The coin game worth infinity

Question

A casino offers this game. A fair coin is flipped repeatedly until it first lands tails. If the first tails appears on flip number $k$ , you are paid $2^k$ dollars. So tails on flip 1 pays $2, tails on flip 2 pays $4, tails on flip 3 pays $8, and so on.

What is the expected payout of a single game?

Accepted Answer

The first tails lands on flip k exactly when the first k-1 flips are heads and flip k is tails, which has probability \left(	frac12ight)^{k}. That outcome pays 2^k. So the expected payout is E[	ext{payout}] = \sum{k=1}^{\infty} \left(	frac12ight)^{k}\cdot 2^{k} = \sum{k=1}^{\infty} 1 = \infty. Each term contributes exactly \1 to the average, and there are infinitely many of them, so the expected payout is infinite. Yet almost nobody would pay even \20 to play, and that gap is the paradox. The infinite average is built entirely from enormous prizes that essentially never occur: you win …

The coin game worth infinity

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