Trilateral Escalation in the Dollar Auction∗

Fredrik Ødegaard† Charles Z. Zheng‡

July 9, 2020

Abstract

We find a new set of subgame perfect equilibria in a Dollar Auction that involves three active bidders. The player who falls to the third place continues making efforts to catch up until his lag from the frontrunner widens to a critical distance beyond which the catchup efforts become unprofitable. At that juncture the second-place player pauses bidding thereby bettering the chance for the third-place one to leapfrog to the front so as to perpetuate the trilateral rivalry. Once two players have emerged as the top two rivals, any such trilateral rivalry equilibrium produces larger total surplus for the three players than its bilateral rivalry counterpart does, where anyone who falls to the third place immediately drops out. Similar multilateral rivalry equilibria remain valid when we enlarge the number of active bidders from three, or complicate the tie-breaking rule that regulates the case where multiple players bid simultaneously.

Keywords: dollar auction, three-player bidding dynamics, leapfrog

∗We thank seminar participants at the 2019 Econometric Society North American Winter Meetings, the University of British Columbia, the University of Montreal, National University of Singapore, Singapore Management University and Lingnan University College at Sun Yatsen University for their comments. The ﬁnancial support from the Social Science and Humanities Research Council of Canada is gratefully acknowl- edged, with Discovery Grant R4181A05 for Ødegaard, and Insight Grant R4809A04 for Zheng. †Ivey Business School, The University of Western Ontario, London, ON, Canada, N6A 3K7, fode- [email protected],, https://www.ivey.uwo.ca/faculty/directory/fredrik-odegaard/. ‡Department of Economics, The University of Western Ontario, London, ON, Canada, N6A 5C2, [email protected], https://sites.google.com/site/charleszhenggametheorist/.

1 1 Introduction

Are there situations where having three competitors turns out to be more advantageous to the competitors themselves than having only two competitors? On one hand, having more rivals may intensify the competition thereby worsening the total welfare of all contenders. On the other hand, having multiple rivals may give a contender breathing room to disengage without conceding, watching the other rivals fight for a while before returning to the race. It is not obvious a priori which of the two opposite effects will prevail. The advantage of three-sided over two-sided competition has been suggested in the evolutionary biology literature. Frean and Abraham [15] and Sinervo and Lively [35] propose a trilateral rivalry dynamics to explain perpetual coexistence of competing species, which these authors suggest a bilateral rivalry model cannot explain. Their models assume a pecking order among the three rivaling species in a cyclic manner like paper-scissors-rock and so allow for little strategic interaction. Other game-theoretic literature on dynamic trilateral rivalry includes Hunter [22] on three-nation arms race, Kress et al. [25] on three- side combats, and a few on three-player duels—or truels—such as Kilgour and Brams [24], Bossert et al. [5], Brams et al. [7], Amengual and Toral [1], and Toral and Amengual [36]. Given the complication of interactive reasoning across more than two players in dynamic games, the main interest of these studies is to illustrate dynamic patterns of three-side conflicts given specific games, and the set of games that have been found tractable for such purpose is still tiny. Furthermore, the three-sidedness of a conflict has been assumed exogenous in these studies. The interesting idea that trilateral conflicts might sustain longer coexistence among rivals than bilateral ones, as suggested in the above-cited biology studies, has not been formally examined. There has not been even an example to illustrate how a third party would join an otherwise bilateral conflict, let alone why trilateral rivalry might have an advantage—be it modeling or normative—over bilateral rivalry. Meanwhile, studies on R&D races, such as Aoki [2], Budd et al. [8], Doraszelski [14], Fudenberg et al. [16], Harris and Vickers [18] and Hörner[20], have tried to explain perpetual coexistence of duopolists despite one’s innovation advantage over the other. Such coexistence is observed to be perpetuated by a cyclic pattern wherein the firm that has fallen behind makes a leapfrog occasionally, and the leading firm despite its advantage continues to make preemptive investments to widen its lead. Assuming to have only two firms, these works

