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This is more of a curiosity to me, but I'm sure I don't have the mathematical skills to answer it. That said... I took a look at several other posts with questions that relate to this one, but I haven't seen this specifically addressed...

Given a uniform solid rhombicosidodecahedron, what are the odds of rolling each of the faces: triangle, square and pentagon?

I'm interested in a theoretical answer here, I'm not concerned with reality... like a roller trying to "cheat" the roll for a specific outcome. And I'm obviously not concerned with the die being "fair". So let's assume that the roll is truly random.

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    $\begingroup$ related: mathoverflow.net/questions/46684/… $\endgroup$
    – Carlo Beenakker
    Commented Apr 26, 2018 at 21:04
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    $\begingroup$ This can be calculated by finding the solid angles, e.g. with the tetrahedral formulas at en.wikipedia.org/wiki/Solid_angle#Tetrahedron, using the Cartesian coordinates in the article linked in the question. $\endgroup$
    – user44143
    Commented Apr 26, 2018 at 21:22
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    $\begingroup$ I am doubtful that simple geometry will resolve this issue; here is an article from the physics literature that addresses some of the complicating factors. $\endgroup$
    – Carlo Beenakker
    Commented Apr 26, 2018 at 21:57
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    $\begingroup$ I agree with Carlo Beenakker; it is unclear what is meant by a "theoretical answer." As soon as you talk about "rolling," you must at minimum bring physics into the picture, and then lots of complicated considerations immediately arise. See for example this paper: researchgate.net/publication/… $\endgroup$
    – Timothy Chow
    Commented Apr 27, 2018 at 20:15
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    $\begingroup$ @MattF. : Your "spherical approximation" sounds like Simpson's model, which, as the paper cited by Carlo Beenakker explains in detail, is a poor fit to experimental data. It might be okay for an "adhesive surface" if by that you mean a surface that brings the die's motion to an abrupt halt the instant it first touches the surface, but this is not what most people think of as "rolling" a die. $\endgroup$
    – Timothy Chow
    Commented Apr 27, 2018 at 23:38

1 Answer 1

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As a first pass, we can approximate the odds of landing on a face by projecting the polyhedron on a sphere, and taking the fraction of the sphere which it covers.

polyhedronprojection

Then the odds of a triangular, square or pentagonal roll overall are $$14.4\%,\ 50.4\%,\ 35.1\%$$ and the odds for an individual triangular, square or pentagonal face are $$0.72\%,\ 1.68\%,\ 2.93\%.$$

This is Simpson's method. It also represents the result of throwing the die high above an adhesive surface, so that the die is well-randomized in the air, and then after touching the surface falls onto the nearest side.

The angles have exact formulas which are easy enough to calculate in Mathematica:

data = PolyhedronData["SmallRhombicosidodecahedron"];
faces = Map[data[[1, 1]][[#]] &, data[[1, 2, 1]]];
angle[a_, b_, c_] := 2 ArcTan[Abs[a.Cross[b, c]]/
     (Norm[a] Norm[b] Norm[c] + a.b Norm[c] + b.c Norm[a] + c.a Norm[b])];
Map[Length[#] angle[# // Mean // Simplify, #[[1]], #[[2]]] &, faces]
     // FullSimplify // Union

This gives the following solid angle measures for each triangular, square or pentagonal face: $$6 \cot^{-1}\left(2 \sqrt{3} u+\sqrt{124 u-61}\right),\\ 8 \cot^{-1}\left(2u+\sqrt{40 u-21}\right),\\ 10 \cot^{-1}\left(2 \sqrt{5u}+3 \sqrt{2u+1}\right)$$ where $u=5+2\sqrt{5}$.

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