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?
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.