A confusion about my solution A caterpillar is moving on the edges of a tetrahedron $ABCD$ on whose top there is glue. In a unit of time the caterpillar goes from any vertex (except D) to any other vertex with the same probability $1/3$. Suppose the caterpillar at time $t=0$ is on $A$.
(a) Pr(The caterpillar finally gets stuck)
(b) Pr(The caterpillar finally gets stuck coming from B)
What I did :
See if we do first one and it has probability $p$ then the second one has probability $p/3$ due to symmetry. So we try to find the first one. Let us denote $p_n=\Pr($there are $n$ vertices in between the time the insect reaches D from A $:=A_n)$. Now $p_0=1/3$. We induct
$P(A_n)=P(A_n|A \rightarrow B)P(A\rightarrow B)+P(A_n|A \rightarrow C)P(A\rightarrow C)=1/3p_{n-1}+1/3p_{n-1}=\dfrac{2p_{n-1}}{3}$
Now our required probability is $\sum_{n=0}^{\infty}p_n=1$. This seems correct to me but the answer of $1$ makes me believe I am wrong. Can someone help me out? If I am wrong then please give me the right solution. Thanks.
 A: It's almost certain that the caterpillar will get stuck if this scenario continues long enough; it's only a question of when it makes the terminal move, and from where.  
While it might seem intuitive that there is an equal chance that the caterpillar can approach $D$ from each of the three other vertices, the situation is not actually symmetrical.   We must consider the initial state.
At $t=0$ the caterpillar starts at node $A$.   There is a $1/3$ chance of immediately crawling to node $D$, and a $1/3$ chance each of crawling to node $B$ or $C$.   From either $B$ or $C$ it can continue to move back and forth between these nodes an indefinite amount of times until it eventually either returns to node $A$ or moves off to node $D$ with equal chance.
So the probability that it will ultimately move to $D$ from node $A$ is:
$$\mathsf P(A) = \frac 1 3 + \frac 2 3 \frac 1 2 \mathsf P(A)
\\ \therefore \mathsf P(A) = \frac 1 2$$
Now at time $t=1$ it will either be stuck at node $D$, or else at either node $B$ or $C$ with equal chance.   Now we can use symmetry to state that the conditional probability that the caterpillar moves to $D$ from $B$ when given that it does not do so from $A$ is: $1/2$
$$\mathsf P(B\mid\neg A) = \frac 12 \\ \mathsf P(B) = \mathsf P(B\mid \neg A)\mathsf P(\neg A) + \underbrace{\mathsf P(B\mid A)}_{0}\mathsf P(A) \implies \mathsf P(B) = \frac 1 4$$
Likewise for $C$. So thus:$$\mathsf P(A) = \frac 1 2, \mathsf P(B) = \frac 1 4, \mathsf P(C) = \frac 1 4$$

Alternatively.
At $t=0$ the caterpillar is not on node $B$.  It can move back and forth between nodes $A$ and $C$ an indefinite number of times until it eventually moves to either node $D$ or $B$ with an equal chance.  Once at node $B$ it may either move to node $D$ or return to either node $A$ or $C$.
Thus the probability of ultimately moving to $D$ from $B$ is:
$$\mathsf P(B) = \frac 12 \left(\frac 1 3 + \frac 2 3 \mathsf P(B)\right) \\ \therefore \mathsf P(B) = \frac 1 4$$
Since a symmetrical argument can be made for the probability of ultimately moving to $D$ from node $C$, $\mathsf P(C)=\mathsf P(B) = \frac 1 4$, and hence $\mathsf P(A) = 1 -\mathsf P(B\cup C) = \frac 1 2$
