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View Full Version : Math problem...how long until I win?

Josh W
11-05-2003, 04:32 AM
No, this isn't a bad beat/streak story....I'm just curious about a hypothetical situation.

If I'm playing a game where I have exactly a 5% edge per game, how many games would I need to play before there is a 90% chance of me being up?

More importantly, is there a simple formula for this (i.e., what if I wanted to know how many games until there was a 95% chance of winning, or 60% chance of winning)?

Josh

BruceZ
11-05-2003, 05:30 AM
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.

BruceZ
11-05-2003, 06:23 AM
(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.

I'm equating flips to hours here. For blackjack, the ratio of SD to EV is 20-40+ for an hour, with the low end being hard to maintain, and for poker it is around 10. Here the ratio is 21, so it's like a very good blackjack game if flips are hours. If you compare a flip to a hand, then this is much better than either game. It would be about like a BJ game with a 5% edge.

BruceZ
11-05-2003, 12:48 PM
SD = sqrt [ E(x)^2 - E(x)^2 ], where E(x) is expected value. I didn't subtract E(x)^2 = .05^2. This only makes a difference of 1 flip. Here are the corrected examples.

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 - .05^2 ] = 1.05 for 1 flip. Then there is a 90% chance that you will be ahead after (1.28*1.05/.05)^2 = 723 flips. More than you thought, eh? This is actually much faster than both blackjack and poker if flips are like hands.

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 - .05^2 ] = .9987 per flip. Then there is a 90% chance that you will be ahead after (1.28*.9987/.05)^2 = 654 flips. So although both games have a 5% edge, you are more likely to break even a little faster in this one.