Take a random $n \times n$ nonnegative symmetric matrix $D$ with zero diagonal. What is the probability that it is an abstract distance matrix, i.e. satisfies $D_{xy}+D_{yz} \geq D_{xz}$ for all index triples $x,y,z$?
In other words: what is the probability that a nonnegative function on $V \times V$, where $V$ is some finite set, defines a finite metric space?
Since you haven't given a distribution, let me make an observation giving the right form of the answer in the case where the $D_{xy}$ are independent uniform $[0,1]$ random variables.
I want to claim that the probability, $p$, that the matrix defines a metric satisfies $\alpha^{n^2}\le p\le \beta^{n^2}$ for constants $\alpha$ and $\beta$.
First, notice that if all the $\binom n2$ edge lengths are in $[\frac12,1]$, then the triangle inequality is satisfied, so that $p\ge 2^{-\binom n2}$.
Conversely, consider the triples $(b,b+a,b+2a)$ where $a$ is an odd number in the range $1$ to $\frac n4$ and $1\le b\le a$. Notice that no two of these triples contain two elements in common. There are $\Theta(n^2)$ such triples. For such a triple, consider the event $E_{a,b}$ that $D_{b,b+2a}\le D_{b,b+a}+D_{b+a,b+2a}$. These events all have the same probability that is strictly less than 1. They are also independent, since no edge occurs in two events. Hence we obtain the upper bound.
