A set of nontransitive dice is a set of dice whose face numbers are such that the relation "is more likely to roll a higher number than" is not transitive. (See wikipedia)

For some sets, the deviation from transitivity is small in the sense that A beats B beats C beats A with probabilities $p_{ij}$ only slightly greater than $0.5$ . Efron's dice (there are 4 of them) beat each other nontransitively with probability $2/3$.

Can we make a strictly better set of 4 six-sided dice? That is, a set of 4 six-sided dice such that they beat each other nontransitively with all probabilities $> 2/3$ ?

Can we make a strictly better set of 4 $n$-sided dice for some small $n$ which one can conveniently make a die out of, e.g. $n = 4, 8, 12, 20 $ ?

Can we make a strictly better set of 5 $n$-sided dice for some small $n$ which one can conveniently make a die out of, e.g. $n = 4, 6, 8, 12, 20 $ ?

Can we make a strictly better set of 3, 4 or 5 dice, each having potentially a different number of sides ($4, 6, 8, 12$ or $20$) ?

Ideally I would like to find a fairly small set of fairly easy-to-make, preferably platonic-solid dice which beat each other nontransitively with probabilities > 80%. They would make an excellent teaching aid and magic trick. There is another answer on math.stackexchange which claims that the best you can do with 3 dice is $p = 0.58$, which is disappointingly close to $0.5$; for a teaching aid you need to be able to beat students almost every time for them to spot the pattern quickly. Efron's dice are substantially better at $2/3$, but is that really the best we can do?

EDIT: I missed this answer which argues that the probability cannot be > than 0.75 irrespective of the details of the dice. Still, it would be nice to know what the "simplest"/smallest set of "simple" dice is that gets you above, say, 70%, 72%, etc.

Also, assuming that the other player doesn't yet understand what's going on, a uniform nontransitive probability of $75%$ - $\epsilon$ can still be improved by making some dice lose with probability > $75$%, so that if the other player chooses randomly from the dice, the will get beaten on average much more than $75$% of the time. The "worse" choices can be encouraged by making the highest number on them very high. As far as I understand the proof in this answer, this possibility is not excluded.

  • $\begingroup$ With 4 dice, I suppose you would like to forbid A > B > C > A; D > A, D > B, D > C, and insist A > B > C > D > A? Or what, exactly? $\endgroup$ – MJD Jan 24 '13 at 23:15
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    $\begingroup$ If there is a nontransitive 3-chain in a set of 4 dice, then the 4th dice is superfluous, and there is another answer here on math.stackexchange which claims that the best you can do with 3 dice is $p = 0.58$ $\endgroup$ – Rationalist Jan 25 '13 at 17:33
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    $\begingroup$ You're not going to get $p > 80$% no matter what you do. As [this question](math.stackexchange.com/questions/57338/…) notes, there's a strict upper bound of $p<3/4$ for any intransitive set of random variables, and for 6-sided dice considerations of discreteness reduce that to $p<2/3$ (so Efron's dice are optimal for a set of 6-sided dice). $\endgroup$ – Micah Jan 25 '13 at 18:03
  • $\begingroup$ Thanks, Micah, I just caught that. Presumably there might be a set of 3 20-sided dice which achieve the bound for n=20 sides, 72.5%. This would still be fairly impressive; although the constructed solution in the answer we linked to uses n n-sided dice, Efron shows that you can hit the bound with less than n dice. How good does it get? $\endgroup$ – Rationalist Jan 25 '13 at 18:35
  • $\begingroup$ Note that the question for which I determined the optimum $p=21/36\approx0.58$ required the win probability to be independent of which die you pick. The code I posted there is easily modified not to impose that requirement, and then it finds this set of three dice: $(0,3,3,3,3,3)$, $(2,2,2,2,2,5)$, $(1,1,1,4,4,4)$, with win probabilities $25/36$, $21/36$, $21/36$, for which the expected win probability upon uniformly randomly choosing a die is $67/108\approx0.62$, and the worst choice has the highest number. I hope to be posting an answer with more results when I find the time. $\endgroup$ – joriki Jan 26 '13 at 13:01

A great visual aid for developing a set of four transitive dice is this one:

enter image description here

In the left image, each quadrant shows the probability diagram of each of the four duels. In the right image you see that these diagrams form a spiral.

See also: https://www.quora.com/What-are-some-mathematical-paradoxes/answer/Job-Bouwman


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