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Take a bunch of identical shapes and stick them together to form a convex 3-d solid. Of course, this won't be possible for most shapes. But when it is possible, one should always get a fair die. All platonic solids are examples of this, so are two tetrahedrons pasted together along a face. So is a Tetratoid where the faces are not even regular polygons. This result seems to make intuitive sense. But is it possible to prove it beyond doubt?

Rohit Pandey
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    I assume you want a convex solid. (Otherwise, just glue a tetrahedron to an icosahedron or octahedron to get a counterexample - some faces have a $0%$ chance of being rolled!) – anon Oct 29 '20 at 02:54
  • I'm not sure what you think a fair die is. But even a cube will give a non-uniform distribution of outcomes depending on how it is thrown. –  Oct 29 '20 at 03:28
  • @runway44 - yes, edited the question. Hadn't thought of that. – Rohit Pandey Oct 29 '20 at 03:31
  • @MatthewDaly - how do you toss a coin? – Rohit Pandey Oct 29 '20 at 03:40
  • If a magician borrows a coin from you and flips twenty heads in a row, do you conclude your coin is magical, or do you accept that a magician can flip a coin so that she knows how many times it spins in the air before it is caught? –  Oct 29 '20 at 04:31
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    @Matthew, the question is clear enough to anybody who wants to understand it -- why do you need to confuse the issue like this? – TonyK Oct 30 '20 at 00:42

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This statement is not true. At least, there are shapes for which it no longer makes "intuitive sense" that they would be fair dice, though rigorously proving unfairness would require a more formal specification of what it means to roll a die.

As an example, consider the snub disphenoid:

enter image description here

The general issue here is that intuitive "fairness" comes from a polyhedron being face-transitive, or isohedral, so that we cannot distinguish any two faces from one another. Thus, if spun with enough random noise that any orientation of the shape is equally likely, we shouldn't expect any one face to be favored over another, since we can't even write down a property possessed by one face and not another without making reference to a fixed orientation or location on the solid.

But "having all congruent faces", or being monohedral, is not in general enough to guarantee face-transitivity; one can find other counterexamples among the non-uniform convex deltahedra, like the triaugmented triangular prism and the gyroelongated square bipyramid. I believe these are the only counterexamples with regular polygons as faces, but using non-regular faces we can find examples like the pseudo-deltoidal icositetrahedron, the rhombic icosahedron, and the Belinski dodecahedron.

RavenclawPrefect
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    Thanks! The Wikipedia article on the snub disphenoid: https://en.wikipedia.org/wiki/Snub_disphenoid has a mesh image. Is it obvious to you which faces in that mesh will be more likely for it to rest on if we rolled it randomly (with enough white noise)? – Rohit Pandey Oct 29 '20 at 20:48
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    I think the red faces should all have a higher chance of being landed on than the yellow faces; an example where it's more obvious (to me, anyway) is the triaugmented triangular prism. If you view it as a triangular prism with three square pyramids glued onto the walls of the prism, then the faces that are part of the triangular prism and not the pyramids seem intuitively much more "stable" - the solid would like to balance on them if possible. – RavenclawPrefect Oct 29 '20 at 20:59