[AaronBrown]

I'll start with a simpler version. You and your opponent both get independent numbers drawn from a uniform distribution from 0 to 1. There are no antes or blinds. You go all in every time, your opponent calls with every number greater than X.

Your chance of winning is X*(1 - X)/2 and your chance of losing is X*(1 + X)/2. The other 1 - X of the time your opponent folds and no money changes hands.

Given that your opponent calls, your chance of losing is (1 + X)/2. There is a 50% chance that you will lose 1/[1 + ln(1+X)/ln(0.5)] hands in a row. So you will have a 50% chance of winning if you have 2^N - 1 times as much money as your opponent, and she sets X = 2^(1 - 1/N) - 1.

The table below shows how much money you have relative to your opponent, and what X gives you each a 50% chance of winning.

1 0.0000
3 0.4142
7 0.5874
15 0.6818
31 0.7411
63 0.7818
127 0.8114
255 0.8340
511 0.8517
1,023 0.8661

Your game is made more complex due to the blind and ante, and because the odds of winning at hold'em are not uniform from zero to one. They're approximately Normal with a mean of 50% and a standard deviation of 10%.