= Return probability from effective resistance
{title2=$P_x(T_x^+<T_y)=1-(d(x)R(x,y))^{-1}$}
For the conductance-weighted <random walk on a graph>, set $d(x)=\sum_vc_{xv}$. The voltage $h(u)=P_u(T_x<T_y)$ has boundary values one and zero. The total <electric current> is $\sum_vc_{xv}(1-h(v))=1/R(x,y)$. The <first-step analysis> therefore gives the displayed positive-return-time formula. Wiring the exterior of finite sets and taking limits gives $P_x(T_x^+=\infty)=1/(d(x)R(x,\infty))$.
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