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Don't worry to much about winrate at low limits. You normally don't stay there long enough to have a large enough sample for it to be significant. But over a large sample 12bb/100 is still quite decent. (and congratz arvyzalo but 2.7k is not a sample
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If your standard deviation is about 90 bb/100, then about 95% of the time, after n hands your win rate will be within 1800/squareroot(n) bb/100 of your true win rate. So, even after 10k hands this would be a very large +- 18 bb/100 interval.

It is important to look at more than the results. Look at how you got there, and try to identify edges you have over your opponents. Once you have a solid understanding of poker, you can see whether you have a big advantage or a small one long before your win rate settles down.

Originally posted by pzhon
If your standard deviation is about 90 bb/100, then about 95% of the time, after n hands your win rate will be within 1800/squareroot(n) bb/100 of your true win rate. So, even after 10k hands this would be a very large +- 18 bb/100 interval.

That was a normal approximation. Perhaps I should have said that those figures assume you don't tilt much.

The standard deviation per 100 hands is reported by some software. The standard deviation of your win rate decreases by 1/sqrt(# hands). So, if your standard deviation per 100 hands is 90 bb (so that playing 100 hands is about as risky as a coin flip for 90 big blinds), then the standard deviation for your win rate after 100 x H hands is 90/sqrt(H) bb/100, or 900/sqrt(# hands) bb/100.

95% of a normal distribution (an ideal bell curve) is within 2 standard deviations of the mean, and if you play a lot under the same conditions (without tilting) then your results should be roughly normally distributed. Two standard deviations would be 1800/sqrt(# hands) bb/100.

I'd say 50K gives you a good idea, but even then u might have moved up already, luck plays such a big factor. I have had a 100K hands breakeven stretch and am a solid winner.