11. (1973 USA Mathematical Olympiad) In a given regular $(2n+1)$-sided polygon, three different vertices are chosen at random. If all such choices are equally likely, find the probability that the center of the polygon lies in the interior of the triangle formed by the randomly chosen points.
11. First fix a vertex of the polygon, 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:
$\triangle p_{0} p_{k} p_{-(n-k+1)}, \triangle p_{0} p_{k} p_{-(n-k+2)}, \cdots, \triangle p_{0} p_{k} p_{-n}$. ( $k$ takes $1,2, \cdots, n$ ), 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
$$
C_{2 n+1}=\frac{(2 n+1) \cdot 2 n \cdot(2 n-1)}{6},
$$

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