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CHAPTER XVIII

PROBABILITY
1. Suppose that a bag contains a hundred balls of exactly the same size and shape, of which one is coloured and the remaining ninety-nine white. If one of these balls be drawn at random it is safe to say that there is a greater probability of drawing a white ball than of drawing the coloured ball. Again, of a number of men in a community, all subject to the same conditions, the probability that one aged 20 will survive 10 years is obviously greater than the probability that one aged 90 will survive the same period. These examples serve to illustrate what is understood by the term " probaMany more such simple examples could be given, from which we could say that the probability that event A would happen is greater or less than that event B would happen. A difficulty arises, however, when we attempt to assign a measure to the probability that one of the events A or B might happen.
In the first of the above examples it is not unreasonable to argue that since there is one coloured ball among a hundred balls, the probability that this coloured ball would be drawn is one in a hundred. Whether this be so or not, we have used the data at our command and have given a measure to the probability. In the second example, however, we have no data immediately available to enable us to assign a numerical value to the probability in question. It would be necessary to collect statistics of the mortality of the men in the community, and to supplement the collection of these statistics by more or less elaborate calculations. In the application of the theory of probability to most actuarial problems the aggregation of statistics is necessary before the data are avail-able, and it is not often that we are immediately in possession of such simple facts as are given in the first type of question. A comanswer to a problem in -actuarial work would involve the collection and interpretation of the relevant statistics : these are matters which are outside the scope of this book, and in dealing with questions on probability it will be sufficient to assume that

314    PROBABILITY
we have sufficient data to enable us to proceed without further analysis.
2. Numerical definitions of probability.
A simple definition of the measure of the probability that an event may happen is as follows :
(i) If an event can happen in a ways and fail to happen in b ways, all of these ways being equally likely and such that not more than one of them can occur, then the probability of the event happening is —
a
a+b
The probability that the event does not happen is a b b , which
a
is equal to 1 —
a    b'
Although this definition enables us to solve arithmetical problems in probability, it has, however, certain defects which render it unsatisin general. In the first place we have to define the phrase "all of which are equally likely." It is difficult to say what this means without rendering the definition unintelligible, for if we define the words as equivalent to "there being no reason why one way should occur more than another" we should be involved in a further definition. Moreover, however we explain the phrase it may not be possible to say that the happenings of an event are equally likely: the probability of its occur
rence cannot then be measured by the simple ratio a + b. The attempt
that has been made to define "equally likely" events as those which occur equally often in the long run, i.e. when the number of occasions is very large, does not seem to solve the difficulty.
While the above definition—which is sometimes called the "unitary" definition—is sufficient for solving the questions inprobability that will be dealt with in this chapter, it will not be out of place to quote other definitions.
NUMERICAL DEFINITIONS OF PROBABILITY    315
be associated with one of two or more mutually exclusive events E, E', E", ... , then if E would occur on pN occasions, E' on p' N occasions, ... and so on, p, p', p", ... are said to be the probaof the event M being associated with E, E', E", ... , and p + p' + p" + ... = 1. (Sheppard, Ency. Brit., "Probability.")
The term" mutually exclusive "in the above definition means that on each occasion one and only one of the events E, E', E", ... can occur.
We can arrive at an approximate agreement between these deand the unitary definition thus. If an event occur on a occasions out of (a + b), we may say that the frequency of its occuris a, and the relative frequency will be a/(a + b). Then the probability of the event is the relative frequency when the number of occasions is very large.
For example, we may revert to an illustration similar to the first illustration in paragraph 1. If there be five exactly similar balls in a bag, two coloured and three white, then by the unitary definition the probability that a coloured ball will be drawn (supposing one ball be drawn at random) is 5. This does not mean that if a ball were drawn five times, being replaced each time, two of the draws would necessarily give coloured balls. The true interpretaof the fraction 4 is that if the experiment were repeated a large number of times a coloured ball would be drawn about twice in five times. After many trials we should find that twice in five times was approximately correct, and we might be said to have confirmed our unitary result by experiment.
There are thus two distinct methods of ascertaining simple probabilities, one from a priori considerations, and the other by experiment. Usually only one of these methods is applicable to any particular problem, and for the present purpose it will be unto consider statistical methods.
3. The basis for the solution of the problems that will be dealt with later is, as has been remarked, the unitary definition. We may amplify this definition by certain others.
(a) The measure of certainty is unity. In other words, if an event is certain to happen the probability of its happening is 1 ; if it is certain not to happen the probability of its happening is o.

