Can a game with negative expectation still have a positive utility? Intuitively, I think not. But I can't clearly prove why. Specifically, I've been thinking about lottery games, where the expectation is obviously negative. But can the utility of hitting the jackpot ever be enough to justify playing?
 A: Imagine the following game: You toss a fair coin, if heads you get a buck, if tails you pay $2$ bucks. Clearly the expectation is $-0.5\ $bucks . We now define the following utility function: $u(x)=(x+2)^3$. Notice if you don't play the expected  utility is $8$, on the other hand the expected utility if you play is $\frac{0+3^3}{2}=\frac{27}{2}=13.5$. So this gives us a positive utility.
A: A common assumption about a utility function is that it is concave down: $U(tx + (1-t)y) \geq tU(x) + (1-t)U(y)$ for any $x,y$ and $0 \leq t \leq 1$. One can prove that any concave down function also satisfies the condition that, if $p_i$, $1 \leq i \leq n$ are nonnegative numbers with $\sum_i p_i = 1$, then for any $x_i$ we have:
$$U(\sum_ip_ix_i) \geq \sum_i p_i U(x_i)$$
In other words, if $E$ is the expectation of the game and $E_U$ the expected utility:
$$U(E) \geq E_U$$
It is also generally assumed that $U$ is increasing, and $U(0) = 0$. Hence if $E < 0$ one has $E_U \leq U(E) < U(0) = 0$, so the game has negative expected utility.
If one does not assume that utility is concave down this result does not hold, as shown in Jorge's answer.
