limited the chance of
winning the whole number (2r - 1) is not aF+T» ^ut is ~fp » tne same
as the chance of winning 2r~ l - 1 florins.
THE DISADVANTAGE OF GAMBLING. 225
mathematical expectation is not the price which a man
of limited means ought to pay for the prospect. It
expresses the value of the expectation to a man who is
able to repeat the venture indefinitely without the risk
of his operations being ever terminated by lack of
means. The speculator's fund, to begin with, must be
infinite in comparison with the stakes involved, before
he may venture to give the absolute value of the
mathematical expectation for any contingent prospect
which he may desire to purchase. In the Petersburg
problem the mathematical expectation is infinite : but
if one is to give an infinite sum for the venture one
must take care to hold funds infinite in comparison with
this infinity. In other words, the speculation on these
terms is only proper for one with respect to whose
funds the infinite stake is inconsiderable. The stake
which he lays down may be 00 , provided that his funds
are 00 x QO .
But to find the sum which a man of limited means
may pay for the expectation defined in our statement
of the problem we may apply the result of Proposition
XXXVII. Thus if m + 1 be the number of florins
which the man possesses, the formula to determine
the sum which he may pay may be written as
follows :
2(m + l)
226 APPENDIX IV.
For example. If the speculator possess nine florins
we have ??i=8. The numerator is "2137 ... and the
denominator '0966 .... Hence X = 2'212.
If he possess 33 florins we have ?7i=32. The nume
rator is now '0807... and the denominator '0278... .
Hence he may pay nearly three florins.
If he possess 1025 florins we have m=1024. The
numerator is now '00488 ... and the denominator
•00097... . Hence he may pay ahout five florins for
the venture.
The result at which we have arrived is not to be
classed with the arbitrary methods which have been
again and again propounded to evade the difficulty of
the Petersburg problem and other problems of a similar
character. Formulae have often been proposed, which
have possessed the one virtue of presenting a finite
result in the case of this famous problem, but they have
often had no intelligible basis to rest upon, or, if they
have been established on sound principles, sufficient
care has not been taken to draw a distinguishing line
between the significance of the result obtained, and the
different result arrived at when the mathematical expec
tation is calculated.
We have not assigned any new value to the mathe
matical expectation ; we have not substituted a new
expression for the old ; but we have deduced a separate
result, which without disturbing the mathematical ex
pectation h^s a definite meaning of its own. We
THE DISADVANTAGE OF GAMBLING. 227
have found not the fair price at which a contingent
prospect may be transferred from one man to another,
but the value which such a prospect has to a man in
given circumstances. We have simply determined the
terms at which a man may purchase a contingent pros
pect of advantage, so that by repeating the operation —
each time on a scale proportionate to his funds at that
time — he may be left neither richer nor poorer when
each issue of the venture shall have occurred its own
average number of times. By continuing the operation
indefinitely, the recurrences of each issue will tend to be
proportional to their respective probabilities, and, there
fore, the condition we have taken is equivalent to the
condition that in the long run the man may expect to
be neither richer nor poorer.
It would be a great mistake to suppose that the
price which one man may prudently give for a venture
is the price which the man with whom he is dealing
may prudently take for it, or that it is a fair price at
which to make the compact. The price which the man
may prudently give, is not even the price which the
same man may prudently take if he change sides with
his fellow gambler. The sum in consideration of which
a man possessed of n pounds may accept a position in
which p± is the chance of his having to pay Pv pz the
chance of his having to pay p.2, and so on f where
2p = lJ must be obtained by changing the algebraical
signs of X, jP1? P2 . . . in the formulae of Propositions
XXXVI. and XXXVII.
228 APPENDIX IV.
Thus on the hypothesis of Proposition XXXVI.
(Cor.), we shall have
and in the more general case dealt with in Proposition
XXXVII. we shall have
/ PVi/ P\P>/ P\P*
1-fl--1 I--2 I--3) ...
X= ^ n J \ nj\ n )
^^l+n-P,+n^Ps+'''
The historical notes which follow are mainly derived
from Mr. Todhunter's History of the Mathematical
Theory of Probability.
