A set of dice with a loop
Ordinary reasoning says that if A is better than B and B is better than C, then A is better than C. Rankings of heights, speeds and scores work that way. The relation is called transitive.
“Wins more often when rolled against” does not have to be transitive. It is possible to build three dice where die A beats die B more often than not, B beats C more often than not, and C beats A more often than not. Such dice are called non-transitive. Nothing is rigged: each die is fair, with six equally likely faces. They are a standard curiosity in recreational mathematics, and several well-known sets exist. Here is one small set that is easy to check.
The illustrative set
Each die has six faces, with each value appearing twice. This is an illustration chosen for easy checking, not a set sold or used in any particular game.
| Die | Faces |
|---|---|
| A | 2, 2, 4, 4, 9, 9 |
| B | 1, 1, 6, 6, 8, 8 |
| C | 3, 3, 5, 5, 7, 7 |
Because every value appears exactly twice, each die effectively has three equally likely values. That means any pairing has 3 × 3 = 9 equally likely value combinations, which keeps the counting small. The doubling does not change any probability: it only means each value has probability 1/3.
A beats B
Compare every value of A with every value of B and count the cases where A is higher.
- A shows 2: beats 1 only. That is 1 win (loses to 6 and 8).
- A shows 4: beats 1 only. That is 1 win.
- A shows 9: beats 1, 6 and 8. That is 3 wins.
Total: 1 + 1 + 3 = 5 wins out of 9. Die A beats die B with probability 5/9, about 55.6 percent. There are no ties, since no value appears on both dice.
B beats C
- B shows 1: beats nothing in C (3, 5, 7). That is 0 wins.
- B shows 6: beats 3 and 5. That is 2 wins.
- B shows 8: beats 3, 5 and 7. That is 3 wins.
Total: 0 + 2 + 3 = 5 out of 9. Die B beats die C with probability 5/9.
C beats A
- C shows 3: beats 2 only. That is 1 win.
- C shows 5: beats 2 and 4. That is 2 wins.
- C shows 7: beats 2 and 4. That is 2 wins.
Total: 1 + 2 + 2 = 5 out of 9. Die C beats die A with probability 5/9.
So the loop is complete: A beats B, B beats C, and C beats A, each by 5 to 4 in the nine pairings.
Why this is not a contradiction
The apparent paradox comes from treating “beats” as if it measured a single quality of the die. It does not. It measures how two distributions overlap, and each matchup involves a different overlap.
Die A has a mix of low and very high faces. Against B, its 9s win every time, which makes up for the low faces. Die C has no very high or very low faces, so it steadily beats A’s low and middle values but loses to A’s 9s. Die B has the lowest faces and the second-highest ones. Each die has a different strength, and what counts depends on the opponent.
It also helps to compare the averages. Each die’s average face is the same: A gives (2 + 4 + 9) ÷ 3 = 5, B gives (1 + 6 + 8) ÷ 3 = 5, and C gives (3 + 5 + 7) ÷ 3 = 5. The expected value of each roll is identical, yet the dice still beat each other in a loop. That shows why a comparison of averages cannot predict who wins a head-to-head contest. The question “which is higher on average” and the question “which is higher more often” are different questions.
Playing a game with these dice
The structure gives the second chooser an advantage. If one player picks a die first and the other picks afterwards from the remaining two, the second player can always choose the die that beats the first player’s, and then wins about 5 times in 9 over many rolls. A single game is still uncertain; 5/9 is an edge in the long run, not a guaranteed result, and a short run of rolls can easily go the other way. This is the same kind of reversal as in Penneys game, where the order of choosing decides who is favoured.
Where the idea appears
Non-transitive dice are mainly a teaching tool, but the idea carries over. In game design, it is the logic behind rock-paper-scissors balance, where no option is best in every matchup, and it is one reason designers avoid ranking pieces on a single number. It also shows why dice distributions need to be examined beyond their averages, since two dice with the same mean can behave very differently.
If you want to verify the result without trusting the arithmetic, a short Python dice simulation that rolls each pair many thousand times will show each win rate settling near 55.6 percent. Any short run will wobble around that figure, which is a useful reminder that a 5/9 edge is modest and takes many rolls to show clearly.