**St. Petersburg Paradox**

Here's a shocker! It turns out that your Average Winnings are infinite.
Let's resort to a little math. If P is your probability of losing on any particular roll of the die (P = 2/3 in our original problem), and Q = 1 - P, then your expected winnings are given by

E = 2^{0}P + 2^{1}QP + 2^{2}Q^{2}P + 2^{3}Q^{3}P + ...= P/(1 - 2Q)

since this results in a geometric series for which there is a closed form representation. In our original problem, P = 2/3, Q = 1/3, and the Average Winnings, E, computes to a value of 2 dollars.

But look what happens when when P = 1/2, as is the case for the second problem posed. Now the expected winnings have no bound. Even when the player's chance of losing on any particular roll is 0.51 instead of 1/2, E still takes on the fairly respectable value of $25.50. So things go to pot very quickly in the tight neighborhood of P = 0.50.

**This is one of those games for which there is a very low probability of winning a very high amount of money, and Expected, or Average Winnings are generally not a particularly good criterion upon which to base one's action. I wouldn't pay $10 on a one-in-a-million chance of winning 20 million dollars, even though the expectation, in this case, makes it seem like a good bet.
**