同弧问题

对圆周上均匀随机选出的 $N$ 个点，当 $0\le x\le\pi$ 时，它们全部落在某一段长度为 $x$ 的圆弧内的概率为

$$P = N\left(\frac{x}{2\pi}\right)^{N-1}.$$

直观理解是：任选一个点作为“起点”，其余 $N-1$ 个点都必须落在从该点开始、长度为 $x$ 的那段圆弧内；这样的起点一共有 $N$ 个可选。

代入 $x=\frac{\pi}{2}$、$N=4$：

$$P = 4\left(\frac{\pi/2}{2\pi}\right)^3 = 4\left(\frac14\right)^3 = 4\cdot\frac1{64} = \boxed{\frac1{16}}.$$

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Original Explanation

Step #1: General Formula

For $N$ points uniformly at random on the circumference, the probability that all points lie within an arc of length $x$ radians ($0\le x \le \pi$) is

$$\Pr(\text{all points in arc of length } x) = N \left(\frac{x}{2\pi}\right)^{N-1}.$$

This formula comes from the fact that we can "anchor" the first point anywhere, and the remaining $N−1$ points must lie within an arc of length $x$ starting from that first point. For $x≤\pi$, this formula is valid.

Step #2: Plug in the numbers

For $N=4$ and $x=\frac{\pi}{2}$:

$$\Pr = 4 \left( \frac{\pi/2}{2\pi} \right)^3 = 4 \left( \frac{1}{4} \right)^3 = 4 \cdot \frac{1}{64} = \boxed{\frac{1}{16}}.$$