Question

A player $X$ has a biased coin whose probability of showing heads is $p$ and a player $Y$ has a fair coin. They start playing a game with their own coins and play alternately. The player who throws a head first is a winner. If $X$ starts the game, and the probability of winning the game by both the players is equal, then the value of ${}^{\prime }{p}^{\prime }$ is

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

Answer 1

The correct option is A $\frac{1}{3}$

$X$ wins when the outcome is one of the following set of outcomes: $H,TTH,TTTTH,....$

Since subsequent tosses are independent, the probability that $X$ wins is $p+\frac{p}{4}+\frac{p}{16}+...=\frac{4p}{3}$

Similarly $Y$ wins if the outcome is one of the following: $TH,TTTH,TTTTTH,...$

So, the probability that $Y$ wins is $\frac{1-p}{2}+\frac{1-p}{8}+\frac{1-p}{32}=\frac{2(1-p)}{3}$

Since $X$ and $Y$ win with equal probability, we have $\frac{4p}{3}=\frac{2(1-p)}{3}\Rightarrow p=\frac{1}{3}$

So, option A is the correct answer.