[BruceZ]

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Let me reword and and try to answer:
Q. What is the probability that a player playing against a specific set of opponents with a playing style that yields a true winrate of 2BB/100hands (a good winrate) and a standard deviation of 13BB/100 will be a 2BB/100 loser over a 10k hand stretch?
A. Less than one in five hundred

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Actually 1 in 956 that he will be a -2 bb/100 loser or worse in 10K hands. I believe that you found the probability for being +/- (4/1.3) standard deviations from the mean, rather than at least this many standard deviations strictly below the mean, so your answer is off by a factor of 2.

The standard deviation of the win rate, or standard error, is 13/sqrt(10,000/100) = 1.3. A win rate of -2 bb/100 would be 4/1.3 standard errors below the average win rate of +2 bb/100. From Excel =NORMSDIST(-4/1.3) =~ 1 in 956.

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Actually, I did a much rougher calc, which is why I went conservative and added "less than"
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With the nice numbers that you chose, you should be able to do this quickly in your head and see that the answer is very close to 1 in 1000. That's how I noticed the problem.

In case anyone's interested in this thought process, it went like this. The SD is given in units of bb/100 hands, so to get the standard error, we need to divide the SD of 13 by the square root of the number of 100 hand units in 10K hands. 10K/100 is 100, and the square root of 100 is 10, so we divide the SD of 13 by 10 to get a standard error of 1.3. Now -2 bb/100 is 4 bb/100 below the average of 2 bb/100, and 4 bb would correspond to 4/1.3 standard deviations below the mean, which is a little more than 3 standard deviations. I remember that 3 standard deviations one sided is about 99.9%, so the probability of being more than this many standard deviations below average is about 0.1% or 1 in 1000.