Question

Two players A and B throw a die alternately for a prize of Rs 11, which is to be won by a player who first throws a six. If A starts the game, their respective expectations are

• A. Rs 6; Rs 5
• B. Rs 7; Rs 4
• C. Rs 5.50; Rs 5.50
• D. Rs 5.75; Rs 5.25

Answer 1

The correct option is A Rs 6; Rs 5

Probability through a six = $\frac{1}{6}$
$P\left(A\right)=\frac{1}{6},P\left(\overline{A}\right)=\frac{5}{6},P\left(B\right)=\frac{1}{6},P\left(\overline{B}\right)=\frac{5}{6}$
Probability of A winning
$=P\left(A\right)+P\left(\overline{A}\right)P\left(\overline{B}\right)P\left(A\right)+P\left(\overline{A}\right)P\left(\overline{B}\right)P\left(\overline{A}\right)P\left(\overline{B}\right)P\left(A\right)+...$
$=\frac{1}{6}+\frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}+\frac{5}{6}\times \frac{5}{6}\times \frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}+...$
$=\frac{\frac{1}{6}}{1-\frac{25}{36}}=\frac{6}{11}$
Probability of B winning $=1-\frac{6}{11}=\frac{5}{11}$
$\therefore$ Expectations of A and B are
$\frac{6}{11}\times 11=$ Rs 6 and $\frac{5}{11}\times 11$ =Rs 5