The volume of the Commentarii of the Petersburg
Academy for the years 1730 and 1731, was published in
1738. It contained a memoir by Daniel Bernoulli,
entitled, Specimen Theoriae novae de mensura sortis,
expounding a theory of moral expectation as distin
guished from mathematical expectation. The author
estimates that if a man's fund is increased by a small
increment, the value of the increment to that man
varies directly as the increment, and inversely as his
original fund. But while he assumes this as a mathe
matical measure of what he regards as a moral value,
Bernoulli does not attempt to give any proof of his
assumption : and rightly, for it is beyond the province
of mathematics to deal with such a subjective value as
he speaks of. Mathematics can only be applied to the
measure of such a quantity by some such arbitrary con-
THE DISADVANTAGE OF GAMBLING. 229
nection as that which he assumes. But that which he
takes to represent his moral expectation is substantially
identical with the quantity which we have heen inves
tigating, viz., the price to be paid for a venture, in
order that repetitions indefinitely multiplied may tend
to neutralise one another. Bernoulli draws from his
theory the inference which we have established at the
beginning of this Appendix, that even a fair chance is
disadvantageous. The Petersburg problem, as he deals
with it, is somewhat simpler than the modern variety
of it, which we have enunciated above. A is to receive
a florin if head falls the first time ; two florins if it falls
the second time, and not before ; four florins if it falls
the third time, and not before, and so on. The mathe
matical expectation is infinite. For the moral expecta
tion, Bernoulli gives an equation equivalent to that
which we should write down in accordance with Pro
position XXXVII.
Daniel Bernoulli's memoir contains a letter addressed
to Nicolas Bernoulli, by Cramer, in which two methods
are suggested of explaining the paradox of the Peters
burg problem. One suggestion is, that all sums greater
than 224 are practically equal; the other (which is
equally arbitrary), that the pleasure derivable from a
sum of money varies as the square root of the sum.
On one of these suppositions the expectation in the
Petersburg problem, as enunciated by Bernoulli, would
be 13 ; according to the other, it would be about 2'9.
D'Alembert (in the year 1754) maintained that a
very small chance was to be regarded as absolutely
230 APPENDIX IV.
zero. He does not suggest a limit to the smallness,
but he gives an example in which the chance is (J-)100.
In another place he suggests that, in the Petersburg
problem, we should take (/3 being a constant)
-. instead of — ,
as the chance that the head will not appear before the
nth throw. From time to time he seems to have pro
posed a variety of arbitrary assumptions, for none of
which any better reason can be assigned than that they
lead to finite results.
Beguelin, in 1767, gave six different solutions of the
Petersburg problem, with different results.
In 1777, Buffon, the Naturalist, published his
Essai d'Arithmetique Morale, in which he speaks against
gambling in language singularly resembling that which
we have employed in the earlier pages of this Appendix.
" Je dis qu'en general le jeu est un pacte mal-entendu,
un contrat desavantageux aux deux parties, dont 1'effet
est de rendre la perte toujours plus grande que le gain ;
et d'oter au bien pour ajouter au mal." But, among
other arbitrary assumptions, this writer maintains that
any chance less than 10000 is to be considered abso
lutely zero.
Laplace, whose great work, the Theorie Analytique
des Probabilites, was published in 1812, has developed
many of Daniel Bernoulli's ideas on this subject.
MISCELLANEOUS EXAMPLES.
1. — There are five routes to the top of a mountain, in how
many ways can a person go up and down ?
2. — Out .of 20 knives and 24 forks, in how many ways
can a man choose a knife and fork ? And then, in how many
ways can another man take another knife and fork ?
3. — In how many ways can the letters a b c d be arranged
without letting b and c come together ?
4. — A has 7 different books, B has 9 different books, in
how many ways can one of A's books be exchanged for one
of B s ?
5. — In the case of the last question, in how many ways
can two books be exchanged for two ?
6. — Five men, A, B, C, D, E, are going to speak at a
meeting, in how many ways can they take their turns without
B speaking before A ?
7. — In how many ways, so that A speaks immediately
before B ?
232 CHOICE AND CHANCE.
8. — Five ladies and three gentlemen are going to play at
croquet, in how many ways can they divide themselves into
sides of four each, so that the gentlemen may not be all on
one side ?
9. — The number of ways of selecting n things out of 2?i + 2
is to the number of ways of selecting n things out of 2n — 2
as 99 to 7. Find n.
10. — One man has 4 books, another man has 6. In how
many ways can they exchange books, each keeping the
number he had at first ?
