Question

Two players $A$ and $B$ toss a coin alternatively, with $A$ beginning the game. The players who first throw a head is deemed to be the winner. $B$'s coin is fair and $A$'s is biased and has a probability $p$ showing a head. Find the value of $p$ so that the game is equiprobable to both the players.

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $1$
• D. $\frac{1}{6}$

Answer 1

Answer: (B)

$\frac{1}{3}$

Player "A' wins if he gets head in the first trial or in the third {if B does not get head in his first trial] and so on.

$P\left(A\right)=p+(1-p)\times \frac{1}{2}\times p+(1-p)\times \frac{1}{2}\times (1-p)\times 12\times p+...$

This is an infinite G.P of the first term p and common ratio $\frac{(1-p)}{2}=\frac{p}{1-\left(\frac{(1-p)}{2}\right)}=\frac{2p}{(p+1)}$

According to the question,

$P\left(A\right)=P\left(B\right)⟹\frac{2p}{(1+p)}=1-\frac{2p}{(p+1)}⟹\frac{4p}{(1+p)}=1⟹p=\frac{1}{3}$