Probability of Exiting A Roundabout

By symmetry we can merge states $(b,d)$. Let $P[a], P[b], P[c]$ be probabilities of entering state $A$, given we start in states $a,b,c$, respectively. Then

• $P[a] = (1/3) + (2/3)P[b]$

• $P[b] = (1/3)P[a] + (1/3)P[c]$

• $P[c] = (2/3)P[b]$

So $P[a] = 7/15$, $P[b] =1/5$, $P[c] = 2/15$.

Here we have implicitly used the Markov chain structure by using the transition probabilities $P_{ij}$ and the fact that future states are conditionally independent of the past given the current state.

One way to short circuit the complications of the Markov chains is to work recursively. The situation is so very symmetric that this ends up simplifying the vast majority of the calculations.

Let the desired probabilities be $P_A,P_B, P_C, P_D$. Of course $P_B=P_C$ and the four variables sum to $1$. Thus there are really only two unknowns here. Let $x=P_A, y=P_B$. Of course, $P_C=y$ and $P_D=1-x-2y$.

Consider what happens once one makes that first choice. With probability $\frac 13$ the game is over (and you exit at $A$). Otherwise you move to either $B$ or $C$. Note that which ever of those two states you move to, you are now in the position $B,C$ started in. It follows that $$x=\frac 13\times 1 + \frac 23\times y$$

Similarly, still considering the start, let's analyze what happens to $P_B$. If you exit at $A$, you can't eventually exit at $B$. If you move to $B$, then the probability that you eventually exit at $B$ is $x$. And if you move to $C$ the probability that you eventually exit at $B$ is now $1-x-2y$. Thus $$y=\frac 13\times 0 +\frac 13\times x+\frac 13\times (1-x-2y)$$

This is easily solved and we get $$P_A=\frac 7{15}\quad P_B=P_C=\frac 15 \quad P_D=\frac 2{15}$$

We remark that this aligns well with your simulated results.