11. — One man has 4 books, another has 6, and a third
has 3. In how many ways can they exchange books, each
keeping the number he had at first, but every one's set being
altered ?
12. — Four digits are arranged at random so as to form a
number in the ordinary scale of notation. Two cyphers are
then associated with them, and they are re-arranged at random
so as again to form a number. Prove that the average value
of the first number is to the average value of the second as
101 to 6734.
13. — A ferry-boat which can carry n people has to convey
m people across a river. It takes a full load every time
except the last : find the number of ways in which the work
can be done.
14. — There are 2n guests at a dinner party ; supposing
that the host and hostess have fixed seats opposite to one
MISCELLANEOUS EXAMPLES. 233
another, and that there are two specified guests who must not
be placed next to one another, find the number of ways in
which the company can be placed.
15. — Out of three consonants and two vowels, how many
words can be formed containing 2, 3, 4, or 5 letters, words
being excluded in which two consonants or two vowels come
together.
16. — How many five-lettered words can be made out of 26
letters, repetitions being allowed, but no consecutive repeti
tions (i. <?., no letter must follow itself in the same word).
17. — A boat's crew consists of eight men, of whom two can
only row on the stroke side of the boat, and three only on the
bow side. In how many ways can the crew be arranged ?
18. — There are m parcels, of which the first contains n
things ; the second 2n things ; the third 3n things ; and so
on. Shew that the number of ways of taking n things out
of each parcel ia \jnn -f- S\n \ m
19. — How many different rectangular parallelepipeds can
be constructed, the length of each edge being an integral
number of inches not exceeding 10?
20. — The number of ways of dividing 2» different things
into two equal parts, is to the number of ways of similarly
dividing 4n different things, as the continued product of the
first n odd numbers to the continued product of the n odd
numbers succeeding.
234 CHOICE AND CHANCE.
21. — In how many ways can the letters of the word
facetious be deranged without deranging the order of
the vowels ?
22. — In how many ways can the letters of the word
abstemiously be deranged without deranging the
order of the vowels ?
23. — In how many ways can the letters of the word
parallelism be deranged without deranging the order
of the vowels ?
24. — How many solutions can be given to the following
problem ? " Find two numbers whose greatest common
measure shall be G and their least common multiple M
— GraabPcyd8 ; a, 6, c, d being prime numbers."
25. — How many solutions can be given to the following
problem? "Find two numbers of which G shall be a common
measure, and M (as in the last question) a common
multiple."
26. — Prove that the number of ways in which p positive
signs and n negative signs may be placed in a row, so that no
two negative signs shall be together, is equal to the number
of combinations of p + 1 things taken n together.
27.— In the expansion of (^ + az + ... + ap)n where n is
an integer not greater than p, there are
MISCELLANEOUS EXAMPLES. 235
terms, in none of which any one of the quantities % az...ap
occurs more than once as a factor; and the coefficient of
each of these terms is [n.
28. — Out of 20 consecutive numbers, in how many ways
can two be selected whose sum shall be odd ?
29. — Out of 30 consecutive integers, in how many ways can
three be selected whose sum shall be even ?
30. — Out of 3n consecutive integers, in how many ways
can three be selected whose sum shall be divisible by 3 ?
81. — If four straight lines be drawn in a plane and pro
duced indefinitely, how many points of intersection will there
generally be ?
82. — If n straight lines be drawn in a plane, no two being
parallel and no three concurrent, how many points of inter
section will there be ?
33. — If n straight lines be drawn in a plane, no two being
parallel, and no three concurrent except p which meet in one
point, and q which meet in another point, how many other
points of intersection will there be ?
34. — A square is divided into 16 equal squares by vertical
and horizontal lines. In how many ways can 4 of these be
painted white, 4 black, 4 red, and 4 blue, without repeating
the same colour in the same vertical or horizontal row ?
35. — Find the number of combinations that can be formed
236 CHOICE AND CHANCE.
out of the letters of the following line (Soph. Philoct. 746) :
a7ra7T7ra7ra» -TraTraTTTraTraTrTraTraTTTraTrai,
taking them (1) 5 together, and (2) 25 together.
86. — In the case of the preceding question, if the number
of combinations r together is to the number r — 1 together
as 9 : 10, find ?*, it being known that it lies between 17
and ^4.
