Expected normalized maximum: expectation of a ratio $\ell_\infty/\ell_1$. 
Let $\alpha_1,\dots,\alpha_k > 0$, and $1\leq s\leq k$ be an integer. Suppose $S\subseteq[k]$ is a random subset chosen uniformly among all subsets of $[k]$ of size $s$.
  Is there anything known about the quantity
  $$
\mathbb{E}_S \frac{\max_{i\in S} \alpha_i}{\sum_{i\in S}\alpha_i}
$$
  as a function of $s$ and say the various $\ell_p$ norms of the vector $\alpha$?

Equivalently, if one renormalizes $\alpha$ to get a probability distribution $p$ over $[k]=\{1,2,\dots,k\}$, and denote by $p_S$ the conditional distribution induced by $p$ on $S$, this is asking about $\mathbb{E}_S \lVert p_S\rVert_\infty$.
I have tried to analyze this, but the denominator makes every attempt either go haywire or seem way too contrived to continue. I feel it should either be known or have an elegant solution, however.
A positive answer (depending on the answer, of course, but things are what they are) would possibly greatly simplify a proof I am working on.
 A: Assuming (without loss of generality) that $\sum_{i=1}^n \alpha_i=1$, you can define $V= Y/X$ with $Y = \max_{i \in S} \alpha_i$ and $X = \sum_{i=1}^k \alpha_i I_i$ with $I_i$ an indicator function that is $1$ if $i \in S$, zero else.  Then $1\geq V\geq 1/s$ always, and by Jensen's inequality for the concave function $\log(\cdot)$: 
\begin{align}\log(E[V]) &\geq E[\log(V)] \\
&= E[\log(Y)] - E\left[\mbox{$\log\left(\sum_{i=1}^k \alpha_i I_i\right)$}\right]\\ 
&\geq E[\log(Y)] - \mbox{$\log(\sum_{i=1}^k \alpha_i E[I_i])$} \\
&= E[\log(Y)] - \log(s/k)
\end{align} 
So 
$$1\geq E[V] \geq \max\left\{(k/s)\exp(E[\log(Y)]), 1/s\right\}$$

For a simple example with $k=4, s=2$: 
$$ \{\alpha_i\} = \{1/8, 1/8, 1/4, 1/2\}$$
$$E[\log(Y)] = (1/6)\log(1/8) + (2/6)\log(1/4) + (3/6)\log(1/2) $$
So 
$$ E[V]\geq \max\left\{(k/s)\exp(E[\log(Y)]), \underbrace{1/s}_{0.5}\right\} = 0.6299605249474366$$ 
But the exact answer is 
\begin{align}
E[V] &= \frac{1}{6}\left[\frac{1/8}{1/8+1/8} + \frac{1/4}{1/8+1/4} + \frac{1/2}{1/8+1/2}\right] \\
&\quad +\frac{1}{6}\left[ \frac{1/4}{1/8+1/4}+\frac{1/2}{1/8+1/2}+\frac{1/2}{1/4+1/2}\right]\\
&=41/60 \\
&= 0.6833333333
\end{align}

Assuming $\alpha_1 \leq \alpha_2 \leq ... \leq \alpha_k$ we get
$$ E[\log(Y)]= \frac{1}{{k}\choose{s}}\sum_{i=s}^k \log(\alpha_i){{i-1}\choose{s-1}} $$
and so 
$$ \exp(E[\log(Y)]) = \prod_{i=s}^k \alpha_i^{{{i-1}\choose{s-i}}/{{k}\choose{s}}}$$

Other simple bounds are: Assume $\alpha_1\leq \alpha_2\leq...\leq \alpha_k$ and define $z = \alpha_1 + ... + \alpha_{s-1}$. Then:
$$ \frac{1}{{k}\choose{s}}\sum_{i=s}^k \left(\frac{\alpha_i}{\sum_{j=i-s+1}^i \alpha_j}\right) {{i-1}\choose{s-1}}\leq   E[V] \leq \frac{1}{{k}\choose{s}}\sum_{i=s}^k \left(\frac{\alpha_i}{\alpha_i+z}\right) {{i-1}\choose{s-1}} $$
For the example $\{1/8, 1/8, 1/4, 1/2\}$ these bounds give
$$ 0.63888888888 \leq E[V]\leq 0.705555555 $$
