Question

Two players $A$ and $B$ toss a coin alternately. A begins the game and the player who first throws heads in the winner. $B^{\prime }s$ coin is fair but $A^{\prime }s$ is biased and has probability $p$ of showing heads. The value of $p$ so that the game is equiprobable to both players

• A. $\frac{2}{3}$
• B. $\frac{1}{3}$
• C. $\frac{2}{5}$
• D. None of these

Answer 1

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

Let $E_i$ and $F_i$ denote the events that the ith toss results in heads and tails, respectively.
Then if $q=1-p$, the probability that $a$ wins the game is
$P(E_1$ or $F_1{F}_2{E}_3$ or $F_1{F}_2{F}_3{F}_4{E}_5$ or $...)$
$P\left(E_1\right)+P\left(F_1{F}_2{E}_3\right)+P\left(F_1{F}_2{F}_3{F}_4{E}_5\right)+...$
$P\left(E_1\right)+P\left(F_1\right)P\left(F_2\right)P\left(E_3\right)+P\left(F_1\right)P\left(F_2\right)P\left(F_3\right)P\left(F_4\right)P\left(E_5\right)+...$

$=p+q\left(\frac{1}{2}\right)p+{\left(\frac{q}{2}\right)}^{2}p+...=\frac{p}{1-\left(\frac{q}{2}\right)}=\frac{2p}{2-q}=\frac{2p}{1+p}$
Subtracting this form $1$, the probability that $B$ wins the game is
$1-\frac{2p}{1+p}=\frac{1-p}{1+p}$
so the game will be equiprobable is
$\frac{2p}{1+p}=\frac{1-p}{1+p}\Rightarrow 3p=1\Rightarrow p=\frac{1}{3}$