What is the equilibrium location of the six numbers on the dodecahedron?

There is a dodecahedron, it is necessary to have on its sides the numbers 1 to 6 so that the probability of their loss was the same.

I will be glad to see or just a ready answer, or algorithm, how to the answer is to come. Or, at least, the names of the chapters in the theory of probability that will help me to understand this.
September 19th 19 at 12:16
September 19th 19 at 12:18
Solution
To arrange as you like, if each figure is met exactly two faces. It is believed that the regular polyhedron, the probability of each face is the same.
If you are afraid that the numbers will be different weight and the heavier will be the bottom often place the same numbers on the opposite faces. Then they are each other's balance.
> the same numbers on the opposite faces
but there is no opposite dodecahedron faces. - Carrol commented on September 19th 19 at 12:21
: Why would it? There are opposite faces, and edges, and vertices. This is only the tetrahedron (from regular polyhedra) they are not. And some antiprisms from semiregular. - Pansy.Marquar commented on September 19th 19 at 12:24
oops, blunted. - Carrol commented on September 19th 19 at 12:27
> If you are afraid that the numbers will be different weight and the heavier will be the bottom often place the same numbers on the opposite faces. Then they are each other's balance.

If we assume that the more severe will be down more often then Your accommodation will only aggravate the deviation from the statistics. - Jarvis96 commented on September 19th 19 at 12:30
September 19th 19 at 12:20
Solution
The dodecahedron 12 faces. I understand the meaning of the task only to the number of occurrences of every digit from 1 to 6 was equal, ie - 2. The relative placement of the digits relative to each other will not affect the result.

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