We know a coin is a fair die with a 50-50 probability for two alternatives. Similarly, all five Platonic solids are fair dice. That makes six solids that can be fair dice, but can there be more? One example could be a two tetrahedra pasted together along one face. The resulting solid is not platonic since two vertices have three faces meeting at them while three of them have four faces meeting at them. However, this too can be a regular die as far as I can tell since all faces are identical. The question is, how many solids can exist that can be used as fair dice?

**Answer**

Infinitely many. For example, I think that by taking the intersection of $n $ identical spheres with adequate radius and centres on the vertices of a regular $n $-sided polygon, you create a fair die with $n $ outcomes.

An example with 3 sides would look similar to the following:

[Chkhartishvili, Levan & Suryamurthy, Gokul. (2015). Volume of intersection of six spheres: A special case of practical interest. Nano Studies. 11.]

If we restrict ourselves to polyhedra, one could build a generalisation of your two-tetrahedra example: simply build two identical regular pyramids with $n$-sided bases and paste them together to make the two bases coincide. This gives a fair die with $2n$ outcomes.

Finally, while all answers (including this one) focus on solids with sides of equal surface, this need not be the case. There can be additional sides with arbitrary surface as long as the die can never land and stay on that side, for example when the projection of the centre of mass falls outside the convex hull of the side. For example a pencil sharpened on both sides:

(Some symmetry of these “impossible sides” is required so that they do not alter the probabilities of the other sides.)

**Attribution***Source : Link , Question Author : Rohit Pandey , Answer Author : Luca Citi*