A player X has a biased coin whose probabilit | vedclass

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Question

(A) Let $P(X)$ be the probability that player $X$ wins and $P(Y)$ be the probability that player $Y$ wins.Player $X$ uses a biased coin with $P(H) = p

Vedclass · Accepted Answer

$\frac{1}{3}$

Vedclass · Answer

(A) Let $P(X)$ be the probability that player $X$ wins and $P(Y)$ be the probability that player $Y$ wins.Player $X$ uses a biased coin with $P(H) = p$ and $P(T) = 1-p$. Player $Y$ uses a fair coin with $P(H) = 1/2$ and $P(T) = 1/2$.$X$ wins if $X$ gets $H$ on the $1^{st}$ throw,or $X$ gets $T$,$Y$ gets $T$,and $X$ gets $H$ on the $3^{rd}$ throw,and so on.$P(X) = p + (1-p)(1/2)p + (1-p)^2(1/2)^2p + \dots = p \sum_{n=0}^{\infty} (\frac{1-p}{2})^n = \frac{p}{1 - \frac{1-p}{2}} = \frac{2p}{1+p}$.Since the total probability is $1$,$P(Y) = 1 - P(X) = 1 - \frac{2p}{1+p} = \frac{1-p}{1+p}$.Given $P(X) = P(Y)$,we have $\frac{2p}{1+p} = \frac{1-p}{1+p}$.$2p = 1 - p \Rightarrow 3p = 1 \Rightarrow p = \frac{1}{3}$.

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