[ncray]

[ QUOTE ]

I am no guru in statistics, but I think you can multiply your std by the swing hand amount to get std of profit for the amount of hands you are talking about.


[/ QUOTE ]
It's the square root of the sample size if the samples are iid. Take the session S to be h_1 + ... + h_n where the h_i's are iid with standard deviation sigma. var(S) = var(h_1+...+h_n) = by independence var(h_1)+...+var(h_n) = by identical variance sigma^2+...+sigma^2 = n*sigma^2. So std(S) = sigma*n^(1/2)

[ QUOTE ]

If you have a large sample with winrate 0$/hand(to simplify) and sd of 1$, the chance of making a 1000$ loss over 1000 hands should be 1 sd from average which means you'd do better about (100-68%)/2 of the time. So the chance would be 16% to lose 1000$ or more during 1000 hands. I divided by two, since the result is 68% likely to be within one sd of average, this is just for the negative half (remember 0$ profit average).

[/ QUOTE ]
This is correct way to do the problem (but incorrect answer because your sd is too large) if the question is "What is the probability that by the end of 1000 hands you have lost more than $1000?" I think OP is asking for the solution to a more difficult question: What is the probability that, at any point over the course of 1000 hands, you find yourself down $1000 from some previous max (not necessarily your starting $)? It's been a while since I've done problems like that. Sound similar to brownian motion reflection principle problems anyone? We have a clear lower bound since the event down $1k from starting point at the end of 1k hands is a subset of the event down $1k from any point.