316    PROBABILITY
(b) If p is the probability that an event might happen, the probability of its failing is 1 — p. This follows directly from the unitary definition.
(c) If the odds in favour of an event are a : b, the probability of its happening is a'(a + b).
(d) If an event can happen in more than one way, all ways being mutually exclusive, the probability of its hapat all is the sum of the probabilities of its happening in the several ways.
This last proposition is known as the "addition" rule, and is fundamental in the application of the theory of probability. The rule may otherwise be stated thus:
If p and p' are the probabilities of two mutually exclusive events, the probability that one of these events happens on any occasion in which both events are in question is p + p'.
The words "mutually exclusive" in this definition are of the utmost importance. A proof of the addition rule depending on Chrystal's definition of probability will show that unless the events are mutually exclusive we cannot add the simple probabilities p and p'. Thus, if there be a large number N out of a series in which the two events are in question, the first will happen on pN occaand the second on p'N occasions respectively. Since the events are mutually exclusive, one and only one of the events can occur on any one occasion. Therefore out of N occasions one or other of the events will happen on pN + p'N occasions, i.e. on (p + p') N occasions. The probability that one of the events hap-pens is therefore p + p'. It will be seen that this proof breaks down if the second event happens on any of the pN occasions that the first happens, i.e. if the events are not mutually exclusive.

4. The simplest problems in probability are those in which both the number of favourable ways and the total number of ways in which the event may happen can be counted, either arithmetically or by the aid of elementary rules. The enumeration is often best performed by the application of the theorems of permutations and combinations, and a thorough knowledge of this branch of algebra is essential for the speedy solution of many of these questions.

ILLUSTRATIVE EXAMPLES Example 1.
An ordinary pack of cards contains 13 cards of each of four suits; spades, hearts, diamonds and clubs. Ten cards of each suit are numbered i (the ace), 2, 3, ... 10, the remaining three being "court cards," namely, king, queen, knave.
Find the chance that if a card be drawn at random from an ordinary pack it will be a heart.
Since there are 52 cards in a pack, the number of ways in which a card can be drawn (i.e. the total number of ways in which the event can happen) is 52.
The number of ways favourable to the event is 13, since there are 13 hearts in the pack.
The required probability is therefore 1'z or } .
Notes: (i) The above result means that over a long series of trials the proportion of favourable draws would approximate to one-fourth of the whole number of trials. It is convenient, however, to think in terms of a single trial, although the results of our calculations merely epitomize the experience of a long series of trials.
(ii) The word "chance " may be taken to be the same as the word "probability." There is a shade of difference between the two ideas, "probability" referring to past, present and future events, and "chance" to future events only. This distinction can be ignored.
Example 2.
In a given race, the odds against three horses A, B, C are 2 : I ; 5 : 2; to : i respectively. Assuming that a dead-heat is impossible, find the chance that one of them will win the race.
The events are mutually exclusive, and we may therefore use the "addition" rule.
p' = probability that A wins = p„=    B    ,
p, _
,5    C    73
The probability that one of the horses wins is therefore
p, + p„ + p,,, = 1~
Example 3.
Out of 211 + I tickets consecutively numbered, three are drawn at random. Find the chance that the numbers on them are in arithmetical progression.

