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#1 | ||||
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In reference to this thread a couple of people have asked how to properly calculate the variance on ternary outcomes events, specifically those that may end only in a win, lose, or push (so this wouldn't apply to Asian handicap).
Without further ado: Let pw = win probability
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So here's an example of how to practically use the second formulation above to test for significance.
Observed: 260 observations (N = 260) all bets at -110 (f = 10/11) wins = 143 (pw = 55%) losses = 104 (pl = 40%) pushes = 13 (pd = 5%) Assuming results were properly obtained out-of-sample, what would be our 95% confidence lower bound for true strategy ROI (using the Central Limit Theorem)? Calculating ROI and ROI variance we find: absolute ROI = 143 * 10/11 - 104 = +26 units (+10%) σ2 = 260 * (100/121*55%*45% + 40%*60% + 2*10/11*55%*40%) ≈ 219.58 units2 or σ = 14.8183 units. So taking into account the ternary outcome set, our (frequentist) 95% confidence lower bound on ROI would be 26/260 - 1.6261*14.8183/260 ≈ +0.6254%. Let's compare this to the results we'd find using conditional probability with a binary outcome set. Let p* = win frequency conditioned on not pushing = 140/(260-13) ≈ 57.895% Which would give us a conditional ROI and variance of conditional ROI of: conditional ROI = +26 units (that's 26 / (260-13) = +10.526%) σ2 = (260-13)*(100/121 + 1 + 2*10/11) * 57.895%*(1-57.895%) ≈ 14.8137 units2 So taking into account the binary outcome set, our (frequentist) 95% confidence lower bound on conditional returns would be 26/(260-13) - 1.6261*14.8137/(260-13) ≈ +0.6614%. Removing the conditioning would yield a lower bound of +0.6614% * (260-13)/260 ≈ 0.6283% on absolute returns. Pretty close to what we had found above (and the error is certainly less than that introduced by using the Central Limit Theorem in the first place). But here's real issue: Taking into account the ternary outcome set, how would one go about testing the null hypothesis that the strategy is in reality no better than breakeven? Anyone? This is straightforward if we condition the results and use binomial variance, but not quite so otherwise. It's exactly this difficulty that leads most practitioners to simply condition out pushes rather than deal with the vagaries of the trinomial distribution.
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But yeah, you got the idea. It's a pain in the ass.
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This spreadsheet illustrates what I had in mind. It implicitly assumes a uniform prior distribution of pd and uses the t-distribution to impute the final p-value (in both the binomial and trinomial cases).
Assuming I didn't make an error, the trinomial p-value of the null-hypothesis (that population EV ≤ 0%) actually turns out to be slightly lower than the binomial p-value (4.168% vs. 4.199%), and coming up with the latter is a whole lot easier.
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I like the MLE because the essence of hypo-testing is to ask, "How objectionable does our data find these probabilities?". Since we are presumably a lot more interested in the win/loss ratio than the draw probability, I chose the draw probability the data finds least "objectionable", which is the essence of the MLE. So if we reject, we know the data has no issues with our pd parameter, and thus is telling us the null-hypo win/loss ratio is the part of the null that is causing our result to be improbable under the null. |
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![]() Regardless, I think we can both readily agree that it's not going to change the results much at all one way or the other and that the best easiest solution is probably clearly to go with binomial modeling coupled with conditional-form model parameters.
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