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Suppose $A$ and $B$ are playing the following game.

They each possess a fair coin, and individually flip their own coins until they get a tail. Whomever had a longer sequence of consecutive tosses that came up heads wins, and the loser has to pay the winner an amount that depends on the number of heads the loser flipped. If they tossed an equal number of heads before their respective first tail, there is a tie.

That is, denote the number of heads flipped by $A$ before the first tail by $a$, and call the number of heads flipped by $B$ before the first tail $b$. Then $A$'s payoffs in monetary terms from this game is given by

$$ p(a,b) = \begin{cases} 4^b & \text{if }a>b \\ 0 &\text{if } a=b \\ -4^a &\text{if }a

Straightforward calculations show that

$$ \operatorname{E}[p(a,b)] = \frac{1}{2} $$

How can a zero-sum game have positive expected payoff? In other words, how can both players expect to gain from playing this game when the amount that one player wins is exactly the amount that the other loses?

To see the expected value calculation, note that

$$ \operatorname E\left[4^b \vert a>b, b \right] \cdot \Pr (a>b \vert b) = 4^b \cdot \frac{1}{2^{b+1}} =\frac{1}{2}\cdot2^b $$ $$ \operatorname E\left[4^a \vert a

Which means that $\operatorname E [p(a,b)\vert b] = \frac{1}{2}$. The conditional expectation is independent of $b$, so it must equal the unconditional expectation.

I suspect there is some connection to the St. Petersburg paradox, but I'm not quite certain what exactly the relation is.

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    It's clear that something is wrong with your calculations. The game cannot create money. So it is a zero sum game, the sum of the expected values of the players must be zero. Because of symmetry, the expected value of each player must be zero. – 2017-02-18
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    @Paul I thought that as well, but I've gone over the calculations several times. If you can spot the mistake in the calculations, that would be helpful. The calculations are very simple, so I'm sure it should take you no more than a couple of minutes to look them over. – 2017-02-18
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    A discussion of this paradox: http://www.thebigquestions.com/alice.pdf – 2017-02-18
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    @Julian Aha. Thanks. – 2017-02-18
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    The short version of the discussion @Julian links to: http://mathoverflow.net/a/261563 – 2017-02-18
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    Note that the chance to get a certain $b$ changes if you introduce the assumption $a > b$. Maybe that's where it went wrong. One thing you could do is try a different approach that gives less problems with conditions and such. For example, calculate for each combination of $a$ and $b$ the probability that it occurs, together with the expected value for player A. – 2017-02-18
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    @Paul That is not the problem. The change in probabilities has been accounted for correctly upon conditioning on that event; see the calculations. The problem is the (implicit but) invalid appeal to Fubini-Tonelli. See the link posted by Julian or the MO post I link to above. – 2017-02-18
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    @JulianRosen if you write a brief answer explaining the problem, I'll happily accept it. – 2017-02-18

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The problem is that, while this is a symmetric zero-sum game, the expressions for each player's expected return are not absolutely convergent, making it impossible to say these expectations are $0$ or $\frac12$ or anything else

Looking at each particular result when Alice wins or loses using indicator functions, her expected return might be $$\sum\limits_{a,b \in \mathbb N_{0}} \frac{1}{2^{a+1}}\frac{1}{2^{b+1}} \left(4^b I[a>b] - 4^a I[a

If we were to try to separate these terms into Alice's wins and losses, we might get something apparently like $$\sum\limits_{b=0}^{\infty} \sum\limits_{a=b+1}^\infty {4^b}\frac{1}{2^{a+1}}\frac{1}{2^{b+1}} - \sum\limits_{b=1}^\infty \sum\limits_{a=0}^{b-1} {4^a}\frac{1}{2^{a+1}}\frac{1}{2^{b+1}}$$

and thus find your conditional calculations in the superficially less symmetric $$\sum\limits_{b=0}^\infty \frac12 2^b \frac{1}{2^{b+1}} - \sum\limits_{b=0}^\infty \frac12 (2^b-1) \frac{1}{2^{b+1}}$$

It should be easy to see that both the sums are infinite. This implies that Alice's expected gain if she wins is infinite, and Alice's expected loss if she loses is also infinite; the probability of each is $\frac13$ so we cannot find an overall expected value

Her net gains and losses do have a distribution: the probability of her gaining $4^n$ is $\frac{1}{4^{n+1}}$ for non-negative integer $n$; so too is the probability of her losing $4^n$; and the probability of her tying with Bob is $\frac13$. So $0$ is the median of this symmetric distribution, as well as being the most likely outcome. The issues are not far away in concept from a St Petersburg Game which started with Alice and Bob flipping a coin to decide which one of them would get to play

The apparent paradox comes from attempting to simplify the original sum in different ways and getting different results. Such combinations cannot be justified: for example the precondition of Fubini's theorem is not met

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    Did you inadvertently submit your answer too early? It sounds like you had more to say there. In any case, thanks for writing this up! I meant to write an answer based on the comments above, but I never got around to it. – 2018-07-31
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    @TheoreticalEconomist - there is often more to say, which is why I rarely use full stops at the ends of paragraphs – 2018-07-31