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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 situa

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).

hammock 10y ago
Mean EV doesn't work because the dice have different variances? Trying to wrap my head around the rationale here
- hawkice 10y ago
  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.
  - hammock 10y ago
    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 :)
    - hawkice 10y ago
      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.
- ouid 10y ago
  EV of what? The faces of dice cannot be added together, except in a very formal sense.
Double_Cast 10y ago
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).

hammock 10y ago

Mean EV doesn't work because the dice have different variances? Trying to wrap my head around the rationale here

hawkice 10y ago
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.
- hammock 10y ago
  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 :)
  - hawkice 10y ago
    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.
ouid 10y ago
EV of what? The faces of dice cannot be added together, except in a very formal sense.

Double_Cast 10y ago

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).