Re: 10,000 hand of bad beats, whats the odds?
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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 probabilty 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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