Expected value of largest integer in a draw Suppose I pick $k$ integers without replacement from $\{1, \ldots, n\}$. Let $I$ be the value of the highest integer. A calculation with binomials reveals $$E[I] = \frac{k}{k+1}(n+1)$$  This is a very simple formula - does it have a simple calculation-free proof? 

Picking $k$ integers without replacement from $\{1, \ldots ,n\}$ involves breaking the interval $[0,n+1]$ into $k+1$ pieces where the length of each piece has the same distribution.  
So the expected length of each piece (including the piece from the highest sampled integer to $n+1$) is $\dfrac{n+1}{k+1}$ and so the expected value of the highest sampled integer is $n+1 - \dfrac{n+1}{k+1}$.  
Simple calculation will give your result. 
So in general, the expected value of the $j$th sampled integer (counting from the bottom) is $j \dfrac{n+1}{k+1} $.
