O fairest of randomizers

On most numbered dice, opposite sides are complementary; on a cube, for example, they add to 7. As a result, if you have the skill to throw a die so that the {1,2,3} corner lands on the table, the upward face must be at least 4.

I would prefer to design dice so that, if numbers are considered as masses, the center of mass coincides with the geometric center. I think this is equivalent to saying: the sum of the numbers in any hemisphere must be equal.

You can’t do that with a tetrahedron, cube, D10 or regular dodecahedron; but I found three solutions for the octahedron, and 876 for the rhombic dodecahedron. (At least I see no obvious way to reduce that number further with symmetries.)

For a cube the best you can do is put pairs of adjacent numbers on opposite faces.
The twelve best arrangements on the regular dodecahedron all have 0 and 11 opposite each other, 1–5 around 11, and 6–10 around 0.
The two best arrangements for D10 are 0285364179 and 0582367149 (reading around the fivefold axis, alternating upper and lower faces).
(Add 1 to each number if you don’t like zero-based indexing; it doesn’t affect the math.)

I’ll update here if I come up with an approach to the icosahedron problem that won’t take thousands of years.

3 thoughts on “O fairest of randomizers”

  1. List of fair dice (in the sense that the faces are equivalent by symmetry).
    It would be interesting to see whether a less regular (but still fair) pentagonal dodacahedron can meet my balance criterion. I’ll think about that later.

  2. I have been measuring the distance of the center of mass from the geometric center, but it would be better (or at least more complete) to show the difference between the sums of faces in the two hemispheres defined by the plane perpendicular to the offset vector.

  3. Lacking a better idea, I started the “naïve” permutation search for D20 with an initial arrangement that puts pairs of adjacent numbers roughly opposite each other. Permuting from this, I got seven solutions in the first ~millionth of the range. Here is one:

     0   4   5   7   9
    19  11  18  12  10
      17  14  16   8  15
       1   3   2  13   6

    Unless I have miscounted, hemisphere sums are not all equal!

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