Multi-Battle Contests, Finite Automata, and the Tug-of-War

This paper examines multi-battle contests whose extensive form can be represented in terms of a finite state machine. We start by showing that any contest that satisfies our assumptions decomposes into two phases, a principal phase (in which states cannot be revisited) and a concluding tie-breaking phase (in which all non-terminal states can be revisited). Degenerate cases are the finite-horizon contests on the one hand (e.g., the match race), and the tug-of-war on the other. Next, assuming a probabilistic technology in each battle, we show that any contest satisfying our assumptions, with either finite or infinite horizon, admits a unique symmetric and interior Markov perfect equilibrium. This entails, in particular, a complete characterization of the equilibrium in the tug-of-war. Finally, we explore, both analytically and numerically, the intricate problem of a contest designer that maximizes expected total effort. In the absence of a complexity constraint, the revenue-maximizing contest is always a match race, where the optimal length of the race increases as the technology of the component contest becomes more noisy. If, however, the complexity constraint is binding, then the optimal contest is typically (but not always) a tug-of-war.


Introduction
In a large variety of environments, dynamic contests arise as a natural form of interaction between two competing parties (Fudenberg et al., 1983;Harris and Vickers, 1987;Roberts and Samuelson, 1988;Hörner, 2004;Konrad andKovenock, 2005, 2009;Agastya and McAfee, 2006;Klumpp and Polborn, 2006;Doraszelski and Markovich, 2007;Fu et al., 2015;Häfner, 2017;Karagözoglu et al., 2018).Given the temporal dimension of the interaction, the winner of the bilateral competition is not determined immediately, but through a potentially inde…nite sequence of individual battles.While the literature has focused mainly on the important special cases of the match race and the tug-of-war, a dynamic contest may in general be governed by a more elaborate set of rules.For example, while initially, a start-up may be able to use its seed capital for cost-intensive R&D, it may be forced to give up at a later stage if lagging behind too much.Similar forms of dynamic contests, in which the conditions for winning change over time, can be found, for instance, in prolonged political con ‡ict (e.g., Organski and Lust-Okar, 1997), biological struggles (e.g., Reeve et al., 1998), and warfare (e.g., Edwards, 2000).
Still, a ‡exible way to model general dynamical con ‡icts has, to our knowledge, been lacking in the literature.Moreover, the way in which contestants'incentives for e¤ort provision depend on the "rules" of the con ‡ict is, overall, not particularly well-understood.
In this paper, we consider a class of dynamic games in which two players repeatedly face each other in the same type of contest.Following the literature, we will refer to the individual encounter as a component contest (or battle), and to the overall interaction as a (multi-battle) contest.It will be assumed that, at each stage of the contest, the history of victories in prior battles determines if the contest ends and is won by one of the players, or if the contest continues.If the contest ends, then the prize is allocated.Extending the before-mentioned literature on dynamic contests, we assume that the contest rules can be de…ned in terms of a …nite state machine (Moore, 1956;Rubinstein, 1986;Salant, 2011).Since the class of such rules is still quite ‡exible (in particular, too ‡exible to guarantee the existence of an equilibrium), we impose three assumptions.These assumptions, referred to as Exchangeability, Monotonicity, and Centeredness, actually impose some structure on the multi-battle contest.Indeed, as our …rst main result shows, any contest that satis…es our assumptions decomposes naturally into two phases.In the principal phase that is entered into at the start of the contest, every state is visited at most once.In the subsequent tie-breaking phase, however, every non-terminal state
Following the ground-breaking work by Fudenberg et al. (1983), a key contribution to the literature has been Harris and Vickers (1987) who studied the match race and the tug-of-war assuming that each battle is a probabilistic contest as in Lee and Wilde (1980).They showed existence of a Markov Perfect Equilibrium for the tug-of-war, and also o¤er conditions su¢ cient for uniqueness. 1 Below, we build on their insights, yet using a di¤erent contest technology.Grossman andShapiro (1987, …rst draft 1985) considered a related model of a race with uncertainty, yet restricted attention to the case of twostages.Judd (2003Judd ( , likewise, …rst draft 1985) ) o¤ers an interesting model of a multi-stage race in which stochastic progress is measured on a continuous scale.Taylor (1995) allows for incomplete information in a dynamic research tournament.
After this initial period, both the match race and the tug-of-war have been studied quite thoroughly under the assumption of a deterministic contest technology.In this case, tie-breaking rules may matter for the equilibrium outcome.Stipulating that the player with the positive continuation value wins, Konrad and Kovenock (2005) identi…ed an intuitive equilibrium outcome in which positive levels of e¤ort are expended only at the initial state.In contrast, Agastya and McAfee (2006), see also McAfee (2000), assumed that a state is revisited if both e¤orts are zero, and showed that a stalemate of this type may occur in a Markov perfect equilibrium.In an applied variant of the tug-of-war, McBride and Skaperdas (2006) assumed that an additional bias supports the initial victor.Fu et al. (2015) point out that dynamic team contests behave entirely di¤erent than contests between individuals.Häfner (2017) considered a tug-of-war between groups, where in each stage, a pair of agents is matched under incomplete information.He …nds a unique Markov-perfect equilibrium with interesting comparative statics.Karagözoglu et al. (2018) were …rst in proving existence and uniqueness of a Markov perfect equilibrium of a tug-of-war with a lottery technology and quadratic costs, which corresponds in our setting to the special case of a Tullock contest with parameter R = 1 2 . 2 Correspondingly, our Theorem 2, including also the similarity principle for continuation payo¤s, extends their equilibrium characterization 1 See also the working paper version (Harris and Vickers, 1986), as well as the stochastic-process di¤erence-form formulation of the tug-of-war (Budd et al., 1993).
2 Karagözoglu et al. (2018) multiply the …rst-order conditions with each other and solve for the equilibrium e¤ort.As they note, however, that trick works only for the case of quadratic costs.Karagözoglu et al. (2018, p. 22) conclude: "Whether our results generalize to an impact function of the form, f (x) = x r , where r < 1 or 1 < r 2 is far from trivial, and hence left as an open question."Sincetheir r corresponds to 2R in our notation, Section 4 of the present paper addresses precisely this open research question.in a number of ways.Klumpp and Polborn (2006) considered races with Tullock technology.Assuming a deterministic contest technology, Konrad and Kovenock (2009) studied races with heterogeneous deadlines, a potential headstart for one player, and positive intermediate prizes.Multi-player races have been considered by Do¼ gan et al. (2018).None of these contributions, however, uses …nite automata to de…ne a ‡exible class of dynamic multi-battle contest structures.Neither does any of these contributions characterize the Markov perfect equilibrium in the tug-of-war for a general class of contest technologies. 3  In our setting, the e¤ort provision is the lower the more unbalanced the state.This is in line with contest models in the tradition of Dixit (1987), for instance.However, in a framework with in…nitely many stages and no premature end, Hörner (2004) shows that the leader in a highly unbalanced state may exert excessive e¤ort so as to keep the follower at a safe distance.
The type of …nite state machine introduced below is known as Moore (1956) machines. Rubinstein (1986) used …nite automata to model the strategy choice of boundedly rational players in in…nitely repeated non-cooperative games.Salant (2011) employed …nite automata to model boundedly rational choice.In contrast to those papers, we use the Moore machine to model a system of rules for a multibattle contest.

