Three dice, no best die

Imagine three dice with three equally likely outcomes each. Roll independently; the higher number wins. These are mathematical dice, without a claim that a particular three-sided object would roll fairly.

• A: 1, 5, 9
• B: 3, 4, 8
• C: 2, 6, 7

Every die has an average of 5. Yet A beats B more often than it loses, B beats C, and C beats A. Each advantage is exactly 5 chances in 9.

You can check A against B without simulation. A's 1 beats none of B's faces; its 5 beats 3 and 4; its 9 beats all three. That's five winning pairs among nine equally likely pairs. The same count works around the cycle:

• A against B: 0 + 2 + 3 = 5 wins.
• B against C: 1 + 1 + 3 = 5 wins.
• C against A: 1 + 2 + 2 = 5 wins.

The average answers a question about the number rolled. It doesn't, by itself, answer the question of beating a particular opponent. Here, knowing which die the other player chose is more useful than comparing averages.

I liked finding a mechanism small enough to hold in view. There is no hidden bias in the rolls, and no need to run a thousand games and hope the frequencies settle down. The nine pairs give the answer exactly.

Could every advantage be stronger?

I checked every way of dividing the numbers 1 through 9 among three such dice, using each number once. There are 280 partitions when the order of the dice and the order of their faces don't count. Five partitions have a strict winning cycle. None makes its weakest advantage greater than 5/9.

The search found the bound; this counting argument explains it.

Roll all three dice together. In each of the 27 equally likely triples, count how many of these comparisons hold: A beats B, B beats C, C beats A. At most two can hold, because ordinary numbers cannot go round a strict cycle.

Call the three pairwise win counts u, v and w, each out of nine. Across all 27 triples, each pair is counted three times, once for each face of the third die. So three times their sum is at most 54: u + v + w is at most 18.

If every edge had at least six wins, all three would have exactly six, and every triple would have to satisfy exactly two comparisons.

Now rotate the names so A contains 1. In any triple where A rolls 1, A cannot beat B, while C always beats A. To get two true comparisons, B must beat C for every possible choice of their faces. But that gives B nine wins against C, contradicting the six it was supposed to have.

Some edge therefore has at most five wins. Our equal-average example reaches that limit on all three edges.

This bound belongs to the stated rules: three equally likely outcomes per die, with the distinct labels 1 through 9 used once. It says nothing about the best cycle with more faces, repeated labels or unequal probabilities.

Calculation and illustration by Ansel. This is an exploration of a mathematical example, not a claim to a new theorem. The finite search can be reproduced by forming each three-element subset A, each three-element subset B of the remaining labels, and letting C contain the rest; retain only partitions whose smallest labels are in increasing order, then check both cyclic orientations.