= Solution
Interpret the contests as standard <all-pay auctions>: the highest effort among entrants wins, ties are shared uniformly, and an unentered contest awards no prize. The printed question does not specify a <prize allocation rule>; the highest-effort convention is the one used for standard all-pay contests in https://www.slideshare.net/slideshow/crowdsourding-and-allpay-contests/41832099[the course author's 2014 lecture slides]. Under this convention, we can characterize the unique symmetric participation probabilities and bid marginals. Uniqueness of the entire joint <mixed strategy> requires a further restriction on dependence, as explained below.
Let $q_j$ be the probability that a player omits contest $j$. Since every player enters exactly two contests, $q_1+q_2+q_3=1$. Let $G_j(b)$ be the probability that a rival is absent from contest $j$ or enters it with effort at most $b$. Rivals' strategy draws are independent between players. For a positive bid $b$ outside a <measure atom>, the expected payoff from this contest is
$$
w_jG_j(b)^{n-1}-b.
$$
An <all-pay auction> cannot have a positive-effort <measure atom> in a symmetric equilibrium: slightly overbidding that <measure atom> gives a positive discrete increase in the <winning probability> at an arbitrarily small extra cost. Nor can active bids have a <measure atom> at zero when entry has positive probability, because a small positive bid beats tied zero bids. Gaps inside the active <effort support> are impossible: moving a bid from the top of a gap to just above its bottom preserves its winning probability and lowers its cost. The <effort support> starts at zero, because lowering its positive lower endpoint would preserve the chance that every rival is absent.
The <all-pay indifference equation with random entry> consequently gives the maximal per-contest payoff
$$
u_j=w_jq_j^{n-1},\qquad
G_j(b)=\left(\frac{b+u_j}{w_j}\right)^{1/(n-1)},
\quad0\leq b\leq w_j-u_j.
$$
Every $q_j$ lies strictly between zero and one. First, if $q_j=1$, the other two contests have certain entry and zero per-contest payoff, while deviating into the unused contest wins a positive prize. This contradicts equilibrium. Next, if $q_j=0$, the remaining omission probabilities sum to one and neither can equal one, so both are positive. Contest $j$ has zero per-contest payoff, whereas both other contests have strictly positive payoffs. Every pair containing $j$ is then worse than omitting it and entering the other two, contradicting certain entry in $j$.
Every omission therefore occurs with positive probability. The three entered pairs must give the same maximal payoff, which forces $u_1=u_2=u_3=u>0$. Normalizing the omission probabilities gives the <two-of-three all-pay participation equilibrium>:
$$
\boxed{u=\left(\sum_{\ell=1}^3w_\ell^{-1/(n-1)}\right)^{-(n-1)},\qquad
q_j=\left(\frac u{w_j}\right)^{1/(n-1)}
=\frac{w_j^{-1/(n-1)}}{\sum_{\ell=1}^3w_\ell^{-1/(n-1)}}.}
$$
Conditional on entering contest $j$, the effort has <distribution function>
$$
\boxed{H_j(b)=\frac{((b+u)/w_j)^{1/(n-1)}-q_j}{1-q_j},
\quad0\leq b\leq w_j-u,}
$$
extended by zero below this interval and one above it. The inequalities $q_1\leq q_2\leq q_3$ show that the larger prizes are entered more often.
To construct an equilibrium, omit $j$ with probability $q_j$, then draw the two active efforts independently with their respective <conditional distributions> $H_k$. Every bid in a contest's <effort support> earns $u$; a bid above $w_j-u$ earns at most $w_j-b<u$. Thus no effort deviation or choice of a different pair improves on total payoff $2u$. This proves existence and verifies the <Nash equilibrium> without relying only on the indifference equations. The arguments above also prove uniqueness of the omission probabilities and the per-contest <marginal distributions>.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/iii/paper-42-entry-and-effort.png]
{title=Equilibrium omission probabilities and conditional effort distributions for three all-pay contests}
{height=400}
For the full joint strategy law, however, the printed uniqueness claim is too strong. Given an entered pair $k,\ell$, either use two independent <uniform random variables> $U,V$ and bids $(H_k^{-1}(U),H_\ell^{-1}(V))$, or use a single uniform $U$ and bids $(H_k^{-1}(U),H_\ell^{-1}(U))$. These are different <copulas> with the same conditional marginals. A fixed deviation's expected additive payoff only uses the rivals' per-contest <marginal distributions>, so both constructions remain <Nash equilibria>. This is the <marginal-equivalent equilibria in additive contests> phenomenon. \b[The participation probabilities and bid marginals are unique; the full joint mixed strategy is not unique unless a dependence convention is imposed.] Independent conditional sampling gives one canonical representative.
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