4. Find the maximum real number $c$, such that the following conclusion holds for all integers $n \geqslant 3$: Let $A_{1}, A_{2}, \cdots, A_{n}$ be $n$ arcs on a circle (each arc includes its endpoints). If there are at least $\frac{1}{2} \mathrm{C}_{n}^{3}$ triples $(i, j, k)$, satisfying
$$
1 \leqslant i < j < k \leqslant n$, and $\cap_{i \in I} A_{i} \neq \varnothing$. (Provided by Ai Ehua)
4. The sought maximum real number $c$ is $\frac{\sqrt{6}}{6}$.

First, we prove that the conclusion in the problem holds when $c=\frac{\sqrt{6}}{6}$.
Orient the circle clockwise, so each arc has a starting point and an endpoint according to this orientation.

Notice that if $A_{i} \cap A_{j} \neq \varnothing$, then the starting point of $A_{i}$ must be contained in $A_{j}$, or the starting point of $A_{j}$ must be contained in $A_{i}$, or both.

Construct a directed graph $G$ with the vertex set $\left\{A_{1}, A_{2}, \cdots, A_{n}\right\}$. For $i \neq j$, if the starting point of $A_{i}$ is in $A_{j}$, then draw a directed edge $A_{j} \rightarrow A_{i}$.
For a pair of directed edges with the same endpoint
$$
\left\{A_{q} \rightarrow A_{p}, A_{r} \rightarrow A_{p}\right\}(q \neq r),
$$

let $p, q, r$ be arranged in ascending order as $i\frac{3}{2}+\frac{n-2}{\sqrt{6}}>\frac{n}{\sqrt{6}}$,

This proves that $c=\frac{\sqrt{6}}{6}$ satisfies the requirement.
Next, we prove that $c \leqslant \frac{\sqrt{6}}{6}$.
Consider the following construction: for a large integer $n$, let
$$
a=\left\lceil\frac{1}{2}+\sqrt{\frac{1}{4}+\frac{(n-1)(n-2)}{6}}\right\rceil<\frac{n}{2},
$$

where $\lceil x\rceil$ denotes the smallest integer not less than the real number $x$.
Take $n$ equally spaced points on the circle, and label them in clockwise order as $P_{1}, P_{2}, \cdots, P_{n}$. Let $A_{i}$ be the minor arc $\overparen{P_{i} P_{i+a}}$ (with $P_{i}$ as the starting point and $P_{i+a}$ as the endpoint), where the indices are understood modulo $n$.

Since the starting points of the arcs are distinct and the union of any two arcs is not the entire circle, $A_{q} \rightarrow A_{p}$ and $A_{p} \rightarrow A_{q}$ cannot both hold. Therefore, the mapping $\Phi$ : $S \rightarrow T$ defined earlier is also injective.
Thus, $|T|=|S|$.
Also note that in graph $G$, each $A_{i}$ has exactly $a$ incoming edges, i.e., $d\left(A_{i}\right)=a$.
Then $|T|=|S|=\sum_{i=1}^{n} \mathrm{C}_{d\left(A_{i}\right)}^{2}=n \mathrm{C}_{a}^{2} \geqslant \frac{1}{2} \mathrm{C}_{n}^{3}$.
If $I \subset\{1,2, \cdots, n\}$ satisfies $\bigcap_{i \in I} A_{i} \neq \varnothing$, take a point $X$ in the intersection. Let the starting point of the first $A_{i}(i \in I)$ that $X$ encounters when moving counterclockwise along the circle be the starting point of $A_{p}$, where $p \in I$. Then for any $q \in I, q \neq p$,
$A_{q} \rightarrow A_{p}$. Thus,
$$
\begin{array}{l}
c n<|I| \leqslant d\left(A_{p}\right)+1=a+1 . \\
\text { Hence } c \leqslant \lim _{n \rightarrow \infty} \frac{1+\left\lceil\frac{1}{2}+\sqrt{\frac{1}{4}+\frac{(n-1)(n-2)}{6}}\right]}{n} \\
=\frac{\sqrt{6}}{6} .
\end{array}
$$

In summary, the maximum $c$ is $\frac{\sqrt{6}}{6}$.