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A game involves flipping a coin until the first head appears and winning $2^n$ dollars if the first head appears on the $\mathrm{n^{th}}$ coin flip. We want to determine the expected winnings for this game.

Based on my understanding on the problem,

$$\begin{align}&X=\{ \mathrm{Coin\ Flips}\} \sim\mathrm{Geo}(p=0.5) \\ &W=\{\mathrm{Winnings}\}=2^X\\ &E[W]=E[2^X]=\sum_{n=1}^\infty 2^nP(X=n)=\sum_{n=1}^\infty (2^n)(0.5^n)=\sum_{n=1}^\infty 1=\infty \end{align}$$

However, this doesn't sound right to me because we also know that

$$E[X]=\frac{1}{p}=\frac{1}{0.5}=2$$

Rob
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    This is known as the "St Petersburg Paradox" https://en.wikipedia.org/wiki/St._Petersburg_paradox – Doug M Apr 22 '17 at 03:10
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    It shouldn't bother you that $E[2^X] = \infty$ while $2^{E[X]} = 4$; in general, taking expected values doesn't commute with arbitrary operations. – Misha Lavrov Apr 22 '17 at 03:58

1 Answers1

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As @Doug M said in the comments, this is the St. Petersburg Paradox. To summarize the wikipedia page for people's convenience:

That expected value assumes infinite resources on both ends. One common solution is Utility Theory which accounts for the demand and satisfaction of consumers and is more Game Theory than probability. Another solution is to simply weight more unlikely events to be smaller and more negligent, giving a convergent/finite series. Others look only at the scenario of finite resources of the gambler, and others still reject the notion of expectation rigorously applying to real world scenarios. This paradox is by no means solved yet, so even today people discuss it at length.

Vedvart1
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