A and B alternately throw a pair of dice. A wins if he throws a sum of 5 before B throws a sum of 8, and B wins if he th…

\(P(A)=\frac{4}{36}=\frac{1}{9},P(B)=\frac{5}{36}\\ P(\text{ A wins })=\frac{4}{36}+\frac{32}{36}\cdot \frac{31}{36}\cdot \frac{4}{36}+\ldots ...\infty \\ P(\text{ A wins })=\frac{\frac{4}{36}}{1-\frac{32}{36}\cdot \frac{31}{36}}\\ =\frac{\frac{1}{9}}{1-\frac{8}{9}\cdot \frac{31}{36}}\\ =\frac{\frac{1}{9}}{1-\frac{62}{81}}=\frac{9}{19}\)