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If the base of a right prism is a regular $n$-gon of side 1, what height makes it a fair dice? The $n=4$ case is obvious by symmetry. Assume constant density, constant downwards gravity, throwing on a flat surface with no friction in a perfect vacuum, and that Newton's laws apply.

If the mass were all concentrated in the geometric center the height which made the solid angle of each of the faces viewed from the center the same would make a fair dice. I don't think this applies when the mass is distributed because if $n>4$ you could exert a lot of torque, say shelling die with artillery, to avoid them landing on the the top and bottom faces.

Furthermore fore large $n$ the ratio of area to solid angle of the top face is smaller than that of the side faces, so that might cause weird behavior.

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    Given the significant physics considerations, this question might be more appropriate for the [Physics StackExchange](http://physics.stackexchange.com). – 2017-01-02
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    Without friction the dice would never stop. I would assume that for any such die with $n\neq 4$ there would be ways to throw the die to bias it towards or away from top and bottom faces. A vertical drop is probably different than one with a lot of rolling. For large *n* you'd probably get a long rod, to make top and bottom face sufficiently unlikely, but throwing that in a random way would be tricky, you'd most likely roll it along its long axis while throwing it. I agree with Blue that this is probably more of a Physics question. – 2017-01-02
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    For some hint of the complexities mentioned by @Blue, see the amazing response by Bill Thurston to the question, "[Fair but irregular polyhedral dice](http://mathoverflow.net/q/46684/6094). – 2017-01-02

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