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Nontransitive, sometimes intransitive or non-transitive, dice are a fascinating concept in probability. It concerns dice such that, in head to head matches, instead of having a neat ranking of "Die A will beat Die B which will beat Die C" and so on when rolled against each other, loops occur. For instance, this post concerns the three set problem, where when paired, A rolls higher then B, B rolls higher than C, and C rolls higher than A.

What is the most unfair set of three nontransitive dice?

But not only does it ask that, it asks a very specific extension: how favorable can you make the odds? The answer turns out to be 7:5; in the highest voted answer, eight dice sets are given, four with symmetry, such that each beats the next one seven times out of twelve.

My question is this: suppose I want a set of four nontransitive dice with six faces each, A, B, C, D, such that A beats B, B beats C, C beats D, and D beats A. Among those matchings, there should be no ties possible. However, if the pairings are A v C or B v D, it should be evenly matched. If the odds for the loop have to be equal and maximized, how far from even can they get?

A further question might be, for five dice, can you have two loops, A>B>C>D>E>A and A>C>E>B>D>A such that all probabilities are the same? And again, how unfair can you make it?

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  • $\begingroup$ You have a great question, but here is a guide post on MathJax as you seem to not have used any. Thanks for joining! $\endgroup$
    – Тyma Gaidash٠
    Jul 20, 2021 at 20:38
  • $\begingroup$ So your question is the linked one, but with one more die and an additional constrain that "non-beating" pairs are evenly matched? It seems that the same answer from there also applies here in the case of 4 dice, i.e. it should be feasible to use a computer to check all possibilities for 4 (or less) dice. $\endgroup$
    – Vepir
    Jul 28, 2021 at 15:48
  • $\begingroup$ To Tyma: Thank you for your concern, but I don't really see any place where using proper notation would help. To Vepir: This may be true, even though with 4 dice it goes from 17 million possibilities to 2 trillion, and for 5 dice it would be roughly a quintillion arrangements. However, i am not knowledgeable with coding, so such a task would be both daunting and grueling for me. $\endgroup$
    – Math Hatter
    Jul 29, 2021 at 20:53
  • $\begingroup$ There is an earlier similar question: math.stackexchange.com/questions/57338/… $\endgroup$
    – Dylan Thurston
    Jan 18 at 12:25

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As you suspect in the comments, going through all possibilities doesn't work great. The answer for 4 dice turns out to be $2/3$ (for any number of faces), so the original set of dice Efron gave were optimal. For $n$ dice the answer is $$ 1-\frac{1}{4 \cos^2 \frac{\pi}{n+2}} $$ For $n=5$ this is around 69.2%; it is always less than $3/4$.

The best reference is this 2021 paper:

It has some very nice geoemtric pictures.

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  • $\begingroup$ Unfortunately, while this does give the maximum probability for any n, I was asking about a set of four dice, A, B, C, and D, so that A>B>C>D>A but A and C tie and B and D tie. $\endgroup$
    – Math Hatter
    Jan 23 at 21:33
  • $\begingroup$ Oops, sorry! That answer should be deducible from the inequalities found by Trybuła in this paper: Trybuła, S. On the paradox of n random variables. Zastos. Mat. 8 (1965), 143–156. This is relevant to some current research, so I'll see if I can work out the explicit answer and get back to this. There are a number of questions in this general area that are (surprisingly) still open. $\endgroup$
    – Dylan Thurston
    Jan 24 at 16:36

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