318    PROBABILITY
This problem can be solved by virtually writing down the sets of numbers that can be in A.P. Consider, for example, the sets of three numbers in A.P. beginning with 4: they will be
45;6    4;6;8    4;7;10 ...4;71+I;2n—2    4;n+2;211,
where the highest number that can enter into any set is 2n. There will evidently be n — 2 such sets of numbers (being the number of numbers from 5 to n + 2, both numbers inclusive). We can therefore write down the following schedule :
Lowest number of the three Number of favourable ways
I
 n
2
 n — I
3
 71 — I
4
 n—2
5
 n — 2
6
 71—3
2n—2 I
271— I I

The total number of favourable ways is 2n (n + I) — nor n=, and since 2

the total number of ways of drawing three tickets is (2n    I) 2n (211 — I) 6
it is easily seen that the required probability is    371    
471— I Example 4.
If the letters of the word REGULATIONS be arranged at random, what is the chance that there will be exactly four letters between the R and the E?
The conditions are satisfied if
the R is in the first place and the E in the sixth,
or „ R second    „ E „  seventh,
„ 
 R third    „ E ,, eighth,
„    „ 
 R fourth    „ E „  ninth,
„ 
 R    „ fifth    „ E „  tenth,
„ 
 R sixth E „  eleventh,

or if the R and the E are transposed in any of the above.

THE MULTIPLICATION RULE    319
Now there are ii letters in the word; therefore the letters R and E can jointly occupy I I x to positions. Since there are 6 x 2 favourable positions (as above) the chance that the R and the E are in the favourable position is Tyr or A.
Example 5.
Find the probability that the number 8 will be thrown in a single throw with two dice.
The total number of different numbers that can be thrown with two dice is 36, since each of the 6 ways of throwing the first die can be associated with each of the 6 ways of throwing the second.
The number of ways favourable to the event can be found by simple enumeration, thus:
First die
Second die
2
6
3
5
4
4
5
3
6
2

Total: 5 ways. The required probability is therefore A.
5. The multiplication rule.
If there be a series of events such that the happening of any one of them in no way affects the happening of any other of them, the events are said to be independent.
The probability that two independent events happen on any one occasion on which they are both in question is the product of the chances of their happening severally.
For consider two independent events C and D, the probabilities of which are p and p' respectively. Out of a large number (N) in which the events are in question C will happen on pN occasions, and out of these pN occasions D will happen on p' (pN) or pp'N occasions. The probability that they both happen is therefore pp'.
This rule is known as the "multiplication" rule, and its applicato the probabilities of happening or otherwise of two indeevents may be seen from a consideration of the following schedule.
Let the probabilities of happening of two independent events be p and p' respectively. Then

3 20    PROBABILITY
 ILLUSTRATIVE EXAMPLES    32I
 322    PROBABILITY
Example 7.
Find the chance of drawing a king, a queen and a knave in that order from an ordinary pack in three consecutive draws, the cards drawn not being replaced.
This is an example of the application of the multiplication rule to dependent probabilities.
If pi be the chance of drawing a king;
    p2    a queen, a king having been drawn;
    p3    a knave, a king and a queen having been drawn,
then the chance of drawing a king, a queen and a knave in succession
is pip2p3
Now there are 4 kings in the pack, so that the chance of drawing a  king is When the king has been drawn there are 51 cards left of which 4 are queens. The chance of drawing a queen is therefore 5 . Similarly, the chance of drawing a knave from the remaining 5o cards is
P1 = 52; P2= r' P3= ~1.
3
The required chance = plp2p3 = 52 5 3 5° .
Example 8.
A bag contains three balls, one red, one white and one blue. X and Y draw a ball at random alternately. If X draws the red ball or Y the white ball it is retained. Otherwise the ball drawn is immediately replaced. Find the chance that just before the fifth draw is made the blue ball only is in the bag.
Let R, W, B denote the red, white and blue ball respectively. Now the blue ball is the only one in the bag just before the fifth draw if any of the following have happened :
  X Y X Y
  first draw
first draw
 second draw second draw
(i)
 R B B W
(ii)
 R B W IV
(iii)
 B W R B
(iv)
 W W R B
(v)
 R W B B
(vi)
 B R R W
(vii)
 B B R W
(viii)
 W R R W
(ix)
 W B R W



ILLUSTRATIVE EXAMPLES    323
The chances of these events happening are:    Product
    (1)    2    2    2
(lam
    
