Example 8 Randomly select three points $A$, $B$, and $C$ on the unit circle. Find the probability that the side lengths of $\triangle ABC$ are all no more than $\sqrt{3}$.
【Analysis】Let's set point $C(1,0)$, which means point $C$ corresponds to an argument of 0, point $A$ corresponds to an argument $\alpha \in[0, \pi]$, and point $B$ corresponds to an argument $\beta \in[-\pi, \pi)$.

Thus, $(\alpha, \beta)$ is uniformly distributed in the rectangular region $[0, \pi] \times[-\pi, \pi)$. Clearly,
$$
\begin{array}{l}
A C \leqslant \sqrt{3} \Leftrightarrow \alpha \in\left[0, \frac{2 \pi}{3}\right], \\
B C \leqslant \sqrt{3} \Leftrightarrow \beta \in\left[-\frac{2 \pi}{3}, \frac{2 \pi}{3}\right], \\
A B \leqslant \sqrt{3} \\
\Leftrightarrow \beta \in\left[\alpha-\frac{2 \pi}{3}, \frac{2 \pi}{3}\right] \text { or } \alpha=\frac{2 \pi}{3}, \beta=-\frac{2 \pi}{3} .
\end{array}
$$

Therefore, the area of the region corresponding to $(\alpha, \beta)$ that satisfies the problem (as shown in Figure 1) is $\frac{2 \pi^{2}}{3}$.

The area of the rectangular region $[0, \pi] \times[-\pi, \pi)$ is $2 \pi^{2}$, so the required probability is $\frac{1}{3}$.