37. — The number of ways of selecting 4 things out of n
different things is one-sixth of the number of ways of select
ing 4 things out of 2n things which are two and two alike of
n sorts : find n.
38. — If pq-\-r different things are to be divided as equally
as possible among p persons, in how many ways can it be
done ? ( r < p )
39. — In how many ways can a pack of cards be dealt to
four players, subject to the condition that each player shall
have three cards of each of three suits and four cards of the
remaining suit ?
40. — Into how many parts is an infinite plane divided by
n straight lines, of which no three are concurrent ?
41. — Into how many parts is infinite space divided by n
planes, of which no four meet in a point ?
42. — In how many ways can three numbers in arithmetical
progression be selected from the series 1, 2, 3 ... 2«, and in
how many ways from the series 1, 2, 3 ... ( 2?i-f 1 ) ?
MISCELLANEOUS EXAMPLES. 237
43. — If there be n straight lines in one plane, no three of
which meet in a point, the number of groups of n of their
points of intersection, in each of which no three points lie in
one of the straight lines, is % In — 1.
44. — 120 men are to be formed at random into a solid
rectangle of 12 men by 10; all sides are equally likely to be
in front. What is the chance that an assigned man is in
the front ?
45. — If the letters of the alphabet are written down in a
ring so that no two vowels come together, what is the chance
that a is next to b ?
46. — If the letters of the alphabet are written down in a
row so that no two vowels come together, what is the chance
that a is next to b ?
47. — A, B, C have equal claims for a prize. A says to B,
let us two draw lots, let the loser withdraw and the winner
draw lots with C for the prize. Is this fair ?
48. — Five men, A, B, C, D, E, speak at a meeting, and it
is known that A speaks before B, what is the chance that A
speaks immediately before B ?
49. — if n things (a, /3, y, &c.) be arranged in a row, subject
to the condition that a comes before /3, what is the chance
that a comes next before /3 ?
238 CHOICE AND CHANCE.
50. — Two numbers are chosen at random, find the chance
that their sum is even.
51. — There are n counters marked with odd numbers, and
n more marked with even numbers ; if two are drawn at
random shew that the odds are n to n — 1 against the sum of
the numbers drawn being even.
52. — The figures 142857 are arranged at random as the
period of a circulating decimal, which is then reduced to a
vulgar fraction in lowest terms. Shew that the odds are
119 : 1 against the denominator being 7.
53. — There are ten counters in a bag marked with num
bers. A person is allowed to draw two of them. If the
sum of the numbers drawn is an odd number, he receives that
number of shillings ; if it is an even number, he pays that
number of shillings. Is the value of his expectation greater
when the counters are numbered from 0 to 9 or from 1 to 10 ?
54. — If a head counts for one and a tail for two, shew that
Sn is the most likely number to throw when 2/? coins are
tossed. Also shew that the chance of throwing 3(n + l)
with 2(?i + 1) coins is less than the chance of throwing 3n
with 2rc coins in the ratio 2w + 1 : 2n + 2.
55. — A bag contains 2» counters, of which half are marked
with odd numbers and half with even numbers, the sum
of all the numbers being S. A man is to draw two counters.
If the sum of the numbers drawn be an odd number, he is to
receive that number of shillings ; if an even number, he is to
MISCELLANEOUS EXAMPLES. 239
pay that number of shillings. Shew that his expectation is
worth (in shillings)
S
56. — If in the case of the last question there be ra+n
counters, of which m are marked with odd numbers, amount
ing to M, and n with even numbers amounting to N, the
man's expectation is worth
M+N-(m-n)(M-N)
|- (m+n) (ra-fn — 1)
57. — What are the odds against throwing 7 twice at least
in three throws with two dice ?
58. — Two persons play for a stake, each throwing two
dice. They throw in turn, A commencing. A wins if he
throws 6, B if he throws 7 : the game ceasing as soon as
either event happens. Shew that A's chance is to J5's as
80 to 31.
59. — Four persons draw each a card from an ordinary pack.
Find the chance (i) that one card is of each suit : (ii) that
no two cards are of equal value : (iii) that one card is of each
suit and no two of equal value.
60. — Each of four persons draws a card from an ordinary
pack. Find the chance that one card is of each suit, and
that in addition, on a second drawing, each person shall draw
a card of the same suit as before.
