Cover Story: Suppose there are $n$ balls in an urn that are numbered from $1$ to $n$. You draw $k \leq n$ balls from this urn without putting them back. Let $X_1, ..., X_k$ be the numbers of the $k$ balls that are drawn and $Y =\min \{X_1, ... , X_n\}$.
I'm interested in (a nice expression for) $\mathbb{E}[Y]$.
If I use the brute-force standard approach, I get
$\mathbb{P}[Y > i] = \mathbb{P}[ X_1 > i, ..., X_k > i] = \frac{n-i}{n} \cdot \frac{n-i-1}{n-1}\cdot ... \cdot\frac{n-i-k+1}{n-k+1} = \frac{(n-i)!(n-k)!}{n!(n-i-k)!}$
for $i \leq n-k$ and otherwise $0$. This determines the law of $Y$ but I'm not able to simply the sum afterwards to obtain a nice formula for the expectation.
But I suppose that there is one because a reasonable source stated (without proof) that
$\mathbb{E}[Y] = \frac{n + 1}{k+1}$
which can be easily checked for $n \leq 2$.
We have $$\begin{align} E(Y) &= \sum_{i \ge 0} P(Y>i) \tag{1}\\ &= \sum_{i=0}^{n-k} \frac{\binom{n-i}{k}}{\binom{n}{k}} \\ &= \frac{1}{\binom{n}{k}} \sum_{i=0}^{n-k} \binom{n-i}{k} \\ &= \frac{1}{\binom{n}{k}} \sum_{j=k}^n \binom{j}{k} \\ &= \frac{1}{\binom{n}{k}} \binom{n+1}{k+1} \tag{2}\\ &= \frac{n+1}{k+1} \end{align}$$ $(1)$ is true for any discrete random variable $Y$ which is always non-negative.
$(2)$ is by the Hockey-Stick Identity.