Nontransitive dice
BoingBoing had an interesting post about nontransitive dice.
Say you have four dice, A, B, C, and D, which are labeled as follows:
A = {1, 1, 1, 5, 5, 5}
B = {2, 2, 2, 2, 6, 6}
C = {3, 3, 3, 3, 3, 3}
D = {0, 0, 4, 4, 4, 4}
Then, 2/3 of the time, B beats A, C beats B, D beats C, and A beats D. Whatever one of the four dice you pick, you can pick another die that will have a 2/3 chance of beating it!
At first glance, this seems counterintuitive. For instance, the mean outcome of C is the higher than for any other die — so how can D beat C? The answer is that the mean outcome of a die is completely irrelevant to the problem at hand. You see, in this scenario, when comparing two dice, you do not care about the magnitude of the difference between the two numbers that come up. 2 beating 1 is the same result as 6 beating 1. The result you care about is a boolean function — whether one number is greater than the other, or not. And this boolean function, averaged over the 36 possible outcomes which result from throwing a pair of dice, does indeed give us a nontransitive ordering.
I find this rather amusing.