Zero-sum game with positive expectation? 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
$$ p(a,b) =
\begin{cases}
4^b & \text{if }a>b \\
0 &\text{if } a=b \\
-4^a &\text{if }a<b 
\end{cases} $$
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
$$ \operatorname E\left[4^b \vert a>b, b \right] \cdot \Pr (a>b \vert b) = 4^b \cdot \frac{1}{2^{b+1}} =\frac{1}{2}\cdot2^b $$
$$ \operatorname E\left[4^a \vert a<b, b \right] \cdot \Pr (a<b \vert b) = \sum_{k=0}^{b-1}4^k\cdot \frac{1}{2^{k+1}} = \frac{1}{2}\left(2^b - 1 \right) $$
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.
 A: 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 $$\sum\limits_{a,b \in \mathbb N_{0}} \frac{1}{2^{a+1}}\frac{1}{2^{b+1}} \left(4^b I[a>b] - 4^a I[a <b]\right)$$ 
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  
