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## math oddity with dice

Here is an old math oddity that involves dice. Even though I'm posting it, I'm still not sure I totally understand it. If you walk into a casino, the dice they use have 6 faces, each with a number from 1 to 6 ( 1, 2, 3, 4, 5, 6 ). Let's say we were to make up 3 new dice with faces as follows:

die A : 3, 3, 5, 5, 7, 7

die B : 2, 2, 4, 4, 9, 9

die C : 1, 1, 6, 6, 8, 8

OK, so give one gambler die A and a second gambler die B. Who would win the most rolls? The possible outcomes with equal probabilities are as follows:

3 and 2, die A wins
3 and 4, die B wins
3 and 9, die B wins
5 and 2, die A wins
5 and 4, die A wins
5 and 9, die B wins
7 and 2, die A wins
7 and 4, die A wins
7 and 9, die B wins

As you can see, the gambler with die A would win over the gambler with die B 5 out of every 9 rolls. So let's compare dice B and C.

2 and 1, die B wins
2 and 6, die C wins
2 and 8, die C wins
4 and 1, die B wins
4 and 6, die C wins
4 and 8, die C wins
9 and 1, die B wins
9 and 6, die B wins
9 and 8, die B wins

So a gambler with die B would beat a gambler with die C 5 out of 9 rolls. OK, so if we have proven die A is superior to die B, and die B is superior to die C, then die A MUST be superior to die C since die A is the best. Right? Well, let's find out by comparing die A to die C.

3 and 1, die A wins ( GO A !!! RA RA RA !! )
3 and 6, die C wins
3 and 8, die C wins
5 and 1, die A wins
5 and 6, die C wins
5 and 8, die C wins
7 and 1, die A wins
7 and 6, die A wins
7 and 8, die C wins

So die C, which should be the weakest, beats die A, the strongest, 5 out of 9 rolls.