Real life motivation. In my younger son's class, there are $18$ students. His teacher provided $18$ time slots for the parents of each child to have a 30-minute conversation of their kid's progress in school and asked the parents to provide $4$ possible time slots that would be convenient for them. I wondered how probable it is that given all the choices, the teacher can select a convenient time slot for the parents of every child so that everyone can have the "progress conversation."

Formal version. We regard every $n\in\omega = \mathbb{N}$ as an ordinal, that is $0 = \emptyset$ and $n = \{0, \ldots, n-1\}$ for $n > 0$. For $n, k \in\mathbb{N}$ we let $[n]^k$ be the collection of $k$-element subsets of $n$ (so that $[n]^k = \emptyset$ whenever $k > n$). We say that ${\cal S}\subseteq [n]^k$ has the satisfiability property if $|{\cal S}|=n$ and there is a bijection $\varphi:{\cal S}\to n$ such that for all $T\in{\cal S}$ we have $$\varphi(T) \in T.$$ We denote the selection of $n$-element collections of $[n]^k$ by $\big[[n]^k\big]^n$.

Given an integer $n>1$, what is the smallest $k$ (in terms of $n$) such that at least half of the elements of $\big[[n]^k\big]^n$ have the satisfiability property? If no exact formula can be given, is there an approximation for when $n\to\infty$?