Re:Nontransitive dice
by kybernetikos on Saturday April 29, 2017 @06:32PM
(#14229137)
I think the numeric and gambling aspect helps too. If you're used to probability, you probably start thinking about expected value automatically in these situations, but it isn't the case that the die with the highest expected value wins most of the time against a die with lower expected value. (e.g. 2,2,2,2,2,2s vs 1,1,1,1,1,100).
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Re:Nontransitive dice
by Double_Cast on Sunday April 30, 2017 @01:39AM
(#14230288)
Seasoned gamblers count "outs" (winning outcomes), convert to "hand odds" (out:loss ratio), then weigh hand odds against "pot odds" (weigh expected profit vs investment).
Mean EV doesn't work because the dice have different variances? Trying to wrap my head around the rationale here
Variance isn't the best lens here, better to look at the distributions. It's also a discrete problem, so EV isn't as precise as looking at all the possible combinations.
Variance describes distribution...
Discreteness is another layer. What are the exact defining characteristics of this problem though? You could take two dice 1-2-3-4-5-6 and 2-3-4-5-6-7 and the typical mean EV calculation would work fine...
1-2-3-4-5-7 and 2-3-4-5-6-8 have different variances but are probably not intransitive (haven't checked the math) since they are translations of each other
I'm just thinking out loud here and trying to narrow it down...maybe someone reading wants to help me :)
Variance does describe the distribution, but not well enough, in this case. It reduces the vector of possibilities to a scalar, and hides the most important aspect, the non-transitivity. Numbers have transitive comparisons, so there's essentially no way a scalar value _could_ capture that in any straightforward way. Basically, abandon all single-value summaries. They'll do you no good in understanding this phenomenon. But the detailed analysis isn't too complicated, so it's okay.