111)    3    2    2    2
(iii)
3    2    18
(iv)
3    2    11_7
The probability that the bag contains (a) is -. 6 . or for the chance that a green ball is drawn is obviously , the chance that a second green ball is drawn is and the chance of a third green ball is then l . Similarly,
the chances under (b), (c), (d) are o, A,    respectively.
Nov three different colours on the second draw can only be obtained if the six balls come under headings (b) and (c).
Under (b) the probability at the second draw of drawing three different coloured balls is A, and therefore the compound probability that this will happen is A. ,r. Under (c) the probability is the same.
The total chance is the sum of these and amounts to    Therefore,
since the odds against the favourable happening are 73 to 27, the person
2I-2

324    PROBABILITY
drawing the balls may lay 8 to 3 against his drawing three of different colours and obtain a slight advantage.
Example 10.
Q is the probability that a man aged x will die in a year. Find the probability that out of five men, A, B, C, D, E, each aged x, A will die in the year and be the first to die.
The chance that a given man dies in the year    = Q.
The chance that a given man does not die in the year =
 1 — Q.
The chance that none of the five dies in the year    =
 (1 — Q)5.
The chance that at least one man dies in the year    =
— (1 — Q)5.

Since the chance that A is the first to die is obviously , and since this is independent of the chance that at least one man dies in the year, the required chance = {1 — (i — Q)5}.
8. The methods for the solution of the above examples depend largely on the simple application of the formulae for permutations and combinations. It is safer as a general rule to use permutations rather than combinations. If repetitions are not allowed this is not very material, as each combination of r things forms r! permutaIf, however, repetitions are allowed then in most instances it is essential to use arrangements.
Other algebraic devices are often useful. For example, in dealing with questions involving the sum of the numbers that can be thrown with dice and kindred problems dealing with homoneous products, it is of advantage to employ the binomial theorem.
The following examples illustrate the use of this method.
Example 11.
Four dice are thrown. Find the chance that the sum of the numbers appearing will be 18.
Regard being had to the different ways of making up the same total, the number of numbers that can be thrown with four dice is the sum of the coefficients in the expansion
(x + x` + x3 + x4 + x5 + x6)4.
The sum of the coefficients will be found by putting x = I. The total number of possible numbers is therefore 64.
The number of ways of throwing 18 is the coefficient of xis in the above expansion.

    
ILLUSTRATIVE EXAMPLES    325 The coefficient of x18 in (x + x2 + x3 + x4 + x5 + x6)4
x18inx4(I +x+x2+x3+x4+x5)4
x14 in (I + x + x2 + x3 + x4 + x5)4
x14 In (1 -    x6)4
(I — x)4
x14 In (I — x6)4 (1 — x)-4
x14in(I -4x6+6x12...) (I+4x+ lox' +...
9.10.II 8    15.16.17 14
+ 6      x +...+    6    x" +
— 15.16.17 — 4.9.10.11
6    6      + 6. lo = Ho.
The required chance is therefore 680
Note. We are here dealing with arrangements which would be disif one die were red, one blue, one yellow and one green. Selections in an example of this type would give quite wrong results.
Example 12.
Nine cards are drawn at random from a set of cards. Each card is marked with one of the numbers 1, o or — 1, and it is equally likely that any of the threg numbers will be drawn. Find the chance that the sum of the numbers on the cards thus drawn is zero.
The number of favourable drawings will be the coefficient of x° in the expansion of (x-1 + x° + x1)9; the total number of possible drawings will be the sum of all.the coefficients in the same expansion.
The coefficient of x° in (x-1 + x° + x')9
x° in x-9 (1 + x + x2)9
=    x9 in (1+x+x2)9
x9 in (I — x3)9 (I — x)-9.
Proceeding on the same lines as in Example II, the required proba
bility is found to be 3199. 3
9. The theory of probability was evolved from a consideration of games of chance, and many problems dealing with these games can be solved by the elementary methods outlined above. No new
,,

