Mathematical_Recreations-Kraitchik-2e

1~4 Mathematical Recreations
to throw in the effort to make the point, that is, to throw this
number again before he throws a 7. If he is successful, he
wins and retains the dice. But if not, that is, if the point
never reappears and a 7 is thrown, the thrower loses on the
throw of 7 and the dice pass to the next player. It is possible
that neither the point nor the 7 might come up, but, as we
shall see, the probability that play could continue in this way
for an appreciably long time is so slight that it may be disregarded.
Consider first the probability of throwing a particular total
on a single throw. There are 6·6 = 36 equally likely ways in
which one of the 6 faces of die A and one of the 6 faces of die
B may come up. If we denote by (a, b) the respective numbers
of spots exposed on the two dice, we see that the total 2
can appear only as (1,1), the total 3 as (1, 2) or (2,1),···,
the total 6 as (1,5), (2,4), (3,3), (4,2), (5, 1),···, the total
12 as (6, 6), The results of such an analysis are listed in
the following table, in which the number of ways in which
each total may be thrown (the frequency of that total) is
given.
Total.... 2
Frequency 1
3 4
2 3
5
4
6
5
7
6
8 9 10 11 12
5 4 3 2 1
From this table we can draw all the conclusions needed in
the theory of the game. First, the probability of throwing
a particular total, say 5, is the ratio of its frequency to 36, in
this case -Is = t. Again, the probability of throwing some
one of several totals, say of throwing a 2, a 3, or a 12, is the
sum of the probabilities of the separate totals, * + :h +
:h = -Is = t· The probability of making a point, say of
throwing a 5 before throwing a 7, is the ratio of the frequency
of the point to the sum of the frequency of the point plus the
4 4 2
frequency of 7: in the case of making 5, 4 + 6 = 10 = 5·