There are three dice, with different numbers on each side.

A: 1-1-5-5-5-5
B: 2-2-2-2-6-6
C: 3-3-3-4-4-4

Two players each choose one of the three dice, and roll once; high roll wins. You get to / have the misfortune to go first. Which of the three dice should you choose?

Follow-up; is it possible to make a perfect set of dice like these, so that

* P(A defeats B) = P(B defeats C) = P(C defeats A) = P > 1/2
* all 18 numbers are natural numbers
* no ties can happen; and
* the average of the numbers on the six sides of each die is the same?

If there are multiple such solutions, how high can P be?

If it is not possible, why not? Perhaps it could be done with dice with a number of sides other than six?

Originally posted by talzamir There are three dice, with different numbers on each side.

A: 1-1-5-5-5-5
B: 2-2-2-2-6-6
C: 3-3-3-4-4-4

Two players each choose one of the three dice, and roll once; high roll wins. You get to / have the misfortune to go first. Which of the three dice should you choose?

Follow-up; is it possible to make a perfect set of dice like these, so that ...[text shortened]... not possible, why not? Perhaps it could be done with dice with a number of sides other than six?

Originally posted by forkedknight C vs A is identical to B vs C, why would you pick either one?

If i pick C they're going to pick A and if i pick B they're going to pick C. Picking A is the only logical choice to go against their B. I have 4 sides that'll beat 4 B sides, granted B has the same but it's a lot better odds in a one off roll than any of the other possibilities for one roll.

A vs B: A wins with a five over two probability 2/3 x 2/3 = 4/9
B vs C: B wins with a six over any, probability 1/3 x 3/3 = 3/9
C vs A: C wins with any over one, probability 3/3 x 1/3 = 3/9

so A is the best choice. I would like to have a set like this where it doesn't matter what the first player picks, the odds are always worse than 50% and always the same. With these exact dice, the odds are not identical.