Return probability from effective resistance ($P_x(T_x^{+}<T_y)=1-(d(x)R(x,y){)}^{-1}$)

For the conductance-weighted random walk on a graph, set $d(x)={\sum }_v{c}_{xv}$. The voltage $h(u)=P_u(T_x<T_y)$ has boundary values one and zero. The total electric current is ${\sum }_v{c}_{xv}(1-h(v))=1/R(x,y)$. The first-step analysis therefore gives the displayed positive-return-time formula. Wiring the exterior of finite sets and taking limits gives $P_x(T_x^{+}=\infty )=1/(d(x)R(x,\infty ))$.