Overview over the paper
The remainder of this paper is structured as follows.Multi-battle contests are introduced in Section 2. Section 3 presents the characterization result for multi-battle contests that satisfy our de…nitions.
The tug-of-war is discussed in Section 4. Section 5 deals with the existence of a unique Markov perfect equilibrium.Revenue-maximizing contests are studied in Section 6. Section 7 concludes.The analysis of the tug-of-war and other technical derivations have been relegated to an Appendix.

Finite state machines
Two risk-neutral contestants i 2 f1; 2g repeatedly face each other in the same type of component contest (or battle), until the rules of the game determine that the contest ends, in which case one of the players is declared the winner and the other is declared the loser.The winner of the multi-battle contest receives 3 Dynamic contests are nice objects for experimental analysis as well.Zizzo (2002) tests the race (best-of-19) in the lab.Deck and Sheremata (2015) o¤er an experimental analysis of the tug-of-war with all-pay technology.Mago and Sheremeta (2017) implement the race (best-of-three) as an experiment with the all-pay auction as CSF.Dechenaux et al. (2015) survey the literature.a prize of value V > 0, while the loser obtains no prize.Without loss of generality, the prize will be normalized to one, i.e., V = 1.As mentioned in the Introduction, we will consider multi-battle contest rules that may be expressed in terms of a …nite deterministic state machine.
De…nition 1.A multi-battle contest M is composed of: (i) a …nite set of states , including an initial state !init 2 ; (ii) a partition = act [ 1 [ 2 of into three pairwise disjoint and nonempty subsets act , 1 , and 2 ; and (iii) a transition map : act f1; 2g ! .
Any state ! 2 act will be referred to as an active state, and any state ! 2 i , with i 2 f1; 2g, will be called a winning state for player i.The interpretation is that the multi-battle contest starts (i.e., the machine begins to run) at time t = 1 at state ! 1 = !init .Then, for any t 1 such that !t 2 act , there is a battle at time t.Depending on the winner i t 2 f1; 2g of the battle at time t, the machine transits into a new state !t+1 = (! t ; i t ).The multi-battle contest ends (i.e., the machine stops) when ! t 2 i for some i 2 f1; 2g.Then, player i (player j 2 f1; 2g, j 6 = i) is declared the winner (the loser) of the multi-battle contest.
The following two examples illustrate the de…nition.
Thus, in a match race of length M , a player having won M battles wins the multi-battle contest, where the improper case M = 1 corresponds to the one-shot contest.We will refer to a race of length M alternatively as a win-with-M .It is easy to see that the maximum number of consecutive battles in Electronic copy available at: https://ssrn.com/abstract=3343709a race of length M is (2M 1).Therefore, for instance, a race of length M = 2 (of length M = 3) corresponds to what is known in the literature as a best-of-three (best-of-…ve) multi-battle contest.In the tug-of-war of distance N , however, a player needs to win N battles more than the opponent.Again, the improper case N = 1 corresponds to the one-shot contest.Note that, for N 2, the tug-of-war has a potentially in…nite horizon.We will refer to a tug-of-war of distance N alternatively as a win-by-N .

Symmetric and minimal contests
It is not di¢ cult to see that two distinct automata may describe multi-battle contests that are "isomorphic" for the contestants.For example, there could be states in the Moore machine that are never reached in …nite play.Even if all states are reachable, there may be redundancies in the representation.
To make this precise, we introduce the following de…nitions.Given a multi-battle contest M, let H i (M) denote the set of …nite histories for each player i 2 f1; 2g that end in a state !T +1 2 i .We say that two multi-battle contests M and M 0 are equivalent For a given multi-battle contest M with state space , we call the number of elements of the complexity (M) of M. Note that the complexity counts not only active, but also terminal states.For example, the win-with-two (i.e., a race of length two) has a complexity of eight states, because there are four active and four terminal states.Clearly, by merging terminal states, an equivalent contest can be found with no more than = 6 states.This motivates the following de…nition.
De…nition 2. A multi-battle contest M is minimal if, for any multi-battle contest M 0 equivalent to M, we have (M) (M 0 ).
Assumption 1.The contest is minimal.
This assumption will be tacitly used in the proofs.However, in our illustrations of speci…c contests, it proves useful to work with non-minimal contests.Obviously, Assumption 1 is not a limitation at all in the sense because it is not structural but only about the representation of contests.Note also that, in a minimal contest M in which both contestants have the opportunity to win, the number of active states equals (M) 2.
We will focus on multi-battle contests that are ex-ante symmetric for the players.
The interpretation of the symmetry condition is straightforward.Since, by Assumption 1, every state is reachable from the initial state !init , a straightforward induction argument may be used to show that, if a multi-battle contest is symmetric, then the bijection is unique.We will call a state ! 2 balanced For instance, in Example 1, the mapping maps any state ! (m 1 ;m 2 ) 2 to ! (m 2 ;m 1 ) .In particular, Assumption 2. The contest is symmetric.

