Ants and Vertices of 3-D Figures

[Lokito]

This question is entirely for my own enrichment.

Imagine that there is an ant at each vertex of a 3-D figure, and that all the ants simultaneously crawl along an edge to the next vertex. The chosen paths are random, and each ant has total control of its own locomotive processes. What is the probability that no ant will encounter another, either en route or at the next vertex, for:

a cube

a tetrahedron

etcetera.

I know of no way to do this type of problem other than listing out every possibility and counting up the number of combinations in the set that work and the total number of combinations. As the number of vertices and the complexity of the shape increase, this becomes really hard to figure out, and it takes forever. What's the right way to do this?

 

[wjm11]

Imagine that there is an ant at each vertex of a 3-D figure, and that all the ants simultaneously crawl along an edge to the next vertex. The chosen paths are random, and each ant has total control of its own locomotive processes. What is the probability that no ant will encounter another, either en route or at the next vertex, for:

a cube

I do not have a general solution. I believe (but cannot prove) that the problem may be simplified by examining what movements are necessary around the perimeter of a single face of any 3-D object: a circular rotation. The object needs to be broken down into discrete faces around which movement takes place.

A cube has 6 faces, 8 vertices, 12 edges. Each vertex has 3 edges connected to it.
Each ant could take three different routes, so number of routes is 3^8 = 6561.
The number of routes that have no ants passing each other or having more than one ant arrive at a vertex is only 12 (I think -- can someone find more?). Probability of meeting conditions is approximately

12/6561 = .001829

 

[galactus]

That's what I came up with also, wjm. But I hesitated posting because I was a little uncertain.

There appears to be a general formula which can be derived. For a triangle there would be 8 different paths but only two where they don't meet, so the probability they wouldn't meet would be 2/8=1/4.
That is 3 vertices and 2 choices to go at each, so 2^3=8. The ants wouldn't meet if they just went clockwise or counterclockwise. The same idea applies to the cube. There are 8 vertices and 3 choices to go at each. So 3^8.
But since there are 6 faces and they can go clockwise or counterclockwise, there are 12 options where they would not meet.

One could use the same principle on any other polyhedra.