[stevepa]

Ok, let's try to do some math (Disclaimer: I make no guarantees I do this right)

Say we have a player with a winrate of 1bb/100. We'll use the previously quoted 15bb/100 SD for full ring games. Now, we want to know the probability of a 300bb downswing (so #of100hands * observed win rate over those hands = -300)

So, starting from Z = (E - o)/(SD/sqrt(n)),
where Z = # of standard deviations away from expected
E = expected win rate
o = observed win rate
SD = standard deviation
n = # of 100 hands played
Writing out the rest isn't worth the effort, but substitute o=300/n, then set dz/dn = 0 and solve for n. Then substitute that to get z. I get that the most likely 300bb downswing occurs over ~30,000 hands with probability 1.04%.

So, for someone having played 300k hands of no-tilt, constant 1bb/100 poker, the probability of having had a 300bb downswing is ~ 1-(0.9896^10) = 10%.

Again, no guarantees that this is right.

Edit: I messed up the shorthanded calculation...Fixed now.

Also, if you're interested, for the short handed # quoted (17bb/100 SD), the same player would have a 300bb downswing over 30,000 hands about 2.08% of the time. So after 300k hands, prob of at least 1 300bb downswing = 19.0%.

Steve