Component contests
Given e¤ort levels x 1 0 for player 1 and x 2 0 for player 2, consider a battle with stage payo¤s where V 1 and V 2 , respectively, are player 1's and player 2'valuations, and p : R + R + ![0; 1] is the contest success function (CSF), capturing the probability that contestant 1 wins the battle.Constant unit costs are assumed for expositional simplicity.Indeed, as usual, the consideration of continuously di¤erentiable, convex and strictly increasing cost functions that vanish at zero e¤ort can be accomplished by a simple transformation of the e¤ort variable.
Lemma 1. Impose Assumption 3. Then the probabilistic contest admits a pure-strategy Nash equilibrium if and only if either (i) V 1 > 0 and V 2 > 0, or (ii) V 1 0 and V 2 0. If the equilibrium exists, it is unique.In case (i), equilibrium e¤ ort levels are given by and equilibrium payo¤ s are given by ). (5) In particular, equilibrium payo¤ s are positive.In case (ii), There is a wide variety of probabilistic speci…cations that are consistent with Assumption 3. Some examples are listed below.
Tullock CSF.Following Tullock (1980), one may assume that player 1's probability of winning is given 2 ) for some R 2 (0; 1], with the convention that the ratio is read as Serial CSF .Let 2 (0; 1).Player 1's probability of winning is given as Again, we have the convention that the ratio x 1 =x 2 is read as 1=2 if x 1 = x 2 = 0.This speci…cation has been introduced by Alcalde and Dahm (2007).
, where 2 (0; 1].This speci…cation is a special case of a functional form introduced by Beviá and Corchón (2015).It corresponds to a 4 Indeed, Electronic copy available at: https://ssrn.com/abstract=3343709convex combination of a Tullock contest with parameter R = 1, which receives a weight of , and a coin toss, which receives a weight of 1 .
However, the di¤erentiability condition in Assumption 3 excludes the all-pay auction, which has been studied before by Agastya and McAfee (2006) and Konrad and Kovenock (2005).The homogeneity assumption excludes lotteries with noise, as studied by Blavatskyy (2010), for instance.Homogeneity is crucial for our analysis of the tug-or-war, so we have presently no way of including such cases, even though it seems plausible that our results should extend.The main reason for Assumption 3 is that we can thereby ensure existence of a unique pure-strategy Nash equilibrium.

Histories, strategies, and Markov perfect equilibrium
By a …nite history h, we mean a …nite sequence (i 1 ; i 2 ; :::; i T ) of players in f1; 2g (where i t is interpreted as the winner of the t-th battle), such that there is a …nite sequence (! 1 ; ! 2 ; :::; !T +1 ) of states in satisfying (i) ! 1 = !init , and (ii) !t 2 act with !t+1 = (! t ; i t ), for any t 2 f1; :::; T g.In this case, we will say that h is of length T 0 and ends in state !T +1 .Denote by H the set of all …nite histories h.Similarly, an in…nite history h consists of a sequence fi t g 1 t=1 , such that there is a sequence f! t g 1 t=1 satisfying (i) ! 1 = !init , and (ii) !t 2 act with !t+1 = (! t ; i t ), for any t 2 N f1; 2; :::g.A terminal history h is either a …nite history of length T such that !T +1 2 1 [ 2 , or an in…nite history.A …nite history h 2 H given by …nite sequences (i 1 ; i 2 ; :::; i T ) is a subhistory of some history h 0 of length T 0 2 fT; T + 1; :::g [ f1g given by (…nite or in…nite) sequences (i 0 1 ; i 0 2 ; :::) if i t = i 0 t for any t 2 f1; :::; T g.
We call a state ! 2 reachable if there is a …nite history h 2 H ending in != !T +1 .A recurrent state ! 2 is a state that can be visited more than once during play.
Let H act H denote the set of histories h 2 H that end in an active state !T +1 2 act .A pure We denote by S i the set of pure strategies for contestant i 2 f1; 2g.Given a …nite history h 2 H, we consider the set H term (h) of terminal histories that have h as a subhistory.Then, given a pair of pure strategies ( 1 ; 2 ) 2 S 1 S 2 , and given a contest success function fp i ( ; )g i=1;2 , we may de…ne the outcome of the game as the probability distribution over H term (h) induced by ( 1 ; 2 ).6 Given a …nite terminal history and a …nite sequence of e¤ort choice we assign payo¤ 1 P T t=1 x i t to a multi-battle contest-winning player i, and similarly, P T t=1 x i t to a multi-battle contest-loser player i, at any …nite terminal history of length T .Further, payo¤ 1 2 P 1 t=1 x i t is assigned to player i from any in…nite terminal history.
Given a …nite history h 2 H, and a pair of pure strategies ( 1 ; 2 ), player i's continuation payo¤ at h is the expectation of terminal payo¤s over the resulting outcome. 7A pair of pure strategies ( 1 ; 2 ) is a subgame-perfect equilibrium if, for any …nite history h ending in !T +1 2 act , each player i 2 f1; 2g may depend on the …nite history h only through the most recent state !T +1 (Ericson and Pakes, 1995;Maskin and Tirole, 2001).A subgame-perfect equilibrium is a Markov perfect equilibrium (MPE) when it employs Markovian strategies.We call an MPE interior if players exert a positive e¤ort at all active states.

Assumptions
It turns out that, for general multi-battle contests, a MPE need not exist (for examples, see Section 5).
To avoid this non-existence problem, and also because we had the impression that not all multi-battle contests make sense (i.e., the de…nition appeared to us as being still too ‡exible), we will impose the following three assumptions.
7 This expected payo¤ exists in the extended real line R [ f 1g because a contestant's expected revenues are …nite.Indeed, a contestant may not collect more prize money than V = 1.However, since the contest is repeated, and the outcome of each battle uncertain, there is no upper limit to the contestant's total of ex-post delivered e¤orts.Economically, it should be obvious that no player ever …nds it optimal to choose a pure strategy that leads to a negative expected payo¤.We may therefore assume, without loss of generality, that the continuation payo¤ is …nite.

Assumption (C for Centeredness).
There is at most one state ! 2 that is both recurrent and balanced.
Exchangeability captures the requirement that any two …nite histories that di¤er only in the order in which the last two contests have been won end up in the same state.For instance, Exchangeability would hold within a tennis game, but it would fail to hold in a tennis match considering points as battles.To understand why, suppose that, at a score of 40:15 in the …rst game of the match, the server makes a point and wins the game, but loses the …rst point of the next game, so that the score is 1:0 0:15.This is certainly a situation that strategically di¤ers from the one reached if the server …rst loses and then wins a point, resulting in the score 1:0 0:0.The possibility that a tennis match can be won with less than half of the total points is well-known, of course. 8ext, Monotonicity says that, starting from any active state, winning all battles in a row will let you win the multi-battle contest.Monotonicity implies, in particular, that in order to win the multi-battle contest, a player must necessarily win the last battle.We will later see that this assumption is crucial for equilibrium existence.It may also be noted that Monotonicity combined with Exchangeability excludes multi-battle contests in which the outcome is determined by simple majority.Indeed, this would be pointless because the winner may be determined before the last battle.9 Finally, Centeredness says that there is at most one balanced state that may be revisited after a sequence of battles.As we show below, both Exchangeability and Monotonicity are indispensable for equilibrium existence.Centeredness, however, does not seem crucial for existence but dropping it leads, as will become clear below, to a less intuitive class of multi-battle contests.

