Probability of Exiting A Roundabout I am piggybacking on this question: probability of leaving
The question was closed, but I found it interesting and I would like feedback on what I have done with it, and I have more questions about it. I'm not sure what proper etiquette is for piggybacking on closed questions.
I will rephrase the question as I understand it:
In the diagram below, you start at the node marked with lowercase $a$.

From $a$, you move to one of the adjacent nodes, chosen uniformly at random. That is, you move to $b$, $d$, or $A$.
If you are at a lowercase node, you continue in the same fashion, moving to an adjacent node uniformly at random. Once you reach an uppercase node, you have left the roundabout, and the journey stops. The question is, what are the probabilities of the journey ending at $A$, $B$, $C$, and $D$?
In the original question, there was an answer that used Markov chains, but I am wondering if there is a way to do it without that technique.
I modeled this experiment in Excel, and after $100,000$ trials, the probabilities appear to be roughly:
$P(A)=46.5\%$
$P(B)=P(D)=20\%$ 
$P(C)=13.5\%$
I have no way of knowing if these are the exact answers, or even if the exact answers are rational numbers, but they make intuitive sense to me because of the structure and symmetry in the diagram.
I wonder if there is a way to "juggle" conditional probabilities to find exact answers to this question, without having to use Markov chains. 
I would start by calculating $P(A|a)$, the probability of leaving at $A$ given you started at $a$. By symmetry, $P(B|b)$, $P(C|c)$, and $P(D|d)$ would be the same as $P(A|a)$.
To get $P(A|a)$, notice that the total length of the walk must be odd. Either you leave immediately ($1$ step), or you take an even number of steps in the roundabout to return to $a$, and then leave at $A$ ($2m+1$, for some integer $m$, steps).
For a given walk of $k$ steps, the probability of taking that walk is simply $(\frac13)^k$.
Let $N_k$ be the number of walks of $k$ steps that leave the roundabout at $A$.
Then $P(A|a)=\frac13+N_3\cdot(\frac13)^3+N_5\cdot(\frac13)^5+...$
I'm unable to find a systematic way of calculating $N_k$. It is easy enough to do for $k=3$ or $5$, but I can't be sure of what pattern is emerging. 
Beyond that, assuming I did have an exact answer for $P(A|a)$, I would still need to figure out $P(B|a)$. Would that just be $\frac13\cdot P(A|a)$ since there is a $\frac13$ probability of going to $b$ on the first step?
By symmetry, $P(D|a)=P(B|a)$, so if I had $P(A|a)$ and $P(B|a)$, I could easily figure out all the probabilites.
I would appreciate any input on this!
 A: 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.
A: 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.
