I'm interested in making a set of 3 strongly non-transitive dice. For anyone not aware, this means if you have dice A, B, and C you can design them in such a way that A usually rolls higher than B, B usually rolls higher than C, and C usually rolls higher than A.

This question has been answered before (What is the most unfair set of three nontransitive dice?), but what I'm curious is if there is a way to make the dice more unfair (Probability being larger than 7/12). The way I would hope to achieve this would be to incorporate different n-sided die. I would prefer to keep these to realistic and simple dice, so things like a d4, d6, d8, or d12 would be best. Aside from brute force simulation as mentioned in the previous question I cannot think of a clever way to approach this problem.

Would more faces help increase the unfair-ness of the dice, or less? Maybe a mixture? Does anyone have any thoughts on this, or a method I could use to approach this?

  • $\begingroup$ Have you tried some d4, d8, and/or d12 to see what you can come up with experimentally? Physicists love experiments, no? Have you tried modifying Joriki's code (from the problem you link to) to test out d4 cases (more than 6 sides might take too long)? Cheers! $\endgroup$ May 16, 2017 at 19:25
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    $\begingroup$ I'm actually running that code right now. I'm just expecting it to take far too long when I get to d8, d12, and combinations. $\endgroup$ May 16, 2017 at 20:03
  • $\begingroup$ Great! Let us know your results! Cheers! $\endgroup$ May 16, 2017 at 20:08
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    $\begingroup$ The code showed that using three 4-sided die has a maximum probability of 9/16 for each die if you want everything to be even (can be stretched to 10/16 for one of the die but then it's not balanced). I've found other proofs that 3-sided die would have a limit of 5/9, so it looks like to make the dice more strongly non-transitive you need to ADD faces. Any clever ideas to quickly simulate 8-sided dice would be much appreciated, as it would take too long the way I'm currently doing it. $\endgroup$ May 17, 2017 at 4:25

1 Answer 1


You might be able to construct lower bounds of losing probabilities.

Example: If you have 1 sided dice, and the first dice has a value on that one side, if the second wins against the first it has to be higher than the first, and if the thrid wants to win against the second it has to be higher than the second. But as numbers are transitive the thrid will also always win against the first. So there is at least one dice that has probability 1 to lose.

If you have 2 sided dice, you can chose only one side of those three dice for now. Again there is at least one dice that loses against both of the other dice if this side shows. Now you would give the other side of the losing dice a winning side against both dices to balance it out. But you can't lower the probability of losing below 1/2 (probability of throwing the first side on your dice) * 1/2 (probability of your opponent throwing the first side) so in total $1/2^2$.

Continuing on, you can already tell that there will always exist at least one dice of three n-sided dice that loses with at least $1/n^2$ probability. If you assume that your solution will results in 3 dice such that you will have the same probability of winning whatever dice your opponent picks this results in the fact that you will lose with at least $1/n^2$ probability with either dice.

And this was a very bad lower estimate. But you get the idea. If you don't just throw away all the interactions from the following sides you might get a better estimate. Look for repeating patterns and then maybe try induction.

Personally I would try to prove the hypothesis that the sum of losing probabilites over all dice must be greater than 1

Intuition: Given that you have the probability of 1 on one dice in the one sided case you already get that.

On the two sided dice you only get 1/4 probability on the second dice by assuming the first side will lose against both other first sides but always win against the second side - which would imply that the second side consists only of smaller numbers (because this dice has the smallest number of the first side). So if you then take the second side there is also a dice that is the lowest on the second side. So this dice loses against the two other dice if they throw the first side (because the first side is always higher than the second side - assumption). And it always loses when they throw the second side. So it has the probability of losing of 1. $1+1/4\ge1$. But if you construct the second side in such a way that it is not completely lower than the first side you will increase the propability of 1/4 to something higher. And again there must be a mirror dice for the second dice so I think you will get something higher than 1 always. But that is just my hypothesis.

If that hypothesis would be true, then 1/3 would be a lower bound to the losing probability of the dice if you distribute it evenly. You could try to prove that. Then you would also know that you can't get over a winning probability of 2/3.

  • $\begingroup$ I understand what you're suggesting. I have found other sources that also would suggest that the maximal winning probability on 4-sided dice would be 2/3 (mathoverflow.net/questions/119867/… has an equation that gives 2/3 for n=4). I'm just struggling to find the actual dice that would create this set. $\endgroup$ May 16, 2017 at 20:11

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