Question

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

Answer 1

Answer on Question #81884 — Math — Statistics and Probability

Question

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

Solution

Let A get the head in Nth trial to win the game.

Since he is flipping the coin in odd trials,

$$\mathrm{P}(N=1) = \mathrm{p},$$ $$\mathrm{P}(N=3) = (1 - p) * 0.5 * p,$$ $$\mathrm{P}(N=5) = (1 - p)^2 * 0.5^2 * p, \text{ and so on.}$$

Thus, $\mathrm{P}(A \text{ wins}) = p + (1 - p) * 0.5 * p + (1 - p)^2 * 0.5^2 * p + \cdots$

$$= \sum (0.5^{i-1} * (1 - p)^{i-1} * p) = \frac{p}{1 - (0.5 * (1 - p))} = \frac{p}{0.5 + 0.5p}$$ $$\mathrm{P}(\text{wins}) = 0.5$$ $$\frac{p}{0.5 + 0.5p} = 0.5$$ $$p = 0.25 + 0.25 * p$$ $$0.75 * p = 0.25$$ $$p = 1/3 \approx 0.33.$$

Answer: $p = 1/3$.

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