Waiting for the slower of two coin-flippers

Question

Two people each flip their own fair coin, once per round, until they get their first head; then they stop. They flip simultaneously each round.

On average, how many rounds until both people have gotten a head (i.e., the later of the two finishers)?

Accepted Answer

Let G1, G2 be the number of rounds each person needs, independent geometric variables with p = 	frac12, each with mean 1/p = 2. We want E[\max(G1, G2)]. Use the identity \max + \min = G1 + G2, so E[\max] = E[G1] + E[G2] - E[\min]. The minimum of the two waits is easy: both people are still flipping until at least one gets a head. In each round the chance that neither gets a head is q^2 = 	frac14, so the first finisher's round is geometric with success probability 1 - q^2 = 	frac34: E[\min] = \frac{1}{1 - q^{2}} = \frac{1}{3/4} = \frac{4}{3}. Therefore E[\max] = 2 + 2 - \frac{4}{3} = 4 - \f…

Waiting for the slower of two coin-flippers

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