Non-transitive cubes

Imagine you have three six-sided cubes that have arbitrary numbers on them. Could it be that the dice A is more likely to have a higher number than the dice B, the dice B is more likely to have a higher number than the dice C, and the dice C is more likely to have a higher number than the dice A?

Transitivity

Transitivity is a concept from binary relations. We say that a session R on M is transitive if for all a, b, c ∈ M it holds that if [a, b] ∈ R and also [b, c] ∈ R, then [a, c] is in R.

In the case of dice, we could have a "be a better die" session in the sense that a A die is better than a B die if A rolls a higher number more often than B. Then transitivity will look like this: if A is better than B and at the same time B is better than C, then A must also be better than C. Is this true?

Let's try to find a counterexample, that is, three cubes that are cyclically better. The cube A will be better than the cube B, the cube B will be better than the cube C, and the cube C will be better than the cube A. We always consider only six-sided cubes.

Example

These three dice are an example:

• 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 the dice A rolls a larger number than the dice B? There are a total of 36 different pairs of ways the dice can fall. If the B die rolls a 1 first, the A die will always roll a larger number. That is, we have 6 pairs where a higher number is rolled on the A die. If a second 1 is rolled on B, we have more 6 pairs where a higher number is rolled on A. If one of the sixes is hit, one of the nines must be hit on A. These are the other 4 possibilities. If one of the eights falls, one of the nines must fall on A. Again, 4 possibilities. In total, we have 6 + 6 + 4 + 4 = 20 possibilities. The probability that a higher number falls on A is 20/36 = 5/9, which is approximately 55 %.

Now B vs. C. If one of the threes falls on C, one of the sixes or one of the eights must fall on B. That's a total of 2 · 4 = 8 possibilities. If a five is rolled, again B must roll 6 or 8, so again 8 possibilities. If a seven is rolled, an eight must be rolled at B, 4 options. The total is 8 + 8 + 4 = 20. The probability is again 5/9.

Finally, C vs. A. If a two falls on A, everything from C is higher. That's 12 possibilities. If a four hits, C can hit 5 or 7. That's 8 possibilities. If a nine hits, we're out of luck. All in all, that makes 12 + 8 = 20 possibilities, and we get the probability 5/9 again.

Efron's dice

Efron cubes are another case of a set of cubes that are not transitive with respect to the session being a better cube. They have the following 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

In doing so, a die beats the next die always with probability 2/3. For example, the moment a four is rolled on dice A, it wins over dice B. The moment a zero is rolled, it loses. And on the dice A a four is rolled with probability 2/3. Similarly for the other dice.

There are other variations of Efron's dice. Those that have the same average value per roll (for example, the A die rolls the number 8/3 on average, while C has a higher average, 10/3) or those that divide among themselves all the natural numbers from 1 to 24.

Other resources

• English Wikipedia