2 explain such cyclic patterns through introducing complicated structures to the various as- pects of technological innovation. An alternative modeling choice, introducing a third firm thereby considering trilateral rivalry, has not been considered. Aimed at inspiring future research that might eventually fill in the void, this paper uncovers a cyclic dynamic pattern of a specific game where trilateral rivalry emerges and perpetuates at equilibrium. The game is Shubik’s [33] dollar auction. Albeit a classroom favorite for decades, the general case of this game with more than two bidders has not been analyzed before. We find a class of subgame perfect equilibria (SPE) where three active bidders perpetuate the competition in a cyclic pattern, wherein the third-place bidder occasionally manages to leapfrog to the front, and the second-place bidder stays put to accommodate the leapfrog. Moreover, such trilateral rivalry arises endogenously in a model where bilateral rivalry is also equilibrium-feasible, and we find that the trilateral equilibrium generates larger total surplus for all the bidders than its bilateral rivalry counterpart does. This finding is largely due to a dynamic aspect of the dollar auction that has not been considered in previous studies of the game: Different from the conventional clock model of ascending auctions, the dollar auction allows a bidder to refrain from raising his bid while others are raising theirs and to join the race later through making a leapfrogging bid that tops the frontrunner’s, as long as the race has not ended. Theoretical studies of the dollar auction, O’Neill [30], Leininger [27], Demange [13], Hörnerand Sahuguet [21], and Dekel et al. [11, 12], all assume only two bidders. Thus leapfrogging bid is impossible because if one bidder does not top the other player’s bid immediately, the other player wins right away.1 Another kind of dynamic auctions that has been found to also exhibit leapfrogging bidding patterns is online penny auctions, considered by Augenblick [3], Hinnosaar [19], Kakhbod [23] and Ødegaard and Anderson [28]. As in the dollar auction, bidders in a penny auction incur a sunk cost for each bid increment they submit. The main difference between the two games is that the sunk cost in the dollar auction is counted as part of a bidder’s eventual payment but not so in a penny auction, where the sunk bidding cost is only a fee and a winner still needs to pay the entire price for the good. Another difference is with

1The same applies to two-player models in the literature of all-pay auctions and wars of attrition. Although n-player models in the all-pay and war-of-attrition literature have been considered by Baye, Kovenock and de Vries [4], Bulow and Klemperer [9], Krishna and Morgan [26], Seel and Strack [31], and Siegel [34], they are either static games where all players simultaneously choose their one-shot bids, or stopping games where anyone who stops raising his bid quits irrevocably, without possibility of leapfrogging.

3 regard to their tie-breaking rules when multiple players bid simultaneously. In both games a random frontrunner is selected from the multiple bidders, but the difference is that while all bidders incur the sunk bidding cost in online penny auctions, only the chosen frontrunner incurs the sunk cost in the dollar auction. Given these differences the bidding incentive in a penny auction is different from that in the dollar auction. However, in our section that extends our analysis, we also consider an alternative tie-breaking rule similar to the one in online penny auctions. With three players, the dollar auction game still has bilateral rivalry as some of its SPEs, where the third player quits forever once the other two have emerged as the top two rivals. This paper, differently, presents a new kind of SPEs, featuring trilateral rivalry, where a player who happens to fall behind tries to leapfrog after he failed to raise his bid in the past while his rivals have emerged as the top two. Since leapfrog is possible, a player’s equilibrium action depends on the current distances among the players and also on whether a player who falls behind still wants to leapfrog. Since leapfrogging takes sunk costs, a player’s willingness to leapfrog in turn depends not only on the endogenous value of being in the frontrunner position but also on the current distance between the player’s currently committed bid and the frontrunner’s bid. Thus, construction of such an equilibrium requires a recursive method more complicated than steady state dynamics. We obtain explicit characterization of the strategy profiles of these equilibria (Theorem1). In any equilibrium of such there is an endogenous node at which the third-place bidder falls so far behind that he is indifferent between leapfrog and dropout. At that node, the second-place bidder pauses raising her bid to better the chance for the third-place bidder to leapfrog into the front. Thus, our equilibrium construction shows that it is a best response for a rival near the top to accommodate, rather than deter, the leapfrogging efforts of an underdog. Furthermore, we compare the bidders’ welfare in our trilateral rivalry equilibria with that in bilateral rivalry equilibria. In each bilateral rivalry equilibrium, once the game has started so that the highest and second-highest bidders have emerged, the subgame equilibrium takes away almost the entire surplus from the bidders. In each trilateral rivalry equilibrium constructed here, by contrast, once the top two bidders have emerged, the subgame equilibrium yields a larger expected payoff for each of the three players than the bilateral rivalry counterpart does (Theorem3). Thus, should the top two bidders have a chance to choose which equilibrium to play thereafter, each would prefer trilateral rivalry to bilateral

