Markov Chain Contest Problem Five cards labeled 1, 3, 5, 7, 9 are laid in a row in that order, forming the five-digit number 13579 when
read from left to right. A swap consists of picking two distinct cards, and then swapping them. After
three swaps, the cards form a new five-digit number n when read from left to right. Compute the
expected value of n.
I have shown my attempt to solve the problem in an answer.  Could someone vet the solution.

Here's a slightly simpler way of calculating the exact expected value of $\ n\ $.  After the OP's correction to my original calculation (in which I made an arithmetical error), it gives the same answer as his.
After $3$ swaps, there will be a probability $\ p\ \ \big($which turns out to be $\ \frac{3}{10}\ \big)
 $ that the digit in any given place will be the same as it was at the start, and a probability of $ \frac{1-p}{4}\ \left(=\frac{7}{40}\right)\ $ that it will be any one of the other $4$ digits.  The expected value of $\ n\ $ is therefore
\begin{align}
10^4\Big(&p\cdot1+\frac{(1-p)(25-1)}{4}\Big)+10^3\left(p\cdot3+\frac{(1-p)(25-3)}{4}\right)\\
&+10^2\Big(p\cdot5+\frac{(1-p)(25-5)}{4}\Big)+10\left(p\cdot7+\frac{(1-p)(25-7)}{4}\right)\\
&\hspace{1.5em}+p\cdot9+\frac{(1-p)(25-9)}{4}\\
&\hspace{3em}=\left(\frac{5p-1}{4}\right)13579+\left(\frac{25(1-p)}{4}\right)11111\\
&\hspace{3em}=\frac{13579+35\cdot11111}{8}\\
&\hspace{3em}=50308
\end{align}
You can calculate the value of $\ p\ $ by treating the presence or absence of the original digit in any place as a two-state Markov chain with transition matrix
$$
\pmatrix{\frac{3}{5}&\frac{2}{5}\\\frac{1}{10}&\frac{9}{10}}\ ,
$$
since the probability is $\ \frac{1}{10}\ $ that a single swap will return any given digit back to its original place once it has been swapped out of it.  The initial state of the chain is $1$, and there are $4$ sequences of states that end with with the digit originally in the place being back there after $3$ swaps.  Those sequences and their probabilities are:
$$
\begin{array}{cc}
\text{state sequence}&\text{probability}\\
1111&\left(\frac{3}{5}\right)^3=\frac{27}{125}\\
1121&\left(\frac{3}{5}\right)\left(\frac{2}{5}\right)\left(\frac{1}{10}\right)=\frac{3}{125}\\
1211&\left(\frac{2}{5}\right)\left(\frac{1}{10}\right)\left(\frac{3}{5}\right)=\frac{3}{125}\\
1221&\ \left(\frac{2}{5}\right)\left(\frac{9}{10}\right)\left(\frac{1}{10}\right)=\frac{9}{250}\ ,
\end{array}
$$
and the sum of these probabilities is $\ \frac{3}{10}\ $, as stated above.
