Two-of-three all-pay participation equilibrium ($q_j=\frac{w_j^{-1/(n-1)}}{{\sum }_{\ell =1}^3{w}_{\ell }^{-1/(n-1)}}$)

Consider $n\ge 2$ 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 $0\le b\le w_j-u$ 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.

If contest rewards and costs add, a fixed deviation's expected payoff depends on each rival's per-contest marginal distributions, rather than on dependence between that rival's efforts across contests. Replacing the conditional independent sampling of two active efforts by another copula with the same conditional marginals therefore preserves every deviation payoff. Consequently it preserves the Nash equilibrium, even when the new joint strategy distribution is different. This explains why unique participation probabilities and bid marginals need not imply a unique full mixed strategy.