Multiple choice

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.

  1. $\dfrac {1}{2}$
  2. $\dfrac {1}{3}$
  3. $1$
  4. $\dfrac {1}{6}$
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B Correct answer
Explanation

Let p be the probability of A getting a head and 1/2 be the probability of B getting a head. The probability of A winning is p + (1-p)(1/2)p + (1-p)(1/2)(1-p)(1/2)p + ... which is a geometric series summing to p / (1 - (1-p)/2) = 2p / (1+p). Setting this equal to 1/2 gives 4p = 1+p, so 3p = 1, p = 1/3.

AI explanation

For the game to be fair, the sum of the probabilities of A winning on his first and subsequent turns must equal the probability of B winning. The probability of A winning is p plus (1-p) times (1/2) times p, which forms the sum p + 0.5p - 0.5p^2. The probability of B winning is (1-p) times (1/2) plus (1-p)^2 times (1/2)^2 times p, and solving the equality 1.5p - 0.5p^2 = 0.5 - 0.25p yields 4p^2 - 7p + 2 = 0. Factoring this quadratic equation gives two roots, 2 and 1/3, but since p cannot exceed 1, the only valid solution is p = 1/3. The result is 1/3.

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