A and B alternately throw a pair of dice.

Question

$A$ 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 throws a sum of 8 before $A$ throws a sum of 5 . The probability, that A wins if A makes the first throw, is

Accepted Answer

$$\begin{aligned}
& \mathrm{p}\left(\mathrm{~S}_5\right)=\frac{1}{9} \
& \mathrm{p}\left(\mathrm{~S}_5\right)=\frac{5}{36} \
& \text { required prob }=\frac{1}{9}+\frac{8}{9} \cdot \frac{31}{36} \cdot \frac{1}{9}+\left(\frac{8}{9} \cdot \frac{31}{36}\right)^2 \cdot \frac{1}{9}+\ldots \infty \
& =\frac{\frac{1}{9}}{1-\frac{62}{81}}=\frac{9}{19}
\end{aligned}$$

Solutions · 1