52, *pn* for 13, *qn* for 39 (where *p* + *q* = 1; *pn* and *qn* are integers).
A formula thus generalized is proposed by Professor Karl Pearson^{[1]}
as proper to represent the frequency with which different values
are assumed by a quantity depending on causes which are *not*
independent.

29. *Miscellaneous Examples: Games of Chance*.—The majority of
the problems under this heading cannot, like the preceding two, be
regarded as conducing directly to statistical methods which are
required in investigating some parts of nature. They are at best
elegant exercises in a kind of mathematical reasoning which is
required in most of such methods. Games of chance present some
of the best examples. We may begin with one of the oldest, the
problem which the Chevalier de Méré put to Pascal when he questioned:
How many times must a pair of dice be thrown in order that
it may be an even chance that double *six*—the event called *sonnez*—may
oocur at least once?^{[2]} The answer may be obtained by finding
a general expression for the rob ability that the event will occur
at least once in *n* trials; and then determining *n* so that this expression = ½.
The probability of the event occurring is the difference
between unity and the probability of its failing. Now the probability
of “sonnez” failing at a single throw (of two dice) is 3536. Therefore
the probability of its failing in n throws is . Whence we
obtain, to determine *n*, the equation , which gives
*n* = 24.605 nearly.

30. In the preceding problem the *quaesitum* was (unity *minus*)
the probability that out of all the possible events an assigned one
(“sonnez”) should fail to occur in the course of *n* trials. In the
following problem the *quaesitum* is the probability that out of all
the possible events one or other should fail—that they should not
all be represented in the course of *n* trials. A die being thrown *n*
times, what is the probability that all three of the following events
will not be represented (that one or other of the three will not occur at
least once); viz. (*a*) either ace or deuce turning up, (*b*) either 3 or 4,
(*c*) either 5 or 6. The number of cases in which one at least of these
events fails to occur is equal to the number of cases in which (*a*)
fails, *plus* the number in which (*b*) fails, *plus* the number in which (*c*)
fails, minus the number of cases in which two of the events fail
concurrently (which cases without this subtraction would be counted
twice).^{[3]} Now the number of cases in which (*a*) fails to occur in the
course of the *n* trials is of all the possible cases numbering 3^{n}.
Like propositions are true of (*b*) and (*c*). The number of cases in
which both (*a*) and (*b*) fail is of the total;^{[4]} and the like is true
of the cases in which both (*a*) and (*c*) fail and the cases in which both
(*b*) and (*c*) fail. Accordingly the probability that one at least of
the events will fail to occur in the course of *n* trials is

.

31. One more step is required by the following problem: If *n*
cards are dealt from a pack, each card after it has been dealt being
returned to the pack, which is then reshuffled, what is the probability
that one or other of the four suits will not be represented? The
probability that hearts will fail to occur in the course of the *n* deals
is ; and the like is true of the three other suits. From the sum
of these probabilities is to be subtracted the sum of the probabilities
that there will be concurrent failures of any two suits; but from this
subtrahend are to be subtracted the proportional number of cases
in which there are concurrent failures of any three suits (otherwise
cases such as that in which *e.g.* hearts, diamonds and clubs concurrently
failed^{[5]} would not be represented at all). Now the probability
of any assigned two suits failing is ; the probability of
any assigned three suits failing is . Accordingly the required
probability is

.

The analogy of the Binomial Theorem supplies the clue to the solution
of the general problem of which the following is an example.
If a die is thrown *n* times the probability that every face will have
turned up at least once is^{[6]}

.

32. If in the (first) problem stated in paragraph 31 the cards are dealt in the ordinary way (without replacement), we must substitute for , the continued product; for the continued product , and so on.

33. Still considering miscellaneous examples relating to games of
chance let us inquire what is the probability that at whist each of
the two parties should have two honours?^{[7]} If the turned-up card
is an honour, the probability that of the three other honours an
assigned one is among the twenty-five which are in the hands of
the dealer or his partner, while the remaining two honours are in
the hands of the other party, is 2551⋅2650⋅2549. But the assigned card
may with equal probability be any one of three honours; and
accordingly the above written probability is to be multiplied by 3.
If the turned-up card is not an honour then the probability that an
assigned pair of honours is in the hands of the dealer or his partner,
while the remaining two honours are in the hands of their adversaries,
is 2551⋅2450⋅2649⋅2548; this probability is to be multiplied by *six*, as
the assigned pair may be any of the six binary combinations formed
by the four honours. Now the probability of the alternative first
considered—the turned-up card being an honour—is 413; and the
probability of the second alternative, 913. Accordingly the required
probability is

.

34. The probability that each of the four players should have an
honour may be calculated thus.^{[8]} If the card turned up is an honour
then *ipso facto* the dealer has one honour and the probability that the
remaining players have each an assigned one of the three remaining
honours, is 1351⋅1350⋅1349. Which probability is to be multiplied by 3!,
as there are that number of ways in which the three cards may be
assigned. If the card turned up is not an honour the probability
that each player has an assigned honour is 1351⋅1350⋅1349⋅1248. Which
probability is to be multiplied by 4!. Accordingly the required
probability is

.

(the chance not being affected by the character of the card turned up).

35. The probability of all the trumps being held by the dealer is

or , which being calculated by means of
tables for (logarithms of) factorials^{[9]} or directly,^{[10]} is 158,753,389,900.

36. There is a set of dominoes which goes from double blank to double nine (each domino presenting either a combination—which occurs only once—of two digits, or a repetition of the same digit). What is the probability that a domino drawn from the set will prove to be one assigned beforehand? The probability is the reciprocal of the number of dominoes: which is 10 × 9/2 (the number of combinations of different digits) + 10 (the number of doubles) = 55.

37. *Choice and Chance*.—When we leave the sphere of games of
chance and frame questions relating to ordinary life there is a danger
of assuming distributions of probability which are far from probable.
For example, let this be the question. The House of Commons
formerly consisting of 489 English members, 60 Scottish and 103
Irish, what was the probability that a committee of three members
should represent the three nationalities? An assumption of indifference
where it does not exist is involved in the answer that the required
probability is the ratio of the number of favourable triplets, viz.
489 × 60 × 103 to the total number of triplets, viz. 652 × 651 × 650 × 3!
A similar absence of selection is postulated by the ordinary
treatment of a question like the following. There being *s* candidates

- ↑
*Trans. Roy. Soc.*(1895). See below, par. 165. - ↑ Todhunter,
*History . . . of Probability*, and Bertrand,*Calcul**des probabilités*, p. 9. - ↑
*All three*events cannot fail. - ↑ (
*c*) occurring*n*times. - ↑ The reasoning may be illustrated by using the area of a circle to represent the frequency with which hearts fail, another (equal) circle for diamonds; for the case in which both hearts and diamonds fail the area common to the circles interlapping, and so on.
- ↑ See Whitworth,
*Exercises in Choice and Chance*, No. 502 (p. 125); referring to prop. xiv. of the same author's*Choice and Chance*. - ↑ Cf. Whitworth,
*Choice and Chance*, question 143, p. 183, ed. 4. - ↑ Ibid.
- ↑ There is such a table at the end of De Morgan's article in the
Calculus of Probabilities in the
*Ency. Brit.*“Pure Sciences,” vol. ii. - ↑ Cancelling factors common to the numerator and denominator.