Probability of traversing all other states and finally landing on one state This is a cross-post from math.stackexchange.com. There has been no response there.
Given a Markov chain of finite states with constant transition probabilities, what is the method to compute the probability of a path finally landing in a particular state for the first time having traversed all other states? Are there any references? The following is a puzzle as an example.
$12$ people sit on a round table to play a variation of the telephone game. They are numbered from $1$ to $12$ in clockwise order, i.e. the people with adjacent numbers (including $1$ and $12$) sit next to each other. Person #1 chooses a secret word and starts the game by randomly selecting one of the two neighbors and whispering the word to that person. Upon hearing the word, each person continues the game by randomly selecting one of the neighbors and whispering the word. The game ends when everyone knows the secret word.
What is the probability that the last person whispering the word is numbered $6$?
 A: For the example, the considered event $E$ is the intersection of two events $A$ and $B$, namely that both 5 and 7 are reached before 6. The union of the two events has probability $1$. So the probability of $E$ is $P(A)+P(B)-1$. By this answer:
https://math.stackexchange.com/questions/725996/reaching-a-level-before-another-for-a-random-walk
$P(A)=7/11$ and $P(B)=5/11$ so the sought probability is $1/11$. The argument can be adapted to any position on the circle : they all have the same probability to be reached last.
A: Here I am exploring an approach that may be slightly different from Rob Pratt.  I will use a less complex example.
a
Instead of $12$ people, we seat $4$ labeled clockwise as $(0,1,2,-1)$. Let $A:=\{0,1,2,-1\}$. Define $p(i,S),\,i\in A,\,S\subset\{-1,1\}$ as the probability of paths starting from $i$ and satisfying the following property. The subset of $\{-1,1\}$ each of these path passes before for the first time landing on $2$ is $S$. 
By the Markov property, we have
\begin{align}
p(0,\{-1,1\}) &= \frac12\big(p(-1,\{-1,1\})+p(1,\{-1,1\})\big) \\
p(-1,\{-1,1\}) &= \frac12\big(p(0,\{-1,1\})+p(0,\{1\})\big) \\
p(1,\{1,-1\}) &= \frac12\big(p(0,\{1,-1\})+p(0,\{-1\})\big)  \\
p(-1,\{-1,1\}) &= p(1,\{1,-1\}) \\
p(0,\{1\}) &= p(0,\{-1\}))
\end{align}
All these $p(i,S)$ are equal. But then I can not proceed further.

Rob Pratt, please feel free to add your comment here.
A: Here is the insightful solution of user100927 in the top comment section. I transcribe his answer.
Let us define $(,)$ to be the probability of the word starting from  and traversing all other persons before landing on $$. $(,)=\frac12((−1,)+(+1,)),\forall |−|>1$. For, say, $j=i+1$, $p(i,i+1)=\frac12\big(p(i-1,i+1)+p(i-1,i)\big)$. The second term in the parenthesis comes from the equivalence of the following two events. A path starts from $i-1$ and hits $i+1$ before $i$ and one starts from $i-1$ and hits $i$ last. The equivalence of the two is implied by the fact that the starting point $i-1$ is the neighbor of $i$ opposite of $i+1$. Moreover, obviously $\sum\limits_jp(i,j)=1$.
Now the symmetry in the problem implies that $p(i,j)$ depends only on $|j-i|$. Without the risk of notational confusion, we denote $p(|j-i|):=p(i,j)$. We have 
$p(i)=\frac12\big(p(i-1)+p(i+1)\big),\,\forall i>1;\   p(1)=\frac12\big(p(2)+p(1)\big);\,\sum\limits_{i\ne0}p(i)=1$. With $n$ nodes, we obtain $\displaystyle p(i)=\frac1{n-1}$.
