# random walk along edges of tetrahedron --- which face gets hit last?

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.

• I assume you mean "(i.e. in this case any edge other than cd)" and not "(i.e. in this case any edge other than bc)"
– user17762
Dec 9, 2011 at 22:02
• Yes, fixed, thanks. Dec 9, 2011 at 22:03
• Faces $abc$ and $abd$ are likely to be hit earlier than $acd$ or $bcd$, since you will hit one of them from the first step, and the other is then in no worse a position than one of the other two. So sadly your wish will not come true. Dec 10, 2011 at 0:06

Doing this the hard way by actually constructing a transition matrix for $90$ states [i.e. $6 \times(2^4-1)$ as we do not want to reach all faces hit], the answer is that when starting at the edge $ab$ the probability that $abc$ is hit last is $0.2$, $abd$ also $0.2$, $acd$ $0.3$, and $bcd$ also $0.3$. This looks plausible.