4 rivalry. That is, each would include rather than exclude the third player. The dynamic pattern of our trilateral rivalry equilibrium complements the common theme “survival of the weakest” in the above-cited three-player-duel literature. There, the duelist with the weakest marksmanship survives because he poses less a threat to each of the other two opponents. Here, an underdog is accommodated by another rival to keep the underdog from dropping out. Regarding the above-cited R&D race literature aimed at explaining perpetual coexistence of oligopolists, the trilateral rivalry equilibrium, generating a stochastic process in which a frontrunner’s lead sometimes widens and sometimes narrows, suggests a new way to formalize leapfrog dynamics. Instead of making the assumption about innovation more complicated than simply increasing the investment, one could simply introduce a third firm. Furthermore, the normative advantage of trilateral rivalry over bilateral rivalry suggests that the incumbents might want to accommodate a third entrant. Although the main results are based on the specific model originally formulated by Shubik [33], we find similar dynamic patterns in some variants of the model. We consider two directions to extend the model. One is alternative tie-breaking rules in the event where multiple players bid simultaneously. The other is to have more than three active bidders. Our main model, following Shubik’s tie-breaking rule, assumes in the event where multiple players bid simultaneously that only one bidder is randomly selected to increment his bid and become the frontrunner. Alternatively, we change the rule to have all simultaneous bidders increment their bids—thereby all incurring the sunk costs—and become tying frontrunners with the proviso that the prize is not allocated unless only a single frontrunner remains and is not outbid immediately. While this alternative rule complicates the analysis, given a nondegenerate range of parameter values there exists a trilateral rivalry equilibrium that exhibits cyclic patterns of leapfrog and accommodation. Not only is there an event where the second-place bidder pauses bidding to accommodate the leapfrog efforts from the lowest bidder, but there is also an event where the lowest bidder pauses bidding—to see the top two rivals bid against each other—and then makes a leapfrog bid (Proposition1). The other alternative tie-breaking rule we consider is to have all simultaneous bidders increment their bids—thereby incurring the sunk costs—but designate only one of them, selected randomly, as the provisional winner, who would win the prize if not outbid immediately. While this rule imposes a sunk cost on a bidder who ties with others and hence may discourage an underdog from making a leapfrog effort, we find that players can avoid

5 such costly ties in equilibrium. With the sunspot coordination idea àla Shell [32] and Cass and Shell [10], one can, under the new tie-breaking rule, replicate all the trilateral rivalry equilibria characterized in the main model (Proposition2). Towards the direction of having more than three active bidders in an equilibrium, we first extend the basic model to have four players and construct an equilibrium of quadrilateral rivalry analogous to a trilateral rivalry equilibrium (Proposition3). Next we consider an n- player model where m of them (m = 2, . . . , n) are involved in a perpetual loop of competition and accommodation. If the top m bidders are positioned consecutively, the subgame admits an equilibrium where only the player in the mth-position bids at all and all the other players stay put to accommodate his leapfrog efforts (Proposition4). We shall define the game in Section2. Then Section3 sets up the notation and equilibrium refinement condition for the recursive analysis necessary to construct trilateral rivalry equilibria. Section4 presents the main result, which explicitly derives the strategy profile for trilateral rivalry. The exposition there shows our method in this paper. Section5 verifies that such strategy profiles do constitute subgame perfect equilibria. Section6 points out their superiority over bilateral rivalry equilibria. Section7 presents the four extensions mentioned previously. Section8 concludes. Proofs are in the appendix, in the order of appearance of the corresponding claims.