326    PROBABILITY
principles are involved; all that is required in attacking these problems is a clear understanding of the particular happenings that may arise. The following examples are illustrative of the methods : the first of these examples is analysed fully, and the student should be able to solve problems of the same type by the application of similar reasoning.
Example 13.
A and B throw alternately with a pair of ordinary dice. A wins if he throws six before B throws seven, and B if he throws seven before
A throws six. If A begins, show that his chance of winning is R. (Huyghens' Problem.)
First, let us consider the chances of throwing six or seven with two dice. We can find the number of ways of throwing these numbers either by actually counting the number of ways (as in Example 5) or by finding the coefficients of x6 and x7 in the expansion of (x + x2 + x3 + x4 + x5 + x6)2 (as in Example I I). Since the figures are small, the first way is simpler, and it is easily seen that six can be thrown in 5 ways, and seven in 6 ways. The chance of throwing six in one throw with the two dice is therefore
and of throwing seven A.
A can win if he throws six the first time. His chance of throwing this number is A.
He may fail to throw six the first time. He can then win if B fails to throw seven at his first throw and A throws six with his second throw. The chance that A fails to throw six is I —    = N.
The chance that B throws seven is    and the chance that he fails
to throw this number is I    =
Therefore the chance that A wins at the second throw is R.N.-A.
If A fails to win at the second throw, he can win at the third throw if
B has not thrown seven at his second throw. The chance of this is
s1 30 :'1 ro    and so on.
A's chance of winning in the long run, i.e. after a very large number of trials, is therefore
5    31 30 5    31 :30 31 30 5
_ S (I + r + r2 + r3 + ...),
where r =
3a•:;c•
If we sum this geometrical progression to infinity, we find that A's chance
_ 5 I _ 5 1296 30 36 1 — r 36' 366 61 '

ILLUSTRATIVE EXAMPLES    327 Example 14.
A and B play a match to be decided as soon as either has won two games. The chance of either winning a game is 1-, and of its being drawn 1"u. What is the chance that the match is finished in lo or less games?


If the match is not finished in lo games, the following must occur:
since this result may occur in to different ways and either A or B may win ;
since the number of orders in which this may occur is 10P2
The chance that the match is not finished in to games is, by the addition rule,
Example 15.
A, B, C, D each throw two dice for a prize. The highest throw wins, but if equal highest throws are made by two or more players, those players continue. A throws q, B throws 7. Find C's chance of winning the prize.
Consider the following scheme, showing the total number of possible ways in which C may win:
2. Ioth'
i.e. the chance that the match is finished in to games or less
98.387
= I — 22.'13010 = • 17 approximately.

328
 PROBABILITY      
 
C may win if
 Probability
C throws Iz, D less than 12
I
 35  
  62 36  
C throws 12, D 12, and C wins later I
62		 I
62		 I*
2
C throws I1, D less than I1 2 33  
  62 36  
  2 2 1*
C throws I1, D It, and C wins later 6 62 2
C throws to, D less than to 3 30  
  62 36  
C throws lo, D Io, and C wins later 3 3 I*
  62' 62 2
  4 26 I*
C throws 9, D less than 9 and C wins later 62 36 2
(since A has thrown 9)
C throws 9, D 9, and C wins later		 4 4 1*
  62' 62 3

The total of these chances is ;T.
10. Most probable value.
It is essential to distinguish between the absolute probability of an event, the average number of times that it may happen over a series of trials and the most probable number of times that it will occur. The most probable value is the value that occurs with greatest frequency, and in simple examples this value can be easily deter-mined by considering separately the contingency of each event. Thus, in a single throw with two dice the numbers that may turn up and the chances of occurrence of these throws are given in the following table :
Possible numbers:    2 3 4 5 6 7 8 9 10 1 1 12 Chances of occurrence : 1 2 3 4 5 6 5 4 3 2 I (- 36)
The most probable number to be thrown is therefore 7.
Again, in Example 17 below, a hasty conclusion would be that half the pack (i.e. 26 cards) would have to be turned up before two aces out of the four would appear. This is not so, however.
* The final factor in each of these terms expresses the chance that C will
win after having equalled the number thrown by D and/or A.

