You need to design a pair of dice where each number one 1-12 has an equal chance of being rolled. You have been given the task of desi

At last, Nauseated

The die is cast

As Sisyphus hath said

And found

Such problems do not lend themselves

To Numbers

"Round"

Try this:  

Label one die with the numbers:          2, 4, 6, 8, 10, and 12.

Label the other die with the numbers: -1, -1, -1, 0, 0, and 0.

When the second die shows a -1, the sum of the two dice will be 1, 3, 5, 7, 9, or 11.

When the second die shows a 0, the sum of the two dice will be 2, 4, 6, 8, 10, or 12.

Since there is an equal chance for the first second die to show a -1 or a 0,

there will be equal chances for the two dice to sum to 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, or 12.

Very nice, geno....I hadn't considered negative numbers.....!!!!

That's brilliant Geno, I never would have thought of that.  

Could also have one die with faces numbered 1 to 6 as normal, and the other die with half the faces numbered 6 and half numbered zero.

Or one die with faces numbered 7 to 12, and the other die with half numbered -6 and half zero.

Or one die with faces numbered 13 to 18 and the other with half numbered -12 and half numbered -6.

Or one die with ...!!!

.

So.....by Alan's reasoning ....the general solution would be for one die to have six faces labeled  n, n+1, n+ 2, n+3, n + 4  and n+ 5  for n ≥ 1. And the other die would have three faces labeled  -n + 1  and three labeled -n + 7

A long with being true that is quite funny, CPhill. You should get a professional troll’s license. It’s to die for!

Now, for a very important question: How could we do it with roman numerals?