2 The Dollar Auction

There is one indivisible good and three risk-neutral players. The value of the good, commonly known, is equal to v for every player. The good is to be auctioned oﬀ via an ascending-bid procedure with bid increment ﬁxed at a positive constant δ such that 2δ < v. In the initial round, all players simultaneously choose whether to bid or stay put; if all stay put then the game ends with the good not sold, else one among those who bid is chosen randomly, with equal probability, to be the frontrunner, whose committed payment becomes δ, with everyone else’s committed payment being zero, and the current price of the good becomes δ. Suppose that the game continues to any subsequent round, with p being the current price

and bi player i’s committed payment (bi ≤ p and strictly so unless i is the frontrunner), all players but the frontrunner simultaneously choose whether to bid or stay put. If all stay put then the game ends, the good is sold to the frontrunner, who pays the price p, and every other

6 player i pays bi; else the current price becomes p + δ and one among those who bid in this round is chosen randomly, with equal probability, to be the frontrunner, whose committed payment becomes p + δ, with the committed payments of others unchanged. Then the game continues to the next round. If the game never ends, then each bidder pays the supremum of his committed payment, and the good is randomly assigned to one of those whose supremum committed payments equal infinity. Our assumption that price ascends in a fixed increment δ, dating back to Shubik’s [33] original formulation, idealizes the notion that a player cannot easily preempt competition through making the price jump to a level that renders rivals’ entry unprofitable. The anticom- petitive effect of such price jumping has been observed by Dekel, Jackson and Wolinsky [11]. Our tie-breaking rule for the case of multiple bids being submitted within a round again follows Shubik’s formulation. We extend our finding to two other tie-breaking rules in Sec- tion 7.1, where all those who bid within a round have to bear their sunk costs . Section 7.2 presents extensions involving more than three active bidders.

3 Preliminary Analysis

3.1 The State Variable and Equilibrium Concept

Denote α for the frontrunner, whose committed payment is the current price p (bα = p), and β the follower, whose committed payment, by the rule of the game, is always just δ below the frontrunner’s (bβ = p−δ); denote γ for the underdog, whose committed payment bγ is the lowest. A feature that sets the dollar auction apart from the stopping game models of war of attrition is that the active bidders’ committed prices are not necessarily close to one another: bα − bγ is not only larger than the constant bα − bβ but also is variable. Thus, let the state variable of the game be represented by the frontrunner-underdog lag

s := (bα − bγ) /δ,

i.e. bγ = p − sδ. Note that s ≥ 2 once a third-place bidder has emerged (or once the game has entered the third round). From then on, s ≥ 2 as long as the game continues. Consider subgame perfect equilibria in the form of

∞ π0, π1, (πβ,s, πγ,s)s=2

7 such that π0 is every bidder’s probability of bidding at the initial round, π1 the probability of bidding at the second round for everyone but the current α player and, for every s ≥ 2 and each i ∈ {β, γ}, πi,s is the probability with which the current i player bids at state s. Implicit in the above formulation are two conditions that we use in the main part of the paper to restrict the set of subgame perfect equilibria. First is symmetry in the sense that a player’s equilibrium strategy depends not on his name but rather on his relative position with respect to other players. The second one is history-independence, that each player’s equilibrium strategy depends only on the current state (s ∈ {2, 3,..., }) when the game has entered the third round or thereafter, or on the current round (t ∈ {0, 1}) if it is the initial or the second round. In addition, we restrict the set of equilibria with a third condition:

0 [∀s ≥ s∗ : πγ,s = 0] ⇒ [∀s, s ≥ s∗ : πβ,s = πβ,s0 ] . (1)

That is, once the third-place bidder has become inactive from now on, the other two bidders play a steady state subgame equilibrium. One may think of (1) as an independence condition of non-participants. It rules out the equilibria where the active bidders choose their actions based on their distance from the player who is no longer active.

3.2 The Surplus-Dissipating Subgame Equilibrium

Our ﬁrst observation is that bilateral rivalry is always equilibrium-feasible. Consider any subgame with state variable s ≥ 2. That is, the price p has risen to at least 2δ, with the frontrunner having committed a payment p, the follower having committed p − δ, and the underdog having committed at most p − 2δ.

Lemma 1 In any subgame starting with s ≥ 2:

i. there is an equilibrium where, in each round, the current follower bids with probability 1 − 2δ/v, the current underdog stays put, the continuation value is equal to 2δ for the current frontrunner, and zero for the other two players;

ii. this is the only equilibrium where any player in the underdog position stays put forever.

This equilibrium we shall call surplus-dissipating subgame equilibrium, because a player in climbing up to the frontrunner position has already incurred a sunk cost no less than 2δ.

8 Due to Lemma1.ii and condition (1), we consider only equilibria that, to any subgame where any player in the underdog position stays put forever, prescribe the surplus-dissipating subgame equilibrium. That provides a basis to compare one equilibrium with another.

