Let we fix $0<E<1$ and an integer $t>0$.for any $n$ and for any vector $\bar{c} \in [0,1]^n$ such that $\sum_{i\in [n]} c_i \geq E \times n$$A_{\bar{c}} :=|\{ S \subseteq [n] : \sum_{i \in S}~ c_i \geq E \times t \}| \geq \binom{ E \times n}{ t }$I don't know if the statament is true o false. I think it's true.My intuition come from the observation that for the vectors $\bar{c} \in \{0,1\}^n$ (with the desidered property about the sum) we have $A_{\bar{c}}= \binom{ E \times n}{t}$; in this case we can only select subset from the set $\{ i ~|~ c_i = 1 \}$.In the others case we can create good subset (s.t. the sum is greater then $E \times t$) using the coordinate in $\{ i ~|~ c_i > E \}$ but also, maybe, using few coordinate from the set $\{ i ~|~ c_i \leq E \}$ we could create other good set!So,Prove it or find the bug! hoping that it could be a funny game for you!Motivation of the question:Suppose you have a random variable $X\in \{0,1\}^n$, a typical measure of how much randomness there is in $X$ is the min-entropy$H_{\infty}(X) = min_{x} \{ -log(Pr[X = x]) \}$In some intuitive sense the min-entropy is the worst case of the famous Shannon Entropy (that is the average case).We are interested to lowerbound the min-entropy of the random variable $(Z=X \wedge Y| Y)$ where $Y$ is uniformely distributed over the set $\{ y ~|~ \sum_i y_i = t\}$.Loosely speaking if we are lucky we can catch the bits of $X$ that have good entropy and so we if $H_{\infty}(X)\geq En$ then  $H_{\infty}(Z|Y)\geq Et$What is the probability that we are lucky?The problem is well-studied one and there exists a lot of literature,for example see Lemma A.3. in Leakage-Resilient Public-Key Cryptography in the Bounded-Retrieval Model
A simple(?) funny combinatorial problem!
co.combinatorics;it.information theory