math oddity with dice | Golden Skate

math oddity with dice

PolymerBob

Record Breaker
Joined
Feb 17, 2007
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 !! ) :yay:
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. :confused:
 
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