MOST PROBABLE VALUE    329
The four aces divide the remaining 48 cards into 5 groups. If a large number of such divisions were made the average number in each of the 5 groups would be 48/5. The average number of cards which must be turned up before two aces appear is therefore 2 of these groups plus 2 aces, i.e. 96/5 + 2 or zi•z, as given below. The solution on p. 330 shows that z6 is neither the average number nor the most probable number, and, further, that there is a definite probability associated with each number of cards turned up. At the risk of labouring the point, it must be emphasized that these separate probabilities are only to be realized over a long series of trials.
Example 16.
A purse contains two half-crowns and three shillings; a second purse one half-crown and four shillings. A coin is drawn from the first and placed in the second, and then a coin is drawn from the second and placed in the first. Assuming that the chance of drawing a half-crown is twice that of drawing a shilling, find the most probable value of the coins in the first purse after the second operation.
At the end of the second operation the first purse may contain:
 330    PROBABILITY
The greatest of these values is
The most probable value of the purse after the second draw is the value associated with this fraction, namely 8 shillings.
Example 17.
Cards are dealt one by one from an ordinary pack (without replaceuntil two aces have appeared. Find (i) how many cards (on the average) will be turned up, (ii) the most probable number of cards to be turned up.
(i) Since two aces cannot appear until the second trial at the earliest, we have
Chance of success at the second trial = -5-4,. A.
third fourth
= 2 48 0 ,3 er    •F48    1 =~•
48 47 4    •3
3    s i ~o s
and generally
Chance of success at the xth trial = ('C — 1) (52 — x) (51 - x) 4.3 52.51.50.49
The average number of cards to be turned up will be the sum of the series whose xth term is
x l(x — 1) (52 — x) (51 — x) 52.51 4 .30.49
for all values of x from 2 to 50 (since there must be two aces at least in the first 5o cards).
Summing this by ordinary algebraic or finite difference methods, it is found that the average number of cards to be turned up is 21.2.
(ii) The most probable number to be turned up will be the value of x which gives
(x— 1)(52—x)(51 —x)4.3
52.51.50.49
its greatest value.
The value for x = r > the value for x = r — 1 so long as
(r—1)(52— r)(51—r)>(r—2)(53—r) (52—r), i.e. so long as    — r2 + 52r — 51 > — r2 + 55r — 1o6,
i.e. so long as    55 > 3r,
i.e. r is not greater than 18 (since r must be integral).
The most probable number of cards to be turned up is therefore 18.
EXPECTATION AND PROBABLE VALUE
11. Expectation and probable value.
The probability of happening of an event may be combined with a measure of the quantity depending on the event, thus producing an expectation. In other words, expectation is the product of an expected gain in actual value and the mathematical probability of obtaining such a gain. No special difficulty arises in this type of question, and the methods outlined above may equally well be applied to the solution of problems involving expectation. All that is necessary is to combine the respective chances with the gains in value should the events occur. Thus, if a player throwing
an ordinary die is to receive £2n, where n is the number of throws that he takes to throw the first 6, his expectation would be
I    I    5 I    I    5 5 I    I    r5 r I    I
6£2+6 6£22+66.6£23+...+\6) •6£27+1+..
the sum of which is £?-.
Probable value is the same as expectation, except that it refers to the things in question and not to the person. For example, in Example 16, the probable value of the coins in the purse at the end of the second operation (i.e. the expectation of the person drawing the coins) is
x 8 shillings +    x 9z shillings + x 62 shillings = 74? shillings.
In other words, it would be worth while to give about 7s. 9d. for the purse after the second draw.
The following example is instructive :
Example 18.
A table is divided into six squares numbered I to 6. A player places a coin on a certain square. Three dice are thrown. If the number thus backed appears once, twice or thrice, the player receives back his own coin together with one, two or three others respectively of the same value. In any other event he loses his stake. Does the advantage in the long run lie with the player or the "banker"?
(a) The chance that no die shows the number backed = (8)3.
(b) one die = 3 (s)z 1
(c) two dice    „ = 34.G-)2
(d) three dice = (03.