Strongly connected components
A state ! 2 is called …nite if there exists T = T (!) 0 such that, for any …nite history h ending in != !T +1 , we have T T We will say that a multi-battle contest has a …nite horizon if each state is …nite.Conversely, a multi-battle contest that does not possess a …nite horizon will be referred to as having an in…nite horizon.A strongly connected component (SCC) is a maximal subset C act with the properties that (i) C has at least two elements, and (ii) any state ! 2 C can be reached from any other state ! 02 C. Clearly, an SCC consists of recurrent states only.
Lemma 2. A multi-battle contest has an in…nite horizon if and only if it has an SCC.

Statement of the result
The following is the …rst main result of this paper.It is crucial for our analysis of general multi-battle contest structures.
Theorem 1. Impose Assumptions (E), (M), and (C).Then, either the multi-battle contest has a …nite horizon, or there is precisely one multi-battle contest-ending tug-of-war that is reached after a maximum …nite number of steps.
Two simple classes of contests that are consistent with the prediction of Theorem 1 are the following.
Example 3 ("Win-with-M -or-by-N ") This …nite-horizon contest may be thought of as a match race of length M 2 that is ended prematurely if one of the contestants has an advantage of at least N points over the opponent.Note that for N M , this is equivalent to a race of length M .Conversely, if N = 1, then we have a one-shot contest.Therefore, we will assume that N 2 f2; :::; M 1g.The simplest non-trivial case is the win-with-three-or-by-two, which is illustrated in Figure 1 In the remainder of this subsection, we prove Theorem 1.The proof is based on a sequence of lemmas.
Suppose …rst that the multi-battle contest has an in…nite horizon.Then, by Lemma 2, there exists an SCC.
For states !, ! 02 and a player i 2 f1; 2g such that (!; i) = ! 0, we say that the move from ! to ! 0 is irreversible if there is no …nite sequence (i 1 ; i 2 ; :::; i T ) of players in f1; 2g (where i t is the winner of the t-th battle) and a …nite sequence (! 1 ; ! 2 ; :::; !T +1 ) of states in such that (i) and (iii) !t 2 act with !t+1 = (! t ; i t ), for any t 2 f1; :::; T g.The following lemma shows that any irreversible move leading away from any state in an SCC leads to a terminal state.
Lemma 3. Impose Assumptions (E) and (M).Suppose that ! 2 is recurrent and that the move to Proof.Since ! is recurrent, it is contained in some SCC C (see Figure 1 for illustration).We claim that ! 0= (!; 1) is terminal.Suppose not.Then ! 02 act .Moreover, since ! is recurrent, and the move to ! 0 is irreversible, ! 00= (!; 2) 2 C must be recurrent.In particular, ! 002 act .By Exchangeability, (! 0 ; 2) = (! 00 ; 1) ! 000.Further, by Monotonicity, a player winning a battle and ending the multi-battle contest must win the multi-battle contest.Hence, it is impossible that ! 000is a terminal state.Hence, ! 0002 act .Next, since the move from ! to ! 0 is irreversible, so is the move from ! to ! 000 .But, ! and ! 00lie in the same SCC.Hence, the move from the recurrent state ! 00to ! 000 = (! 00 ; 1) is irreversible.By induction, and the …niteness of the state space, there is an iteration, at which a repeated win for player 2 leads back to !. This, however, is in con ‡ict with Monotonicity.
The claim follows.Thus, any state that is reached by an irreversible move from a recurrent state is terminal.To prove the Theorem, it remains to be shown that (i) there cannot be more than one SCC in the multi-battle contest, and (ii) any symmetric SCC is a tug-of-war.

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Electronic copy available at: https://ssrn.com/abstract=3343709 Lemma 4. Impose Assumptions (E) and (M).Let C be an SCC.Then C can be reached from any active state ! 2 act .In particular, there is at most one SCC in the multi-battle contest.
Proof.Note that Monotonicity implies that ! 02 act .To provoke a contradiction, suppose that there is some active state ! 2 act such that C cannot be reached from !.By Assumption 1, however, !can be reached from ! 0 , i.e., there is a …nite sequence of states ! 1 ; :::; !K and of players i 0 ; ::; i K 1 2 f1; 2g such that (i) !k 2 act for k = 1; :::; K 1, (ii) !k+1 = (! k ; i k ) for k = 0; :::; K 1, and (iii) Since C can be reached from ! 0 , but cannot be reached from !, there must be some index k 2 f0; :::; K 1g such that C can be reached from ! k but not from ! k+1 .Without loss of generality, i k = 1.Since !k 2 act , and invoking Monotonicity, there is a …nite sequence b ! 1 ; :::; b 2) for l 2 f1; :::; L 1g, and (iii) b !L 2 2 .Note now that 2) is terminal, then we have a contradiction (because there is no SCC at all reachable from ! k ).Therefore, b ! 2 = (! k ; 2) is active, and we have by Exchangeability a state (b ! 2 ; 1).By Monotonicity, the state (b ! 2 ; 1) is active.From (b ! 2 ; 1), we cannot reach C, because this would imply we can reach C from ! k+1 , which is not true.However, we can reach C from b ! 2 .By  The following lemma, which crucially exploits the symmetry of the contest, identi…es an important structural property of the tie-breaking phase.
Lemma 5. Impose Assumptions (E) and (M).Then any SCC contains a balanced state.If, in addition, Assumption (C) holds, then the SCC is a tug-of-war.
This concludes the proof of Theorem 1.

Tightness
We conclude this section with two examples that illustrate the tightness of our assumptions.
Example 5.The multi-battle contest shown in Figure 5 below satis…es Exchangeability and Monotonicity.Lemma 5 shows, however, that the SCC appearing in Figure 5 cannot be a SCC.This SCC is actually symmetric, but it does not contain a balanced state.Thus, Assumption 1 (symmetry) is crucial for obtaining the conclusion of Lemma 5.

