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Calculate your winrate

  
 
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TylerK
Old 06-29-2005, 11:41 PM     Post subject: Calculate your winrate #1 (permalink)  
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It's not pretty but it works. Based on a conversation with Phyl and this thread.

http://www.zerodivide.cx/~tyler/winrate.php
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pokerfanatic
Old 06-30-2005, 02:17 AM #2 (permalink)  
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What’s the source code in? I'm just wondering if you coded it or it was just out there. If you could send me that source code, that would kick ass.
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TylerK
Old 06-30-2005, 02:31 AM #3 (permalink)  
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It's in PHP. It's not pretty, I just knocked it together in 10 mins. Send me your email address if you want me to send it to you tho.
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euphoricism
Old 06-30-2005, 02:36 AM #4 (permalink)  
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M +- (1.96X)

Where M is the mean, and X is the standard error (which is Standard Deviation / sqrt(n)?

So you take the standard deviation (input by user), devide it by the squareroot of the number of hands (input by user), and multiply it by the z score (which is 1.96, and you don't need to care why), then add it to (and subtract it from) the average winrate

Ugh I'm leaving out a step... *thinks thinks thinks*
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euphoricism
Old 06-30-2005, 02:37 AM #5 (permalink)  
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And im a dipshit because its all explained on the 2+2 link.
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euphoricism
Old 06-30-2005, 02:38 AM #6 (permalink)  
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Oh, and note that your hand-size should be above 10k for it to be reasonably accurate. Small sample sizes don't fit the central limit theorem
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TylerK
Old 06-30-2005, 03:16 AM #7 (permalink)  
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Quote:
Originally Posted by euphoricism
M +- (1.96X)

Where M is the mean, and X is the standard error (which is Standard Deviation / sqrt(n)?

So you take the standard deviation (input by user), devide it by the squareroot of the number of hands (input by user), and multiply it by the z score (which is 1.96, and you don't need to care why), then add it to (and subtract it from) the average winrate

Ugh I'm leaving out a step... *thinks thinks thinks*
I'm multiplying by 2*STDERR instead of 1.96 but I'm bad at math and I don't know why. Is this ok y/n
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euphoricism
Old 06-30-2005, 03:41 AM #8 (permalink)  
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Definitely change it to 1.96. Thats the z score for .05 (corrosponding to 95% conf)
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TylerK
Old 06-30-2005, 03:56 AM #9 (permalink)  
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Quote:
Originally Posted by euphoricism
Definitely change it to 1.96. Thats the z score for .05 (corrosponding to 95% conf)
Though I won't likely understand it, can you tell me why? Or give me some kind of link?
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Nehmer
Old 06-30-2005, 04:05 AM #10 (permalink)  
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Quote:
Originally Posted by TylerK
Quote:
Originally Posted by euphoricism
Definitely change it to 1.96. Thats the z score for .05 (corrosponding to 95% conf)
Though I won't likely understand it, can you tell me why? Or give me some kind of link?
It just means that 95% of the normal distribution is within 1.96 Standard Deviations of the mean. Unless you are doing work for a project where accuracy is extremely important the difference between 2 and 1.96 will be hardly noticable and would just give us like the 95.4% confidence interval(So in our case the 2 is probably fine).
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euphoricism
Old 06-30-2005, 04:12 PM #11 (permalink)  
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Yeah, thats pretty much right. Its not vital, but theres no reason not to change it and increase accuracy

Basically, to get a z-score you take your percentage of confidence wanted (as a decimal) and subtract it from 1. so 1-.95 = .05, look that up on a ztable and it corrosponds to 1.96 (95% is basically used all the time, so everyones memorized it). You could do 99% the same way, just google a z table and do 1-.99 = .01, and look up .01
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elipsesjeff
Old 07-01-2005, 12:02 AM #12 (permalink)  
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With a 95% confidence interval, your true winrate is between -0.9 BB/100 and 5.56 BB/100.


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elipsesjeff
Old 07-01-2005, 12:03 AM #13 (permalink)  
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You know, that is probably the most useless thing I've done today.


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