How to get rank-in probabilities? Suppose that we have n instances $X_1,\dots,X_n$ $(X_i \in \mathbb{R})$ uniformly sampled from the population without replacement.
We do not know the distribution of $X_i$ but know that the population size is N.
How to estimate the probabilities $Pr(pos(X_i) \geq T)$?
where $1\leq pos(X) \leq N$ indicates the position of $X_i$ in the ordered list of the entire instances in the population; the ordered list is obtained through sorting by the values of $X_i$ in descending order.
Here, $T$ $(1 \leq T \leq N)$ is a certain position in the ordered list.
 A: I'm going to rewrite the problem just a little bit, please let me know if I've changed the intent of your question.
We have a collection of $N$ random variables $X_1,\ldots,X_N$. We do not know the distribution of the $X_i$s but we do know the collection is independent and identically distributed, meaning all the $X_i$ follow the same distribution and the $X_i$ are mutually independent.
I'm going to make one assumption that I think you intended: the probability distribution of the $X_i$s is continuous which will imply $P(X_i=X_j)=0$ if $i\neq j$. This will give us that with probability $1$ all the $X_i$ will be different values. This helps avoid some extra complications with ties.
Here's the part I want you to reason out for yourself: for an arbitrary $i$ and $j$ such that $i\neq j$ what are $P(X_i< X_j)$ and $P(X_i > X_j)$? Take a pause here.

Once you accept that $P(X_i< X_j)=P(X_i > X_j)$, then you can see that they both must be $1/2$. Extending out this line of reasoning further, you'll see that all $N!$ orderings of the $N$ random variables are equally likely with probability $\frac{1}{N!}.$
With this insight, you can deduce that the probability that a specific random variable, say $X_i$, is ranked exactly $s$th is $\frac{(N-1)!}{N!} = \frac{(N-1)!}{N(N-1)!} = \frac{1}{N}$. The probability that $X_i$ is in the top $T$ can be found by summing these, so you get $\frac{1}{N}+\cdots + \frac{1}{N} = \frac{T}{N}$.
