Non-transitive dice

Imagine you have three six-sided dice with any numbers on them. Is it possible that dice A rolls a higher number more often than dice B, dice B rolls a higher number more often than dice C and dice C rolls a higher number more often than dice A?

Transitivity

Transitivity is a concept from binary relations. We say that a relation R on M is transitive if for all a, b, c ∈ M it holds that if (a, b) ∈ R and at the same time (b, c) ∈ R, then (a, c) is also in R.

For dice we could have the relation “is a better dice”, meaning that dice A is better than dice B if dice A rolls a higher number more often than dice B. Then transitivity looks like this: if dice A is better than dice B and at the same time dice B is better than dice C, then dice A must also be better than dice C. Is that true?

Let’s try to find a counter-example, that is, three dice that are better in a cycle. Dice A will be better than dice B, dice B will be better than dice C and dice C will be better than dice A. We only ever consider six-sided dice.

Example

An example is these three dice:

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

What is the probability that dice A rolls a higher number than dice B? There are 36 different pairs of outcomes in total. If the first 1 comes up on B, then A always rolls a higher number. So we have 6 pairs in which A rolls the higher number. If the second 1 comes up on B, we have another 6 pairs in which A rolls the higher number. If one of the 6s comes up, one of the 9s must come up on A. That gives another 4 possibilities. If one of the 8s comes up, one of the 9s must come up on A. Again 4 possibilities. In total we have 6 + 6 + 4 + 4 = 20 possibilities. So the probability that A rolls the higher number is 20/36 = 5/9, which is approximately 56%.

Now B against C. If one of the 3s comes up on C, one of the 6s or one of the 8s must come up on B. That is 2 · 4 = 8 possibilities in total. If a 5 comes up, again a 6 or an 8 must come up on B, so again 8 possibilities. If a 7 comes up, an 8 must come up on B, giving 4 possibilities. In total 8 + 8 + 4 = 20. The probability is again 5/9.

Finally C against A. If a 2 comes up on A, everything on C is higher. That is 12 possibilities. If a 4 comes up, a 5 or a 7 can come up on C. That is 8 possibilities. If a 9 comes up, we are out of luck. In total that makes 12 + 8 = 20 possibilities, and again we get a probability of 5/9.

Efron’s dice

Efron’s dice are another example of a set of dice that are not transitive with respect to the relation “is a better dice”. They have these numbers:

• A: 4, 4, 4, 4, 0, 0,
• B: 3, 3, 3, 3, 3, 3,
• C: 6, 6, 2, 2, 2, 2,
• D: 5, 5, 5, 1, 1, 1.

Each dice beats the next one with probability 2/3. For example, when a 4 comes up on dice A, it beats dice B. When a 0 comes up, it loses. And a 4 comes up on dice A with probability 2/3. The same goes for the other dice.

There are also other variants of Efron’s dice. Some have the same mean value per roll (for example, dice A rolls 8/3 on average, whereas C has a higher mean of 10/3), and some share out all the positive integers from 1 to 24 among themselves.

Further sources

• English Wikipedia