You have 4 dice with the following numbers on them:
3,3,3,3,3,3
2,2,2,2,6,6
1,1,1,5,5,5
4,4,4,4,0,0
your challenges is to choose one of the dice, and then I will take another. We will roll our dice together, trying to get the highest roll, and the winner is the one who rolls the best out of ten. Which die should you choose?
I think 2 is "best" in sense of having highest probability of winning against the other die. But Alex is correct in saying that for any die that you pick, I can select one that has a probability = 2/3 of winning against you on one roll of the dice.
1 wins against 2 with Prob=2/3, against 3 with P=1/2, against 4 with P=1/3
2 wins against 1 with P=1/3, against 3 with P=2/3, against 4 with P=5/9
3 wins against 1 with P=1/2, against 2 with P=1/3, against 4 with P=2/3
4 wins against 1 with P=2/3, against 2 with P=4/9, against 3 with P=1/3
And since the second person gets to pick another die (as I read the puzzle, at least), I'll stick with my "it doesn't matter" answer. I agree that if the second person takes another die at random, 2 is the best die to start with, giving you slightly better than 50-50 odds, 14/27 chance of winning if I calculated correctly.
Sorry Alex, I misread your answer. You are correct. This is kind of a trick question kind of like the rock, paper , scissors game. For any choice, there is another choice with a higher probability.
No problem. My answer was rather cryptic, as I was trying to type it out fast on my phone before having to turn it off for a red-eye flight.
This is interesting, taking me back to sophomore abstract algebra. You expect a relationship like "is more probable than" to be transitive, but the relationship here is not transitive.