# Karger's min-cut (contraction): Combinatorial argument for success probability?

The contraction algorithm for min-cut is: pick an edge $$(u,v)$$ uniformly at random, and "contract" it by merging $$u$$ and $$v$$ into a single vertex, deleting self-loops. Continue until two vertices remain; return this cut.

The probability of finding the min cut can be shown to be at least

$$\frac{1}{{n \choose 2}} .$$

This feels like it should have some clever combinatorial interpretation or proof. (Something that sounds like e.g. "we place the vertices in a uniformly random order and if the last pair is correct, the algorithm succeeds.") But every proof I have found involves brute-forcing the product $$\frac{n-2}{n} \frac{n-3}{n-1} \cdots \frac{1}{3} = \frac{2}{n(n-1)} .$$

Is there any direct combinatorial argument that the probability exceeds $${n \choose 2}^{-1}$$?