Related: St. Petersburg Paradox
I was reading today the Wikipedia page on the St. Petersburg Paradox.
The posted expected value is: $ 1/2 * 1 + 1/4*2 + 1/8*4 ... $
This seems very wrong to me. Here is a game which would lead to the same expected value calculation:
Flip a coin. If it's heads, you win \$1.
Regardless of result of previous game: Flip 2 coins. If both are heads, you win \$2.
Regardless of result of previous game: Flip 3 coins. If all three are heads, you win \$4.
...
Regardless of result of previous game: Flip N coins. If all are heads, you win $\$2^{N-1}$.
And you take this as N goes to infinity
Now this game obviously has an infinite expected value, and I would pay any amount of money to play it. It's hard to believe that the original game has the same expected value as this one.
Is there a flaw in my reasoning?
I don't think your constructed game and St. Petersburg are equivalent even though they have the same expected value. Consider this: in SP, your payout is always a power of 2; in your game you can have any whole number payout. That is, if you look at the probability of winning \$3 in SP, it's obviously $0$, whereas in your game, it's some finite non-zero value.
And we can't even claim that the expected payout of both the games is same. Sure both tend to infinity, but you can't say that their expected payout is the same on that basis. Consider $E_1$ to be the payout of SP and $E_2$ the payout of your game. $$E_1= 1 P_1(1)+2 P_1(2) + 3P_1(3) + \cdots $$ $$E_2= 1 P_2(1)+2 P_2(2) + 3P_2(3) + \cdots$$
Clearly, $P_1(3)=0$ but $P_2(3) \neq0$ and is some finite value. Same goes for the other terms. So you can see that $E_2>E_1$, hence the expected value of your game is more than SP.