As background: there are $k!$ strict orders on the set $K = \left\{ 1, \ldots k \right\}$. Under symmetry, each order occurs with probability $\tfrac{1}{k!}$.
Now consider the set of weak orders on $K$. When $k=2$, this has three elements: $\left( 1|2 \right)$, $\left( 2|1 \right)$ and $\left( 12 \right)$. For general $k$, its cardinality is the $k^{\text{th}}$ Fubini or ordered Bell number, and is indexed by the OEIS as A000670.
I want a notion of symmetry for this set of weak orders. Specifically, I want it to have the property that the probability factor associated with each term in an order depends only on the number of elements in that term (e.g. one or two in the $k=2$ example above).
Concretely: let each $i \in K$ be uniformly and independently distributed over a set of $n$ 'slots'. Thus:
- when $n=1$, the element $\left\{ \left\{ 1, \ldots, k \right\} \right\}$ has probability one: all elements of $K$ are assigned to the same slot.
- when $n \rightarrow \infty$, we collapse to the set of strict orders, each of which with probability $\tfrac{1}{k!}$: each element of $K$ is assigned to a unique slot.
I would like to know these probabilities for any positive integers $k$ and $n$.
For example:
when $k=2$, $p_{(12)} = \tfrac{1}{n}$, while $p_{(1|2)} = p_{(2|1)} = \tfrac{1}{2n} (n-1)^+$, where $x^+ = \max \left\{ 0, x \right\}$; and
when $k=3$, $p_{(123)} = \tfrac{1}{n^2}$, $p_{(1|23)} = p_{(12|3)} = \tfrac{1}{2n^2} (n-1)^+$ and $p_{(1|2|3)} = \tfrac{1}{6n^2} (k-1)^+ (k-2)^+$.