Sicherman dice are pairs of dice numbered [1, 2, 2, 3, 3, 4] and [1, 3, 4, 5, 6, 8] respectively. They have the nice feature that the distribution of results when you roll them and add them together is the same as for a normal pair of six sided dice. But some games care about rolling doubles, so they don't work the same as regular dice for that purpose.

This suggests a puzzle: color the faces of a pair of Sicherman dice so that rolling a matching color pair works the same as rolling doubles on normal dice. (That is, there is exactly one matching color pair for each even sum: 2, 4, 6, 8, 10, and 12.) How many distinct ways are there to do this? (Up to isomorphism, in other words, the choice of particular colors isn't what's important.)

Some faces might not match any face on the other die. It's convenient to designate one color to be the color that never matches, so we don't have spurious extra solutions where we choose various ways of splitting the never-matching faces into different colors. So let's use black for that. All faces that match no others are required to be black, and black faces are considered to not match each other, in case there happen to be non-matching faces on both dice.

Given this formulation of the problem, I have found four solutions. I think I have them all, but I'm not entirely certain. I hope to get physical copies of some of these solutions made; last year I backed a Kickstarter to produce custom engraved dice, and some of these will probably make it into my order.

This suggests a puzzle: color the faces of a pair of Sicherman dice so that rolling a matching color pair works the same as rolling doubles on normal dice. (That is, there is exactly one matching color pair for each even sum: 2, 4, 6, 8, 10, and 12.) How many distinct ways are there to do this? (Up to isomorphism, in other words, the choice of particular colors isn't what's important.)

Some faces might not match any face on the other die. It's convenient to designate one color to be the color that never matches, so we don't have spurious extra solutions where we choose various ways of splitting the never-matching faces into different colors. So let's use black for that. All faces that match no others are required to be black, and black faces are considered to not match each other, in case there happen to be non-matching faces on both dice.

Given this formulation of the problem, I have found four solutions. I think I have them all, but I'm not entirely certain. I hope to get physical copies of some of these solutions made; last year I backed a Kickstarter to produce custom engraved dice, and some of these will probably make it into my order.

- There's more than that.

Clearly, the two 1s must be the same color, and the first die's 4 must match the second die's 8, as these are the only ways to get 2 and 12.

All of the ways to get 4 are 1+3, so one of the 3s must be the same color as the 1s. The ways to get 10 are 2+8 (twice) and 4+6, so either the 6 or one of the 2s must be the same color as the 4+8. This provides 4 distinct partial solutions for the 2, 4, 10, and 12 doubles. I'm arbitrarily using letters to denote colors here:

[1A, 2B, 2, 3A, 3, 4B], [1A, 3, 4, 5, 6, 8B]

[1A, 2, 2, 3A, 3, 4B], [1A, 3, 4, 5, 6B, 8B]

[1A, 2B, 2, 3, 3, 4B], [1A, 3A, 4, 5, 6, 8B]

[1A, 2, 2, 3, 3, 4B], [1A, 3A, 4, 5, 6B, 8B]

6 is made from 1+5, 2+4 (twice), or 3+3 (twice). 8 is made from 4+4, 3+5 (twice), or 2+6 (twice).

SOLUTIONS WITH 6=2+4

For each partial solution, it's possible to color a 2 like a 4 and a 3 like a 5 in new colors to get a complete solution:

[1A, 2B, 2C, 3A, 3D, 4B], [1A, 3, 4C, 5D, 6, 8B]

[1A, 2C, 2, 3A, 3D, 4B], [1A, 3, 4C, 5D, 6B, 8B]

[1A, 2B, 2C, 3, 3D, 4B], [1A, 3A, 4C, 5D, 6, 8B]

[1A, 2C, 2, 3, 3D, 4B], [1A, 3A, 4C, 5D, 6B, 8B]

In the cases where 6 is not yet colored, we can color that the same as 2+4, instead of making 3+5, to make more solutions.

[1A, 2B, 2C, 3A, 3, 4B], [1A, 3, 4C, 5, 6C, 8B]

[1A, 2B, 2C, 3, 3, 4B], [1A, 3A, 4C, 5, 6C, 8B]

In the cases with 2B, it's also possible to color 4B in the second die to make 2+4 and 4+4 both in B:

[1A, 2B, 2, 3A, 3, 4B], [1A, 3, 4B, 5, 6, 8B]

[1A, 2B, 2, 3, 3, 4B], [1A, 3A, 4B, 5, 6, 8B]

SOLUTIONS WITH 6=1+5

In the cases where the 3A is on the second die, it is possible to color 5A to make the 6 without making any more numbers. Then 3+5 cannot be made without making extra numbers, but in one case we can make 8 with 4+4 in B and in the other case we can make 2+6 in a new color.

[1A, 2B, 2C, 3, 3, 4B], [1A, 3A, 4, 5A, 6C, 8B]

[1A, 2, 2, 3, 3, 4B], [1A, 3A, 4B, 5A, 6B, 8B]

In the cases where 3A is on the first die, we can color 5A to make both 6 and 8, and no more coloring is needed.

[1A, 2B, 2, 3A, 3, 4B], [1A, 3, 4, 5A, 6, 8B]

[1A, 2, 2, 3A, 3, 4B], [1A, 3, 4, 5A, 6B, 8B]

SOLUTIONS WITH 6=3+3

We can only ever make 6 as 3+3 if we use the two previously unmarked 3s in a new color (otherwise, we'd get two color-matched ways to make 4). This is possible when the 3A is on the first die. Then we can make 8 either by coloring the 5 in the same colors as these 3s in both of those cases, and in one case we can also do it by making 4+4 in B, and in the other by making 2+6 in a new color:

[1A, 2B, 2, 3A, 3C, 4B], [1A, 3C, 4, 5C, 6, 8B]

[1A, 2, 2, 3A, 3C, 4B], [1A, 3C, 4, 5C, 6B, 8B]

[1A, 2B, 2D, 3A, 3C, 4B], [1A, 3C, 4, 5, 6D, 8B]

[1A, 2, 2, 3A, 3C, 4B], [1A, 3C, 4B, 5, 6B, 8B]

So, 16 distinct solutions. Of of these, I think

[1A, 2, 2, 3, 3, 4B], [1A, 3A, 4B, 5A, 6B, 8B]

is the best, since it only needs 3 colors including black, it does not need the "black doesn't match any other color" rule since all the black is on one die, and each of the colors appears on 4 faces.Mar 3, 2015 - Ah. Just goes to show you; I lack the focus to be a diligent puzzle solver. I did find the one you consider best. It has a nice mnemonic: 1 on the low die, odd on the high die; 4 on the low die, even on the high die. You could easily use it without even coloring the dice.

The other one I like is:

[1A, 2B, 2_, 3_, 3C, 4D], [1A, 3A, 4B, 5C, 6D, 8D]

(Your 4th solution, but with letters swapped.)

In addition to only having non-matching faces on one die, this has a couple nice properties:

· Each of the 36 rolls is distinct: No roll has a pair of faces with the same numbers and colors as another roll. As far as I can tell, only your first four solutions have this property, of which this and the first are the ones with non-matching faces on one die only.

· The colors (excluding black) can be given a strict ordering that follows the numerical order of the faces on both dice.

I'll probably make a set of these and a set of the one you thought was best.Mar 4, 2015