How many hands to achieve 95% winning days?

[kjander]

Lets assume you are a 5.26BB/100 winner over 200K hands with a standard deviation of 44bb/100. How many hands do you have to play to win over that stretch 95% of the time.

[timex]

To be 95%, a negative result has to be at least 1.65 standard deviations. 5.26n > 1.65* 44 *root(n)
root(n) > 13.8
So about 19000 hands to be 95% sure you are up.

[Phil153]

A 95% confidence interval occurs at very close to 2 standard deviations from the mean. For 95% of your days to end in the positive, your daily mean must be > 2 Standard Deviations, or 88bb/100. That corresponds to 88/5.26 =~ 1673 hands/day.

[kjander]

that sounds abotu righ but..... I didn't really follow the math. Can you baby step it through to me?

[AaronBrown]

This is almost right. 2 standard deviations is a 2-sided 95% confidence interval, meaning that there would be a 2.5% chance of losing money and a 2.5% chance of making more than double the expected profit. 1.64, as timex used, is correct for a one-sided 95% interval, with a 5% chance of losing money.

Second, you have to square the ratio, also as timex did. Using your numbers it's (2*44/5.26)^2 = 280 hundred hands instead of (2*44/5.26) = 17 hundred hands.

[The_Bankroll]

just out of curiosity, if you did play enough hands to win 95% of the time, what would you're average win rate/day be?

[timex]

Number of hands played * average win/hand