Challenge: design a polyhedral die that will always give one of five outcomes, each with equal probability.
While achieving that, minimize the total number of faces of the die. Think of the die as convex and made of a uniform density material.
More specifically, design a polyhedron that has 5 symmetric faces that are stable, where the die won't tip over if sitting on those faces, and any number of unstable faces, where the die will tip over. There should be no stable, asymmetric faces. Minimize the number of unstable faces.
By "symmetric", I mean that there should exist symmetries of the polyhedron mapping each of the five stable faces to each other.
By a "stable" face, I mean that the center of mass of the polyhedron, projected onto the plane of the face, lies within the face. If placed face-down on a table, the center of mass should lie above a stable face, and off to the side of an unstable face. Avoid having the center of mass exactly above the boundary of the face.
As a starting point, this double pencil construction by Julian Rosen achieves 15 total faces (5 stable, 10 unstable). Can you do better?
Edit: A die of 10 faces, as posted by @Bass, is the best solution I know. Can you prove it is optimal, or find a better one?
Bonus question: Can you find a construction with less than $3n$ faces for arbitrary $n$?