= Edge-inclusion formula for a uniform spanning tree
{title2=$\mathbb P(e\in T)=c_eR_{\mathrm{eff}}(a,b)$}
For a finite <electrical network> with positive <electrical conductances> $c_e$, choose a <spanning tree> with law
$$
\Pr(T)=\frac{\prod_{e\in T}c_e}{\sum_{S\text{ spanning tree}}\prod_{e\in S}c_e}.
$$
The one-edge case of the <transfer-current theorem> gives $\Pr(e\in T)=c_eR_{\mathrm{eff}}(a,b)$ for $e=\{a,b\}$. A unit injected <electric current> produces voltage $R_{\mathrm{eff}}(a,b)$, and <Ohm's law> makes the <electric current> along $e$ equal to $c_eR_{\mathrm{eff}}(a,b)$.
In a triangle of unit resistors, one resistor is parallel to a series pair, so $R_{\mathrm{eff}}=1\cdot2/(1+2)=2/3$. Each <edge> consequently has inclusion <probability> $2/3$, also clear from the three equally likely <spanning trees>. For a <wired uniform spanning forest>, take the finite wired limit and use the limiting wired <effective resistance>.
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