4. Find the largest 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
$$
1 \leqslant i < j < k \leqslant n \text{, and } \bigcap_{i \in I} A_{i} \neq \varnothing \text{. }
$$
4. The sought maximum real number $c$ is $\frac{\sqrt{6}}{6}$.

First, we prove that when $c=\frac{\sqrt{6}}{6}$, the conclusion in the problem holds.
Orient the circle in a clockwise direction, 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 either the starting point of $A_{i}$ is contained in $A_{j}$, or the starting point of $A_{j}$ is contained in $A_{i}$, or both.

Construct a directed graph $G$ with 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 $i < j < k$. Then, the edge $A_{j} \rightarrow A_{i}$ must exist, and the edge $A_{k} \rightarrow A_{j}$ must exist. Therefore, the edge $A_{k} \rightarrow A_{i}$ must also exist. Thus, the graph $G$ is transitive.

Let $S$ be the set of all pairs of arcs with a common endpoint, and let $T$ be the set of all pairs of arcs with a common starting point. Define a mapping $\Phi: S \rightarrow T$ as follows: for each pair $\left\{A_{q} \rightarrow A_{p}, A_{r} \rightarrow A_{p}\right\} \in S$, map it to $\left\{A_{k} \rightarrow A_{i}, A_{k} \rightarrow A_{j}\right\} \in T$.

Since the graph $G$ is transitive, $\Phi$ is well-defined. Moreover, $\Phi$ is injective. Therefore, $|T| \geq |S|$.

Since $|S| = \frac{1}{2} \sum_{i=1}^{n} d(A_{i})(d(A_{i})-1) = \frac{1}{2} \sum_{i=1}^{n} \left(\frac{n-2}{\sqrt{6}}\right)\left(\frac{n-2}{\sqrt{6}}-1\right) = \frac{n(n-2)(n-3)}{2\sqrt{6}}$,

and $|T| = \frac{1}{2} \sum_{i=1}^{n} d(A_{i})(d(A_{i})-1) = \frac{1}{2} \sum_{i=1}^{n} \left(\frac{n-2}{\sqrt{6}}\right)\left(\frac{n-2}{\sqrt{6}}-1\right) = \frac{n(n-2)(n-3)}{2\sqrt{6}}$,

we have $\frac{1}{2} \sum_{i=1}^{n} \left(\frac{n-2}{\sqrt{6}}\right)\left(\frac{n-2}{\sqrt{6}}-1\right) > \frac{n(n-2)(n-3)}{2\sqrt{6}}$,

which implies $\frac{3}{2}+\frac{n-2}{\sqrt{6}}>\frac{n}{\sqrt{6}}$,

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

Take $n$ equally spaced points on the circle, and label them in a clockwise direction as $P_{1}, P_{2}, \cdots, P_{n}$. Let $A_{i}$ be the minor arc $P_{i} P_{i+a}$ (with $P_{i}$ as the starting point and $P_{i+a}$ as the endpoint), where the indices are taken 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 exist. Therefore, the mapping $\Phi: S \rightarrow T$ defined earlier is also injective.
Thus, $|T|=|S|$.
Also, note that in the graph $G$, each $A_{i}$ has exactly $a$ incoming edges, i.e., $d(A_{i})=a$.
Then, $|T|=|S|=\sum_{i=1}^{n} \binom{d(A_{i})}{2}=n \binom{a}{2} \geqslant \frac{1}{2} \binom{n}{3}$.
If $I \subset\{1,2, \cdots, n\}$ satisfies $\cap_{i \in I} A_{i} \neq \varnothing$, then take a point $X$ in the intersection. Let $A_{p}$ be the arc whose starting point is the first point $A_{i}(i \in I)$ that $X$ encounters when moving counterclockwise along the circle, with $p \in I$. Then, for any $q \in I, q \neq p$, we have $A_{q} \rightarrow A_{p}$. Thus,
$$
\begin{array}{l}
c n<|I| \leqslant d(A_{p})+1=a+1 . \\
\text { Hence } c \leqslant \lim _{n \rightarrow \infty} \frac{1+\left[\frac{1}{2}+\sqrt{\frac{1}{4}+\frac{(n-1)(n-2)}{6}}\right]}{n} \\
=\frac{\sqrt{6}}{6} .
\end{array}
$$

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