Two flies sits on the plane. First one sits at $(0,0)$ and the second one at $(0,8)$, where first coordinate indicates time and second indicates location. Then, they start to move randomly and independently. First one goes $1$up with probability $\frac13$ or $1$down with probability $\frac23$. Second one goes $1$up with probability $\frac34$ or $1$down with probability $\frac14$. If a distance between them equals to $12$ - flies fly away. Find the probability of encounter and the average time to encounter or to fly out.
I know that I have to find a proper difference equation. It's a similar exercise to gambler's ruin, but now there are "$2$ people instead of $1$" and I have to find difference equation depending on distance.
My guess is: $\psi_k=\frac13\psi_{k+1}+\frac23\psi_{k-1}$ for the first one and $\psi_k=\frac34\psi_{k+1}+\frac14\psi_{k-1}$ for the second one, then I just compare them. Or do I calculate them independently with different boundary values?
Any help will be much appreciated.
Construct a Markov chain $(X_n)_{n \in \mathbb{N}_0}$ that models the distance between the two flies. The chain has the possible states $\{0,2,4,6,8,10,12\}$. (We just say that once the chain has reached either $0$ or $12$, it stays there.) What are the transition probabilities?
Now let $p_i := P((X_n) \text{ reaches } 0 \text{ before } 12 | X_0 = i)$ for $i \in \{0,2,4,6,8,10,12\}$. (You are looking for the value of $p_8$.) What are the values of $p_0$ and $p_{12}$? Do first step analysis to relate the other $p_i$'s to each other to get a system of linear equations that you can solve.