Conditional probability: reverse probability, Markov chain There are two urns A and B. Initially, urn A contains $2$ red marbles and urn B contains $2$ white marbles. Consider a sequence of trials each consisting in simultaneously drawing one marble from each urn and switching them. Let $X_n$ be the number of red marbles in urn A after $n$ trials, $n ∈ \Bbb N$.
Suppose that, at some point, a long time after the process started, urn A contains at least one red marble. What is the probability that it contained exactly one red marble before the previous trial? I.e., what is $P (X_{n−1} = 1\mid X_n ≥ 1)$, as $n →∞$
I know that the stationary distribution is $[1/6, 2/3, 1/6]$ and the probability matrix is \begin{bmatrix}0&1&0\\1/4&1/2&1/4\\0&1&0\end{bmatrix}
but I am not too sure how to solve for the conditional probability in reverse.
 A: Fix a positive integer $n$.

The forward and reverse transition probabilities are the same, hence
$$P(X_{n-1}=1\mid X_n \ge 1)=P(X_n=1\mid X_{n-1} \ge 1)$$
Let events $F,G$ be defined by


*

*$F$ is the event $X_{n-1} \ge 1$.$\\[4pt]$

*$G$ is the event $X_n=1$.


Then $F$ is the disjoint union of events $F_1,F_2$ defined by


*

*$F_1$ is the event $X_{n-1} =1$.$\\[4pt]$

*$F_2$ is the event $X_{n-1} =2$.


Letting $n$ approach infinity, we get
\begin{align*}
&\lim_{n\to\infty} P(X_{n-1}=1\mid X_n \ge 1)\\[4pt]
=\;&\lim_{n\to\infty} P(X_n=1\mid X_{n-1} \ge 1)\\[4pt]
=\;&\lim_{n\to\infty} P(G\mid F)\\[4pt]
=\;&\lim_{n\to\infty} \frac{P(G\cap F)}{P(F)}\\[4pt]
=\;&\lim_{n\to\infty} \frac{P(G\cap (F_1\cup F_2))}{P(F_1\cup F_2)}\\[4pt]
=\;&\lim_{n\to\infty} \frac{P(G\cap F_1)+P(G\cap F_2)}{P(F_1)+P(F_2)}\\[4pt]
=\;&\lim_{n\to\infty} \frac{P(G\mid F_1){\,\cdot\,}P(F_1)+P(G\mid F_2){\,\cdot\,}P(F_2)}{P(F_1)+P(F_2)}\\[4pt]
=\;&\lim_{n\to\infty} \frac{\left(\frac{1}{2}\right)\,\!{\cdot}\,P(F_1)+(1){\,\cdot}\,P(F_2)}{P(F_1)+P(F_2)}\\[4pt]
=\;&{\large{\frac{\left(\frac{1}{2}\right)\!{\cdot}\!\left(\frac{2}{3}\right)+(1){\,\cdot}\!\left(\frac{1}{6}\right)}{\frac{2}{3}+\frac{1}{6}}}}\\[4pt]
=\;&\frac{3}{5}\\[4pt]
\end{align*}
