Suppose I have 100 integers and I sample 10 without repetition. What is the expected rank of the lowest out of 10 samples? Suppose I have 100 integers and I sample 10 without replacement. What is the expected rank of the lowest out of 10 samples? i.e. my lowest integer in the 10-sample is kth smallest out of 100. what is k?
How would the answer change if I sampled with replacement?
 A: To obtain an answer we must know how many ties there are among these $100$ integers and where they occur: that's too complicated and likely is not the intent of the question.  (Nevertheless, the techniques applied below continue to work.)  Henceforth, then, suppose all these integers are unique.  Without any loss of generality they might as well be the ordered set of their ranks, $1 \lt 2 \lt \cdots \lt 100.$
Let $S(k)$ be the chance that the smallest element of the sample is $k$ or larger. These events correspond to samples of $\{k, k+1, k+2, \ldots, 100\},$ a set with $101-k$ elements.

*

*When sampling without replacement, these constitute $\binom{101-k}{10}$ out of the $\binom{100}{10}$ equiprobable samples.


*When sampling with replacement, they constitute $(101-k)^{10}$ out of the $100^{10}$ equiprobable samples.
The expectation is the sum of the $S(k)$ beginning with $k=1.$  The two answers therefore are
$$\sum_{k=1}^\infty \frac{\binom{101-k}{10}}{\binom{100}{10}} = \frac{101}{11}$$
(without replacement) and
$$\begin{aligned}
&\sum_{k=1}^{100} \frac{(101-k)^{10}}{100^{10}} \\&=\frac{6(100)^{10} + 33(100)^9 + 55(100)^8 - 66(100)^6 + 66(100)^4 - 33(100)^2 + 5}{66(100)^9}
\end{aligned}$$
(with replacement).
As decimals they equal $9.\overline{18}$  and $9.599241\ldots,$ respectively.
