= Total-effort formula for a proportional contest with outside effort
{title2=$H_k(R+\delta)^2-(k-1)(R+\delta)-\delta=0$}
With $k>0$ active players of <harmonic mean> valuation $\bar v_k$, unit-cost <Nash equilibrium> effort satisfies $R+\delta=\frac{\bar v_k}{2k}[(k-1)+\sqrt{(k-1)^2+4k\delta/\bar v_k}]$. Sum the active-player first-order equations $b_i=(R+\delta)(1-(R+\delta)/v_i)$ to obtain the quadratic. If $\delta$ is at least the largest valuation, no one is active and $R=0$. With no outside effort, at least two players must be active.
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