#11
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Re: Question for statisticians
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[ QUOTE ] [ QUOTE ] Supposing that one has a "true" win-rate of n bb/100. This is not actual win-rate but just an assumed number based on one's own (fluctuating) quality of play relative to the field. Now, the law of large numbers says that your actual win-rate will approach n the more hands you play. It seems like there should be some statistical formula involving standard deviation and number of hands played for just how fast actual win-rate (call it a) approachs n. I would think this would actually be more like a probability function of the form: if n is your true win-rate and s is your standard deviation, and you've played h hands (in 100s), then the probability that a lies in the range [n-s/10, n+s/10] is p. Anyone have a good formula (or formulas) for this? [/ QUOTE ] You want the standard error, which is simply s/sqrt(h), where s has units of bb for 100 hands. Then your actual win rate will lie within [n-s/sqrt(h), n+s/sqrt(h)] about 68% of the time, [n-2*s/sqrt(h), n+2*s/sqrt(h)] about 95% of the time, etc. [/ QUOTE ] OK, I'm taking a risk trying to correct Bruce here, but ... let, S1 = SD per hand s = SD/100 hands Sh = SD per "h" hands general equation: (Sh)^2 = h*S1^2 Therefore, s^2 = 100*S1^2 (Sh)^2 = h*S1^2 so, s^2/100 = (Sh)^2/h Sh = sqrt(h)*s/10 if, WR = your win rate/100 hands after "h" hands then, Total winnings = h*WR/100 95% confidence interval is, 2*Sh = 2*sqrt(h)*s/10 Total winnings with confidence interval is: h*WR/100 +/- 2*sqrt(h)*s/10 multiply this by 100/h to get it in terms of per 100 hands, WR +/- 2*s/sqrt(h/100) = WR +/- 20*s/sqrt(h) [/ QUOTE ] I'm not sure what you are "correcting". My 95% confidence interval is WR +/- 2*s/sqrt(h) not WR +/- 20*s/sqrt(h) because he stated that h is already in units of 100 hands. If you are addressing the units of s, this is a common source of confusion caused by the way standard deviation is typically reported by Poker Tracker and others with units of "bb/100 hands". Technically variance would have units of bb^2/hand, so standard deviation would have units of bb/sqrt(hands), but what is meant is that they are evaluating the standard deviation in units of bb FOR a 100 hand interval, and thus giving you the result of bb/sqrt(hands) * sqrt(100 hands). I always state this as bb FOR 100 hands to make sure people use the correct numbers while still being technically accurate. |
#12
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Re: Question for statisticians
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I'm not sure what you are "correcting". My 95% confidence interval is WR +/- 2*s/sqrt(h) not WR +/- 20*s/sqrt(h) because he stated that h is already in units of 100 hands. [/ QUOTE ] Oops, I didn't see that part. I never imagined someone would express it this way. |
#13
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Re: Question for statisticians
i havent read all the responses but i make an intuitive approach:
if you double the sample size and divide the result by 2 your standard deviation will be 1/2^0.5*(the former standarddeviation) ig you double the sample size twice the new standard deviation will be 1/4^0.5=1/2*(the former standarddeviation) so you double your accuracy by taking a 4x bigger sample size. |
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