probability winning / losing

[meriwether]

Given a winrate and standard deviation (BB/100), how do you calculate the probability of winning/losing q BB in r hands?

[BruceZ]

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Given a winrate and standard deviation (BB/100), how do you calculate the probability of winning/losing q BB in r hands?

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Compute your average win for r hands. Compute your SD for r hands as the SD for 100 hands times sqrt(r/100). Now q will lie some number of SDs above the mean (a negative number if it lies below the mean). Determine this number of SDs, and look up the probability in a table of the standard normal distribution, or plug it into the Excel function =NORMSDIST(SDs). This will give the probability of winning less than this amount, so subtract this from 1 to get the probability of winning this amount or more. If the number of SDs is negative, then you probably need to make it positive before using a printed table, and then take 1 minus the value the table gives you. The NORMSDIST function can handle negative numbers directly.

For example, say your win rate is 3 bb/100 hands, your SD is 20 bb for 100 hands, and you want to know the probability that you will win or lose various amounts in 10,000 hands. Your average win for 10,000 hands is 300 bb, and your SD is 20*sqrt(100) = 200 bb. Winning 400 bb would be 100 bb above the mean, or 1/2 standard deviation above the mean. The probability that you win less than this amount is =NORMSDIST(1/2) = 69%, so the probability that you win this amount or more is 31%. Winning 200 bb would be 100 bb below the mean, or 1/2 standard deviation below the mean. The probability that you win less than this amount is =NORMSDIST(-1/2) = 31%, so the probability is 69% that you will win more than this.

Now for a loss. Losing 100 bb would be 400 bbs below the mean, or 2 standard deviations below the mean. The probability that you will lose more than this is NORMSDIST(-2) = 2.3%, so there is a 97.7% probability that you will do better than this. If you are using a printed table of the standard normal distribution, then you would look up +2, and the table would give 97.7%. You would then subtract this from 1 to get 2.3%.

[jason_t]

Hi Bruce,

Suppose that one computes the probability p of winning q BB (q could be negative) in r hands. How does one compute the probability f(p) that in s hands (where s >> r), q BB are won over some block of r hands?

Thanks,

Jason.

[Siegmund]

Short reply: if you want an exact answer, with great difficulty.

Slightly longer reply: as s->infinity, f->1. If s>>r, a lower bound for f is 1-(1-p)^(s/r).