probability of winning / losing

meriwether · #1 03-13-2005, 01:17 PM

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

meriwether · #2 03-13-2005, 09:09 PM

bump

Stork · #3 03-13-2005, 09:17 PM

try posting this in the probability forum, you might get more responses there.

Nomad84 · #4 03-13-2005, 09:46 PM

I'm not an expert, but I'll try to explain this in an understandable way.

SD=standard deviation/100 (from PT)
sigma=standard deviation for this particular session
WR=win rate in BB/100 (from PT)
N=# of hands divided by 100

First, you obviously expect to win N*WR BB, on average. So the question becomes "How often am I within X BB of winning exactly N*WR?" To figure this out, take sigma=SD*sqrt(N). Next, take Z=X/sigma. Reference this number to a Z table to figure out the percentages. I use this website. If you like, you can just plug in your "mean result" (N*WR) and sigma into the second applet and play around with the different options to figure out how often you will win more than some amount, lose more than some amount, fall within some range, or fall outside some range. It's a fairly straightforward way of doing what you are asking since you just need to know N*WR and sigma=SD*sqrt(N).

Note: If you prefer, you can use WR and SD in BB/HR, and N=# of hours. Should work just as well.