Let's define a die as a polyhedron that, if rolled over a perfect horizontal plane, ends up being in a physically stable unambiguous state labelled $n$. The die has $N$ states. Each state $n$ has a probability $p_n$ to happen.

The most common dice are the 5 platonic solids: tetrahedron, cube, ... where $p_n=1/N$. However, also others non platonic solids can guarantee uniform probabilities, such as the well known "coin" or, for the sake of the example, a long die with a regular polygon as a base shape (a "long" $N$-regular prism).

Previous solids symmetries allow to avoid to take in account 2 fundamental variables involved in the rigid body dynamics:

  1. center of mass
  2. moment of inertia

The algorithm that allows to generate any discrete probability density function (PDF) with a calculator using the common pseudo-random command rand is well known. If the realisation of rand falls between $\sum_{n=0}^k p_n$ and $\sum_{n=0}^{k+1} p_n$ the resulting state is $k+1$ ($p_0=0$ and $p_{N+1}=1$). That method basically divides the segment from $0$ to $1$ that represent the probability domain in $N$ sub segments each one assigned to every one of the $N$ target output states with pre-defined probability $p_n$.

Analogously, basing on the previous algorithm, it is possible to consider a cylinder with the base divided in $N$ angular sectors each one with angle of $\alpha_n=360 \cdot p_n$. Obviously $\sum_{n=1}^{N} \alpha_n=360$ since $\sum_{n=1}^{N} p_n=1$. Rolling that not-polyhedral die and considering in which angular sector falls the tangent point between the solid and the horizontal landing plane it is possible to define an output state. The output states generated in this way have a realisation PDF that follows the probabilities $p_n$.

However this method does not generate a die as defined before since its states are continuous, the discrete states must be deducted using the angular segments positions and the solid is not a polyhedron.

After all this dissertation the main question is: Does exist a method to design a polyhedral die where its realisations follow a given generic set of probabilities $[p_n]_{n\in [1,...,N]}$?

Is it helpful the reasoning about the sectorized cylinder?


A solution suggested by @Ján Lalinský is to consider a set of probabilities that can be expressed in the form $p_n=k_n/K$ where $K$ and $k_n$ are integers and $\sum_{n=1}^{N} k_n= K$. A long die (prism) with a regular base with $K$ edges can be considered. The $K$ faces of the prism can be grouped in $N$ groups with size $k_n$ and marked with $n$. Every roll of this die will respect the assumptions.

However a solution based on a polyhedron with exactly $N$ faces will be preferable.

  • 1
    $\begingroup$ I did, but your problem was not very clear to me, especially whether by cylinder you mean really cylinder or $N$-gonal prism. If you want $N$-state die where the states have prescribed but nonequal probabilities, I do not think using a prism with irregular polygonal base is a good idea, because it is hard to determine probability for faces of such irregular prism. It is best to use $M\gg N$ -gonal right regular prism and define $N$ groups of faces in such a way that probability of any face from $n$-th group corresponds to the prescribed probability $p_n$. $\endgroup$ – Ján Lalinský Apr 23 '19 at 20:10
  • 1
    $\begingroup$ @Andrea constructing body with irrational probabilities seems a rather mathematical problem - what is the practical difference between probability $1/\pi$ and probability $1/3.1415926535$? From a practical standpoint, there is always some interval of imprecision in every physical quantity that it no longer matters for anything, so one would just use high $M$ to get rational probabilities close enough to the desired ones. $\endgroup$ – Ján Lalinský Apr 24 '19 at 23:12
  • 1
    $\begingroup$ So it's better to have no answer than to have it migrate (which I'm also not in favor of)? $\endgroup$ – noah Apr 25 '19 at 19:12
  • 2
    $\begingroup$ The question is Math when "rolling" bodies are symmetric but without symmetry it is mandatory to take in account the full rigid body dynamic (center of mass and moment of inertia) in the probability calculus. This is Physics. $\endgroup$ – Andrea Apr 25 '19 at 21:19
  • 2
    $\begingroup$ There was no reason to close this question. $\endgroup$ – Andrea May 6 '19 at 12:46