3.3 The Value Functions

∞ Let any equilibrium in the form of π0, π1, (πβ,s, πγ,s)s=2 be given. For any s ≥ 2 and any i ∈ {β, γ}, denote qi,s for the probability with which the current i player becomes the α player in the next round. By the uniform tie-breaking rule,

qi,s = πi,s (1 − π−i,s/2) (2) at any s ≥ 2, with −i being the element of {β, γ}\{i}. Given this equilibrium and any state s, denote Vs for the expected payoff for the current α player, Ms the expected payoff for the current β, and Ls that for the current γ. Denote V0 for everyone’s expected payoff at the initial round, V1 for the initial frontrunner’s the expected payoff, and M1 = L1 for every non-α player’s, at the second round. The law of motion is described below:

 3  0 prob. (1 − π0)  3 2 2 V0 −→ V1 − δ prob. π0/3 + π0(1 − π0) + π0(1 − π0) (3)  3 2 2  M1 prob. 2π0/3 + 2π0(1 − π0) + 2π0(1 − π0) ;  2  v prob. (1 − π1) V −→ (4) 1 2  M2 prob. 1 − (1 − π1) ;

 2  0 prob. (1 − π1)  M1 −→ V2 − 2δ prob. π1 (1 − π1/2) (5)   L2 prob. π1 (1 − π1/2); and, for each s ≥ 2:   v prob. 1 − qβ,s − qγ,s  Vs −→ Ms+1 prob. qβ,s (6)   M2 prob. qγ,s;

9   0 prob. 1 − qβ,s − qγ,s  Ms −→ Vs+1 − 2δ prob. qβ,s (7)   L2 prob. qγ,s;   0 prob. 1 − qβ,s − qγ,s  Ls −→ Ls+1 prob. qβ,s (8)   V2 − (s + 1)δ prob. qγ,s.

3.4 The Dropout State

∞ Since v is ﬁnite, we have, at any equilibrium in the form of π0, π1, (πβ,s, πγ,s)s=2 , that V2 is

ﬁnite and hence V2 < sδ for all suﬃciently large s; thus for the equilibrium

s∗ := max {s ∈ {1, 2, 3,...} : V2 ≥ sδ} (9) exists and is unique. Call s∗ the dropout state of the equilibrium. The next lemma explains the appellation.

Lemma 2 If the dropout state is s∗, then an underdog (γ player) (i) stays put for sure at state s if and only if s ≥ s∗, and (ii) bids for sure at state s if 2 ≤ s < s∗ − 1.

As explained in Section 3.2, we shall construct all equilibria based on the condition that, in any subgame where the underdog never bids, the other two bidders play the surplus- dissipating subgame equilibrium deﬁned there, which yields a continuation value 2δ for the current frontrunner, and zero for the other two. That is,

[∀s ≥ s˜[πγ,s = 0]] =⇒ [Vs˜ = 2δ, Ms˜ = Ls˜ = 0] . (10)

In particular, once the game enters the dropout state s∗, the player currently in the underdog position will stay put forever (Lemma2). Thus (10) implies Vs = 2δ and Ms = Ls = 0 for all s ≥ s∗.

4 Derivation of Trilateral Rivalry Equilibria

By (9) and Lemma2.i, if the dropout state is s∗ = 1, then everyone stays put, thereby ending the game, after the initial round; and if s∗ = 2 then starting from the third round the underdog quits, with the subgame equilibrium being the surplus-dissipating one played by

10 the frontrunner and the follower that emerge at the second round. The focus of this paper, however, is trilateral rivalry, where the underdog does not quit forever after the second

round. In other words, we look for equilibria with dropout states s∗ ≥ 3. Following is our ﬁnding of what such equilibria necessarily entail.

Theorem 1 If the dropout state s∗ ≥ 3 then:

i. s∗ is an even number;

ii. at the initial round everyone bids for sure;

iii. at each state s ∈ {1, 2, . . . , s∗ − 2} every non-α player bids for sure;

iv. at state s∗−1 the β player stays put and the γ player bids with probability πγ,s∗−1 ∈ (0, 1)

such that V2 = s∗δ, where V2 is derived through the law of motion, (3)–(8);

v. at any state s ≥ s∗, the γ player stays put and the β player bids with probability 1 − 2δ/v.

