6 ms·

I find nontransitive dice to be a clear demonstration of the effects of premature rounding. The nontransitivity is only possible because, after each iteration,

• CydeWeys 10y ago
  Negative. Non-transitive dice work even when the expected value of each die is the same. The first example in the Wikipedia article exhibits this: Die A has sides 2, 2, 4, 4, 9, 9. Die B has sides 1, 1, 6, 6, 8, 8. Die C has sides 3, 3, 5, 5, 7, 7. The expected value of each die is 5, yet A beats B, B beats C, and C beats A. Other examples in the article have sets of transitive dice where the expected value is not the same; transitivity still holds.
  • TTPrograms 10y ago
    The "rounding" he's describing is the "beating" you're describing. If the game you play is just long term sums of values then expected value is all you need, and in your example they're all even.
    • CydeWeys 10y ago
      See my other example. Most sets of non-transitive dice do not have the same expected value, and the expected value doesn't tell you which one will win. To use an extreme example, take two dice, (2 2 2 2 2 2) and (1 1 1 1 1 999). The average value of the first die is much smaller than that of the second, yet it wins 5/6ths of the time.
  • amalcon 10y ago
    I'm confused, because this reads to me like exactly what I was trying to say, but you seem to be disagreeing with me. Could you clarify? In the game with rounding, A beats B, B beats C, C beats A. In the game without rounding, where totals are summed, they are evenly matched and it's down to chance. That's exactly the effect I was referring to.
    • dragonwriter 10y ago
      You seem to be using "rounding" to mean "comparing the exact, unmodified results of individual rolls", which is probably the source of the confusion because that doesn't really correspond to any common definition of the word "rounding".
      • jameshart 10y ago
        One way to interpret OP's 'rounding' concept is that in a round of a dice game, the winner's score is rounded up to 1, and the loser's score rounded down to 0. If, on the other hand, each turn each player got a fractional score between 0 and 1 (without 'rounding'), then that would represent a 'truer' evaluation of the dice. The best way I can think of to allocate a fractional score to each die player is to sum all the dice rolled, and give each player their roll divided by the sum. So if players rolled 2 and 5, one player would get 2/7 and the other 5/7 - instead of rounding the scores to 0 and 1. Note that to calculate the EV for a die under this scheme, you have to calculate it versus a particular opposing die, so it would produce a different EV for the transitive A/B/C dice in different combinations. Perhaps that's what the OP was trying to suggest?
        • amalcon 10y ago
          More or less this, except with a simpler way of allocating fractional scores: simply divide the value rolled on that die by the largest such value on any of the dice in the set (or, equivalently, round to either 0 or largest-such-value and skip the division step).
    • CydeWeys 10y ago
      There are many sets of non-transitive dice where the average is not the same though (I should have used one of those examples). Here's one: A: 4, 4, 4, 4, 0, 0 (avg: 8/3) B: 3, 3, 3, 3, 3, 3 (avg: 9/3) C: 6, 6, 2, 2, 2, 2 (avg: 10/3) D: 5, 5, 5, 1, 1, 1 (avg: 9/3) There's no reason that the average values need to be the same to have non-transitive dice. To modify the original three to have the same winning properties, but radically different averages, just imagine doing this: Die A has sides 2, 2, 4, 4, 99, 99. Die B has sides 1, 1, 6, 6, 8, 8. Die C has sides 3, 3, 5, 5, 7, 7. The expected value of an individual die roll doesn't play into it at all.
  • Double_Cast 10y ago
    You two are speaking past each other. CydeWes is using EV in the "first moment" sense. Amalcon is using EV in the "expected utility" sense. E.g. consider rock-paper-scissors. The expected utility of rock is 1/2. But since rock is represented non-numerically, each event has no first moment.