The tug-of-war
The analysis above implies a central role for the tug-of-war in the analysis of multi-battle contests that can be represented by a …nite state machine.In this section, we will prove that the tug-of-war admits a unique symmetric Markov perfect equilibrium.We also characterize the equilibrium.Our analysis builds upon Harris and Vickers (1987), but in contrast to their work, we will admit a di¤erent class of technologies for the component contests.active states ! (N 1) ; :::; !N 1 , and two terminal states !N 2 1 and !N 2 2 .The initial state is !init = ! 0.Let V n;N denote the valuation, i.e., the continuation payo¤, of a state !n 2 for player 1.
Similarly, for player 2, the continuation payo¤ of state !n 2 is denoted by v n;N .
As we show in the Appendix, a Markov perfect equilibrium exists in the tug-of-war.To ensure uniqueness, however, and to characterize the equilibrium, we will impose one more assumption.
The assumption captures a property of the component CSF that is crucial for constructing the Markov perfect equilibrium in the tug-of-war.Speci…cally, Assumption 4 ensures that, from the ratio of di¤erences in continuation payo¤s in the transition between two states state, one may uniquely determine the ratio of valuations in any of the two corresponding component battle.Assumption 4 imposes an upper Electronic copy available at: https://ssrn.com/abstract=3343709 bound on the curvature of the technology, which holds for any of the before-mentioned speci…cations of CSF.
Proof.See the Appendix.
We are now in the position to characterize the equilibrium of the tug-of-war.
Theorem 2. Suppose that Assumptions 3 and 4 hold.Let N 1 be a positive integer.Then, the following holds true for the tug-of-war with 2N 1 active states: (i) the tug-of war has a unique symmetric MPE; (ii) V n;N is strictly increasing in n 2 f N; :::; N g, while v n;N is strictly declining in n 2 f N; :::; N g; (iii) the symmetric equilibrium payo¤ V 0;N = v 0;N is strictly declining in N ; (iv) for any …xed integer n 0, equilibrium e¤ orts X n;N for player 1 and x n;N for player 2 are strictly declining in N 2 fn + 1; n + 2; :::g, and symmetrically, for any …xed integer n < 0, equilibrium e¤ orts X n;N for player 1 and x n;N for player 2 are strictly increasing in N 2 f:::; n 2; n 1g; (v) for i 2 f1; 2g, player i's probability of winning in any active state !n is independent of N 2 fn + 1; n + 2; :::g.
Proof.See the Appendix.
Theorem 2 establishes existence and uniqueness of the symmetric and interior Markov perfect equilibrium for a tug-of-war with an odd number of active states and constant or decreasing returns.Figure 7 illustrates equilibrium continuation payo¤s at active states in a tug-of-war of distance N = 4.These continuation values are identi…ed recursively as the unique solution of a nonlinear equation.For R = 1, each of these nonlinear equations is quadratic, so that an explicit solution is feasible. 11 Example 7.For the case of the lottery technology (i.e., Tullock CSF with parameter R = 1), and a 1 1 The derivations are detailed in the Appendix.

18
Electronic copy available at: https://ssrn.com/abstract=3343709 distance N = 2, we get the following equilibrium continuation payo¤s at active states: The monotonicity of equilibrium continuation values predicted by Theorem 2(ii) is evident from these numbers.
The analysis conducted in the Appendix identi…es an unexpected similarity principle across tug-of-wars with di¤erent N , which is that holds for any n 2 f (N 1); :::; N g.Thus, continuation payo¤s known for the tug-of-war with a larger number of active states may be used to compute continuation payo¤s for a tug-of-war with a smaller number of active states.For instance, scaling Example 7 down to N = 1, relationship (10) implies for and n = 0 that which indeed corresponds to the symmetric equilibrium payo¤ in the one-shot lottery contest.Clearly, this property is inherited from our assumption that the contest success function is homogeneous of degree zero.As a result of the similarity property, the symmetric equilibrium payo¤ V 0;N in the tugof-war is monotone also in N , as stated in Theorem 2(iii).For instance, the rent dissipation in the win-by-2 is always stronger than in one-shot contest, and similarly, the rent dissipation in the win-by-3 is always stronger than in a win-by-2.This is remarkable because the analogous property for the match race does not hold.
The proof of Theorem 2 that we o¤er in the Appendix proceeds algebraically, i.e., we solve the system of Bellman equations in a recursive way.This actually seems necessary because the abstract transformation that maps a vector of continuation payo¤s, via the Bellman equations, into a new vector of continuation payo¤s does not seem to possess any convenient properties.For instance, it is not a contraction.We also found that the Jacobian of this transformation is not a P-matrix, so that uniqueness is not guaranteed by the usual conditions for univalence.To see that the equilibrium strategies predicted 19 Electronic copy available at: https://ssrn.com/abstract=3343709by state-by-state optimization are immune against arbitrary deviations, we verify that the tug-of-war without discounting satis…es a suitable variant of the one-stage deviation principle.

Existence and uniqueness
Our second main result is the following.
Proof.We start by showing that at every active state, the continuation payo¤ for every player at each state lies in the open interval (0; 1).Consider …rst the case where the contest has a …nite horizon.Thus, there is an integer L 0 such that the longest …nite history has length L. Clearly, L 1. Indeed, if L = 0, then !init is terminal, say !init 2 1 .But then, by Assumption 2, (! init ) 2 1 \ 2 = ?, a contradiction.Hence, L 1, as claimed.The proof proceeds now by induction.Suppose …rst that L = 1.Then, there is precisely one active state, which is !init .From Monotonicity, there is an interior equilibrium.Suppose that the claim has been shown for any …nite-horizon contest of length L 1.
Take any !such that the longest …nite history in the subgame starting at ! has length L. There are three cases.Assume …rst that there is another battle at ! 0 = (!; 1) 2 act , but ! 00= (!; 2) is terminal, as illustrated in panel (a) of Figure 8.Since state ! 00is terminal, Monotonicity implies that player 1's payo¤ at ! 00 is zero and it su¢ ces to show that valuation at ! 0 is positive.However, given that the longest …nite history in the subgame starting at ! 0 has length at most L 1, this follows from the induction hypothesis.The situation is similar in a second case, where ! 0= (!; 1) is terminal, while ! 00= (!; 2) 2 act .See as panel (b) of Figure 8 for illustration.Here, by Monotonicity, it must be that ! 02 1 , and contestant 1's continuation payo¤ is one.Conversely, if player 1 loses and there is another battle at ! 00 = (!; 2), then by the induction hypothesis, player 1's continuation payo¤ is strictly below one.In a third case, neither ! 0nor ! 00is terminal, as illustrated in panel (c) of Figure 8.
Then, by Exchangeability, (! 0 ; 2) = (! 00 ; 1), and player 1's continuation payo¤ at ! 0 must be strictly higher than player 1's continuation payo¤ at ! 00 .Therefore, at !, player 1 has a positive valuation of winning.Analogous arguments can be made for player 2, of course, which proves the claim for the case of contests with …nite horizon.Consider next the case where the contest has an in…nite horizon.Then, we may replace the continuation payo¤s at any state in the concluding SCC by the unique and interior equilibrium payo¤s of the corresponding tug-of-war of some distance N 2. By Theorem 2 (ii), player 1's continuation payo¤ V n;N is strictly increasing, and player 2's continuation payo¤ v n;N is strictly declining in n 2 f N; :::; N g.
The proof proceeds now precisely as above by backward induction.This proves the claim.To see that the identi…ed pro…le of Markov strategies constitutes a subgame-perfect equilibrium (i.e., allowing for unilateral deviations that a¤ect e¤ort levels at more than one battle), one may use the same arguments as in the proof of Theorem 2.
The following examples show that existence breaks down in general when the assumptions of Exchangeability and Monotonicity are dropped.
Example 8. Consider the multi-battle contest depicted in panel (a) of Figure 9, and suppose that = 1 with Tullock technology.This contest does not admit a Markov perfect equilibrium.Indeed, in the initial battle, contestants …ght about whether the remaining interaction will be a one-shot contest (if 2 wins the …rst battle) or a win-with-2 (if 1 wins the …rst battle).The one-shot contest is known to yield a payo¤ of 0:25 for each contestant, the win-with-two a payo¤ of 0:18 for each contestant.
Therefore, player 2 has a negative valuation of winning, so she exerts no e¤ort.Player 1, conversely, has a positive valuation of winning, and would not want to leave the outcome to any symmetric tie-breaking rule.So player 1 has no best response.As a result, an equilibrium does not exist.While the example is asymmetric, it is simple to make it symmetric, so it is really assumption (E) that does the job here.Corollary 1.Any multi-battle contest satisfying Assumptions (E), (M), and (C) is pervasive.