Theorem1.iv reveals a striking feature of a trilateral rivalry equilibrium: At the state

s∗ − 1, when the current underdog’s lag is just one step shy of the dropout state of the equilibrium, the currently higher bidder β stays put for sure so that the underdog γ can leapfrog into the front if γ chooses so. This coupled with Claims (ii), (iii) and (v) of the theorem depicts the trajectory of trilateral rivalry: First, as the escalation continues, the gap between the frontrunner and the underdog may collapse or expand, depending on whether the latter manages to leapfrog thereby replacing the frontrunner. Second, the escalation

may end only when this gap reaches s∗ − 1, called critical state, at which the follower lets the underdog decide whether to continue the escalation through leapfrogging. Third, bilateral rivalry never occurs on path, as the game ends if the underdog does not leapfrog at the critical state. Finally, a rather peculiar feature of all these equilibria is, according to Claim (i), that these equilibria support only even numbered dropout states. This section outlines the reasoning for the theorem and leaves the details to Appendices A.3 and A.4.

4.1 Why the Underdog Leapfrogs at All

Here we explain why the γ player at state s∗ − 1 bids at all, i.e., πγ,s∗−1 > 0. First note

s∗ ≥ 4 from the hypothesis s∗ ≥ 3 and Claim (i) of the theorem, which we will explain in

11 Section 4.3. Second, we shall use a Lemma6 in Appendix A.3:

L2 ≤ M2 < V3. (11)

Now suppose, to the contrary of the claim, that πγ,s∗−1 = 0 at equilibrium. Then, since

πγ,s = 0 at all s > s∗ −1 (Lemma2.ii), the γ player quits forever starting from the state s∗ −1.

Thus (10) implies Vs∗−1 = 2δ and Ms∗−1 = 0. Combine (6), (7) and Lemma2.ii to get the following chains of transition:

−2δ −2δ · · · → Ms∗−4 −→ Vs∗−3 → Ms∗−2 −→ Vs∗−1 ↓ ↓ ↓ (12)

L2 M2 L2

−2δ · · · → Vs∗−4 → Ms∗−3 −→ Vs∗−2 → Ms∗−1 ↓ ↓ ↓ (13)

M2 L2 M2

Reason backward along the two chains, starting from Vs∗−1 = 2δ and Ms∗−1 = 0, and repeatedly use the fact L2 ≤ M2 in (11). Then we obtain two chains of inequalities:

Vs∗−1 = 2δ ⇒ Ms∗−2 ≤ max{−2δ + Vs∗−1,L2} = L2 ≤ M2

⇒ Vs∗−3 ≤ max{Ms∗−2,M2} ≤ M2

⇒ Ms∗−4 ≤ max{−2δ + Vs∗−3,L2} ≤ max{M2,L2} ≤ M2 ⇒ · · ·

Ms∗−1 = 0 ⇒ Vs∗−2 ≤ max{Ms∗−1,M2} = M2

⇒ Ms∗−3 ≤ max{−2δ + Vs∗−2,L2} ≤ max{M2,L2} ≤ M2

⇒ Vs∗−4 ≤ max{Ms∗−3,M2} ≤ M2 ⇒ · · ·

The two chains combined imply Vs ≤ M2 for all s ≤ s∗ − 1. But then V3 ≤ M2, contradict-

ing (11). Thus, πγ,s∗−1 > 0.

4.2 Why the Follower Accommodates the Underdog

Here we explain why the β player at state s∗ − 1 stays put for sure, i.e., πβ,s∗−1 = 0. First,

observe that if s∗ ≥ 4 then L2 > 0. That is because the underdog γ at state s = 2 can bid

thereby securing a fraction of the payoﬀ V2 − 3δ, which is strictly positive by s∗ ≥ 4 and (9).

