随机插入排名

因为 101 个独立连续样本的相对次序（全排列）等可能。

Paula 的样本在 1..101 的每一个名次上对称，概率相同。

因此

$$\boxed{\mathbb{P}(\text{名次}=k)=\frac{1}{n+1}},\quad k=1,2,\ldots,n+1.$$

对 $n=100$，每个名次概率都是 $1/101$，所以名次 51 与名次 1 一样可能。

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英文解析

a. Consider first the scenario when Peter draws just two values; let us call these values a and b. Let us call x the value obtained by Paula. Clearly, because the draws are independent and are obtained from the same population, all six orders are equally likely: abx, axb, bax, bxa, xab, xba, where, e.g., bxa stands for the order $\mathbf{b}< \mathbf{x}< \mathbf{a}$ . Two of them are such that Paula's draw (x) is first, second, or third. Thus, all three ranks for Paula's draw are equally likely.

b. In general, there are $(\mathrm{n} + 1)$ ! equally likely orders (i.e., permutations) of the n

values drawn by Peter and the single value drawn by Paula. How many of them are such that Paula's value ends up at rank k? If rank k is reserved for Paula's value then there remain n! orders in which the n values drawn by Peter can be arranged around this rank. For example, if $\mathbf{n} = 3$ , and Paula's value (call it x) is fixed at rank 2, Peter's three values (say, a, b,c) can be arranged as follows: axbc, axcb, bxac, bxca, crab, cxba. These are those sixl outcomes in which Paula's value x ends up at rank no. 2. Obviously, exactly the same generative procedure could be used to put Paula's value x at any rank from 1 to 4. In general, then, of the $(\mathbf{n} + 1)$ ! permutations there are n! such that Paula's value appears at rank no. k. Thus, the probability for each possible rank k of Paula's draw is equal to nty! = n+. Note that this value is independent of the actual rank, k.

Specifically, with $\mathbf{n} = 100$ each rank for Paula's draw from 1 to 101 is equally likely. Intuitively, most people think that it is much more likely that Paula's draw will eventually occupy some middle, intermediate rank, such as $\mathbf{k} = 51$ , rather than an extreme rank such as $\mathbf{k} = 1$ or $\mathbf{k} = 101$ . But that is not correct: any rank is equally likely.