Theoretical results
In this section, we explore the intricate issue of …nding the optimal dynamic structure of multi-battle contests.Of course, the degree to which the extensive form of a contest may be shaped by a third party depends on the application considered.E.g., in a sports contest, the designer has essentially full discretion (maybe up to constraints with respect to timing or players' levels of exhaustion).In contrast, a regulatory body that aims at maximizing the e¤ort expended in R&D activities will likely face a variety of policy constraints.All the more, the regulator's in ‡uence on the extensive form of competition will often be quite indirect.In our analysis below, however, we will abstract from such application-speci…c considerations.
The designer is assumed to maximize expected total e¤orts.Thus, we sum e¤orts both across players and across periods.In particular, no discounting is applied.Similar objective functions have been used in the literature (e.g., Moldovanu and Sela, 2001;Groh et al., 2006, Sec. 3.2).Note that, given that Electronic copy available at: https://ssrn.com/abstract=3343709the prize will be allocated with probability one, we may equivalently assume that the designer aims at minimizing the players'symmetric expected equilibrium payo¤ in the symmetric equilibrium.
Lemma 7. Maximizing expected total e¤ orts is equivalent to minimizing symmetric expected equilibrium payo¤ s in the contest.
Proof.Let = 1 = 2 denote the symmetric equilibrium payo¤.Then, = 1 2 e , where e denotes the total expected payo¤ per player.The claim is now immediate.
We can now show the following.
Theorem 4. The one-shot contest is dominated by any contest in which two consecutive battle wins at the beginning of the contest imply a win of the contest.
Proof.Take any contest in which two consecutive battle wins at the beginning of the contest imply a win of the contest.We denote by the continuation value of the contest after each player has won precisely one battle.Clearly, < 1=2.Then, the continuation payo¤ for a player that has lost the …rst battle is while the continuation payo¤ for the player having won the initial battle is Therefore, in the initial battle of the contest, the symmetric equilibrium payo¤ is given by We claim that Plugging in the expressions for v and V , we obtain , (1 ) 1 By Assumption 4, the left-hand side is strictly increasing in .Moreover, for = 1=2, the left-hand side is identical to the right-hand side.This proves the claim, and hence, the Theorem.
The theorem shows that a very large class of dynamic contests elicits higher expected total e¤orts than the one-shot contest.Intuitively, this is so because even the two-fold repetition of the component contest allows making double-use of the same incentive.The conclusion of Theorem 4 can be extended by recalling that the one-shot contest is a degenerate form of the tug-of-war.Since we have shown that symmetric equilibrium payo¤s in the tug-of-war are strictly declining in the number of active states, it follows that the one-shot contest is dominated, in terms of total e¤ort, also by any non-degenerated tug-of-war.In fact, we conjecture that the one-shot contest minimizes expected total e¤orts across all contests.This is also suggested by the results of our numerical analysis which will be reported below.

Numerical exploration
In the numerical analysis, we employed several speci…cations.For the Tullock case, the well-known equilibrium characterization (Nti, 1999) yields provided that V 1 > 0 and V 2 > 0. The parameter was chosen from the grid R 2 f0:01; :::; 1:00g.In the case of the serial speci…cation, Alcalde and Dahm (2007) have shown that In this case, the parameter was chosen from the grid 2 f0:01; :::; 0:99g, i.e., excluding the case = 1 which implies complete rent dissipation for the contestant with the lower valuation.The relative di¤erence CSF yields equilibrium payo¤s In this case, the parameter was chosen from the grid 2 f0:01; :::; 1:00g.
To compare the expected equilibrium payo¤ across contests, we needed a way to enumerate all designs with a …nite or in…nite horizon.As for the …nite-horizon contests, we considered multi-battle contests for which the longest sequence of consecutive battles feasible has length (2K + 1), where K 1 is an integer.One can convince oneself that the number T K of dynamic contests with longest path (2K + 1) that satisfy our assumptions corresponds to the number of nondecreasing integer sequences 0 x 1 x 2 ::: x K 1 satisfying x k k for any k 2 f1; ::::; K 1g.For instance, for K = 1, there is only the best-of-three (or win-with-two) multi-battle contest.For K = 2, there are two contests, the best-of-…ve (or win-with-three) contest, and the win-with-three-or-by-two contest.For larger values of K, the number of contests increases quickly.In fact, one can show (e.g., Pemantle and Wilf, 2009) that To enumerate all possible multi-battle contests, we chose a recursive programming approach that conveniently solves the issue of an endogenous loop depth.
As for the contests with in…nite horizon, these can likewise be enumerated by replacing the payo¤s in the …nal active states of dynamic contests with …nite horizon by the corresponding continuation payo¤s of a tug-of-war with an odd number of active states.Figure 1(b) illustrates this idea.Note that the minimality assumption implies that the states of the concluding tug-of-war can always be arranged in the inverted L-shape.Altogether, we explored the set of (i) all …nite-horizon contests that end after at most 29 battles, and (ii) all in…nite-horizon contests in which a recurrent state is reached after at most 26 battles.The computations were done using Visual Basic.

