A and B throws alternatively with 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.

Question

A and B throws alternatively with 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. Find their chances of winning.

I did this question but I get 6/11 for A. But the answer is given as 30/61???

Here is how I did it;

Answer 1

From your answer, you seem to have argued that
P(A) throws a $6 =\frac5{36}$, P(B) throws a $7 = \frac6{36}$, thus P(A wins) $=\frac6{6+5} = \frac6{11}$

But that is not correct.
With $A$ throwing first, let $a$ be the probability that $A$ wins.

Then either $A$ wins on first throw, or both $A$ and $B$ fail on their first throw, and we are back to start, so to say.

This gives the equation $a = \frac5{36} + \left(\frac{31}{36}\cdot\frac{30}{36}\right)a$

Solve to get the answer