Ask HN: The St. Petersburg paradox and optimising the geometric mean

I recently asked myself whether the argument behind the Kelly criterion - that optimising the geometric rather than the arithmetic mean is more appropriate for repeated bets - can be applied to the St. Petersburg paradox. Assuming one only has enough money to pay for one round of St. Petersburg with a fee $X, the highest price one should be willing to pay would then be the highest $X still giving a > 1 geometric mean: (2/X)^0.5 * (4/X)^0.25 * (8/X)^0.125 * (16/X)^0.0625 * ... Resulting in a max fee of $4. Now if one had enough money to play two rounds one can again take the probabilities for the possible outcomes ($2+$2, $2+$4, $2+$8, $4+$4, etc.) and calculate the geometric mean. The same applies for any number of rounds, one just needs to swap the geometric distribution with a negative binomial distribution, giving a different max fee $X depending on the money available to one. Of course, when playing sequentially, the money paid out by the game can be used to fund more rounds, but this should only prolong the times one can play before going bankrupt by a constant factor, leaving the argument intact. Any thoughts on this?