Confidence interval calculation has me concerned

[Oblivious]

So assume X is a random variable representing my hourly rate for a particular poker session. Also assume that X has a normal distribution with mean m (also the sample mean of all recorded previous sessions) and variance s^2 (also the sample variance of all recorded previous sessions).

I calculated the probablility of my being a winning player, that is:
P( -m/s < Z < infinity) where Z has a standard normal distribution, m is the mean (of the data set) and s is the standard deviation (of the data set). You can imagine this area pictorially as the standard normal curve shaded from a value slightly negative to positive infinity (assuming m is positive).

So for online games, my win rates in $ per hour had a mean of m = 7.59 and a standard deviation of s = 80.19. According to my method i got 53.78 percent that my true rate is above $0/hour. This seems too low for me. I have over 1000 sessions logged for this data set. Im up almost 7 grand over 1.5 years playing part time playing mostely .50/1 thru 2/4. Pokertracker has me beating .50/1 , 1/2, and 2/4 for over 4BB per 100 hands over 16 thousand hands.

Anyone know how to do this confidence interval correctly?

[BruceZ]

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So assume X is a random variable representing my hourly rate for a particular poker session. Also assume that X has a normal distribution with mean m (also the sample mean of all recorded previous sessions) and variance s^2 (also the sample variance of all recorded previous sessions).

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You are confusing the variance of your winnings with the variance of your hourly rate. The sample variance of your sessions s^2 is the variance of your winnings per session. You must divide this by n total hours to get the variance of your hourly rate s^2/n, and the SD of your hourly rate is s/sqrt(n).


[ QUOTE ]
I calculated the probablility of my being a winning player, that is:
P( -m/s < Z < infinity) where Z has a standard normal distribution, m is the mean (of the data set) and s is the standard deviation (of the data set). You can imagine this area pictorially as the standard normal curve shaded from a value slightly negative to positive infinity (assuming m is positive).

So for online games, my win rates in $ per hour had a mean of m = 7.59 and a standard deviation of s = 80.19. According to my method i got 53.78 percent that my true rate is above $0/hour. This seems too low for me. I have over 1000 sessions logged for this data set. Im up almost 7 grand over 1.5 years playing part time playing mostely .50/1 thru 2/4. Pokertracker has me beating .50/1 , 1/2, and 2/4 for over 4BB per 100 hands over 16 thousand hands.

Anyone know how to do this confidence interval correctly?

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You are confusing the SD of your win rate with the SD of your winnings for 1 hour, and these are two very different things. The SD of your win rate is not $80.19/hr. The SD of your winnings for 1 hour is $80.19. Note that this number does not change much with the number hours you play once you have played enough hours. The SD of your win rate, also called your standard error (SE) measured over 1000 hours is $80.19/sqrt(1000) = $2.54/hour. Note that this number decreases as the number of hours you play increases. Then to compute your confidence interval, take 7.59/2.54 = 3 standard deviations = 99.9% confidence that your win rate is > 0 after 1000 hours. I have assumed that 1000 sessions is 1000 hours. If this is not the case, then divide by the sqrt of the total hours in 1000 sessions.