(In bidding at state s = 2, γ also gets a fraction of L3, which is nonnegative as he can always

12 M2 M2 M2 M2

V2 V3 V4 V5 V6 WON

M2 M3 M4 M5 M6 LOST

L2 L2 L2 L2

L2 L3 L4 L5 L6 LOST

V2 V2 V2 V2

Figure 1: The law of motions and equilibrium winning path if s∗ = 7.

stay put thereafter, cf. (8).) Second, since πγ,s = 0 for all s ≥ s∗ (Lemma2.i), (10) implies

Vs∗ − 2δ = 0. Thus, bidding is worse than staying put for the β player at state s∗ − 1:

Staying put gives him πγ,s∗−1L2, which is strictly positive because L2 > 0 and, as explained in Section 4.1, πγ,s∗−1 > 0; by contrast, bidding gives him (Vs∗ −2δ)πγ,s∗−1/2+L2πγ,s∗−1/2 =

L2πγ,s∗−1/2, strictly less than L2πγ,s∗−1. Hence πβ,s∗−1 = 0.

4.3 Why Dropout States Cannot Be Odd Numbers

The proof of Claim (i) of the theorem, relegated to Appendix A.3, follows a method similar to that illustrated in Section 4.1. Here we provide its intuition. Note that in a trilateral rivalry equilibrium the game ends only when the state is s∗ − 1: When s < s∗ − 1, the underdog bids for sure (Lemma2.ii and Claim (ii) of this theorem); when s = s∗ − 1, the underdog may bid while the follower stays put. If the underdog bids (thereby becoming the next α) then the state returns to s = 2, else the game ends and the current α wins the good. Thus, in order to win, a player needs to be the α player at the critical state s∗ − 1.

Consequently, if s∗ is an odd number then, on the path to winning, a bidder must have in previous rounds been the β player for all odd states s < s∗ − 1, and the α player for all even states s ≤ s∗ − 1. Figure1 illustrates the case of s∗ = 7, where solid lines represent possible transitions if one bids, dashed lines if he stays put, and the thick gray states and arrows indicate the path to winning.

Thus, if s∗ is odd, a player in the β position at any even state s < s∗ − 1 would rather,

13 in order to reach the winning path, become the γ player in state s = 2 (through staying put) than become the superﬂuous α player in the odd state s + 1 at the cost of 2δ (through bidding). In particular, in state s = 2, the β player would never bid while the γ player would always bid; hence the state s = 2 repeats itself, with the players switching roles according to γ → α → β → γ, thereby trapping them in an inﬁnite bidding loop.2

4.4 Why Even Number Dropout States Are Possible

When the dropout state s∗ is an even number, by contrast, a β player is not in the predica-

ment as in the previous case. First, in any even state s < s∗ − 1 the β player wants to bid in order to stay on the winning path and become the α in the odd state s + 1. Second, in

any odd state s < s∗ − 1 the β player would rather bid and become the α in the even state s + 1 than stay put thereby becoming the γ player in state 2. With the former option, it takes a cost of 2δ (to become α in s + 1) and two rounds for the player to have a chance to become the β player in state s = 2 thereby landing on the winning path. With the latter option, it takes a cost of 3δ and three rounds for him to have such a chance of reaching the winning path. In Figure2, with s∗ = 6, the situation of this odd-state β player is illustrated by the node M3, from which the former option (becoming the next α) reaches the winning path state M2 via M3 → V4 → M2, while the latter option (being the next γ) reaches M2 3 via the more roundabout route M3 → L2 → V2 → M2. That is the intuitive reason for the part of Theorem1.iii on the β player’s strategy.

5 Veriﬁcation of Trilateral Rivalry Equilibria

By Theorem1.ii, Eq. (2) and the uniform tie-breaking rule,

2 ≤ s ≤ s∗ − 2 =⇒ qβ,s = qγ,s = 1/2. (14)

2The odd-vs-even contrast in Sections 4.3 and 4.4 echoes an odd-vs-even contrast observed by Kilgour and Brams [24] on three-player duels. An interesting difference is that the odd-vs-even contrast in three-player duels is about the exogenous length of a finite game, whereas our contrast is about the endogenous length of the on-path rivalry of the equilibrium in our potentially infinite game. 3 In the more roundabout route, the last step, from V2 to M2, is preferable to a player because of a nontrivial Lemma 10, saying that in the consecutive configuration it is better-off to be the follower than the frontrunner.

14 M2 M2 M2

V2 V3 V4 V5 WON

M2 M3 M4 M5 LOST

L2 L2 L2

L2 L3 L4 L5 LOST

V2 V2 V2

Figure 2: The law of motions and equilibrium winning path if s∗ = 6.

