A biased coin tames the paradox

Question

The St. Petersburg game pays \2^n if the first head appears on toss n; with a fair coin the expected value is infinite. Now use a biased coin that lands heads with probability p = 	frac23 on each toss (tails with probability 	frac13), keeping the same \2^n payout. What is the expected payout now? For which p is the game worth a finite amount?

Accepted Answer

With head-probability p, the first head lands on toss n when you get n-1 tails then a head: probability (1-p)^{n-1}p. The payout is 2^n, so E[	ext{payout}] = \sum{n=1}^{\infty} (1-p)^{n-1}\,p\,\cdot 2^n. Factor to expose a geometric series. Write 2^n = 2\cdot 2^{n-1}: E = 2p \sum{n=1}^{\infty} \big(2(1-p)\big)^{n-1} = 2p \cdot \frac{1}{1 - 2(1-p)}, valid when the ratio 2(1-p) < 1, i.e. p 	frac12. Plug in p = 	frac23, so 1 - p = 	frac13 and 2(1-p) = 	frac23: E = 2\cdot	frac23 \cdot \frac{1}{1 - 2/3} = \frac{4}{3} \cdot \frac{1}{1/3} = \frac{4}{3}\cdot 3 = \4. A finite, modest \4. The rea…

A biased coin tames the paradox

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