$A$ 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:
Step-by-Step Solution
Key Concept: $P(A\text{ wins})=p+(1-p)(1-q)p+(1-p)^{2}(1-q)^{2}p+\dotsb=\dfrac{p}{1-(1-p)(1-q)}$ where $p=P(\text{sum }5),\,q=P(\text{sum }8).$
$P(\text{sum }5)=\dfrac{4}{36}=\dfrac{1}{9}$ (pairs $(1,4),(2,3),(3,2),(4,1)$).
$P(\text{sum }8)=\dfrac{5}{36}$ (pairs $(2,6),(3,5),(4,4),(5,3),(6,2)$).
$P(A\text{ wins})=\dfrac{p}{1-(1-p)(1-q)}=\dfrac{1/9}{1-\tfrac{8}{9}\cdot\tfrac{31}{36}}=\dfrac{1/9}{76/324}=\dfrac{324}{9\cdot 76}=\dfrac{9}{19}.$
Correct Answer: 2