Suppose we have a tetrahedron $abcd$, and start at edge $ab$. Now walk to any "adjacent" edge (i.e. in this case any edge other than $cd$), each with equal probability $1/4$. This gives a stationary Markov process on the edges.
Walking from one edge to another uniquely determines the face that they share. In this case, we say the face was "hit." For example, if you start at edge $ab$ and walk to edge $bd$ then face $abd$ has been hit. As we continue the random walk indefinitely, the probability approaches $1$ that every face has been hit at least once.
What is the probability that face $abc$ is the last of the four faces to be hit? This tells us the probability for every face, since the probability for $abc$ and $abd$ being the last face hit is the same by symmetry (since we start at edge $ab$), and similarly the probability of $bcd$ and $acd$ being hit last is also the same.
The answer that I'd most like is that the probability of being hit last is the same for every face $1/4$, but I don't know if that's true.