7-15 In a given regular $(2n+1)$-sided polygon, three different vertices are randomly selected. If all such selections are equally likely. Find the probability that the center of the regular polygon lies inside the triangle formed by the randomly chosen three points.
[Solution] First, fix one vertex of the triangle, denoted as $p_{0}$, and the vertices on both sides of $p_{0}$ are sequentially denoted as $p_{1}, p_{2}, \cdots, p_{n}$ and $p_{-1}, p_{-2}, \cdots, p_{-n}$.

For the center of the regular polygon to be inside the triangle, it is necessary and sufficient that the other two vertices of the triangle are one $p_{k}(k>0)$ and one $p_{-m}(m>0)$, and $k+m>n$.

Therefore, for each fixed $k$, there are $k$ such triangles:
$$\Delta p_{0} p_{k} p_{-(n-k+1)}, \Delta p_{0} p_{k} p_{-(n-k+2)}, \cdots,$$
$\triangle p_{0} p_{k} p_{-n} .(k$ taking $1,2, \cdots, n)$, then
$$\sum_{k=1}^{n} k=1+2+\cdots+n=\frac{n(n+1)}{2} .$$

Since there are $2 n+1$ vertices in total, the number of triangles containing the center of the regular polygon is
$$\frac{(2 n+1) n(n+1)}{2 \times 3}=\frac{(2 n+1) \cdot n \cdot(n+1)}{6}$$

The total number of triangles formed by $2 n+1$ vertices is
$$C_{2 n+1}^{3}=\frac{(2 n+1) \cdot 2 n \cdot(2 n-1)}{6}$$

Therefore, the required probability is
$$\frac{\frac{(2 n+1) n(n+1)}{6}}{\frac{(2 n+1) \cdot 2 n \cdot(2 n-1)}{6}}=\frac{n+1}{2(2 n-1)}$$