Question

A fair die is tossed repeatedly. $A$ wins if it is $1$ or $2$ on two consecutive tosses and $B$ wins if it is $3,4,5$ or $6$ on two consecutive tosses. The probability that $A$ wins if the die is tossed indefinitely is

• A. $\frac{1}{3}$
• B. $\frac{5}{21}$
• C. $\frac{1}{4}$
• D. $\frac{2}{5}$

Answer 1

Answer: (B)

$\frac{5}{21}$

Say, $A$ won the first round,

Then for $A$ to win, the possible combinations are :

$AA,ABAA,ABABAA,$ and so on.

The probability of $A$ winning is :

$\frac{1}{3}(\frac{1}{3}+\frac{2}{9}.\frac{1}{3}+...)$ where $\frac{2}{9}$ is the probability of the group $AB$.

$P(A$wins given that $A$ wins the first round$)=\frac{1}{3}.\frac{1}{3}.\frac{9}{7}$ where $\frac{9}{7}=(1+\frac{2}{9}+...)$

$P(A$ wins given that $A$ wins the first round$)=\frac{1}{7}$ ...(1)

If $B$ won the first round , then similarly for $A$ to win we have,

$BAA,BABAA,BABABAA,...$

Hence the total probability is $\frac{2}{3}.\frac{1}{9}.\frac{9}{7}$ $=\frac{2}{21}$

$P(A$ wins given that $B$ wins the first round$)=\frac{2}{21}$ ...(2)

Adding the probabilities (1) and (2), we get $\frac{1}{7}+\frac{2}{21}=\frac{5}{21}$