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Here is an introduction to nontransitive dice. The question is: given $n$-player with a $m$-sided dice each one, the what is the minimum of $m$ for a fixed $n$ to produce nontransitivity?

Here is some related posts:

What is the most extreme set 4 or 5 nontransitive n-sided dice?

What is the most unfair set of three nontransitive dice?

How far can probability intransitivity be stretched?

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  • $\begingroup$ $m=3$ for $n=3$, for example, the three dices are {1,5,9}, {2,6,7},{3,4,8}. $\endgroup$
    – Eden Harder
    Commented Aug 23, 2013 at 14:22
  • $\begingroup$ What is $n$? The number of dice? How nontransitive must they be? $\endgroup$
    – Will Sawin
    Commented Aug 23, 2013 at 17:55
  • $\begingroup$ @WillSawin Yeah $n$ is the number of dice. $\endgroup$
    – Eden Harder
    Commented Aug 23, 2013 at 23:45
  • $\begingroup$ What about the second question? There are many large nontransitive arrangements that are not nontransitive in an interesting way. For instance, take lots of small perturbations of your dice: $\{1.01,5.02,8.99\}$ and so on. This will give a large number of nontransitive dice for $m=3$. But this doesn't seem like the right case to consider. $\endgroup$
    – Will Sawin
    Commented Aug 24, 2013 at 0:38
  • $\begingroup$ @WillSawin It is a cool idea! We can always to do this perturbation, but the nontransitivity is only related to the order of the numbers but not what they exactly are. $\endgroup$
    – Eden Harder
    Commented Aug 24, 2013 at 0:54

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