Say $E_1,...E_n\subset\{1,2,...,k\}= K$, each $|E_i|=4$ and each $j\in K$ appear in at most $3$ sets $E_i$.

Question

Say $E_1,...E_n\subset\{1,2,...,k\}= K$, each $|E_i|=4$ and each $j\in K$ appear in at most $3$ sets $E_i$. We choose from each $E_i$ one number. Prove that we can do that so that a set of all choosen numbers has not more than ${3k\over 7}$ members.

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This was my try but the bound I get is not good and also I'm not even sure if it is correct.

We choose at random from each $E_i$ independently a number with a probability $p=1/4$ (so we can chose the same number more then once) and name this number $c_i$. Let $M$ be a set of choosen numbers and let $X=|M|$. If $X_i$ is indicator random variable for a number $i\in K$ then $$E(X) = E(X_1)+...+E(X_k)$$

Say $i$ is in a sets $E_1,...E_{d_i}$, where $d_i\leq 3$, then \begin{eqnarray}E(X_i) &=& P(X_i=1) \\ &=& P(\{i=c_1\}\cup ...\cup \{i=c_{d_i}\})\\ &=&1-P(\{i\ne c_1\}\cap ...\cap \{i\ne c_{d_i}\})\\ &=&1-P(i\ne c_1)\dots P(i\ne c_{d_i})\\ &=&1-\Big({3\over 4}\Big)^{d_i}\\ \end{eqnarray}

So we have $$E(X)= k-\sum _{i=1}^k\Big({3\over 4}\Big)^{d_i}\leq k-k\Big({3\over 4}\Big)^3$$

So $E(X) \leq {37k\over 64}$ which is not good enough.

Answer 1

I really don't think a probabilistic argument would work. Take $m \ge 1, k = 4m, n = 3m$, and $A_1,A_2,A_3 = \{1,2,3,4\}, A_4,A_5,A_6 = \{5,6,7,8\}$, etc. Then we need at most $\frac{12}{7}m$ elements chosen, so on average we need a bit less than $2$ elements chosen from a batch of $3$. I don't see how a random choice will do this; the choices of elements from $A_2,A_3$ must depend on the choice of element from $A_1$. And once we start having these kinds of dependencies, the proof becomes much more combinatorial/deterministic and falls outside what any reasonable person would call a "probabilistic proof".

Note that the construction just mentioned rules out the probabilistic approach you outlined in the question. Indeed, $E(X)$ will be more than $\lfloor \frac{3k}{7} \rfloor$ ($m=1$ is easy to compute).

In regards to the approach you outlined in an answer, it nearly certainly is just as hard as the original approach. Indeed, it definitely will be true that $P(X=n) > 0$, since a valid choice of elements, one from each $E_i$, with size at most $\frac{3k}{7}$ could be the randomly chosen set $S$. The issue is that $P(X=n)$ will be exponentially small, and thus difficult to prove is nonzero. It will also be exponentially small even if we choose $X$ a bit more wisely, by, for example, choosing $i$ to be in $S$ with probability $\frac{3k}{7}\frac{\#\{1 \le j \le n : i \in A_j\}}{4n}$. I highly doubt there is any natural choice of probabilities that will yield $P(X=n)$ being not exponentially small.

Of course, there could be a completely different approach, that one would consider "probabilistic method" that does well with the construction mentioned at the beginning of my answer. However, I view that as unlikely, but I obviously cannot be sure.

Answer 2

The 500pt belongs to me if the following proof is correct.

We give the bound $\frac{59}{140}k<\frac{3}{7}k$ by using a randomized algorithm introduced in the book ‘Probabilistic Method’ (Section 3.5) by Alon and Spencer.

Let $H=(K,E)$ be a $4$-uniform hypergraph, where $E=\{E_1,...,E_n\}$. For each $v\in K$, let $x_v$ be a uniform random label in the interval $[0,1]$, where all choices are independent. Note that with probability $1$, all labels are distinct.

The algorithm goes over the vertices by an increasing order of their labels, coloring each vertex blue, unless it is the first vertex in some $E_i$.

When the algorithm end, the number (depends on all $x_v$, $v\in K$) of red vertices $X$ is an upper bound. Let's compute $\mathbb E X$.

A vertex $v$ is colored blue if and only if it is not the first vertex in any $E_i$ containing it. A vertex $v'$ lies behind $v$ when its label $x_{v'}>x$, which is of probability $1-x$. So $$\mathbb P(v \text{ is blue })\ge \int_0^1 (1-(1-x)^3)^3dx=\frac{81}{140}.$$

So $\mathbb EX\le \frac{59}{140}k$.