TITLE: The bug "probably" gets stuck!
QUESTION [5 upvotes]: Consider a regular tetrahedron with vertices $A,B,C,D$. A bug starts crawling from $A$. The bug moves from one vertex to the other along the edges continuously until it reaches $D$, where there is glue and hence once the bug reaches $D$, it gets stuck. The bug can move to any one vertex from any other vertex with equal probability. Then, find the probability that the bug reaches $D$ from $B$.
There seems to be varying answers to this question from different people (with $1/4$ and $1/5$ being the commonest responses). What can be the possible equation to be framed? Any hint is appreciated. Thanks!!

REPLY [2 votes]: Using symmetry, you can cut down the number of variables a bit more and extend this to the case of a complete graph with $n$ vertices - the tetrahedron is the case with $n=4$. Assume that the bug starts from vertex $curr$ and that the designated end vertex is $glue(\neq curr)$. Let $p_{v}$ and $p_{other}$ be the probabilities of bug reaching $glue$ from the vertex $v$ and from set of vertices not equal $curr$, respectively. (Note that, as in the question, "from" refers to the last vertex the bug visits before reaching $glue$.)
Since the bug moves from one vertex to another uniformly at random, the probability that it will reach $glue$ is $1$. Therefore,  $\begin{equation} p_{curr} + p_{other} = 1\end{equation}$. Also, $\forall v \notin \{curr, glue\}, p_{v} = \frac{p_{other}}{n-2}$. Lastly, $p_{curr} = \frac{1}{n-1} + \frac{n-2}{n-1}*\frac{p_{other}}{n-2}$. Reducing these equations, we get $p_{curr} = \frac{2}{n}, p_{other} = \frac{n-2}{n}, p_{v} \forall v \notin \{curr, glue\} = \frac{1}{n}$.