61. — A bag contains %n(n+l) counters, one marked 1,
R
240 CHOICE AND CHANCE.
two marked 4, three marked 9, &c. A person draws out a
counter at random, and is to receive as many shillings as the
number marked on it. Prove that the value of his expecta
tion varies as the square of the number of counters in the bag.
62. — A and B throw for a certain stake, A having a die
whose faces are numbered 10, 13, 16, 20, 21, 25 ; and B a
die whose faces are numbered 5, 10, 15, 20, 25, 30. The
highest throw to win, and equal throws to go for nothing.
Prove that the odds are 17 to 16 in favour of A.
63. — A pack of cards consists of p suits of q cards each,
numbered from 1 up to q. A card is drawn and turned up :
and r other cards are drawn at random. Find the chance
that the card first drawn is the highest of its suit among all
the cards drawn.
64. — A and B play for a stake which is to be won by
him who makes the highest score in 4 throws of a die.
After two throws, A has scored 12, and B 9. What is A's
chance of winning ?
65. — A bag contains 6 shillings and 2 sovereigns. What
is the value of one's expectation if one is allowed to draw
till one draws a sovereign ?
66. — There are m white balls and m black ones : in balls
are placed in one bag, and the remaining m in a second bag,
the number of white and black in each being unknown. If
one ball be drawn from each bag, find the chance that they
are of the same colour.
I MISCELLANEOUS EXAMPLES. 241
67. — In the last question, if m== 4, and a ball of the
same colour has been drawn from each, find the chance that
a second drawing will give balls of the same colour : ( i ) if
the balls drawn at first have been replaced, and (ii) if they
have not been replaced.
68. — A player has reckoned his chance of success in a
game to be e, but he considers that there is an even chance
that he has made an error in his calculation affecting the
result by e' (either in excess or defect). Shew that this
consideration does not affect his chance of success in a single
game, but increases his chance of winning a series of games.
69. — Shew that in taking a handful of shot from a bag it
is more probable that an odd number will be drawn than an
p even number.
70. — A bag contains m white and n black balls, and from
it balls are drawn one by one till a white ball is drawn.
A bets B at each drawing, x to y, that a black ball is drawn.
Prove that the value of A's expectation at the beginning of
the drawing is *_.. - x
71. — Counters (n) marked with consecutive numbers are
placed in a bag, from which a number of counters (m) are to
be drawn at random. Shew that the expectation of the sum
of the numbers drawn is the arithmetic mean between the
greatest and least sums which can be indicated by the number
of counters (m) to be drawn.
72. — A and B play a set of games, in which A* a chance
242 CHOICE AND CHANCE.
of winning a single game is p, and B's chance q. Find
(i), the chance that A wins m out of the first w-j-w.
(ii), the chance that when A has won m games, m-\-n
have been played.
(iii), the chance that A wins m games before B wins
n games.
73.— The face of a die, which should have been marked
ace, has been accidentally marked with one of the other five
numbers. A six is thrown twice in two throws. What is
the chance that the third throw will give a six ?
74. — One of two bags contains 10 sovereigns, and the
other 10 shillings. One coin is taken out of each and placed
in the other. This is repeated 10 times. What is now the
expectation of each bag ?
75. — A, B, C are candidates for an office, the election to
which is in the hands of 8m-\-^~L electors. 3m votes, together
with the casting vote if necessary, are promised to A, and %m
votes to B. In how many ways can the remaining votes be
given so that A may be successful ?
76. — A writes to B requiring an answer within n days. It
is known that B will be at the address on some one of these
days, any one equally likely. It is a ^-days' post between
A and B. If one in every q letters is lost in transit, find
the chance that A receives an answer in time. (n>2p.)
77. — What is the probability that a number, consisting of
MISCELLANEOUS EXAMPLES. 243
•7 digits, the sum of which is 59, will be exactly divisible
by 11?
78. — There are n vessels containing wine, and ?r vessels
containing water. Each vessel is known to hold a, a+1,
#+2, ..o or a-\-m — 1 gallons. Find the chance that the
mixture formed from them all will contain just as much wine
as water.
79. — A man has left his umbrella in one of three shops
which he visited in succession. He is in the habit of leaving
it, on an average, once in every four times that he goes to a
shop. Find the chance of his having left it in the first,
second, and third shops respectively.
