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](http://en.wikipedia.org/wiki/Nontransitive_dice))


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](http://en.wikipedia.org/wiki/Nontransitive_dice#Efron.27s_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 a potentially 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 an [answer](https://math.stackexchange.com/questions/260072/what-is-the-most-unfair-set-of-three-nontransitive-dice) 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? ([Crossposted](https://math.stackexchange.com/questions/286193/what-is-the-most-extreme-set-4-or-5-nontransitive-n-sided-dice) from math.stackexchange)


EDIT: I missed [this answer](https://math.stackexchange.com/questions/57338/how-far-can-probability-intransitivity-be-stretched?rq=1) 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" set of "simple" dice is that gets you above, say, 70%, 72%, etc. 



What is the most extreme set 4 or 5 nontransitive n-sided dice?
If you fix only the number of dice, but let the number of faces be arbitrary (or if you simply find a way to make arbitrary probabitities for different faces), then the answer for $n$ faces is $\displaystyle 1-\frac1{4\cos^2(\pi/(n+2))}$. This is proved (that's quite immodest, I know...) in my paper "intransitive roulettes" in *Matematicheskoe prosveschenie, III series*, 2010, Vol, 14, pp.~240--255, but only in Russian, sorry. Similar results (lacking, perhaps, only the explicit constants) can be found in the following papers:


S. Trybula, On the paradox of $n$ random variables, *Zastos. Mat. (Appl. Math.)* 8 (1965), 143--154.


Z. Usiskin, Max--min probabilities in the voting paradox, *Ann. Math. Statist.* 35 (1964), 857--862.


The optimal example is the following. Let $q=\frac1{4\cos^2(\pi/(n+2))}$. Define $r\_n=0$, $r\_{i}=q/(1-r\_{i+1})$. Then one can show that $1-q=r\_1>r\_2>\dots>r\_n=0$. Now let us make the following "dice": $i$th die ($1\leq i\leq n$) makes $i$ with probability $1-r\_i$ and makes $i+n$ with probability $r\_i$. Then each die wins the (cyclically) previous one with probability $1-q$. 


Consequently, $2/3$ is the optimal number for $n=4$ for any number of faces. Pitifully, the answers for $n>4$ are irrational, hence they are not achievable on regular dice.


It seems that the optimal configuration for $n>4$ on regular dice can be made by the corresponding modification of the general optimal example. E.g., for $n=5$ we have $r\_4=q=0.30797\dots$, $r\_3=0.445\dots$, $r\_2=1-r\_3$, $r\_1=1-r\_4$, so we cannot achieve 70%. On the other hand, these values can be approximated to make the following 5 icosahedral dice:
$$(6\times 1, 14\times 6), \quad
 (9\times 2, 11\times 7), \quad
 (11\times 3, 9\times 8), \quad
 (14\times 4, 6\times 9), \quad
 (20\times 5),
$$
where each wins the (cyclically) next one with probability at most $\frac{9\times 14}{20^2}=0.315$. 


Next, there is a bound for the answer when the number of faces is bounded (or fixed, as in our case). If the number of faces is $2k$ for each die, then consider the $k$th maximal numbers on each die. Consider the die which contains the maximal number among them; then it wins the next one with the probability at least $\frac{k+1}{4k}$. Hence for the icosahedral dice the result cannot exceed $\frac{29}{40}=0.725$. THis can be achieved on the following set:
$$
 (5\times 1,15\times 11), \quad
 (7\times 2,13\times 12), \quad
 (8\times 3,12\times 13), \quad
 (9\times 4,11\times 14), \quad
 (10\times 5,10\times 15),
$$
$$
 (11\times 6,9\times 16), \quad
 (12\times 7,8\times 17), \quad
 (13\times 8,7\times 18), \quad
 (15\times 9,5\times 19), \quad
 (20\times 10),
$$
but not on a smaller one. Some less optimal answers with smaller number of dice are
$$
 (5\times 1,15\times 9), \quad
 (7\times 2,13\times 10), \quad
 (9\times 3,11\times 11), \quad
 (10\times 4,10\times 12),
$$
$$
 (11\times 5,9\times 13), \quad
 (13\times 6,7\times 14), \quad
 (15\times 7,5\times 15), \quad
 (20\times 8)
$$
with losing probability at most $\frac{13\times 9}{20^2}=0.2925$, and
$$
 (5\times 1,15\times 7), \quad
 (8\times 2,12\times 8), \quad
 (10\times 3,10\times 9),
 (12\times 4,8\times 10), \quad
 (15\times 5,5\times 11), \quad
 (20\times 6)
$$
with losing probability at most $\frac{15\times 8}{20^2}=\frac{12\times 10}{20^2}=0.3$ --- exactly 30% on 6 dice.


Finally, for dodecahedral dice the bound is $\frac{17}{24}=0.7083\dots$, hence it is also possible to make it more than 70% (but this is impossible for octahedral dice...). The example is as follows:
$$
 (3\times 1,9\times 9), \quad
 (4\times 2,8\times 10), \quad
 (5\times 3,8\times 11), \quad
 (6\times 4,8\times 12),
$$
$$
 (7\times 5,8\times 13), \quad
 (8\times 6,8\times 14), \quad
 (9\times 7,8\times 15), \quad
 (12\times 8).
$$


I cannot claim that the numbers of dice presented above are optimal for these probabilities, but it seems so. 


