I was writing an article about how much bankroll was needed at a certain standard deviation with a certain win rate...
The formula i came across was the following.
Bankroll needed = -(standard deviation^2/2*avg, win per hour)ln(the risk you are willing to accept).
On first sight, this formula looks ok. I found it several times on the internet, so I assumed the general consensus was that this formule did what i wanted to do.
However, someone pointed out to me that this formula accually assumes a normal distribution of the hourly outcomes around the avg per hour! Since my pokertracker doesn't have enough hands in it to be able to say something meaningfull about this i have no way of saying if this is accually the case.
The longer I think about it, the more i think this formula is wrong, the distribution should at least be different for different types of players (loose agressive has bigger wins, bigger losses than tight aggrissive, though problaby larger wins than losses, even when the hourly rate is set to 0, also i could imagine a player with lots of small losses and few Big wins, the formula wouldn't be (entirely)
Would someone with either a lot of math experience or someone who does have the 200000 or so hands in some kind of database be so kind to shed some light on this? This is quite an important concept I think.
Thanks in advance,
Edited by Johan (05/13/04 12:12 PM)
Normal distribution of hourly rate? I, for one, don't think so.....