Conjecture: Let $N$ be a non-empty and finite set. There exists a probability distribution $p$ on the set of all partitions of $N$, $Z(N)$, such that $$\sum_{P\in Z(N):S\in P}p(P)= {1 \over n\cdot \binom{n-1}{s-1}}$$ for all $S\subseteq N:S\ne \emptyset$, where $p(P)$ denotes the probability of partition $P$, $n=|N|$, $s=|S|$.
Trivial? Known? Drops from deeper results? Any idea?
Recall that the number of permutations $\sigma$ of a set $N$ of cardinality $n$
for which a given non-empty subset $S\subset N$ of cardinality $s$ is an orbit (that is a $\sigma$-invariant subset generated by some element $x\in N$, whose complement is also $\sigma$-invariant) is, of course, just $(s-1)!(n-s)! $.
So we may interpret the ratio $$\frac{1} {n {n-1 \choose s-1}}=\frac{(s-1)!(n-s)!}{n!}\qquad\qquad{\bf (1)}$$ as the probability that the subset $S$ be an orbit of a permutation of $N$, uniformly chosen at random.
Therefore, a probability distribution on $Z(N)$ that fulfills your requirement is just the law of the map $\operatorname{Orb}: {\frak S}(N)\to Z(N)$ (that takes a permutation into its orbit partition) seen as a random variable based on ${\frak S}(N)$, with its uniform probability distribution. In particular we have, for $ P\in Z(N)$ $$p(P)=\frac{1}{n!} \prod_{S\in P}(|S|-1)! \qquad\qquad{\bf (2)}$$
Also note that $p$ is unique, under the obvious independence and symmetry assumptions that allow to deduce formula (2) .
