A tough variance problem (at least for me)

[TNixon]

If, given a specific winrate and std.dev, I wanted to know the odds of having an upswing or downswing of size X, over Y hands, how would I go about figuring this out?

For example, what are the chances of having a $500 downswing at some point over 1k hands, with a winrate of $.16/hand, and a std dev of $1.02/hand?

The complicated part of this is that a drop doesn't have to be a continuous loss in every hand, just a drop from point A to point B. So a sequence that went WLLWLWLLW would be considerered to have a downswing of 3, even though the longest consecutive downswing is 2.

[TNixon]

Is this problem difficult enough that only a small number of people here could even give it a go, or so simple that I'm being ignored as a moron?

[img]/images/graemlins/smile.gif[/img]

[LarryLaughs]

So the problem you want to solve is that how likely is it to make a x$ deviation from your long term average during y hands, given winrate and deviation?

I mean it is obvious that your sample for the deviation and the average should be very large in proportion to the "swing" size to start with.

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.

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).

Not sure if sd really scales linearly like that.

Not at all sure if that is correct...

[TNixon]

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So the problem you want to solve is that how likely is it to make a x$ deviation from your long term average during y hands, given winrate and deviation?

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That sounds about right.

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Not sure if sd really scales linearly like that.

Not at all sure if that is correct...


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Can anybody verify this? I think I remember seeing a post dealing specifically with the odds of being a certain number of std. devs from the mean, so I'll try to dig that up again.

[pzhon]

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Is this problem difficult enough that only a small number of people here could even give it a go,

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That's the problem.

Also, just the win rate and standard deviation alone are not enough information to determine the probability of a downswing of a particular size, as Pokey pointed out. Some processes with the same (positive) win rate and standard deviation have no downswings at all. So, it's not enough to set up some model and solve the problem for that model. You may have to match more features with an empirically observed distribution from poker in order to expect the model to have downswings which are as frequent.

[ncray]

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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.


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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)

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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).

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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.

[rufus]

For small runs like 1000 hands, it's probably best to have a computer just grind it out. As the number of hands gets large, risk of ruin models are probably a good approximation.

[pzhon]

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Sound similar to brownian motion reflection principle problems anyone?

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A simplified problem, with 0 win rate and a loss of $1000 from the original forbidden, is easily solved with the reflection method. The density at $x without ever hitting -$1000 is the unrestricted density at $x minus the unrestricted density at -$1000-(x-(-$1000)) = -$2000-x. The probability of avoiding -$1000 is probability of ending up over -$1000 minus the probability of ending up below -$1000.

The reflection method does not apply to the original problem. There is drift, and there isn't a simple forbidden region.

[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)

[pzhon]

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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.

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No, because

[img]/images/graemlins/spade.gif[/img] Having a downswing of a particular size is very different from losing that amount in a fixed number of hands.
[img]/images/graemlins/spade.gif[/img] Losing $x starting at hand 1 is far from independent from losing $x starting at hand 2.