[TNixon]

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As the number of hands gets large, risk of ruin models are probably a good approximation.

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This was mentioned by somebody else in another forum, but it's fairly easy to prove that risk of ruin is an very poor (to the point of being useless) way of approximating the problem.

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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 $)

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Yes, this is the question being asked.

However, I *think* you could get a very good ballpark figure by using the probability of the value in question actually occurring at any given point, given the standard deviation.

Once you know the probability of it occurring at any given point, you can combine that value over whatever number of hands you want, correct? (invert the probability, and multiply it once for each hand)?

So, for example, if the std. deviation is $1/hand, and we want to know how likely it is to be down $3 over 100 hands, then I think the following would be a reasonable ballpark (almost certainly close enough for anybody generally asking this sort of question, even though it doesn't *exactly* answer the question)

$3 is 3 std deviations, we should be within this 99.97% of the time, so the chance of being down $3 from the mean over 100 hands should be .9997^100, or 0.9704? And the probability of being *outside* this would be about 2.96%, and taking half that (because half of the time we're going to be above the mean by more than 3 std. devs), leaves us with a 1.47% chance of being down approximately $3 or more?

This doesn't answer the chance of being down *exactly* $3, but this should be a good approximation of the odds of being down $3 or more? (and a little bit of vagueness in the answer is fine, since there's a fair amount of vagueness in the question itself)