Consider the following set of dice.

• Die A has sides 2, 2, 4, 4, 9, 9.
• Die B has sides 1, 1, 6, 6, 8, 8.
• Die C has sides 3, 3, 5, 5, 7, 7.

The probability that A rolls a higher number than B, the probability that B rolls higher than C, and the probability that C rolls higher than A are all 5/9, so this set of dice is nontransitive. In fact, it has the even stronger property that, for each die in the set, there is another die that rolls a higher number than it more than half the time.

Now, consider the following game, which is played with a set of dice.

1. The first player chooses a die from the set.
2. The second player chooses one die from the remaining dice.
3. Both players roll their die; the player who rolls the higher number wins.

If this game is played with a transitive set of dice, it is either fair or biased in favor of the first player, because the first player can always find a die that will not be beaten by any other dice more than half the time. If it is played with the set of dice described above, however, the game is biased in favor of the second player, because the second player can always find a die that will beat the first player's die with probability 5/9. The following tables show all possible outcomes for all 3 pairs of dice.

 2 4 9 1 6 8 3 C \ A A / B B​ / C Player 1 chooses die A Player 2 chooses die C Player 1 chooses die B Player 2 chooses die A Player 1 chooses die C Player 2 chooses die B C A A A B B C C C C C A A B B B B C C C A A A A B B B
Categories: Math