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I've been looking at polyhedra, incuding Platonic solids as well as Archimedean and Catalan solids. Catalan solids are face-transitive, which I believe implies that they are "fair dice", in the sense that each face is equally likely to land on top if the solid is "rolled" in some suitably non-biased fashion.

There is a solid called a pseudo-deltoidal icositetrahedron, and it is not a Catalan solid: https://en.wikipedia.org/wiki/Pseudo-deltoidal_icositetrahedron. It has the interesting property of being "monohedral" but not face-transitive. In other words, each face is the same polygon, but the symmetry group induces more than one orbit on the faces. In fact, there are two types of faces, symmetry-wise: polar faces (8) and equatorial faces (16). I'm wondering whether this shape is, nevertheless, a fair die.

There are two Catalan solids having 24 faces (the deltoidal icosidetrahedron and the tetrakis hexahedron), so it is possible to make a fair D-24 from either of those. I'm just curious whether this is a third option.

Note: I'm aware that this question comes up in the answers of a question on SO: https://mathoverflow.net/questions/46684/fair-but-irregular-polyhedral-dice. It is conjectured that this shape is a fair die, and someone agrees that they think it is, but nobody proves anything.

I'm very interested in any thoughts anyone can offer on this question. :)

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  • $\begingroup$ I'd just like to add that there's now a 3D model on Wikipedia, which you can download and 3D print, if you're willing to do the experiment. I might 3D print this sometime too. $\endgroup$
    – ViHdzP
    Commented Feb 9, 2020 at 20:37
  • $\begingroup$ actually there is also the triakis octahedron which is a fair die (it's just hard to read the faces). I made a bunch of 3d models here, in case you would be curious. Also: I believe that my post is a duplicate of this post. $\endgroup$
    – ARG
    Commented Nov 25, 2020 at 15:34
  • $\begingroup$ @ViHdzP I also made a bunch of die using OpenSCAD you can find them [there](https://www.thingiverse.com/thing:4145859). Most of the coordinates are from [here $\endgroup$
    – ARG
    Commented Nov 3, 2022 at 12:54

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It is a difficult question to answer, because it depends on physics. Just look at how hard it is to work out how thick a coin needs to be to become a fair three-sided coin.

What you can say about this pseudo-deltoidal icositetrahedron, is that all the faces are congruent, and the dihedral angles (angles between two faces meeting at an edge) are the same everywhere. The latter is maybe not so obvious. The normal deltoidal icositetrahedron has equal dihedral angles. By symmetry, the angle between faces around the equator and the equatorial plane are exactly half the dihedral angle. When you twist the shape to turn it into the pseudo-DI the angle to the equatorial plane does not change, so two faces meeting at the equator still have the same dihedral angle as before.

Does this mean it is a fair die? Probably not.

Imagine that you roll the pseudo-DI from one of the polar faces over two equatorial faces to a polar face at the other pole. This more or less rolls in a straight line, because the four visited faces occur in all four orientations. If you roll a normal DI over any four faces in the same way, it will go in a curved path.

This probably means that once the pseudo-DI is rolling very slowly, it is slightly more likely to end up on a polar face than an equatorial face.

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  • $\begingroup$ The model they used has been shown to be highly flawed in the general case. Here's a nice read on the topic of modeling non-isohedral dice. $\endgroup$
    – ViHdzP
    Commented Feb 9, 2020 at 20:40

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