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?
Construct any solid with identical plane shapes and the resulting solid must be a fair die
geometryplatonic-solidsprobabilitysolid-geometry
Best Answer
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:
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.