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Contest theory studies how prize allocation rules and effort costs influence strategic effort investment, participation, and the resulting Nash equilibria or Bayesian Nash equilibria. Unlike a sale in which only the winner pays, an all-pay auction charges every participant for its effort.

We derive the probabilities for a sequential elimination all-pay contest from a discounted subgame perfect equilibrium, using backward induction, and only then take $\delta \uparrow 1$. This preserves the selection supplied by discounting.

First consider a two-player all-pay auction with effective prizes $A_1\ge A_2>0$, so the incremental payoff is $A_i$ times winning probability minus effort. For $0\le b\le A_2$, the independent mixed strategies have effort cumulative distribution functionsThe second player has an atom $1-A_2/A_1$ at zero. For positive bids on this support, the first player's payoff is $A_1{G}_2(b)-b=A_1-A_2$ and the second's is $A_2{G}_1(b)-b=0$. Bids above $A_2$ cannot improve either payoff. The first has no zero atom, so the second's zero bid also earns zero. If the first deviates to zero, it can win only when the second bids zero; even with every such tie resolved in its favor, its payoff is at most $A_1(1-A_2/A_1)=A_1-A_2$. The two-player complete-information all-pay equilibrium therefore has winning probabilitiesThese follow by integrating $G_2$ against the uniform $G_1$. A third player with effective prize at most $A_2$ cannot profit by entering: for $0<b\le A_2$, its winning probability is $G_1(b)G_2(b)\le b/A_2$, giving payoff at most zero; above $A_2$ its effort exceeds its prize.

For the dynamic induction, relabel any remaining subgame's valuations as $v_1>\cdots >v_k$, with $r<k$ prizes left. Define the backward-induction threshold in an elimination all-pay contestFor $r=1$ the sum is empty and $C_1=v_2$. The discounted continuation value in an elimination contest is obtained from the following net utilities:The base case is the one-prize all-pay auction: only the two highest valuations need positive effort, with prizes $v_1,v_2$ and payoffs $v_1-v_2,0$. Every lower player has a nonprofitable deviation by the preceding calculation.

For $r\ge 2$, if either of the top two wins, the other becomes the highest player in a subgame with $r-1$ prizes. The continuation threshold in either such subgame is the same numberThe remaining top player's continuation payoff is $v_i-D$. Its effective prize in a sequential contest, net of the discounted payoff from losing, is thereforeThus $A_1>A_2$ for $\delta <1$, and the two-player distributions above apply. Adding the losing-state baselines giveswhich agree with the proposed formulas.

For a player of rank $i\ge 3$, the inductive continuation utility after either top player wins is identical: its new rank is $i-1$, and the utility expression depends only on its own value and the lower-valued tail. Thus its current zero-effort payoff is that common continuation value multiplied by $\delta$, exactly the stated $J_i$. Its effective prize for deviating to win now is $A_i=v_i-J_i$. For $3\le i\le r$, direct subtraction givesFor $i\ge r+1$, $A_i=v_i\le v_{r+1}\le C_r$, since the coefficients in $C_r$ are nonnegative and sum to one. Such a player cannot gain by entering against the two active players. There is one additional zero-bid deviation to check for the highest player. If both active bids become zero, the tie could award a prize to any remaining player. Losing to any player below rank $2$ gives no greater continuation utility than losing to rank $2$: deleting rank $2$ makes the ordered remaining rival list componentwise smallest, and its continuation threshold is a nonnegative weighted sum of that list. Thus even resolving every all-zero tie in the highest player's favor gives payoff at most its usual losing baseline plus $A_1(1-A_2/A_1)=A_1-A_2$. No tie rule can improve this deviation. This verifies all best responses and the continuation-utility formulas. Applying the construction to every remaining-player set proves a subgame perfect equilibrium of the discounted contest, not merely an on-path prescription.

Now take the vanishing-discount limit of an elimination all-pay contest. At every nonfinal subgame,Thus the two highest remaining players win with equal probabilities in a nonfinal stage. In the final stage the effective prizes are their actual valuations, so its lower-valued participant beats the higher one with probability equal to half their valuation ratio.

Return to the original ranks. Only the original top $m+1$ can ever win; at most $m-1$ higher players can have left before the final stage, so players of rank $m+2$ or worse never enter its top pair. Player $1$ must lose $m-1$ fair nonfinal contests to remain unawarded until the final stage, where its opponent has value $v_{m+1}$. HenceA player $2\le i\le m$ first enters the top pair at stage $i-1$, after $i-2$ higher-ranked winners have left. To remain unawarded, it must lose the $m-i+1$ nonfinal contests from then on, followed by the final contest against rank $m+1$. ThereforeThere are exactly $m$ distinct winners, so the expected value of their indicator sum is $m$. Combining these calculations givesThis proves the ranked winning probabilities in an undiscounted elimination contest. When $m=1$, it reduces to the ordinary two-player all-pay winning probabilities; when $m>1$, the repeated fair stages explain each power of $1/2$.

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 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 isAn 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 payoffEvery $q_j$ 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:Conditional on entering contest $j$, the effort has distribution functionextended by zero below this interval and one above it. The inequalities $q_1\le q_2\le q_3$ 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.

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. 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.

A prize allocation rule assigns winning probabilities or divisible prize shares to a profile of efforts. In an all-pay auction, the greatest effort wins; a proportional allocation contest instead assigns shares continuously according to relative effort. The rule is essential data of a contest model.

For positive total effort, a proportional allocation contest awards player $i$ the prize with probability $b_i/{\sum }_j{b}_j$, or gives it that proportion of a divisible prize. A convention is needed at zero total effort. This allocation is different from the discontinuous highest-effort rule of an all-pay auction.