at an examination and *r* optional subjects from which each candidate
chooses one (*r* > *s*), what is the probability that no two candidates
should choose the same subject? If the candidates be arranged
in any order, the probability that the second candidate should not
choose the same subject as the first candidate is (*n* - 1)/*n*. The probability
that the third candidate will not choose either of the two
subjects taken by the aforesaid candidates is (*n* - 2)/*n*, and so on.
Thus the required probability is

*n*(*n* − 1)(*n* − 2) ⋅ ⋅ ⋅ {*n* − (*s* − 1)}/*n*^{s}.

38. When as in these cases the interest of the problem lies chiefly
in the application of the theory of combinations, or permutations,
there is a propriety in Whitworth's enunciation of the questions under
the head of choice rather than *chance*. It comes to the same whether
we say that there are *x* ways in which an event may happen, or that
the probability of its happening in an assigned one of those ways is
1/*x*. For example, suppose that there are *n* couples waltzing at a
ball; if the names of the men are arranged in alphabetical order,
what is the probability that the names of their partners will also be
in alphabetical order? The probability that the man who is first in
alphabetical order should have for partner the lady who is first
in that order is 1/*n*. The probability that the man who is second
alphabetical order should have for partner the lady who is second
in that order is 1/(*n* - 1), and so on. Therefore the required probability
is 1/*n*!. Or it may be easier to say that the number of ways,
each consisting of a set of couples in which the party can be arranged,
is *n*!; of which only one is favourable.

39. The same principle governs the following question. For how many days can a family of 10 continue to sit down to dinner in a different order each day; it not being indifferent who sits at the head of the table—what is the absolute, as well as the relative, position of the members? The number of permutations, viz. 10!, is the answer. If we are to attend to the relative position only—as would be natural if the question related to 10 children turning round a flypole—the number of different arrangements would be only 9!

40. *Method of Equations in Finite Differences*.—The last question
may serve to introduce a method which Laplace has applied with
great *éclat* to problems in probabilities. Let *y*_{n} be the number of
ways in which *n* men can take their places at a round table, without
respect to their *absolute* position; and consider how the number
will be increased by introducing an additional man. From every
particular arrangement of the original *n* men can now be obtained *n*
different arrangements of the *n* + 1 men (since the additional man
may sit between any two of the party of *n*). Hence *y*_{n+1} = *ny*_{n}, an
*equation of differences* of which the solution is C (*n* - 1)! The constant
may be determined by considering the case in which *n* is 2.

41. The following example is not quite so simple. If a coin is
thrown *n* times, what is the chance that head occurs at least twice
running? Calling each sequence of *n* throws a “case,” consider
the number of cases in which head never occurs twice running; let
*u _{n}* be this number, then 2

^{n}−

*u*must be the number of cases when head occurs at least twice successively. Consider the value of

_{n}*u*

_{n+1}; if the last or (

*n*+ 2)th throw be tail,

*u*

_{n+2}includes all the cases (

*u*

_{n+1}) of the

*n*+ 1 preceding throws which gave no succession of heads; and if the last be head the last but one must be tail, and these two may be preceded by any one of the

*u*favourable cases for the first

_{n}*n*throws. Consequently,

*u*_{n+2} = *u*_{n+1} + *u _{n}*

If α, β are the roots of the quadratic *x*^{2} − *x* − 1 = 0, this equation
gives^{[1]}

*u _{n}* = Aα

^{n}+ Bβ

^{n}.

Here A and B are easily found from the conditions *u*_{1} = 2, *u*_{2} = 3;
viz.

, ,

whence .

The probability that head never turns up twice running is found
by dividing this by 2^{n}, the whole number of cases. This probability,
of course, becomes smaller and smaller as the number of trials (*n*)
is increased. This is a particular case of a more general problem
solved by Laplace^{[2]} as to the occurrence *i* times running of an event
of which the probability at one trial is *p*.

42. In such problems where we now employ the calculus of finite
difference Laplace employed his method of generating functions.
A distinguished instance is afforded by the problem of points which
was put by the Chevalier de Méré to Pascal and has exercised generations
of mathematicians. It is thus stated by Laplace.^{[3]} Two
players of equal skill have staked equal sums; the stakes to belong
to the player who shall have won a certain number of games.
Suppose they agree to leave off playing when one player, A, wants
*x* “points” (games to be won) in order to complete the assigned
number, while the second player wants *x′* points: how ought they
to divide the stakes? This is a question in *Expectation*, but its
difficulty consists in determining the probability that one of the
players, say A, shall win the stakes. Let that probability be *y*_{x,x′}.
Then, after the next game, if A has won, the probability of his winning
the stakes will be *y*_{x-1,x′}. But if A loses, B winning, the probability
will be *y*_{x,x′-1}. But these alternatives are equally likely.
Accordingly the probability of A winning the stakes may be written

½*y*_{x-1,x′} + ½*y*_{x,x′-1}.

