Question

A and B throw a pair of dice. A wins if he throws $6$ before B throws $7$ and B wins if he throws $7$ before A throws $6$. If A begins, his chance of winning is?

• A. $\frac{5}{16}$
• B. $\frac{30}{61}$
• C. $\frac{35}{61}$
• D. $\frac{60}{61}$

Answer 1

The correct option is B $\frac{30}{61}$

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.