Let $G = (E,V)$ be an undirected graph (which can have multiple edges or loops).

Let $k,l,m\colon E\to \mathbb{R}_{\geq 0}$ be three edge-weight functions that satisfy $2k(e) + l(e) + m(e) = 1$ for all edges $e \in E$. Let $D_G(k,l,m) = (V,A)$ be the random digraph obtained from $G$ in the following manner: independently for each edge $e = \{u,v\} \in E$, remove $e$ and replace it with:

- the arc $(u,v)$ with probability $k(e)$;
- the arc $(v,u)$ with probability $k(e)$;
- nothing with probability $l(e)$;
- both arcs $(u,v)$ and $(v,u)$ with probability $m(e)$.

For a choice of root $q \in V$, we say a digraph $D = (V,A)$ is *$q$-connected* if there is a directed path from $q$ to every vertex $v \in V\setminus \{q\}$.

**Proposition:** $\mathrm{Pr}(D_G(k,l,m) \textrm{ is $q$-connected}) = \mathrm{Pr}(D_G(k',l',m') \textrm{ is $q'$-connected})$ if $k(e) + l(e) = k'(e) + l'(e)$ for every edge $e \in E$.

I know that this proposition is true; it follows from the discussion in Remark 5.4 of this paper. Basically the reason it is true is because both probabilities can be computed via a deletion-contraction recurrence.

But I would like to know if there is a simpler/more conceptual reason why this probability should be independent of the particular distribution $(k,l,m)$ so long as $k(e) + l(e)$ is fixed. (It is pretty easy to see, via ``path-reversals'', why it is independent of the choice of root $q$.)

Here is one heuristic reason why we might expect this probability to depend only on $k(e) + l(e)$. Fix a root $q \in V$. Let us say a cut $C = \{U,W\}$ of $G$ is *bad* for a digraph $D = (V,A)$ if $q \in U$ and there are no arcs from $U$ to $V$ in $D$. Observe that $D$ is $q$-connected iff it has no bad cuts. And the probability a given cut $C$ is bad for $D_G(k,l,m)$ is $\prod_{e \in \mathrm{cutset}(C)} (k(e) + l(e))$. But this is not a rigorous proof that $k(e) + l(e)$ is all that matters because clearly the events of $C$ and $C'$ being bad for $D_G(k,l,m)$ are not independent.