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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 ' $p$ ' is
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Verified Answer
The correct answer is:
$\frac{1}{3}$
$\frac{1}{3}$
If the outcome is one of the following: $H, T T H, T T T T H, \ldots$, then $X$ wins As subsequent tosses are independent, so the probability that $X$ wins is
$$
p+\frac{p}{4}+\frac{p}{16}+\ldots=\frac{4 p}{3} .
$$
Similarly $Y$ wins if the outcome is one of the following: $T H, T T T H, T T T T T H, \ldots$
Therefore, the probability that $Y$ wins is
$$
\frac{1-p}{2}+\frac{1-p}{8}+\frac{1-p}{32}=\frac{2(1-p)}{3}
$$
Since, the probability of winning the game by both the players is equal then, we have
$$
\frac{4 p}{3}=\frac{2(1-p)}{3} \Rightarrow p=\frac{1}{3}
$$
$$
p+\frac{p}{4}+\frac{p}{16}+\ldots=\frac{4 p}{3} .
$$
Similarly $Y$ wins if the outcome is one of the following: $T H, T T T H, T T T T T H, \ldots$
Therefore, the probability that $Y$ wins is
$$
\frac{1-p}{2}+\frac{1-p}{8}+\frac{1-p}{32}=\frac{2(1-p)}{3}
$$
Since, the probability of winning the game by both the players is equal then, we have
$$
\frac{4 p}{3}=\frac{2(1-p)}{3} \Rightarrow p=\frac{1}{3}
$$
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