80. — If mn balls have been distributed into m bags, n into
each, what is the chance that two specified balls will be found
in the same bag ? And what does the chance become when
r bags have been examined and found not to contain either
ball?
81. — One card out of a pack has been lost. From the
remainder of the pack, thirteen cards are drawn at random,
and are found to consist of two spades, three clubs, four
hearts, and four diamonds. What are the respective chances
that the missing card is a spade, a club, a heart, or a
diamond ?
82. — A number of persons A, B, C, D ... play at a game,
their chances of winning any particular game being a, /3, y, 5 . . .
respectively. The match is won by A if he gains a games in
244 CHOICE AND CHANCE.
succession ; by B if he gains b games in succession ; and so
on. The play continues till one of these events happens.
Shew that their chances of winning the match are propor
tional to
(1-«K , (l-0)ff, &c
l-a« ' 1-/35
83. — A goes to hall p times in q consecutive days and sees
B there r times. What is the most probable number of times
that1 B was in hall in the q days ?
Ex. — Suppose p = 4 g = l r=8.
84. — If Mr be the number of permutations of m things
taken r together, and Nr the number of permutations of n
things taken r together, prove that the number of permuta
tions of m-\-n things r together will be obtained by expand
ing (M-\-N)r, and in the result replacing the indices by
suffixes.
85. — Find the number of positive integral solutions of the
equation #+2/+z+ ... (p variables) = m, the variables
being restricted to lie between I and n, both inclusive.
86. — In how many ways can 26 different letters be made
into six words, each letter being used once and only once ?
87. — A body of n members has to elect one member as a
representative of the body. If every member gives a vote,
in how many ways can the votes be given ?
88. — In the case of the last question, how many different
MISCELLANEOUS EXAMPLES. 245
forms may the result of the poll assume, regarding only the
number of votes given to each member and not the names of
his supporters ?
89. — In a company of sixty members, each member votes
for one of the members to fill an office. If the votes be
regarded as given at random, what is the chance that some
member shall get a majority of the whole number of votes ?
Also determine the chance of the same event on the hypo
thesis that every different result of the poll (considered as in
the last question) is equally likely to occur.
90. — In how many ways can 3 sovereigns and 10 shillings
be put into 4 pockets ? (One or more may be left empty.)
91. — In how many ways can 12 sovereigns be distributed
into five pockets, none being left empty ?
92. — In how many ways can 20 books be arranged in a
bookcase containing five shelves, each shelf long enough to
contain all the books ?
93. — In how many ways can a person wear five rings on
the fingers (not the thumb) of one hand ?
94. — A debating society has to select one out of five sub
jects proposed. If thirty members vote, each for one subject,
in how many ways can the votes fall ?
95. — In the last question, what is the chance that upwards
of twenty votes fall to some one subject ?
246 CHOICE AND CHANCE.
96. — A bag contains m counters marked with odd numbers,
and n counters marked with even numbers. If r counters be
drawn at random the chance that the sum of the numbers
drawn be odd is £(1 + p), and that it be even i(l— ju,), where
/x, is the coefficient of xr in the expansion of
m+n
97. — The number of ways in which r things may be distri
buted among n-\-p persons so that certain n of those persons
may have one at least is
98. — Show that for n different things 1 — (number of par
titions into 2 parts) + [2 (number of partitions into 8 parts)
— . . . .± [n — 1 (number of partitions into n parts) = 0.
99. — Find the number of 5-partitions of 21.
100. — Two examiners working simultaneously examine a
class of 12 boys, the one in classics the other in mathema
tics. The boys are examined individually for five minutes
each in each subject. In how many ways can a suitable
arrangement be made so that no boy may be wanted by both
examiners at once ?
101. — If/n denote the number of derangements of n terms
in circular procession so that no term may follow the term
which it followed originally,
MISCELLANEOUS EXAMPLES. 247
102. — A pack of n different cards is laid face downwards
on a table. A person names a certain card. That and all
the cards above it are shewn to him, and removed. He
names another ; and the process is repeated until there are
no cards left. Find the chance that, in the course of the
operation, a card was named which was (at the time) at the
top of the pack.
103. — Three different persons have each to name an
integer not greater than n. Find the chance that the
integers named will be such that every two are together
greater than the third.
104.— A person names a group of three integers (not
necessarily different, but each one not greater than n).
