Isomorphism between tetrahedron and cube paths

Question

Consider the following two problems:

1. A bug is sitting on vertex $A$ of a regular tetrahedron. At the start of each minute, he randomly chooses one of the edges at the vertex he is currently sitting on and crawls along that edge to the adjacent vertex. It takes him one minute to crawl to the next vertex, at which point he chooses another edge (at random) and starts crawling again. What is the probability that, after 6 minutes, he is back at vertex $A$?

2. Same as 1., except the bug is crawling along the edges of a cube.

We can compute the probability of both to find that they are both $\frac{61}{243}$, which probably implies there exists an isomorphism between the paths on the cube and the tetrahedron.

However, I don't quite see what or how the morphism can be built, because there are $4$ states for where the bug can be on the cube, while there are only $2$ states for where the bug can be on the tetrahedron.

I am not necessarily asking for a fully rigorous argument, some solid intuition (preferably geometric) will be accepted as well. (Of course, a rigorous argument will be accepted.)

Answer 1

A handwaving intuitive argument is that in both cases

• each move involves $3$ equally likely choices leaving a vertex

• so after after $6$ moves there are $3^6=729$ paths,

• to a total of four possible places the bug can be (there is a parity restriction on the cube but not the tetrahedron),

• roughly equally likely so about $182.25$ paths each, though the number of paths to each point must each be an integer.

• Three of the places are exactly equally likely by symmetry, so presumably have $182$ paths each,

• and the remaining place (which is vertex $A$ on the tetrahedron and vertex $A$ onm the cube after an even number of moves) is as close as possible to that namely $183$ paths or a probability of $\frac{183}{729}= \frac{61}{243}$.

The same intuitive argument would say that after five moves, the probability of being at vertex $A$ on the tetrahedron or oppositive vertex $A$ on the cube would be $\frac{60}{243}$.

Answer 2

Taking four vertices of the cube, no two of which are adjacent, gives a tetrahedron. Starting from any vertex on this tetrahedron, walking along two random edges on the cube takes you back to this vertex with probability $1/3$, and takes you to each of the other 3 vertices with probability $2/9$ each. The same thing is true if you start at some vertex on a non-cube-derived tetrahedron and walk along two randomly chosen edges on this tetrahedron. Therefore, the two probabilities must be the same.

Addendum: Alternatively, identify each vertex on the cube with its diagonally opposite vertex, and embed the tetrahedron into the cube as above. After doing this, the walks on the tetrahedron and the cube become identical, so that the probability of reaching $A$ or its opposite in 6 steps on the cube is the same as the probability of returning to $A$ in 6 steps on the tetrahedron. But since 6 is even, we can never reach the opposite of $A$ from $A$ on the cube in 6 steps.