Unconstrained contests
In our …rst analysis, we assume that the choice set is the entire set of multi-battle contests satisfying our assumptions.It follows from Theorem 2 that the one-shot contest is always strictly suboptimal.
In fact, any tug-of-war can be strictly improved by replacing it by a longer variant.We now have the following, as we believe, interesting numerical …nding.
Finding 1.The revenue maximizing multi-battle contest is always a match race of length M 2. 12   To understand intuitively why the race is optimal compare the two contests shown in Figure 10.1;0) is reached.Then, to win the contest, player 1 just needs one more win (at least), while player 2 would need three wins in any case.
In contest (b), however, the analogous situation looks less unbalanced.Thus, the continuation value for player 1 at state ! (1;0) is higher in contest (a) than in contest (b).For player 2, the situation is the opposite, i.e., the continuation value for player 2 at state ! (1;0) is lower in contest (a) than in contest (b).Because of this, the net valuation of winning the initial battle is higher in contest (a) than in Electronic copy available at: https://ssrn.com/abstract=3343709For our analysis, the relative size of the changes in the continuation values across players matters.Speci…cally, it turns out that the change in continuation values between contests (a) and (b) is much larger for player 1 than for player 2. In other words, the leading player 1 gains, in relative terms, much more than the lagging player 2 loses.We call this the preemption e¤ ect.As a result, the equilibrium payo¤, i.e., the continuation payo¤ at state !init = ! (0;0) is higher in contest (a) than in contest (b).
In turn, this implies that contest (a) is always suboptimal from the perspective of an e¤ort-maximizing designer.Thus, the intuitive reason for the optimality of the race is that, among all multi-battle contests, it minimizes the preemption e¤ect by making a battle-win as incremental a progress towards winning as feasible.The one-shot multi-battle contest, however, is never optimal.The intuitive reason for this is that a repetition allows a double-use of the same prize incentives, which is impossible in a one-shot contest.

The optimal length of a race
Given that a match race is always optimal, the question is which length is best for a designer that maximize the expected total e¤ort.Figure 11 illustrates that the length of the optimal race is inversely related to the Tullock parameter.Thus, intuitively, if each the contest success function re ‡ects more randomness (i.e., a lower R), then it is preferable for the contest designer to involve contestants into a longer race.We have found the same prediction for the serial and relative di¤erence contest.
Finding 2. The revenue-maximizing length of a match race is monotonically increasing in the noise.In many environments, the scope for changing the rules of the contest will be more limited than assumed above.Suppose, therefore, that the contest designer is restricted to use a multi-battle contest such that the number of active states is no larger than a given number K 1. Below, we report the results for the Tullock contest.
For K 2 f1; 2g, we obtain the one-shot contest.There is no other symmetric multi-battle contest under this complexity constraint.
For K = 3, we need to compare the one-shot contest with a win-by-2.As shown in the observation, however, rent dissipation in the tug-of-war with three active states is always strictly higher than in the one-shot contest.Thus, for K = 3, the win-by-2 is always optimal.
For K = 4, we have the win-with-2 as an additional possibility.Numerically, the win-with-2 strictly dominates the win-by-2 if and only if R 0:89.Intuitively, this …nding captures the trade-o¤ for the contest designer.If randomness is important (R low), then the designer has an incentive to have long multi-battle contest, so as to reward small but continued e¤orts provision.If, however, the constraint of the number of states is binding, then the tug-of-war, with its potentially in…nite horizon, o¤ers a better trade-o¤ than the race.
For K 2 f5; 6g, the designer has the win-by-3 as an additional option.This is numerically optimal if R < 0:95, otherwise the match race is better.This makes sense because the contest designer will make Electronic copy available at: https://ssrn.com/abstract=3343709use of the relaxed complexity constraint.
For K = 7, there are two additional possibilities, the win-by-4, that always dominates smaller tugof-wars.Further, we have a win-with-3-or-by-2 (Example 4).Optimal is the win-by-4 for R < 0:95, and the win-with-2 otherwise.The win-with-3-or-by-2, however, is never optimal.
For K = 8, the win-by-4 remains optimal for R < 0:85, while the win-with-two (i.e., best-of-three) stays optimal for R 0:95.In between, it is optimal to use a win-with-3-and-by-2 (Example 3).
For K 2 f9; 10g, the win-by-5 becomes optimal for R < 0:74, while the win-with-two stays optimal for R > 0:98.In between, it is optimal to use a win-with-three (i.e., a match race of length three).
Thus, for R 2 (0; 1] small, the numerical always analysis …nds a tug-of-war to be optimal, whereas for R 2 (0; 1] large, the complexity constraint relaxes, and we …nd a match race.For intermediate values of R, there are cases (e.g., K = 8 and K = 14) in which we get "sporadic" contests that are optimal but neither a tug-of-war nor a match race.We have continued the analysis to state machines with up to K = 28 active states, but this pattern remained unchanged.Recalling the relationship between complexity and the number of active states to be K = (M) 2, we may summarize our results for complexity-constrained optimal contests as follows.
Finding 3. Let the complexity = (M) be given.Then, for any R not too large, the complexity bound is binding and the optimal multi-battle contest of complexity is the longest tug-of-war that can be realized with complexity :