Find the chance that the integers named will be such that
every two are together greater than the third.
105. — If three numbers be named at random, they are
just as likely as not to be proportional to the sides of a
possible triangle.
106. — A list is to be published in three classes. The odds
are m to 1 that the examiners will decide to arrange each
class in order of merit, but if they are not so arranged, the
names in each will be arranged in alphabetical order. The
list appears, and the names in each class are observed to be
in alphabetical order, the numbers in the several classes
being a, b, and c. What is the chance that the order in each
class is also the order of merit ?
107. — How many different throws can be made with n
248 CHOICE AND CHANCE.
dice, those throws being considered the same in which the
same set of numbers is turned up ?
108. — Prove that the most likely throw in the last
question is one in which the numbers turned up are all
different, if n is not greater than 6. And find the most
likely throw when n is greater than 6.
109. — There are 2>i letters, two and two alike of n
different sorts. Shew that the number of orders in which
they may be arranged, so that no two letters which are
alike may come together, is
110. — From a bag containing m gold and n silver coins, a
coin is drawn at random, and then replaced ; and this ope
ration is performed p times. Find the chance that all the
gold coins will be included in the coins thus drawn.
111. — A train, consisting of p carriages, each of which will
hold q men, contains pq — m men. What is the chance that
another man getting in, and being equally likely to take any
vacant place, will travel in the same carriage with a given
passenger.
112. — If the chance of a trial succeeding is to its chance
of failing as m : n, the most likely e"vent in (m-\-n)r trials
is mr successes and nr failures.
113. — If n witnesses concur in reporting an event of which
they received information from another person, the chance
MISCELLANEOUS EXAMPLES. 249
that the report is true will be (pn+l + gn+l) -f- (pn+(f)
where p~\ — q is the chance of the correctness of a report
made by any single person.
114. — The reserved seats in a concert room are numbered
consecutively from 1 to m + n -\- r. I send for m consecutive
tickets for one concert and n consecutive tickets for another
concert. What is the chance that I shall find no number
common to the two sets of tickets ?
115. — Two persons are known to have passed over the
same route in opposite directions within a period of time
m-\-n-\-r, the one occupying time m, and the other time n ;
find the chance that they will have met.
116. — If p = l — q be the change of success at any trial,
what is the chance that in r-\-n trials there should be at
least r consecutive successes, (i) when n<r and (ii) when
m is the greatest integer in n~-r.
117. — If n numbers be selected at random, what are
the respective chances that their continued product in the
common scale of notation will end with the digits 0, 1, 2, 3,
4, 5, 6, 7, 8, 9 ?
118. — There are n tickets in a bag numbered 1, 2, 3, ...n.
A man draws two tickets at once, and is to receive a number
of sovereigns equal to the product of the numbers drawn.
What is his expectation ?
119. — What would be the expectation in the last question
if three tickets were drawn and their continued product
taken ?
250 CHOICE AND CHANCE.
120. — If a set of dominoes be made from double blank up
to double n, prove that the number of them whose pips are
n — r is the same as the number whose pips are ™+r,
and the number is the coefficient of xn~r in the expansion of
(1 — x — #2+#3)~~1 ; and the total number of dominoes is
121. — If from the dominoes in the last question a man
is to draw one at random, and to receive as many pounds
as there are pips on the domino drawn, what is his expec
tation worth ?
122. — If p be a prime number whose reciprocal in the
decimal scale of notation is expressed by a recurring period
of p — 1 digits, and if the digits of this period be rearranged
at random, the chance that the new period thus formed
will belong to a fraction whose denominator is p, will be
(? + l)r~1([^)10 -T- [p — 2, where q is the quotient and r
the remainder when p is divided by 10.
123. — A vessel is filled with three liquids whose specific
gravities in descending order of magnitude are 8lt S2, S3.
All volumes of the several liquids being equally likely,
prove that the chance of the specific gravity of the mixture
being greater than 8 is
-
' -
according as 8 lies between Sl and £2, or between S2 and SB.
ANSWEES TO THE EXAMPLES.
EXAMPLES ON CHOICE (pages 62-64).
1. 20. 2. 840 ways ; or, considering the arrangement,
20160 ways. 3. 40320, 5040. 4. 1320.
