My thoughts on looking for the proportions of a fair d3: Can we smoothen a d6?

Question

A year or so ago, there was a video by Matt Parker talking about a fair d3, using a cylinder. Matt and his friend then set out to make cylinders with their calculations, and then did a fairly large sample of dice throws. The stats professor in the video said that Matt and his friend were wrong, but they did find a lower and upper bound. When I posted this question in a similar group on other social media, it is said that creating a d3 at this point is just using a d6 then mod 3 the result. Me being a person of results, I wish for a more satisfying conclusion to this problem. Would it be possible to smoothen every other edge of a d6 so as to have a polyhedron with 3 curved "sides"? I suck at draeing so I can't draw what I mean, but just imagine a cube with every other edge being a smooth continuation from one face to the other, as opposed to a hard edge.

Answer 1

You have the right idea, but not with "every other edge". You need to round only three edges out of the 12.

Imagine looking down one body diagonal of the cube. You see three edges aimed towards you, there are three opposing hidden edges meeting at the other end of the diagonal, and the remaining six form a zig-zag band around the "middle". Select either set of three alternating edges in this band and break out your carpenter's tools. The two possible sets of three edges correspond to mirror image configurations.

From a symmetry point of view, the cube starts out with $O_h$ symmetry, having 48 point-group symmetry elements. Your modified die ends up $D_3$, with six elements. This implies you have eight choices for your modification. We just saw two of them given one body diagonal along which to view the cube. You can choose any of the four body diagonal for your vantage point, properly giving $4×2=48/6=8$ possible orientations.