Fancy things involving water and bubbles make things much more difficult/expensive to produce. A perfect sphere, subdivided into equal area “faces” would give a fair die, but reading the number would be difficult - edge cases in which it lands on or near the dividing line between faces would be a problem.
@d_r and @henryseg gave good answers, but I thought I might take a stab at adding in why 120 is the cap.
We want all the faces to be equivalent, where any two are related by a rotation or reflection, so we can do best if we have a type of symmetry with as many of these as possible. One way to do that is to have symmetry about an axis of rotation through a very small angle. This gives the bipyramids with as many faces as you want, but they’re all very acute and so it’s not much like other dice.
The other option is polyhedral symmetry. There the most elaborate type possible is the symmetry of a regular dodecahedron (D12) or the dual icosahedron (D20). In essence, we can look at this as trying to put as many points (where the numbers will go) on one of those in a symmetrical way, and then just fracturing the faces accordingly. It doesn’t matter which you start with, so I’m going to go with the one without pentagons.
An icosahedron has 20 triangular faces; for a D20 we put the points right in the middle, and the dual D12 is equivalent to putting points at the corners. You can have more, though, if you put the points just near the corners. Then each face can have its own three, all related by rotation, for a grand total of 20x3 = 60 points. You get the same number if you put the points near the middle of the edges.
We do even better by putting the points somewhere between the corners and mid-edges. That allows a six points per face, making a little hexagon in the middle of the triangle, like in the top-right of this picture I found:
Since a triangle has 6 different symmetries – the identity, rotating left or right, and three reflections – that’s as many points as you can have and still have symmetries mapping one to each of the others. So that gives us our maximum of 20x6 = 120 equivalent points to put numbers on.
I don’t know, I hope that makes some sense the way I’ve written it. I promise you it could with more pictures, but that’s a lot more work to put together.
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Paging @doctorow - this post deserves to be added to the article on the front page.
(Ars does it with comments all the time!)
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So it’s basically an icosahedron whose faces have been split into six?
You know, now that I’m looking for that, I think I can see it.
If you connect the three closest major points, you make a triangle; that triangle contains six smaller triangles, and if you smoothed each of those larger triangles out, you’d have a d20 (which is as many faces as you can get using regular shapes).
So, in this picture below, one of the faces of the “d20” within the green die would be: 13,81,67,6,76,120.
I think I get it now. Thanks!
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