= Two-of-three all-pay participation equilibrium
{title2=$q_j=\dfrac{w_j^{-1/(n-1)}}{\sum_{\ell=1}^3w_\ell^{-1/(n-1)}}$}
Consider $n\geq2$ identical players, prizes $w_j>0$, unit effort costs, and exactly two entries per player. The symmetric equilibrium omission probabilities $q_j$ sum to one. The common per-contest payoff is $u=(\sum_j w_j^{-1/(n-1)})^{-(n-1)}$, and the unconditional rival distribution on $0\leq b\leq w_j-u$ is $G_j(b)=((b+u)/w_j)^{1/(n-1)}$. Conditional on entry, its <distribution function> is $(G_j(b)-q_j)/(1-q_j)$. Choosing the omitted contest according to $q$ and sampling the two active efforts independently gives a <symmetric equilibrium> with expected payoff $2u$. These formulas determine the marginal laws; they do not determine the dependence between a player's two efforts.
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