Ant going to the vertex

Question

An ant has a $\frac{1}{2}$ chance of moving to the vertex to its left and to its right in a hexagon. What is the probability for the ant to move from a vertex to the opposite vertex

I found this resolution:

I assume you wanted to find the probability that the ant goes to the other vertex, where the ant can go indefinitely.

Solution Let the vertex where the ant is on be $P_1$, and let the two vertices adjacent to it be $P_2$, the two vertices adjacent to these be $P_3$, and let the last vertex be $P_4$.

Using States, we obtain the following equations: $$\begin{cases}P_1=P_2\\ P_2=\frac{P_1+P_3}2\\ P_3=\frac{P_2+P_4}2\\ P_4=1\end{cases}$$By solving these systems of equations, we get: $$(P_1,P_2,P_3,P_4)=(1,1,1,1)$$Therefore, the probability that the ant would get to the opposite vertex is $\boxed1$.

She is right? Or is the answer $ \frac{1}{10}$?

Answer 1

If you want the probability that the ant ever reaches the opposite vertex, that is indeed $1$. All you need to conclude this is that this is a finite irreducible Markov chain. That is:

1. There are a finite number of vertices.
2. At each step, the probability of the ant moving to a given vertex depends only on where it is now.
3. It is possible for the ant to get from any vertex to any other vertex.