Joined: May '09
Location: India
Age: 37 (M)
Posts: 4873
just a basic way to estimate your true win rate
if sample size = n hands average win rate(mean) = x bb/hand standard deviation = s bb/hand
then standard error of mean = e = s / (n^0.5)
now for 95% confidence limits- true win rate = x +/- 2e
****************************************************** 95% confidence limits means - your true win rate will lie within the given range approx 95% of time
****************************************************** most of the tracking software provide data /100 hands so if sample size = n hands effective sample size = N = n/100 average win rate (mean)= X bb/100 hands standard deviation = S bb/100 hands
then standard error of mean = E = S/(N^0.5)
now for 95% confidence limits- true win rate = X +/- 2E
****************************************************** converting /100 hand values to /hand
win rate per 100 hands (X) = 100 * win rate per hand (x) standard deviation per 100 hands (S) = 10 * standard deviation per hand (s)
note- to analyze a data, both mean and standard deviation should have same units.
- standard deviation was calculated by stdevpa/stdeva function in microsoft excel (or equivalent) - sample is representative, meaning you assume your opponents, their skill level, your skill level, your playing style etc will remain same in future. so that standard deviation of sample is equivalent to standard deviation of population. - sample size << population size sample size should not be exceeding 5% of population size for which these results are assumed to be applicable. - sample size should be large enough so that expected sampling distribution of means of sample size n or N follow a normal (or near normal) distribution. how much size is sufficient is kinda debatable. for any kind of distribution without outliers, 40-50 is considered minimum. but outliers are pretty common in poker, so you need a bigger size. i think for /hand, n >500 and for /100 hand analysis N>100