332    PROBABILITY The net expectation of the player
=(b)+2(c)+3(d)—(a)=—A7z-, and that of the banker
=—(b)—2(c)—3(d)±(a)= ?lg,
so that the advantage lies with the banker in the long run.
For a stake of a shilling it is easily seen that this advantage is just under a penny.
Here it might be contended that as the player is to receive back his own coin in addition to the prize, his expectation should be based on respective receipts of two, three or four units instead of one, two or three units, as appears in the above solution. His expectation on this basis might be argued as being 2 (b) + 3 (c) + 4 (d) — (a) or ,, which would show a substantial advantage to the player. Further consideration would show, however, that if the return of the stake is treated as a profit, then the laying of the stake must be treated as a payment to the banker for the privilege of playing and that it is, accordingly, definitely paid away whether the player wins or loses. His expectation then be-comes 2 (b) + 3 (c) + 4 (d) — 1 = — 11,5 as before.

12. In connection with mathematical expectation it is interesting to note the celebrated St Petersburg Problem, which has been a fruitful source of discussion for nearly two hundred years. Briefly stated, the problem is this : A coin is tossed until head turns up. If head turns up first A is to pay B one unit; if head does not turn up till the second throw B is to receive two units ; if not until the third throw four units, and so on. How much must B pay A before the game, in order that the game may be considered fair? That is, what is B's expectation?
The theoretical solution is as follows :
At the first trial B's expectation is z x t    =
second    „    (1)2 x 2    = 2. ..................
15    nth    \].\ 1 n x 2n—1 = 1-.
B's expectation is therefore z + ? + ? + ... to infinity, and as this is a divergent series, it appears that B could afford to pay an infinitely large sum before the game starts for his expectation.
Many explanations of this result have been given by eminent mathematicians, notably by d'Alembert, Bernouilli and de Morgan. An inter-

THE METHOD OF INDUCTION    333
esting account of the problem with an alternative solution depending upon the amount of money that B possesses at the outset is given by Whitworth in his Choice and Chance. This solution depends on a somewhat arbitrary assumption, and a more practical limitation of B's expectation arises from the fact that A's resources, however great, must be limited. Poincare (Calcul des Probabilites, p. 42) shows that if A's total assets are 2', B's expectation is I + ip. For example, if A possesses 230—which is more than a thousand millions—B's expectation is reduced from infinity to i6, and as Poincare drily remarks, this is a considerable reduction!
13. The method of induction.
There is a certain type of problem in probability for which it is not possible to obtain a solution by the direct methods outlined in the preceding paragraphs. In these problems it is necessary to find a relation connecting the chance at any stage with that at sucstages and then to calculate the required probability by adopting an inductive process. The difficulty in these questions is to ascertain the fundamental relation : when this has been established the problem can be solved by the application of algebraic methods.
Since we have to obtain a relation connecting the probability of an event at the nth stage with those at succeeding stages, it is often of advantage to investigate the relation for simple numerical values of n and then to deduce the general result. The following is an excellent example of this method of solution.
Example 19.
Five green balls and sixteen red balls are placed in a bag. A ball is drawn at random n times in succession and replaced after each drawing. Find the chance that no two successive drawings shall have given green balls.
To satisfy the required conditions the events must have taken place as follows :
At the first draw either a red or a green ball must be drawn: R
G
At the second draw one of the following draws must have taken place: R R    R G
G R
i.e. first drawing associated with R; or R G.

