# What is the most unfair set of three nontransitive dice?

In a set nontransitive dice, each die is superior to another die, but is inferior to a third. It is similar to the game of rock-paper-scissors. Here is one example:

die A has sides: 2, 2, 4, 4, 9, 9
die B has sides: 1, 1, 6, 6, 8, 8
die C has sides: 3, 3, 5, 5, 7, 7


Die A has a 5/9ths chance of rolling a higher number than B, which itself has a 5/9ths chance of rolling a higher number than C, which itself has a 5/9ths chance of rolling a higher number than A. It is a circle with no overall winner.

Suppose there is a simple dice game where one person picks a die from the set of three nontransitive dice. A second person then picks another die from the set. The players then roll their dice, and the person who rolls the higher number wins. If there is a tie, the players simply roll again until there is a winner.

If this game is played with the above set of dice, then the second player will always be able to pick a superior die, and will win the game 5/9ths of the time.

What is one possible set of nontransitive dice that maximizes the unfairness of this game? One additional requirement is that the second player's odds of winning must not be affected by the first player's choice of die.

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– anon Dec 16 '12 at 16:52
@anon I don't really understand the second paragraph of that question, or the answer. – PhiNotPi Dec 16 '12 at 16:57
@anon: Great to see that you're still around -- you might be interested in the answer to this one, too :-) – joriki Dec 16 '12 at 21:29

It turns out that you can do very slightly better than the dice in your example. The probability has to be a multiple of $1/36$, your example has $20/36$, and the best possible value is $21/36=7/12$.

From my answer to the question that anon linked to, we know that the probability of one $6$-sided die beating another while having a lower median can be at most $3/4-1/12=2/3=24/36$. Wikipedia gives an example of a cycle of four dice in which each die beats the next one with this probability.

You're looking for a cycle of three dice in which each die beats the next one with the same probability. Without loss of generality we can assume that the dice carry $18$ different numbers. Then there are $18!/6!^3=17153136$ different sets of dice, so it's well within the reach of our electronic friends to try them all out. Here's code that does that (rather inefficiently). The result is that the maximal probability for such a cycle is $21/36=7/12$ and there are eight different sets of dice that attain that probability. Here they are, with consecutive stretches of numbers collapsed into one as in your example:

0 3 3 3 6 6
2 2 2 5 5 5
1 4 4 4 4 4

0 0 3 3 3 6
2 2 2 2 2 5
1 1 1 4 4 4

0 3 5 5 5 5
2 4 4 4 4 7
1 1 1 6 6 6

0 3 5 7 7 7
2 2 4 4 6 9
1 1 1 8 8 8

0 3 3 5 5 5
2 2 2 4 4 7
1 1 1 6 6 6

0 3 5 5 7 7
2 2 2 4 6 9
1 1 1 8 8 8

0 3 3 3 3 5
2 2 2 2 4 7
1 1 1 6 6 6

0 0 3 6 6 6
2 2 2 5 5 5
1 1 1 4 7 7


There are two sets with $7$ different numbers, four sets with $8$ and two with $10$, and your near-optimal example has $9$, whereas the mean number of different numbers after collapsing, averaged over all sets with $18$ different numbers, is $18-17\cdot(5/17)=13$, so consecutive stretches of numbers and thus high collapsibility are a striking feature of these sets.

P.S.: I just realized that there's a symmetry in that you can invert the orders of both the dice and the numbers and get another set with the same probability. The first two sets are related by this symmetry, as are the third and seventh, the fourth and sixth and the fifth and eighth. Thus there are only four really different sets of dice, with $7$, $8$, $10$ and $8$ different numbers, respectively.

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Using numbers 1 4 4 4 4 4, 2 2 2 5 5 5, 3 3 3 3 3 6 has two pairs with 21/36 but one pair has 25/36, so overall has probability is more unfair.

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This should be a comment on joriki's answer – Conrad Turner Jul 3 at 7:09