Let's say we have a finite set $S$ where $|S|=n$ and we take its powerset $X = \mathcal{P}(S)$ to generate a measurable space $(X, \mathcal{P}(X))$. From this we consider the probability space $(X, \mathcal{P}(X), \mathbb{P})$ where $\mathbb{P}$ is a uniform probability measure. What is the probability that a uniformly sampled $A \in \mathcal{P}(X)$ is $\sigma$-algebra on S?
My intuition is that the probability will vanish as $n \rightarrow \infty$, but for finite sets it should be non-zero since there exists members of $\mathcal{P}(X)$ that are measurable.
It is $$ p_n:=\frac{B_n}{2^{2^n}}, $$ where $B_n$ is the $n$-th Bell number. (See this question for details.)
Using Theorem 2.1 in this paper, $$ p_n\le \frac{1}{2^{2^n}}\left(\frac{0.792n}{\ln(n+1)}\right)^n\to 0 $$ as $n\to\infty$.