hm... never finished this thread. Oh well deserves a bump. I just posted this somewhere else, and figured it fit in here:

How many hands must be played in order to accurately know one's winrate within +/- .1BB/100, assuming respective standard variations of 10BB/100 and 20BB/100?



It depends on what level of confidence you want. I will compute this for 95% confidence and 99% confidence, but you probably won’t achieve 0.1 bb/100 accuracy in your lifetime at these levels.

The standard error (SE) of the win rate is always SD/sqrt(N/100), where SD is the standard deviation for 100 hands, and N is the number of hands. For 95% confidence, we need 0.1 bb to be about 1.96 standard errors. For 99% confidence, we need 0.1 bb to be about 2.58 standard errors. These are from a table of the standard normal distribution. (like the one above)

Call this number s, for the number of standard errors. Then we have:

0.1 = s*SD/sqrt(N/100)

N = 100*(s*SD/0.1)^2

We can use this equation to get the following results:

95% confidence:
For SD = 20 bb/100, N =~ 15.4 million hands.
For SD = 10 bb/100, N =~ 3.8 million hands.

99% confidence:
For SD = 20 bb/100, N =~ 26.5 million hands.
For SD = 10 bb/100, N =~ 6.6 million hands.

So "true win rate" is effectively impossible to determine, and therefore should not be relied too heavily upon in determining what your future profits may be.