Sunday, July 29, 2012

Fair Dice

There are three six-sided dice, with the numbers 1 to 18 on their faces.  If you roll any two of the dice against each other, each one is equally likely to "win" by rolling the higher number.  If you roll all three dice, they are equally likely to win.  No ties are possible.

One die has the numbers:

1, 5, 10, 11, 13, 17

Can you figure out the numbers on the other two dice?

This puzzle was inspired by a much harder puzzle submitted by Eric Harshbarger to MathPuzzle in April 2012, where there were four twelve-sided dice.

6 comments:

asexualcuriosities said...

2, 4, 8, 12, 15, 16 and 3, 6, 7, 9, 14, 18?

miller said...

Very close! But the second die (2,4,8,12,15,16) is slightly more likely to win when all three dice are pitted against each other.

asexualcuriosities said...

Damn you, Miller! I ended up just resorting to trial and error, but I think I've got it: 3, 4, 7, 12, 15, 16 and 2, 6, 8, 9, 14, 18?

miller said...

That's correct!

The solution is unique.

Larry, The Barefoot Bum said...

Is it interesting that the solution is unique?

miller said...

No, the solution is unique because I built the puzzle to have a unique solution. I mentioned the uniqueness so that people don't look for additional solutions.

There are, however, more solutions once you relax the constraint that the first dice has 1, 5, 10, 11, 13, 17.