How thick should a cylindrical coin be for it to act as a fair three-sided die? When flipping a coin of radius $r>0$ and thickness $t>0$ in the real world, there is some non-zero probability of getting neither heads nor tails, but instead landing on the thin lateral side. My question is, how thick does this lateral face need to be relative to $r$ so that the coin has equal chance to land on each of its 3 faces? That is, how do we choose $t$ so that the chance of landing on the lateral is $\frac{1}{3}$?
Clearly this problem requires some simplification, so we may assume the coin does not bounce and lands on a flat surface, and, well, any other assumptions implicit in the next paragraph:
I have heard from a friend that there is a method of solving this which entails centering the coin in a sphere, and adjusting $t$ so that a randomly chosen line segment between the center and a point on the sphere has an equal probability of intersecting each face of the coin. Is this a common modeling technique? How should we choose the radius of this sphere? Does it not matter given that the coin is fully contained?
And secondly, how can we proceed to find $t$ as a function of $r$ once this model has been established?

This is an old and charming question. The thick coin is a cylinder. Cut it vertically by its diameter. You get a rectangle by section with sides $e\wedge D\ $. The thickness is $e$ and $D$ its diameter.
When you flip the coin the decision variable are the angles that form the rectangle diagonals. These angles determine if the rest position will be on some head-tail or on the edge. Name $\vartheta_e$ the angle between diagonals having the edge as the opposite side.

The fact is that the angle is a random variable having uniform distribution U[0,2$\pi$]. $\ $Thus $\frac{2\vartheta_e}{2\pi}=\frac{1}{3}$ and $\vartheta_e=\frac{\pi}{3}$. But $\tan(\vartheta_e/2)=e/D$ and $e=D\cdot\tan(\pi/6)=D/\sqrt{3}$.
I've seen some wrong answers having high score for this question. I do not know, but I hope this can be ammended.
