Re: Need help - What\'s the SD for playing starting hands?
Standard deviation for a proportion p is Sqrt(p (1-p)/n), where n is the number of trials (for decent-sized n, like 25ish). So, for p = .1, n = 10, the SD is .09, using the approx. It can be manually found that the SD is really .161, so n=10 isn't quite large enough. More usefully, I guess, for 60 hands, SD is .03878, so you're playing 10% +/- 7.5% of your hands 95% of the time, so being card dead should be actually somewhat commonplace.
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