This is the same probability as that which was before written *y*_{x,x′}.
Equating the two expressions we have, for the function *y*, an equation
of finite difference involving two variables, of which the solution is^{[4]}

.

43. The problem of points is to be distinguished from another
classical problem, relating to a contest in which the winner has not
simply to win a certain number of games, but to win a certain number
of counters from his opponent.^{[5]} Space does not admit even the
enunciation of other complicated problems to which Laplace has
applied the method of generating functions.

44. *Probability of Causes Deduced from Observed Events*.—Problems
relating to the probability of alternative causes, deduced from
observed effects, are usually placed in the separate category of
“inverse” probability, though, as above remarked,^{[6]} they do not
necessarily involve different principles. The difference principally
consists in the need of evidence, other than that which is afforded
by the observed event, as to the probability of the alternative
causes existing and operating. The following is an example free
from the difficulty incident to unverified a priori probabilities, which
commonly besets this kind of problem. A digit having been taken
at random from mathematical tables (or the expansion of an endless
constant such as π); a *second* digit is obtained by taking from a
random succession of digits one that added to the first digit makes
a sum greater than 9. Given a result thus formed, what are the
respective probabilities that the second digit should have been
0, 1, 2, . . . 8 or 9? In the long run the first digit assumes with equal
frequency the values 0, 1, 2 . . . 8, 9. Accordingly the second digit
can never be 0. There is only one chance of its being 1, namely when
the first digit is 9. If the second digit is 2, and the first either 8 or 9,
the observed effect will be produced. And so on. If the second
digit is 9, the effect may occur in nine ways. Accordingly in the
long run of pairs thus formed it will occur that the cases or causes
which are defined by the circumstances that the second digit is
0, 1, 2, . . . 8, 9, respectively, will occur with frequencies in the
following ratios 0 : 1 : 2 . . . 8 : 9. The probability of the observed
event having been caused by a particular (second) digit, *e.g.* 7, is
7/(0 + 1 + 2 + . . + 9) = 7/45.

45. The following example taken from Laplace^{[7]} is of a more
familiar type. An urn is known to contain three balls made up of
white and black balls in some unknown proportion. From this urn
a ball is extracted *m* times (being each time replaced after extraction).
If a white ball is drawn every time, what are the respective probabilities
that the number of white balls in the urn are 3, 2, 1 or 0?
By parity of reasoning it appears that in the first case the result is
certain, its probability 1, in the second case the probability of the
observed event occurring is (⅔)^{m}, in the third case that probability
is (⅓)^{m}, in the fourth case *zero*. Accordingly the respective inverse
probabilities are in the ratios

1 : (⅔)^{m} : (⅓)^{m} : 0;

*provided that* (as in the preceding example, with respect to the second
digits) the alternative causes, the four possible constitutions of the
urn, are (a priori) equally probable. This is rather a bold assumption
with respect to the contents of concrete urns^{[8]} and similar groupings;
but with regard to things in general may perhaps be justified
on the principle of *cross-series*.^{[9]}

46. Often in the investigation of causes we are not thrown back
on unverified a priori probabilities. We have some specific evidence
though of a very rough character. An example has been cited from
Mill in a preceding paragraph.^{[10]} Against the improbabilities calculated
by the methods of the present section there has often to be
balanced an improbability evidenced by common sense, which does
not admit of mathematical calculation. Bertrand^{[11]} puts the following
case. The manager of a gambling house has purchased a roulette
table which is found to give *red* 5300 times, *black* 4700 times, out of
10,000 trials. The purchaser claims an indemnity from the maker.
What can the calculus tell us as to the justice of the claim? Nothing

- ↑ Cf. Boole's
*Finite Differences*, ch. vii. § 5. - ↑
*Op. cit.*liv. II. ch. ii., No. 12. - ↑
*Op. cit.*liv. II. ch. ii., No. 8. - ↑ A clear and corrected version of Laplace's reasoning is given by
Todhunter,
*History. . . of Probability*, art. 973, p. 528, with reference to the more general cases in which the “skills” of each party—their chances of winning a single game—are not equal but respectively*p*and*q*(*p*+*q*= 1). See also Czuber,*Wahrscheinlichkeitstheorie*, pp. 30 seq. - ↑ See Todhunter,
*op. cit.*art. 107, and other articles referring to duration of play. See also Boole,*Finite Diferences*, ch. xiv., art. 7, ex. 6. - ↑ Above, par. 13.
- ↑
*Op. cit.*liv. II. ch. i. No. 1. - ↑ Cf. Bertrand,
*op. cit.*§ 118. - ↑ Above, par. 5.
- ↑ Par. 13.
- ↑
*Op. cit.*§ 134.