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

Answer 1

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

Case1: If $X$ is winner
Outcomes$=H,TTH,TTTTH,...$
$P\left(X\right)=p+(1-p)\times \frac{1}{2}\times p+(1-p{)}^2\times {\left(\frac{1}{2}\right)}^2\times p+...$
$=\frac{p}{1-\frac{1}{2}(1-p)}=\frac{2p}{p+1}$
Case2: If $Y$ is the winner
Outcomes$=TH,TTTH,TTTTTH,...$
$P\left(Y\right)=\frac{1-p}{2}+\frac{(1-p{)}^2}{4}+\frac{(1-p{)}^3}{8}+...$
$=\frac{\frac{(1-p)}{2}}{1-\frac{(1-p)}{2}}=\frac{(1-p)}{p+1}$
Since, the probability of winning the game by both the players is equal
$\Rightarrow \frac{2p}{p+1}=\frac{(1-p)}{p+1}$
$\Rightarrow p=\frac{1}{3}$