Edge-inclusion formula for a uniform spanning tree ($\mathbb{P}(e\in T)=c_e{R}_{\mathrm{eff}}(a,b)$)

For a finite electrical network with positive electrical conductances $c_e$, choose a spanning tree with lawThe one-edge case of the transfer-current theorem gives $\mathrm{Pr}(e\in T)=c_e{R}_{\mathrm{eff}}(a,b)$ 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 $c_e{R}_{\mathrm{eff}}(a,b)$.

In a triangle of unit resistors, one resistor is parallel to a series pair, so $R_{\mathrm{eff}}=1\cdot 2/(1+2)=2/3$. 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.