Question

In a game two players $A$ and $B$ take turns in throwing a pair of fair dice starting with player $A$ and total of scores on the two dice, in each throw is noted. $A$ wins the game if he throws a total of $6$ before $B$ throws a total of $7$ and $B$ wins the game if he throws a total of $7$before $A$ throws a total of six The game stops as soon as either of the players wins. The probability of $A$ winning the game is :

• A. $\frac{31}{61}$
• B. $\frac{5}{6}$
• C. $\frac{5}{31}$
• D. $\frac{30}{61}$

Answer 1

Answer: (D)

$P(6)=\frac{5}{36},P(7)=\frac{1}{6}$
$P(A)=W+FFW+FFFFW+\dots .$
$=\frac{5}{36}+\left(\frac{31}{36}\times \frac{5}{6}\right)\times \frac{5}{36}+{\left(\frac{31}{36}\times \frac{5}{6}\right)}^2\times \frac{5}{36}+\dots$
$=\frac{\frac{5}{36}}{1-\frac{155}{216}}=\frac{5}{36}\times \frac{216}{61}=\frac{30}{61}$