[32 ^ [64
5* [12 [12 [8 6t [32 ([8)2 ([2)6
7. 166320. 8. [15 - 192. 9. 120. 10. 480, 22.
18.[60 8.|60
1L [20 [40' [20 [40
12. 3. 13. 360,120,24. 14. 675675. 15. 1436400.
16. 9849600. 17. — If there be m of one sort, and n of
the other, the number of ways is \m In. 18. 66— 64.
19. 20. 20. 3439,1271. 21. [90 -T- [24 [22 [30 [4 [10
22. 167. 23. 20591. 24. 5040, 75600.
EXAMPLES ON CHANCE (pages 138-141).
2 1
1. q. 2. ~. 3. — The alphabet containing 20 con-
o
sonants and 6 vowels, ^. 4. — The chances are propor
tional to 14256 : 12060 : 10175. 5. JL 6. —
252 CHOICE AND CHANCE.
83 1
7. 3 shillings. 8. . 9. . 10. 10 shillings.
11? ^ 245753125 625 16
243' 729' 387419489' 2187' ' 27
iyi 7 17 11 lr. 0-7, 9nft .- 130 16025
14' 12 '54 '108' 15' 257to208' 16' 837' 17577
17' 19' 5' 20. 31 to 15.
MISCELLANEOUS EXAMPLES (pages 231-250).
1. 25. 2. 480, 437. 3. 12. 4. 63. 5. 756.
6. 60. 7. 24. 8. 30. 9. 5. 10. 209. 11. 59733.
13. Ifw=:gn+r, m-^n3r. 14.
15. 66. 16. 10156250. 17. 1728. 19. 220.
21. 3023. 22. 332629. 23. 277199. 24. 16.
25. in (w-1) where n = (a-f- 1) (/3 + 1) (y +1) (8 +1).
28. 100. 29. 2030. 30, £H (3?i2-3?i+2). 31. 6.
32. *n(n-l). 33. in(«-l)-^(p-l)-H(g--l).
34. 576. 35. 15, 12. 36. 19. 37. 6.
3a 39> - 4a
41. +l)(^-H+6). 42. »i(w-l),wa. 44. .
1 2
45^ _ . 46. — . 47. — Their respective chances
10 21
•* I' I' I- 48- !• 49- * 50' 1-
53.— The latter is better as 11 to 9. 57. 25 : 2.
ANSWERS TO THE EXAMPLES. 253
59
63
74
2197
704
264
60 52728
' 20825 '
P fi I:
4165'
4165'
112559125*
65. 22 shillings.
i- 73-S-
oo 54
2j9\/ 1\8 4
H7\ ~^/ ' '21*
h^+^+.-.+^^v"
r + l( \pq \pq-q
-r-lj
. 21±19(-8)10
78. Middle
*7Q
6'
crowns. '
coefficient
7>~98'
r6/l-
of (1H
(
80.
m J
™* 37' 37'
37*
mn — 1* mn — ?-?z — 1 '
81- WWWW 83.
this and the next lower integer are equally likely. If this be
not an integer, the next lower integer. (Example : 5 or 6.)
1771
86. ^~- x [26. 87. nn. 88. |2re-l -^ \n \n-l.
89. — In the first case, the odds against the event are as the
sum of the first thirty terms to the sum of the remaining
thirty terms, in the expansion of the binomial (59 + 1)59. In
the second case, the chance is 60 [60 [88 -=- [29 [119.
90.— 5720. 91. 330. 92. [24 + [4. 93, 6720.
94. 46376. 99. 105. 100. IS!9^1. 102. I-*,1
.03. ,+. ,04.
m-4-1
106'
254 CHOICE AND CHANCE.
108. — If n lie between 6r and 6(r+l), the most likely throw
will be one in which each number appears either r or r-\-l
times.
(r+l)(r+2) mn+mr+nr
117.— Chance of the final digit being 0 is (10n-8M-5n+4n)
-i-10n. Chance of 5 is (5n-4n)-j-10n. Chancesof 1, 8, 7, 9,
are equal, each being 4n~* — 10n. Chances of 2, 4, 6, 8, are equal,
each being 4n~1(2n-l)-r-10n. 118. (n + 1) (3n + 2) ~ 12.
119. t?rc(?i + l)2. 121. n pounds.
D. MAEPLES, PRINTER, LORD STREET, LIVERPOOL.
HI
7 DAY USE
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