14. The Markov chain is named after the Russian mathematician Andrey Markov, and its process has the property of being "memoryless," meaning the probability distribution of the state at time $n+1$ depends only on the state at time $n$, and is "unrelated" to the states at times $n-1, n-2, n-3, \cdots$. There are two boxes, A and B, each containing 2 red balls and 1 black ball of the same size, shape, and texture. One ball is randomly selected from each box and exchanged, and this operation is repeated $n\left(n \in \mathbf{N}^{*}\right)$ times. Let $X_{n}$ be the number of black balls in box A after $n$ operations, $a_{n}$ be the probability that box A has exactly 1 black ball, and $b_{n}$ be the probability that box A has exactly 2 black balls.
(1) Find the distribution of $X_{1}$;
(2) Find the general term formula for the sequence $\left\{a_{n}\right\}$;
(3) Find the expected value of $X_{n}$.
(1) $X_{1}=0,1,2$, and $P\left(X_{1}=0\right)=\frac{1 \cdot C_{2}^{1}}{C_{3}^{1} \cdot C_{3}^{1}}=\frac{2}{9}$,
$$
P\left(X_{1}=1\right)=\frac{1 \cdot 1+C_{2}^{1} \cdot C_{2}^{1}}{C_{3}^{1} \cdot C_{3}^{1}}=\frac{5}{9}, P\left(X_{1}=2\right)=\frac{C_{2}^{1} \cdot 1}{C_{3}^{1} \cdot C_{3}^{1}}=\frac{2}{9} \text {. }
$$

Therefore, the distribution of $X_{1}$ is
\begin{tabular}{|c|c|c|c|}
\hline$X_{1}$ & 0 & 1 & 2 \\
\hline$P$ & $\frac{2}{9}$ & $\frac{5}{9}$ & $\frac{2}{9}$ \\
\hline
\end{tabular}
(2) When $X_{n}=0$, $a_{n+1}=\frac{C_{3}^{1} \cdot C_{2}^{1}}{C_{3}^{1} \cdot C_{3}^{1}}=\frac{2}{3}$;
When $X_{n}=1$, $a_{n+1}=59$; When $X_{n}=2$, $a_{n+1}=\frac{C_{2}^{1} \cdot C_{3}^{1}}{C_{3}^{1} \cdot C_{3}^{1}}=\frac{2}{3}$,
then $a_{n+1}=\frac{5}{9} a_{n}+\frac{2}{3}\left(1-a_{n}\right)=\frac{2}{3}-\frac{1}{9} a_{n}$.
Thus, $a_{n+1}-\frac{3}{5}=-\frac{1}{9}\left(a_{n}-\frac{3}{5}\right) \Rightarrow a_{n}-\frac{3}{5}=\left(a_{1}-\frac{3}{5}\right)\left(-\frac{1}{9}\right)^{n-1}=-\frac{2}{45}\left(-\frac{1}{9}\right)^{n-1}$ $=\frac{2}{5}\left(-\frac{1}{9}\right)^{n} \Rightarrow a_{n}=\frac{3}{5}+\frac{2}{5}\left(-\frac{1}{9}\right)^{n}\left(n \in \mathbf{N}^{*}\right)$.
(3) Similarly to (2), when $X_{n}=0$, $b_{n+1}=0$;

When $X_{n}=1$, $b_{n+1}=\frac{C_{2}^{1} \cdot 1}{C_{3}^{1} \cdot C_{3}^{1}}=\frac{2}{9}$; When $X_{n}=2$, $b_{n+1}=\frac{1 \cdot C_{3}^{1}}{C_{3}^{1} \cdot C_{3}^{1}}=\frac{1}{3}$, then $b_{n+1}=\frac{1}{3} b_{n}+\frac{2}{9} a_{n}=\frac{1}{3} b_{n}+\frac{2}{9}\left[\frac{2}{5}\left(-\frac{1}{9}\right)^{n}+\frac{3}{5}\right]=\frac{1}{3} b_{n}-\frac{4}{5}\left(-\frac{1}{9}\right)^{n+1}+\frac{2}{15}$ $\Rightarrow b_{n+1}-\lambda+\mu\left(-\frac{1}{9}\right)^{n+1}=\frac{1}{3}\left[b_{n}-\lambda+\mu\left(-\frac{1}{9}\right)^{n}\right] \Rightarrow \lambda=\mu=\frac{1}{5}$, thus $b_{n}-\frac{1}{5}+\frac{1}{5}\left(-\frac{1}{9}\right)^{n}=\left(b_{1}-\frac{2}{9}\right)\left(\frac{1}{3}\right)^{n-1}=0 \Rightarrow b_{n}=\frac{1}{5}-\frac{1}{5}\left(-\frac{1}{9}\right)^{n}$. Therefore, $E\left(X_{n}\right)=a_{n}+2 b_{n}=\frac{3}{5}+\frac{2}{5}\left(-\frac{1}{9}\right)^{n}+2\left[\frac{1}{5}-\frac{1}{5}\left(-\frac{1}{9}\right)^{n}\right]=1$.