A coin lands heads with probability 1/3. Two players alternate flips and the first to throw a head wins. What is the probability the player who goes first wins, and is there any p in (0,1) that makes the duel fair?

A coin lands heads with probability 1/3. Two players alternate flips and the first to throw a head wins. What is the probability the player who goes first wins, and is there any p in (0,1) that makes the duel fair?

Approach: The position after both players miss is identical to the start, so write one equation for the first player's win probability and solve it, then ask what value of p equalises the two sides.

3/5. The position after two misses is identical to the start, so recursion gives w = p + (1-p)^2 w, hence w = p/(1 - (1-p)^2) = 1/(2-p). With p = 1/3 that is 1/(5/3) = 3/5. The same value comes from the geometric series p * sum_{k>=0} (1-p)^{2k}. Setting 1/(2-p) = 1/2 forces p = 0, so no p in (0,1) makes the duel fair. The first mover advantage is strict for every positive p and shrinks toward zero only as p tends to 0.

Follow-up: With three players flipping in a fixed cyclic order and p = 1/3, what are the three win probabilities?

Key concepts: recursion, geometric series, first mover advantage.