Three, (20 points) A town has three teahouses, $A$, $B$, and $C$. The new mayor drinks tea at only one of these three places each day. It is known that on the first day, the probability of him going to each place is $\frac{1}{3}$. If he goes to $A$ on a certain day, the probabilities of him going to $A$, $B$, and $C$ the next day are $\frac{1}{2}$, $\frac{1}{3}$, and $\frac{1}{6}$, respectively. If he goes to $B$ on a certain day, the probabilities of him going to $A$, $B$, and $C$ the next day are $\frac{1}{4}$, $\frac{1}{2}$, and $\frac{1}{4}$, respectively. If he goes to $C$ on a certain day, the probabilities of him going to $A$, $B$, and $C$ the next day are $\frac{1}{4}$, $\frac{1}{4}$, and $\frac{1}{2}$, respectively. Find the probabilities of the mayor going to $A$, $B$, and $C$ on the $n$-th day.
Three, let the probabilities of the mayor going to places $A$, $B$, and $C$ on the $n$-th day be $a_{n}$, $b_{n}$, and $c_{n}$, respectively. Then $a_{n}+b_{n}+c_{n}=1$, and
$$
\left\{\begin{aligned}
a_{n+1} & =\frac{1}{2} a_{n}+\frac{1}{4} b_{n}+\frac{1}{4}\left(1-a_{n}-b_{n}\right) \\
& =\frac{1}{4} a_{n}+\frac{1}{4}, \\
b_{n+1} & =\frac{1}{3} a_{n}+\frac{1}{2} b_{n}+\frac{1}{4}\left(1-a_{n}-b_{n}\right) \\
& =\frac{1}{12} a_{n}+\frac{1}{4} b_{n}+\frac{1}{4} .
\end{aligned}\right.
$$

Equation (1) can be simplified to $a_{n+1}-\frac{1}{3}=\frac{1}{4}\left(a_{n}-\frac{1}{3}\right)$.
Given $a_{1}=\frac{1}{3}$, we have $a_{n}=\frac{1}{3}\left(n \in \mathbf{N}_{+}\right)$.
Substituting $a_{n}=\frac{1}{3}$ into equation (2) yields
$$
b_{n+1}=\frac{1}{4} b_{n}+\frac{5}{18},
$$

which simplifies to
$$
b_{n+1}-\frac{10}{27}=\frac{1}{4}\left(b_{n}-\frac{10}{27}\right).
$$
Given $b_{1}=\frac{1}{3}$, we obtain
$$
b_{n}=\frac{10}{27}-\frac{1}{27}\left(\frac{1}{4}\right)^{n-1}\left(n \in \mathbf{N}_{+}\right).
$$

Finally, from $a_{n}+b_{n}+c_{n}=1$, we get
$$
c_{n}=\frac{8}{27}+\frac{1}{27}\left(\frac{1}{4}\right)^{n-1}\left(n \in \mathbf{N}_{+}\right).
$$

Note: Letting $n \rightarrow \infty$, we have $a_{n}=\frac{1}{3}$, $b_{n} \rightarrow \frac{10}{27}$, and $c_{n} \rightarrow \frac{8}{27}$. The ratio of the three probabilities is $9: 10: 8$.