圆周随机取 N 点都落在某半圆内的概率

答案是

$$\frac{N}{2^{N-1}}.$$

固定其中一点作为半圆起点，其余 $N-1$ 点都落在该半圆内的概率为 $\left(\frac12\right)^{N-1}$。

由于任意一点都可能成为“起点”，乘以 $N$ 得到结论。

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Original Explanation

Label the points $1,2,\ldots,N$ clockwise. First, fix point 1. The probability that the remaining $N-1$ points all lie in the semicircle that starts at point 1 (in the clockwise direction) is

$$\frac{1}{2^{\,N-1}}.$$

Because any point could be the “start” of the containing semicircle, multiply by $N$. Thus, the probability is

$$\frac{N}{2^{\,N-1}}.$$