The bug "probably" gets stuck! 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!!
 A: Let $p_A$ be the probability that the bug reaches $D$ from $B$ given that it is currently at $A$. 
Let $p_B$ be the probability that the bug reaches $D$ from $B$ given that it is currently at $B$. 
Let $p_C$ be the probability that the bug reaches $D$ from $B$ given that it is currently at $C$. 
We want the value of $p_A$.
If the bug is currently at $A$, then the bug has: 
a $\frac{1}{3}$ chance of going to $B$, from which there is a $p_B$ probability of reaching $D$ from $B$,
a $\frac{1}{3}$ chance of going to $C$, from which there is a $p_C$ probability of reaching $D$ from $B$,
a $\frac{1}{3}$ chance of going to $D$, from which there is a no chance of reaching $D$ from $B$.
Therefore, $p_A = \frac{1}{3} \cdot p_B+\frac{1}{3} \cdot p_C+\frac{1}{3} \cdot 0$. 
By similar logic, you can get equations for $p_B$ and $p_C$. 
This gives you a system of 3 linear equations for 3 variables. Solve this to get $p_A$. 
A: The bug will get stuck at $D$ with probability $1$.  Note that this does not say that there are no series that avoid $D$-you could alternate $A$ and $B$ forever, but the chance is zero.  The question really asks, given that the bug travels from $A$ to $D$, what is the chance it visits $B$ on the way. As the chance the first move is to $B$ is $1/3$, it is obvious that $1/4$ and $1/5$ are wrong.  
This is a Markov chain, but you need to distinguish all the states.  Starting you are at $A$ and have not visited $B$ or $C$.  You have two states being at $A$ depending on whether you have visited $B$ or not (for this problem it doesn't matter if you have visited $C$), two states at $C$, and one state at $B,D$, with $D$ absorbing.  Write your transition matrix and use your usual techniques.
A: 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 $p_{curr} = \frac{2}{n}, p_{other} = \frac{n-2}{n}, p_{v} \forall v \notin \{curr, glue\} = \frac{1}{n}$.
