Let $3 \overline{F_{n}}$ be the set of all functions satisfying the following recurrence relation: from $\{1,2, \cdots, n\}$ to $\{1,2, \cdots, n\}$, and
(1) $f(k) \leqslant k+1 \ (k=1,2, \cdots, n)$;
(2) $f(k) \neq k \ (k=2,3, \cdots, n)$.
Find the probability that a function $f$ randomly selected from $\overline{F_{n}}$ satisfies $f(1) \neq 1$.
This problem involves the application of the Fibonacci sequence in counting problems. By enumeration of smaller values of $n$, it is not difficult to find that the result should be $\frac{F_{n-1}}{F_{n}}$, where $F_{i}$ represents the $i$-th term of the Fibonacci sequence.
Below, we prove this using mathematical induction:
$\overline{F_{n}}$ has $F_{n}$ elements, and among them, $F_{n-1}$ satisfy $f(1)=2$ (while $f(1) \neq 1$).
Clearly, when $n=1$, the conclusion holds.
Now, let $n \geqslant 2$, and use the method of constructing a one-to-one correspondence to count.

If $f \in \overline{F_{n}}, f(1)=2$, then a function $g \in \overline{F_{n-1}}$ can be determined.

If $f(k+1)=1$, then $g(k)=1$, and for other $x$, if $f(x+1) \neq 1$, then $g(x)=f(x+1)-1$.

Conversely, for any function $g \in \overline{F_{n-1}}$, there is a unique $f \in \overline{F_{n}}$ such that $f(1)=2$.
Therefore, the number of $f$ is the total number of elements in $\overline{F_{n-1}}$.
By the induction hypothesis, there are $F_{n-1}$ of them.
On the other hand, by setting
$$
g(k)=f(k+1)-1,
$$

we know that the set of elements satisfying $f(1)=1, f \in \overline{F_{n}}$ is in one-to-one correspondence with the set of elements satisfying $g(1)=2, g \in \overline{F_{n-1}}$. Thus, by the induction hypothesis, the number of elements satisfying $f(1)=1, f \in \overline{F_{n}}$ is $F_{n-2}$.

Therefore, the total number of elements in $\overline{F_{n}}$ is $F_{n-1}+F_{n-2}=F_{n}$, and the probability that $f(1) \neq 1$ is $\frac{F_{n-1}}{F_{n}}$.
By induction, the original proposition is proved.