#1
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variance formula question
Let's assume your natural winrate is Xbb per hour,
what is the statistical likelyhood of being +/- YBBs in a Z hour session? Is there a straight forward formula for this? |
#2
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Re: variance formula question
The other thing you need to make this computation is the standard deviation of your win rate. Let's say you win X BB per hour on average, with a standard deviation of 50 BB per hour. After Z hours, you expect to be up Z*X BB, with a standard deviation of 50*SQUARE_ROOT(Z).
If you play 100 hours, Z = 100, the square root of Z is 10. So you expect to earn 100*X, with a standard deviation of 500 BB. About 1 time in 44 you will be up more than 2 standard deviations from your expectation, and 1 time in 44 you will be more than 2 standard deviations below it. About 1 time in 3 you will be above expectation, but within one standard deviation, another 1 time in 3 you will be below expectation but within one standard deviation. The remaining times will be split equally between 1 and 2 standard deviation above or below expectation. |
#3
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Re: variance formula question
[ QUOTE ]
Let's assume your natural winrate is Xbb per hour, what is the statistical likelyhood of being +/- YBBs in a Z hour session? Is there a straight forward formula for this? [/ QUOTE ] You need the standard deviation for 1 hour = sigma. Then the probability is: NORMSDIST[(Y - Z*X)/(sqrt(Z)*sigma)] - NORMSDIST[(-Y - Z*X)/(sqrt(Z)*sigma)] Where NORMSDIST(x) is an Excel function meaning the probability of a variable distributed by the standard normal distribution being less than x. See this post for more info. |
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