Modest winnings are the rule

Question

In the standard doubling game, a fair coin is flipped until the first tails on flip k, paying 2^k dollars (so \2, \4, \8, and so on). What is the probability that you win at least \8 in a single play?

Accepted Answer

Winning at least 8 = 2^3 means the payout 2^k satisfies k \ge 3, that is, the first tails does not arrive until flip 3 or later. That requires the first two flips both to be heads: P(	ext{win} \ge 8) = P(	ext{first two flips heads}) = \left(	frac12ight)^2 = \frac14. So despite the infinite average payout, you take home \8 or more only a quarter of the time. The same logic shows you win at least 1024 = 2^{10} only with probability \left(	frac12ight)^9 = 	frac{1}{512}, and larger prizes are rarer still. This is the heart of the paradox laid bare: the mean is enormous, but the typical g…

Modest winnings are the rule

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