德军坦克

我们需要找到一个 $N$，使得从 $\{1,2,\dots,N\}$ 中无放回地均匀抽取 6 个不同编号时，其最大值 $X_{(6)}$ 的期望为 $120$。

直观地说，如果 $120$ 是第 6 个顺序统计量，那么平均而言，前面还有另外 5 辆坦克，它们会把区间 $\{1,2,\dots,120\}$ 分成 6 段长度大致相等的部分。这意味着除去这 6 个顺序统计量本身（因为这里是离散取值），还剩下 $114$ 个其他序列号。

这些坦克之间的间隔在期望上应当相等，因此平均间隔为 $\frac{114}{6} = 19$。于是，既然 $120$ 是第 6 个顺序统计量，而每个顺序统计量的平均间隔为 $19$，那么最大坦克编号应为 $$120 + 19 = 139.$$

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Original Explanation

We need to find $N$ such that the expected value of the $X_{(6)}$, the maximum among $6$ distinct uniform random draws from $\{1,2,\dots,N\}$ is $120$.

Intuitively, if $120$ is the $6$th order statistic, then in expectation, there are 5 other tanks before it that should partition $\{1,2,\dots,120\}$ into 6 equal length parts. This means that besides all of tanks that are the order statistics themselves (since we have discrete values), there are $114$ other serial numbers.

The spacings between these tanks should be equal in expectation, so the average distance between them should be $\frac{114}{6} = 19$. Therefore, as $120$ is the $6$th order statistic and we know each order statistic on average has a spacing of $19$, the maximum tank number should just be $120 + 19 = 139$.