Determining the distribution of results of a simple algorithm Setting
Consider repeating the following process on the numbers $N=\{1, 2, 3, \ldots, n\}$:

Pick an integer $k \in N$, uniformly at random.
Pick a subset of $k$ elements from $N$, uniformly at random.

After picking $s$ such sets, we repeatedly (until exhaustion) mark each number that is in a set whose other numbers are all marked.
Example
For example, if $N=\{1,2,3,4,5\}$ and $s=3$, we might pick the sets $s_0=\{1,3,5\}$, $s_1=\{1,4\}$ and $s_2=\{4\}$.  In the marking phase, we'd first mark $4$ (because in $s_2$ all 0 other elements are marked), then we'll mark $1$ (from set s_1), and we'll be done.
More generally, if $s=1$, then we know that $\frac{1}{N}$ fraction of times, we will have marked exactly one element with this process, and $\frac{N-1}{N}$ fraction of times we will mark no elements.
Question
Can anyone provide a closed-form expression for the distribution of the number of marked elements after this procedure for arbitrary $s$?
P.S.  I am purposefully leaving out the motivation or reference for this algorithm so as not to bias answers.  Those who recognize the motivation are kindly asked not to mention it, unless it specifically helps with the solution.
I suspect you won't get a closed form solution for the distribution you're looking for. Think of the seemingly easier problem where $k$ is always chosen to be exactly $2$, and where you get the set $\{1\}$ "for free". This problem is just like asking for a closed-form distribution of the cardinality of the connected component in the Erdos-Renyi graph $G(n,s)$ that contains a specific vertex $v_1$. I'm pretty sure that there's no known closed-form description of this distribution, and that in fact the lower order terms are poorly understood, especially around the critical value $s \sim n \log n$.