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

Answer 1

The correct option is C $\frac{30}{61}$

Sum $7=\left\{\right(1,6\left)\right(2,5\left)\right(3,4\left)\right(4,3\left)\right(5,2\left)\right(6,1\left)\right\}$
Probability of getting sum $7$ is
$P\left(7\right)=\frac{6}{36}$

Sum $6=\left\{\right(1,5\left)\right(2,4\left)\right(3,3\left)\right(4,2\left)\right(5,1\left)\right\}$
Probability of getting sum $6$ is
$P\left(6\right)=\frac{5}{36}$

Now, the probability of $A$ winning the game is
$P\left(A_{win}\right)=P\left(6\right)+P\left(\overline{6}\right)\cdot P\left(\overline{7}\right)\cdot P\left(6\right)+\cdots$


$=\frac{5}{36}+\frac{31}{36}\times \frac{30}{36}\times \frac{5}{36}+\cdots$
$=\frac{\frac{5}{36}}{1-\frac{31\times 30}{36\times 36}}$
$=\frac{5\times 36}{36\times 36-31\times 30}$
$=\frac{5\times 36}{6(36\times 6-31\times 5)}$
$=\frac{5\times 6}{61}$
$\therefore P\left(A_{win}\right)=\frac{30}{61}$