Quote:
Originally Posted by kikadell
trying to not make this sound like a moan post, it isnt. but its possible for me to lose 1.22BB/100 over 34k hands and still be a good player right? it can be just variance?
Okay, let's do this in reverse: we assume that you were 1BB/100 winner over the sample. And we ask what's the probability that your actual result is:
-1.22BB/100.
I further assume that your std dev per 100 hands is 18BB, pretty normal for 6-max limit hold'em.
The standard deviation of the sample is:
= std dev of one hand * square root of the number of hands
= 1.8BB * sqrt(34000)
= 331.9BB
The mean result with your assumed 1BB/100 winrate is:
= 34000 / 100 * 1BB
= 340BB
The actual result of the sample is:
= - (34000 / 100) * 1.22BB
= - 414.8BB
Now we can obtain the Z-score of the observed result compared to the assumption:
z = (actual result - sample mean) / std dev of the sample
= (-414.8 - 340) / 331.9
= - 2.27
With the NORMSDIST() Excel/Open Office function we see that the result is 0.0116. (This can be calculated from the Z score tables too.)
So, the probablity that a 1BB/100 winner (over the sample!) would have a result of -1.22BB/100 in 34k hand sample is
1.2%. Draw your own conclusions from that. ;-)
(If my calculations are correct. Leader or someone else please confirm or disconfirm my logic here. :-))
Last edited by JarnoV; 02-22-2010 at 06:01 AM.
Reason: formatting