A and B throw a pair of dice alternately.

Let’s take A1 to be the event of getting a total of 6

A1 = {(2, 4), (4, 2), (1, 5), (5, 1), (3, 3)}

And, B1 be the event of getting a total of 7

B1 = {(2, 5), (5, 2), (1, 6), (6, 1), (3, 4), (4, 3)}

Let P(A1) is the probability, if A wins in a throw = 5/36

And P(B1) is the probability, if B wins in a throw = 1/6

Therefore, the required probability of wining A in his third throw

$=P\left(\bar{A}_{1}\right) \cdot P\left(\bar{B}_{1}\right) \cdot P\left(A_{1}\right)=\frac{31}{36} \cdot \frac{5}{6} \cdot \frac{5}{36}=\frac{775}{7776}$