Question

Two persons A and B are throwing an unbiased six faced die alternatively, with the condition that the person who throws $3$ first wins the game. If A starts the game, the probabilities of A and B to win the same are respectively

• A. $\frac{6}{11},\frac{5}{11}$
• B. $\frac{5}{11},\frac{6}{11}$
• C. $\frac{8}{11},\frac{3}{11}$
• D. $\frac{3}{11},\frac{8}{11}$

Answer 1

The correct option is A $\frac{6}{11},\frac{5}{11}$

$p=\frac{1}{6}$ and $q=\frac{5}{6}$
A wins if he gets $3$ on his 1st turn or he gets $3$ on his second turn but B doesn't get $3$ on his first turn and so on..
$\therefore p\left(A\right)=p+p{q}^2+p{q}^4+......$
$=p(1+q^2+q^4+.....)$
$=\frac{p}{1-q^2}=\frac{\frac{1}{6}}{1-(\frac{5}{6}{)}^2}=\frac{6}{11}$
$\therefore p\left(B\right)=1-p\left(A\right)=\frac{5}{11}$