Two bottles, one with 3 red balls and the other with 3 blue balls. In each trial, one ball is randomly drawn from both bottles and exchanged with each other. What is the distribution of the red ball number in the first bottle after $n$ trials?
Denote the red ball number in the first bottle of trial $n$ is $X_n$. I think this is a Markov process that $X_n$ only depends on $X_{n-1}$: \begin{equation*} \mathrm{Pr}(X_n=k|X_{n-1}=j) = \begin{cases} \frac{j^2}{9} & \textit{if k = j-1}\\ \frac{j^2-6j+9}{9} & \textit{if k = j+1}\\ \frac{6j-2j^2}{9} & \textit{if k = j}\\ \end{cases} \end{equation*}
I wonder, if $\Pr(X_n)$ can be computed, instead of the conditional distribution $\Pr(X_n|X_{n-1})$.
I agree with your formula for the transition probabilities. We can number the states $0, 1, 2, 3$ by number of red balls in the first bottle, and then write the same probabilities as the transition matrix: $$Q = \left(\begin{array}{l}0 & 1 & 0 & 0 \\ \frac 1 9 & \frac 4 9 & \frac 4 9 & 0 \\ 0 & \frac 4 9 & \frac 4 9 & \frac 1 9 \\ 0 & 0 & 1 & 0 \end{array}\right)$$
We start out with 3 red balls in the first bottle, which corresponds to state vector $v_0 = (0, 0, 0, 1)$. To compute the distribution after $k$ timesteps, we need to evaluate $Q^k v_0$.
The point of writing the Markov chain iteration in this matrix form is that we can use linear algebra tricks to compute the matrix power quickly. Find a matrix $T$ such that $A = T^{-1} Q T$ is diagonal. (See Wikipedia for details about how to diagonalize a matrix.) In this case, we can choose $$T = \left(\begin{array}{l} 1 & -1 & -1 & 1 \\ 1 & \frac 1 3 & -\frac 1 3 & -\frac 1 9 \\ 1 & -\frac 1 3 & \frac 1 3 & -\frac 1 9 \\ 1 & 1 & 1 & 1 \end{array}\right) $$ and we end up with $$A = \left(\begin{array}{l} 1 & 0 & 0 & 0 \\ 0 & -\frac 1 3 & 0 & 0 \\ 0 & 0 & \frac 1 3 & 0 \\ 0 & 0 & 0 & -\frac 1 9 \end{array}\right). $$ Then we have $Q = T A T^{-1}$, so $Q^n = (T A T^{-1})^n = T A T^{-1} T A T^{-1} \cdots T A T^{-1} = T A^n T^{-1}$. $A^n$ is easy to compute because $A$ is diagonal, so we can get a final formula for the distribution of $X_n$:
$$X_n = \left(\begin{array}{c} \frac 1 {20} - \frac 1 4 (-\frac 1 3)^n - \frac 1 4 (\frac 1 3)^n + \frac 9 {20} (-\frac 1 9)^n \\ \frac 1 {20} - \frac 1 {12} (\frac 1 3)^n + \frac 1 {12} (-\frac 1 3)^n - \frac 1 {20} (-\frac 1 9)^n \\ \frac 1 {20} - \frac 1 {12} (-\frac 1 3)^n + \frac 1 {12} (\frac 1 3)^n - \frac 1 {20} (\frac 1 9)^n \\ \frac 1 {20} + \frac 1 4 (-\frac 1 3)^n + \frac 1 4 (\frac 1 3)^n + \frac 9 {20} (\frac 1 9)^n \end{array}\right). $$
Note the final distribution is a column vector with 4 entries, corresponding to 0, 1, 2, and 3 red balls respectively in the first bottle.