334
At the third draw:
PROBABILITY
    
R R R    R
    G R1R    G R G R
i.e. second drawing associated with R; or first drawing associated with R G.
At the fourth draw:
    RRRR    RRRG
    GRRR    GRRG
    RGR,R    RGjRG
R R G!R
G R G'R
i.e. third drawing associated with R; or second drawing associated with R G.
It is evident therefore that if we have attained success at the (n — I)th stage, i.e. if no two successive green balls have been drawn, we shall attain success at the nth stage provided that, either
(i) the nth draw gives a red ball;
or (ii) if the nth draw gives a green ball the drawing at the (n — I)th stage gave a red ball.
The chance of drawing a red ball =    and the chance of drawing a green ball = 21.
If therefore un be the required chance after n drawings, we shall have
    — 1 fi    7 fi
    u,z — _711 n_1 +    un_2
This is a relation connecting three successive coefficients of the series
uo + u1x + u2x2 + ... + un-2xn-2 + un_1xn 1 + unxn + ... .
The series is therefore a recurring series with scale of relation
    I — Iex —    iaxs
Also uo = 1 and u1 = 1, since the conditions are satisfied if no ball or one ball is drawn.
Proceeding in the usual way, we find that the generating function is
    1[D + x (111 — 16u0/21)    or    I + 5x/21
(I + 4421) (I — 20x/21)    (I + 4X/21) (I — 20x/21)
25    I    I    I
24 I — 20X/21 24 I + 4x/21
— 25 \I
loxl 1    1    4xl -1
24    21/    24\1+221/
R G R G

ILLUSTRATIVE EXAMPLES    335
The required chance, un, is the coefficient of xn in these two expanThis is easily seen to be
[25 ( i)n — (— 1)n (MI.
Note. As an alternative method of obtaining the coefficients of (? 00)n and (— i41)n we may say that un = a (Mn + b (— 24,)n. a and b can then be found from the conditions uo = I and u1 = 1.
The following examples are illustrative of the same method of attack.
Example 20.
A player tosses a coin and is to score one point for every head turned up and two for every tail. He is to play on until his score reaches or passes n. If pn is his chance of attaining exactly n, show that
P. = (pn—1 + pn—2)
and hence find the value of pn .
There are two ways of reaching n exactly, namely, by throwing
 336    PROBABILITY
By partial fractions this becomes
2    I 3(I—x) 3(I+2x).
Expanding by the binomial theorem, we obtain for the value of pn (the coefficient of xn) the same result as is found by the first method.
Example 21.
A has Io counters and B has 5; their chances of winning a single game are in the ratio 2 : I. The loser in each game is to give a counter to his opponent. The game stops when one or the other has lost all his counters. Find A's chance of winning all B's counters.
Let un be A's chance of winning all B's counters when A has n counters. In the next game A must either win or lose a counter. His chances of these contingencies are iS and * respectively. When he has lost the next game his chance of winning all B's counters is un_1 and when he has won the next game it is un+I
Hence    un = — z
Sun+I + 311n-1•
It is required to find un from the above relation, given that u15 = (since A will then have won) and uo = o (since he will have lost). The required relation for the recurring series is therefore
-x+3x2. Since we may write this relation as
3 (I — x+ix2)= 3 (I — ix)(I — x),
we may put un = a (2)n + b, and then obtain a and b from the values of uD and uls .
This is a particular example of the problem of "duration of play." The problem in its general form gives m counters to A and n counters to B, and states that A's chances of winning, drawing and losing a single game are p, q, r respectively, where p + q + r = I. The method of solution is precisely similar to that above.
The inductive method can be adapted to other types of question in probability. An excellent example will be found in J.I.A. vol. LvI, pp. 102-104, where a problem in direct probability is solved very simply by the inductive process.

14. We will conclude this chapter with some miscellaneous exincluding among them questions involving probabilities of living or dying. These problems require only a careful application of the ordinary methods, and no special comment is necessary.

ILLUSTRATIVE EXAMPLES    337
Example 22.
Three men were known to be alive five years ago when their ages were 31, 48, 69. Assuming that of 98 males born together one dies annually until there are no survivors, find the chances that
We have to find the chance of their dying in a given order. There will be no need therefore to take into consideration the different orders in which the men may die.
(i) P may die at age 96    Chance = z•
Q may die at age 97, i.e. may survive
a year and die in the following year    = (I — 2)
R may die at age 98 or later ; i.e. may
survive 2 years and die in the
following year    = (I — (1 — 3) 1. The total chance under this heading is therefore
n
L x (I&nb