Quote:
Originally Posted by Rodolphe
Because if I compute my winrate as the mean of the profit of the n first hand, then my winrate will converge (not exactly but the more hands I have, the more slowly he will move after a certain number of hands) and then my standard deviation will converge to 0 right ?
So, I don't understand something there.
Your winrate converging shouldn't cause the standard deviation to converge? It sounds like you are trying to use the deviation of the new mean from the previous mean or something similar (that's the only way I can see why your standard deviation would converge to zero).
To work out your standard deviation is pretty easy on a "per session" or "per hand" basis, but I'm not entirely sure how PT does it in terms of BB/100 (the book "Gambling Theory and Other Topics" will tell you IIRC). My guess is that is does something like this:
A) Find your winrate from all your data (ie: the mean).
B) Break up your data into 100 sample "blocks" somehow and find the winrate for each "block" (perhaps with resampling - this is the bit I'm unclear on).
C) Work out the sum of squared deviations of each block's winrate from the mean winrate calculated in (A).
D) Normalize by dividing by the number of blocks and then take the square root.
This is only a guess though and I'm not really 100% sure if it's correct - your best bet is to ask this in the Probability forum (or else buy/borrow a copy of "Gambling Theory and Other Topics").
Juk
PS: I'd be quite interested in seeing any results you get on this. Knowing just how much the luck adjusted version reduces the variance would be pretty interesting (plus if you search the 2+2 archives you'll find some good posts about how to use the info to generate confidence intervals on true winrate estimate and thus you should be able to work out how how many non-luck adjusted hands it takes to get the same level of confidence as the luck adjusted version).
Last edited by jukofyork; 10-15-2008 at 01:49 AM.