11. (20 points) On a straight road, there are $2 n+1$ trees $T_{1}, T_{2}, \cdots, T_{2 n+1}$ (where $n$ is a given positive integer). A drunk man starts from the middle tree $T_{n+1}$ and randomly walks among these trees for $n$ minutes according to the following rules: When he is at tree $T_{i}(2 \leqslant i \leqslant 2 n)$ at the end of a minute, he has a $\frac{1}{4}, \frac{1}{2}, \frac{1}{4}$ probability of being at $T_{i-1}, T_{i}$, $T_{i+1}$ at the end of the next minute.
(1) Find the probability that the drunk man is at each tree $T_{i}(1 \leqslant i \leqslant 2 n+1)$ at the end of the $n$-th minute;
(2) Suppose the distance between adjacent trees is 1. Find the expected distance between the drunk man and his starting position (i.e., tree $T_{n+1}$) at the end of the $n$-th minute (express your answer in the simplest form involving $n$).
11. (1) Suppose $2 n+1$ trees $T_{1}, T_{2}, \cdots, T_{2 n+1}$ are arranged from left to right, with a spacing of 1 unit between each pair of trees. Define a random walk as moving 0.5 units to the left or right with a probability of $\frac{1}{2}$. According to the probability distribution described in the problem, the drunkard's movement status each minute can be decomposed into two random walks. Therefore, the original problem is equivalent to finding the probability $p_{i}$ that the drunkard, starting from $T_{2 n+1}$, is at position $T_{i}$ after $2 n$ steps of random walk.

For some $i(1 \leqslant i \leqslant 2 n+1)$, let the number of times moving 0.5 units to the right and to the left in the entire process of $2 n$ steps be $k$ and $2 n-k$ respectively, then
$$
n+1+\frac{k-(2 n-k)}{2}=i,
$$

Solving for $k$ gives $k=i-1$, meaning in $2 n$ steps, there are $i-1$ steps to the right and $2 n-(i-1)$ steps to the left. The number of such scenarios is $\mathrm{C}_{2 n}^{i-1}$, so the required probability is
$$
p_{i}=\frac{\mathrm{C}_{2 n}^{i-1}}{2^{2 n}} \quad(1 \leqslant i \leqslant 2 n+1) .
$$
(2) For $i=1,2, \cdots, 2 n+1$, the distance between tree $T_{i}$ and $T_{n+1}$ is $|n+1-i|$, and the probability of the drunkard reaching tree $T_{i}$ is $p_{i}$, so the expected value is
$$
E=\sum_{i=1}^{2 n+1}|n+1-i| \frac{\mathrm{C}_{2 n}^{i-1}}{2^{2 n}} .
$$

According to the transformation of summation indices and properties of binomial coefficients, we have
$$
\begin{aligned}
\sum_{i=1}^{2 n+1}|n+1-i| \mathrm{C}_{2 n}^{i-1} & =\sum_{j=0}^{2 n}|n-j| \mathrm{C}_{2 n}^{j}=2 \sum_{j=0}^{n}(n-j) \mathrm{C}_{2 n}^{j}=2 \sum_{j=0}^{n} n \mathrm{C}_{2 n}^{j}-2 \sum_{j=0}^{n} j \mathrm{C}_{2 n}^{j} \\
& =2 n \sum_{j=0}^{n} \mathrm{C}_{2 n}^{j}-2 \sum_{j=1}^{n} 2 n \mathrm{C}_{2 n-1}^{j-1}=2 n \cdot \frac{1}{2}\left(\mathrm{C}_{2 n}^{n}+\sum_{j=0}^{2 n} \mathrm{C}_{2 n}^{j}\right)-4 n \sum_{j=0}^{n-1} \mathrm{C}_{2 n-1}^{j} \\
& =n\left(\mathrm{C}_{2 n}^{n}+2^{2 n}\right)-4 n \cdot \frac{1}{2} \sum_{j=0}^{2 n-1} \mathrm{C}_{2 n-1}^{j}=n\left(\mathrm{C}_{2 n}^{n}+2^{2 n}\right)-2 n \cdot 2^{2 n-1} \\
& =n \mathrm{C}_{2 n}^{n} .
\end{aligned}
$$

Therefore, $E=\frac{n \mathrm{C}_{2 n}^{n}}{2^{2 n}}$.