Permanent as probability of distinct independent choices ($\mathbb{P}(X_1,\dots ,X_n\text{ all distinct})=\mathrm{perm}P$)

Let $X_1,\dots ,X_n$ be independent random variables, each choosing a label from $\{1,\dots ,n\}$. Put their probabilities in the matrix $P=(p_{ij})$, where $p_{ij}=\mathbb{P}(X_i=j)$. Each row sums to one. All choices are distinct precisely when their ordered list is one of the permutations of the $n$ labels. These are mutually exclusive events, and independence makes the probability of one list $\sigma$ equal to ${\prod }_i{p}_{i,\sigma (i)}$. Adding the list probabilities gives the permanent of a matrix:In particular, this permanent lies between zero and one.

For a two-person example,The distinct outcomes $(1,2)$ and $(2,1)$ have probabilities $3/5$ and $1/20$. ThereforeThe colliding outcomes $(1,1)$ and $(2,2)$ have probabilities $3/20$ and $1/5$, totaling $0.35$. All four outcomes are shown below; their probabilities sum to one.

If every choice has the uniform distribution on a finite set of $n$ labels, then $p_{ij}=1/n$ and the distinct-choice probability is $n!/n^n$. For three choices and three labels it is $6/27=2/9$. This counts a subset of all $n^n$ ordered outcomes, not independent pairwise collision events. Independence of the original choices is essential: without it, the individual rows of $P$ do not determine the joint probabilities.