Question

A and B throw alternately a pair of dice. A wins if he throws 6 before B throws 7 and B wins if she throws 7 before A throws 6. If A begins, his chance of winning would be:
A) 30/61
B) 31/61
C) 1/2
D) 6/7

Answer 1

The answer to this would be option (A)30/61

The reason being:

The probability of A throwing ‘6’ is 5/36. So, the probability of A not throwing ‘6’ is 1 – 5/36 = 31/36. The probability of B throwing ‘7’ is 6/36. So, the probability of B not throwing ‘7’ is 1 – 6/36 = 30/36.

The probability that somebody wins is [1-{(31/36)(30/36)}]. So, The probability of winning of A provided that somebody wins = (5/36)/[1-{(31/36)(30/36)}] = 30/61.