棍子的短端

“随机折断”表示断点在木棍上的各个位置等可能出现。断点落在左半段和右半段的概率相同。若断点落在左半段，较短那段就是左边这一段，其平均长度是左半段的一半，也就是整根木棍的四分之一；落在右半段时同理。因此较短那段的平均长度是 $\boxed{\text{全长的 }\frac{1}{4}}$。

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Original Explanation

Breaking "at random" means that all points of the stick are equally likely as a breaking point (uniform distribution). The breaking point is just as likely to be in the left half as the right half. If it is in the left half, the smaller piece is on the left; and its average size is half of that half, or onefourth the length of the stick. The same sort of argument applies when the break is in the right half of the stick, and so the answer is $\boxed{\text{one-fourth of the length}}$

设木棍总长为 1，并假设断点落在右半段，位置为 $x\in[\frac{1}{2},1]$，则短段与长段的长度比为

$$\frac{1-x}{x}$$

由于 $x$ 在区间 $[\frac{1}{2},1]$ 上均匀分布，其平均值为

$$2\int_{\frac{1}{2}}^1 \frac{1-x}{x}\,dx = 2\int_{\frac{1}{2}}^1\left(\frac{1}{x}-1\right)dx = 2\log_e 2 - 1 \approx \boxed{0.386}$$

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Original Explanation

We might suppose that the point fell in the right-second half. Then $(1-x)/x$ is the fraction if the stick is of unit length. Since $x$ is evenly distributed from $1/2$ to $1$, the average value, instead of the intuitive $1/3$ is:

$$2 \int_{\frac{1}{2}}^1\frac{1-x}{x}dx = 2 \int_{\frac{1}{2}}^1(\frac{1}{x} - 1)dx =2\log_e2 - 1 \approx \boxed{0.386}$$