[BruceZ]

You have to know your standard deviation, SD. Then the amount of time it takes to have probability p of being ahead is (x*SD/u)^2, where u is how much you win per game or per unit time, SD is for the same unit time, and x is the number of standard deviations depending on p. You can get x from the Excel function =NORMSINV(p). For 90% it is 1.28. For 60% it is 0.25.

If this is a coin flip type game with a 5% edge by paying 1.1 times your bet when you win (since 0.5*1.1 - 0.5*1 = .05), then your standard deviation is sqrt[ 0.5* 1.1^2 + 0.5*(-1)^2 ] = 1.051 for 1 flip. Then there is a 90% chance that you will be ahead after (1.28*1.051/.05)^2 = 724 flips. More than you thought, eh? This is actually faster than most blackjack games, but not as fast as poker.

Now what about if it's an even money bet, but you know the coin is biased to come up in your favor 52.5% of the time, so again you have a 5% edge. Now your standard deviation is sqrt[ .525*1^2 + .475(-1)^2 ] = 1 per flip. Then there is a 90% chance that you will be ahead after (1.28*1/.05)^2 = 655 flips. So although both games have a 5% edge, you are more likely to break even a little faster in this one.