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 'p' is :
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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Write P(X wins) and P(Y wins) as infinite geometric series and equate them.
Updated On: Nov 11, 2026
- \(\dfrac15\)
- \(\dfrac13\)
- \(\dfrac25\)
- \(\dfrac14\)
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The Correct Option is B
Solution and Explanation
X wins on turn 1, 3, 5, ... Each full round (X tails, Y tails) has probability \((1-p)\cdot\tfrac12\).
\(P(X) = p\left[1 + \tfrac{1-p}{2} + \left(\tfrac{1-p}{2}\right)^2 + \cdots\right] = \dfrac{p}{1 - \frac{1-p}{2}} = \dfrac{2p}{1+p}\).
\(P(Y) = (1-p)\cdot\tfrac12\left[1 + \tfrac{1-p}{2} + \cdots\right] = \dfrac{(1-p)/2}{(1+p)/2} = \dfrac{1-p}{1+p}\).
Equal probabilities: \(2p = 1-p \Rightarrow p = \tfrac13\).
Answer: (2) \(\dfrac13\)
\(P(X) = p\left[1 + \tfrac{1-p}{2} + \left(\tfrac{1-p}{2}\right)^2 + \cdots\right] = \dfrac{p}{1 - \frac{1-p}{2}} = \dfrac{2p}{1+p}\).
\(P(Y) = (1-p)\cdot\tfrac12\left[1 + \tfrac{1-p}{2} + \cdots\right] = \dfrac{(1-p)/2}{(1+p)/2} = \dfrac{1-p}{1+p}\).
Equal probabilities: \(2p = 1-p \Rightarrow p = \tfrac13\).
Answer: (2) \(\dfrac13\)
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