To find the probability of throwing an ace in two throws with a
single die. The probability of throwing an ace the first time is
1/6; whereof 1/ is the first part of the probability required.
If the ace be missed the first time, still it may be thrown on
the second; but the probability of missing it the first time is
5/6, and the probability of throwing it the second time is 1/6;
therefore the probability of missing it the first time and
throwing it the second, is 5/6 X 1/6 = 5/36 and this is the
second part of the probability required, and therefore the
probability required is in all 1/6 + 5/36 = 11/36.
To this case is analogous a question commonly proposed about
throwing with two dice either six or seven in two throws, which
will be easily solved, provided it be known that seven has 6
chances to come up, and six 5 chances, and that the whole number
of chances in two dice is 36; for the number of chances for
throwing six or seven 11, it follows that the probability of
throwing either chance the first time is 11/36, but if both are
missed the first time, still either may be thrown the second
time; but the probability of missing both the first time is
25/36, and the probability of throwing either of them on the
second is 11/36; therefore the probability of missing both of
them the first time, and throwing either of them the second time,
is 25/36 X 11/36 = 275/1296, and therefore the probability
required is 11/36 + 275/1296 = 671/1296, and the probability of
the contrary is 625/1296.
