A and B throw a pair of dice alternately. A wins the game, if he gets a total of 6 and B wins, if she gets a total of 7. If A starts the game, then find the probability of winning the game by A in third throw of the pair of dice.

Let ${A_1} = A$ total of $6 = \{ (2,4),(1,5),(5,1),(4,2),(3,3)\}$

and ${B_1} = A$ total of $7 = \{ (2,5),(1,6),(6,1),(5,2),(3,4),(4,3)\}$

Let $P(A)$ is the probability, if $$A$$ wins in a throw $\Rightarrow$ $P(A) = \frac{5}{{36}}$

and $P(B)$ is the probability, if $$B$$ wins in a throw $\Rightarrow$ $P(B) = \frac{1}{6}$

$\therefore$

Required probability $= P(\bar A) \cdot P(\bar B) \cdot P(A) = \frac{{31}}{{36}} \cdot \frac{5}{6} \cdot \frac{5}{{36}} = \frac{{775}}{{216 \cdot 36}} = \frac{{775}}{{7776}}$