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#1
03-17-2005, 07:22 PM
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?? probability of future results ??
Take a hypothetical situation....
Player Able has an established win rate, WR. Able's WR is expressed as BB/HR or BB/100, whichever is relevant. Able's standard deviation (SD) is also expressed in terms of BB, rather than $$. Given that, I'd like to be able to predict the probability of a ceratin result (or greater) over a specific span of play. In other words, I'd like a generic function to calculate the probability of winning (or losing) X BB's over a specific span Y (HR's or 100's) play, given win (loss) rate parameters WR and SD. Help! Thanks in advance, mosquito. 
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#2
03-18-2005, 11:52 PM
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Re: ?? probability of future results ??
**BUMP**
Someone....anyone..... If a player with a +3.1 BB/100 , SD 4 BB/100 wants to know the chance that he will go -100 BB in the next 500 hands, as an example. Answer this, and it can be generalized...... Thanks.....
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#3
03-19-2005, 12:02 AM
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Re: ?? probability of future results ??
You can just a normal approximation, assuming you have enough hands.
That is, after N hands, the amount won has a normal distribution with mean = win rate (in BB/100) SD = (sample SD)/sqrt(N/100) You can find the chance that he's won between x and y by finding the area under the normal curve between x and y. Cheers, gm
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#4
03-19-2005, 12:10 AM
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Re: ?? probability of future results ??
[ QUOTE ]
Take a hypothetical situation.... Player Able has an established win rate, WR. Able's WR is expressed as BB/HR or BB/100, whichever is relevant. Able's standard deviation (SD) is also expressed in terms of BB, rather than $$. Given that, I'd like to be able to predict the probability of a ceratin result (or greater) over a specific span of play. In other words, I'd like a generic function to calculate the probability of winning (or losing) X BB's over a specific span Y (HR's or 100's) play, given win (loss) rate parameters WR and SD. Help! Thanks in advance, mosquito. [/ QUOTE ] See my recent post.
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#5
03-19-2005, 12:19 AM
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Re: ?? probability of future results ??
[ QUOTE ]
That is, after N hands, the amount won has a normal distribution with mean = win rate (in BB/100) SD = (sample SD)/sqrt(N/100) [/ QUOTE ] You mean that this is the mean and SD for the win rate, not the amount won. The amount won has a mean of (win rate in bb/100)*N/100, and an SD of (SD for 100 hands) * sqrt(N/100).
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#6
03-19-2005, 05:55 PM
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Re: ?? probability of future results ??
[ QUOTE ]
[ QUOTE ] That is, after N hands, the amount won has a normal distribution with mean = win rate (in BB/100) SD = (sample SD)/sqrt(N/100) [/ QUOTE ] You mean that this is the mean and SD for the win rate, not the amount won. The amount won has a mean of (win rate in bb/100)*N/100, and an SD of (SD for 100 hands) * sqrt(N/100). [/ QUOTE ] Yes, I did mean that. thanks.
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