Nontransitive dice

Consider three special dice: A, B, and C. On a fair roll, A is more likely to beat B. B is more likely to beat C. But C is more likely to beat A. These are nontransitive dice.

Mirwin's dice
M Winkelmann, CC BY-SA 2.5, via Wikimedia Commons

Nontransitive dice are an interesting mathematical curiosity. Consider the following configurations:

  • Die A has the numbers 1, 2, 5, 6, 7, and 9 on it.
  • Die B has 1, 3, 4, 5, 8, and 9.
  • Die C has 2, 3, 4, 6, 7, and 8.

If we pit A (top row) against B (left column), the winners are as follows:


Three roll combinations produce a tie. A wins 17 combinations, and B wins 16 combinations. In other words, this is not a fair contest: Die A has a slight edge over Die B.

If we make the same table for B vs. C, we would discover that B has an edge over C. It would be reasonable to conclude that A is the best die, followed by B, with C the clear loser. It would be reasonable, yes, but it’s also wrong. C has the same edge over A that A has over B – and that B has over C. The dice effectively function like rock-paper-scissors, with no one die having an advantage over the other two.

The set of three dice above are known as Mirwin’s dice, but they’re not the only nontransitive dice out there: lots of different number sets produce this effect. Apparently Warren Buffett is a fan: he once tried to scam Bill Gates with a set of nontransitive dice.

One Reply to “Nontransitive dice”

  1. James Grimes (known from numberphiles on YT) came up with a really great set of NTD:

    A: 0,5,5,5,5,5
    B: 1,1,6,6,6,6
    C: 2,2,2,7,7,7
    D: 3,3,3,3,8,8
    E: 4,4,4,4,4,9

    they form two cycles: in one, the dice are winning with about 5/9 and in the second they are winning with nearly 70%. And if your friend insists in letting you pick the color first, you can try to talk them into using your choice of only 1 die each or two dice each. In the latter case one cycle flips and goes the other way round, which gives you the edge again…

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