A set of dice is nontransitive if the binary relation – X rolls a higher number than Y more than half the time – on its elements is not transitive. Different sets of such dice are known.

'Rolling higher' is a stochastic event and 'more than half of the time' an other one.

Most often probabilty is presented as a kind of measure and in trivial cases such as coins or usual dice it is obviously connected to physical/spatial objective properties. For this particular case however it appears to be a kind of (cyclic) ordering.

How would a Frequentist and a Bayesian explain away this fact?

  • Note that if you compare the expected outcomes of the dice, the non-transitivity vanishes. – Lev Jul 11 '16 at 6:28

If you construct the Bayesian network for this problem, like for the three die problem below, you won't find anything in the network's topology that implies any kind of ordering on the "is greater than" nodes, so there is no reason to think that there should be any sort of ordering on those nodes, since we have complete freedom in selecting the distributions over the "dice" nodes (i.e. we could make dice with any sets of numbers on the sides that we like). Although counter-intuitive (like the Monty Hall problem), it is a non-problem when analyzed systematically.

Note, although Bayesian networks are most often applied with an underlying Bayesian interpretation, in this case it is just applied to decompose the overall probability distribution in terms of conditional probabilities. No issues of interpretation arise, we're just given a fully specified game. And all parties can agree that these are the correct conditional relationships.

Three die network structure

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  • thanks but I have even more problems now: the last node should be A>C and the Monty hall paradox is unrelated; nontransitive dice are just a better exemple than the rock-paper-scissors game that is always presented with a directed cyclic graph. – sand1 Apr 11 '16 at 8:51
  • @sand1 It doesn't matter much whether the indicator is A<C or C<A, since P(A<C)=1-P(C<A)-P(C==A). – Dave Apr 11 '16 at 13:24

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