The St. Petersburg Paradox: Is MathWorld’s discussion confused?

MathWorld‘s page on the St. Petersburg paradox is misleading:

This paradox can clearly be resolved by making the distinction between the amount of the final payoff and the net amount won in the game. It is misleading to consider the payoff without taking into account the amount lost on previous bets, as can be shown as follows. At the time the player first wins (say, on the $n$ th toss), he will have lost

$\sum_{k=1}^{n-1}2^k=2^{n-2}$
(3)

dollars. In this toss, however, he wins $2^n$ dollars. This means that the net gain for the player is a whopping $2, no matter how many tosses it takes to finally win. As expected, the large payoff after a long run of tails is exactly balanced by the large amount that the player has to invest.

The fundamental problem here is that the paradox is not mathematical in nature (unlike say this) and to treat it as such leads to confusion. The fundamental issue is not to explain how the unbounded expectation comes about, but how to resolve the unbounded expectation with the intuitive notion of fair value to pay to play the game. The attempt above to explain the unbounded expectation hinges on the idea that by the $n$ th toss, $2^n-2$ dollars have been “lost” by the player and winning in the $n$ th toss results in a net gain of only 2 dollars. This is fallacious because, the “payoff” given up by the player up until the $n$ th toss is $2^{n-1}$ , i.e. what she could have earned by throwing an head on the $(n-1)$ toss. The “net gain” from winning on the the $n$ th toss is therefore $2^{n-1}$ . However, even this reasoning is wrong! The “final payoff” to the player by throwing a head on the $n$ th toss is $2^{n}$ and the “net gain” is $2^{n} -c$ , where $c$ is the amount paid by the player to enter the game. Assuming that the player has reached the $n$ th toss of the St. Petersburg game, she has not won anything on any of the previous tosses and by the rules of the game “lost” only $c$ to enter the game.