Conclusion
In this paper, we have introduced three conditions that jointly de…ne a ‡exible and, as we believe, natural class of symmetric multi-battle contests.This class includes both dynamic contests with …nite horizon, such as the win-with-N contest, also known as the best-of-(2N 1) tournament, as well as contests with in…nite horizon, such as a the tug-of-war.We propose making use of a …nite state machine to obtain a well-de…ned class of dynamic contests that can then be analyzed.It was shown that any multi-battle contest in our class has a very simple structure.Moreover, provided that the contest technology satis…es Electronic copy available at: https://ssrn.com/abstract=3343709 our assumptions, we established that any multi-battle contest admits a unique symmetric and interior Markov perfect equilibrium.
As part of the analysis, we characterized the Markov perfect equilibrium of the tug-of-war.This complements results in the existing literature.While we used the characterization result for the tug-ofwar predominantly as the basis for our numerical analysis of more general contests with in…nite horizon, the result addresses also a long-standing open research question, and therefore may be of independent interest.
Finally, we have studied the optimal design of dynamic contests, where the designer sets the rules so to maximize total aggregate e¤ort.For this type of problem, there is still a regrettable lack of analytical methods, and so we had to resort to numerical methods.Despite this di¢ culty, the optimal design problem could be seen to balance a trade-o¤ between three natural forces in multi-battle contests: (i) lower incentives due to randomness in the contest technology, (ii) a double-use of incentives, and (iii) the discouragement e¤ect in unbalanced states.The standard race was seen to optimally balance this trade-o¤ for a variety of contest success function if the complexity of rules is no concern for the designer.Moreover, the length of the optimal race was shown to increase monotonically in the degree of randomness in the contest technology.However, once there is a positive shadow cost of complexity, the tug-of-war is typically the optimal choice.Taken together, these …ndings shed light on the determinants of optimal dynamic contests.
There are some natural extensions that we chose to not address in this paper.To begin with, one could consider heterogeneous valuations of the multi-battle contest prize.That case, however, would be more demanding.For instance, we conjecture that the Markov perfect equilibrium of the tug-of-war might not be unique in that case.Next, De…nition 1 could be generalized to account for, say, ties in the outcome of an individual contest, or ties in the outcome of the dynamic contest.We did not pursue this line of research because we perceive the multi-battle contest as an instrument to avoid precisely any kind of tied outcomes.Further, there might be random transitions between states.For instance, in Figure 4, an initial coin toss could determine the active state from which interaction starts.We conjecture that, incorporating such possibilities would complicate matters substantially (e.g., because it would be di¢ cult to capture Centeredness) without yielding additional insights.Finally, one could think about having a variety of sets of …nal states, with di¤erent prize allocations in each of those.In that case, our assumptions would appear intuitively too tight, however.For example, in a promotional Electronic copy available at: https://ssrn.com/abstract=3343709 v n 1;N v n+1;N 0. In case (i), the equilibrium characterization of the one-shot contest (cf.Lemma 1) yields the Bellman equations Given that equilibrium payo¤s are positive in the one-shot contest, this implies that V n;N > V n 1;N and v n;N > v n+1;N .Moreover, in this case, X n;N > 0 and x n;N > 0, so that V n;N < V n+1;N and v n;N > v n+1;N , which proves the claim.In case (ii), X n;N = x n;N = 0 by Lemma 1, and consequently, each of the two players wins with equal probability in state !n .Therefore, V n+1;N V n;N V n 1;N and v n+1;N v n;N v n 1;N .By arguments detailed in case (i), this sequence of weak inequalities may be continued to both sides until V N;N ::: V N;N and v N;N ::: v N;N , which is impossible because of V N;N = v N;N = 1 and v N;N = V N;N = 0.This concludes the analysis of case (ii), and proves the lemma.
Given Lemma A.1, we may de…ne the ratio of changes in continuation payo¤s when moving from ! n 1 to ! n , i.e., for n 2 f (N 1); :::; (N 1)g.The following relationship exploits that the contest technology is homogeneous of degree zero, so that the ratio of net valuations at any active state uniquely determines the corresponding ratio of changes in continuation payo¤s.
the mapping (:) is a continuous self-mapping on the unit interval.Hence, there certainly exists a …xed point.This …xed point must be interior because 0 and 1 are not …xed points.Indeed, (0) = > 0, and since < 1 2 .Finally, the …xed point satis…es the relationship and hence, must be unique by Assumption 4. Proof of Lemma 1.If p is constant, then anonymity implies p 0:5, so that x 1 = x 2 = 0.Moreover, 1 = V 1 =2 > 0 and 2 = V 2 =2 > 0, which proves the claim in this case.Assume, therefore, that p is not constant.For that case, Malueg and Yates (2005, Lemma 1) have shown that any pure-strategy equilibrium satis…es x 1 > 0 and x 2 > 0. For > 0, we may de…ne f ( ) = p( ; 1).Then we have by homogeneity that for any x 1 > 0 and x 2 > 0. The necessary …rst-order conditions for an interior optimum read x 2 1 = 0 (98) Combining anonymity with homogeneity, we get for any x 1 > 0 and x 2 > 0. Di¤erentiating w.r.t.x 1 yields so that (a variant of Euler's Lemma) for any x 1 > 0 and x 2 > 0. Evaluating at x 1 = x 1 and x 2 = x 2 , and combining the resulting relationship with ( 98) and (99), we arrive at as in Baik (2004) and Malueg and Yates (2005).Moreover, from (98), Analogously, for player 2, as claimed.Thus, we have shown that, provided that a pure-strategy Nash equilibrium exists, it is unique and given by e¤orts ( 106) and (107).To see that these e¤ort levels constitute a pure-strategy Nash equilibrium, note that the equilibrium payo¤ function for player 1, is concave on R + .Hence, given that the …rst-order condition holds, x 1 is indeed a global maximum of e 1 .Finally, it is obvious that equilibrium pro…ts are positive if the CSF is noncompetitive.To see that equilibrium payo¤s (5) are positive if the CSF is strictly concave, recall that a player can always ensure a nonnegative payo¤ by choosing an e¤ort of zero.Since x 1 > 0 is the unique interior maximum, strict concavity implies that 1 > 0. A similar argument shows that 2 > 0.
induction, we see that only repeated battle wins by player 2 can bring us from ! k to C, and that just one battle win of player 1 on the way leads us to states from which C cannot be reached anymore.But this means that any path in C consists of wins of player 2 exclusively.However, this is inconsistent with Monotonicity!This proves the …rst assertion.The second assertion follows directly from the …rst using Lemma 3.

Figure 7 .
Figure 7. Continuation payo¤s in the tug-of-war

Figure 9 . 0 1 1 n 1 and n 0 2 n 2 .
Figure 9. Multi-battle contests without MPE contest (b), whereas the continuation value for the loser of the initial battle is lower in contest (a) than in contest (b).

Figure 10 .
Figure 10.Optimality of the match race

Figure 11 .
Figure 11.Expected total e¤ort in a Tullock race, as a function of R n 2 f (N 1); :::; N g, as well as the ratio of net valuations at any active state !n ,

Finally
derivation of the continuation payo¤s in the case N = 2. Appendix B. Proofs of Lemmas 1 and 6 balanced if and only if m 1 = m 2 2 f0; :::; M 1g.Similarly, in Example 2, the mapping maps any state !n 2 to ! n .Moreover, state !n is balanced if and only if n = 0.