Given any πγ,s∗−1 ∈ [0, 1], the value functions (Vs,Ms,Ls)s associated to the strategy profile specified by Theorem1 can be calculated based on (14) and the law of motion, (6)–(8). The question is whether such a strategy profile constitutes an equilibrium. The crucial step in answering this question is to verify that, given the strategy profile in Theorem1, bidding is a best response for the β player at every state below s∗ − 1. Verification for all such states might sound cumbersome, but it turns out that we need only to check two inequalities:

Lemma 3 For any even number s∗ ≥ 4 and any strategy proﬁle speciﬁed by Theorem1, bidding is a best response for the β player at state s ∈ {1, 2, . . . , s∗ − 2} if either (i) s is even and V3 − 2δ ≥ L2, or (ii) s is odd and Vs∗−2 − 2δ ≥ L2.

Thus we obtain an equilibrium with an even number dropout state s∗ ≥ 4 if the following conditions all hold: (a) the bidding probability πγ,s∗−1 at the critical state solves the equation V2 = s∗δ (Theorem1.iv), with V2 as well as other value functions derived from the law of motion (6)–(8) according to the strategy proﬁle in Theorem1; (b) V3 − 2δ ≥ L2; and (c) Vs∗−2 − 2δ ≥ L2. Condition (a), by Lemma 18 in the appendix, is equivalent to 3µ v ∗ (1 − π )(2 − µ ) + (2 − µ )2(s − 6 + µ ) (15) δ γ,s∗−1 ∗ ∗ ∗ ∗

= (2(1 + µ∗) − 3µ∗πγ,s∗−1) (3s∗ + 2(1 − 2µ∗) − (s∗ − 4 + µ∗)(1 − 2µ∗ + 3µ∗πγ,s∗−1)) ,

−s∗+3 where µ∗ := 2 . Condition (b), by Lemma 17 in the appendix, turns out to be redundant, implied by Condition (c), which by Lemma 19 in the appendix is equivalent to

3(2 − µ∗) πγ,s∗−1 ≥ 1 − . (16) 2(1 − 2µ∗)(s∗ − 4 + µ∗)

15 When s∗ ≤ 6, (16) is guaranteed as long as (15) admits a solution πγ,s∗−1 ∈ [0, 1], which in turn is guaranteed if v/δ > 35/2 when s∗ = 4, and v/δ > 6801/120 when s∗ = 6 (Lemma 22).

When s∗ > 6, to guarantee both (15) and (16) we need a stronger condition (cf. the end of Appendix A.6), v 1 5 ≥ s2 + s − 8 2s∗−3. (17) δ 3 ∗ 3 ∗ In sum, trilateral rivalry equilibria with even-number dropout states exist provided that the parameter v/δ is suﬃciently large:

Theorem 2 A trilateral rivalry equilibrium exists if v/δ is suﬃciently large. In particular:

i. if v/δ > 35/2 then one exists with dropout state s∗ = 4;

ii. if v/δ > 6801/120 then one exists with dropout state s∗ = 6;

iii. if an even number s∗ ≥ 8 satisﬁes (17), then one exists with dropout state s∗.

Numerical Illustration Consider the case where δ = 1, v varies from 0 to 1000, and

s∗ ∈ {4, 6, 8, 10}. Figure3 shows the equilibrium bidding probability of the underdog (the

γ player) in the critical state s∗ − 1 as a function of the underlying value v. The vertical

lines indicate the point at which additional equilibria for s∗ > 4 are admitted. For instance,

starting at v = 57 (≈ 6801/102) the equilibrium corresponding to the dropout state s∗ = 6 is permissible. Note that within each equilibrium the bidding probability is increasing in v (or v/δ as δ = 1 in this example). Interestingly, when a new equilibrium with a higher dropout state becomes permissible the corresponding equilibrium bidding probability drastically reduces. Furthermore, each additional equilibrium requires an order of magnitude increase in v, consistent with the exponential growth rate of the right-hand side of (17).

6 Welfare Properties

6.1 Bilateral Rivalry Equilibria

Diﬀerent from the equilibria characterized above, where the dropout states are s∗ ≥ 3, the

equilibria with dropout states s∗ = 2 all exhibit bilateral rivalry on path. In each of them, once the game has reached the third round, where two players have emerged as the top

16 1.0

s*=4 0.8

s*=6 0.6

s*=8 0.4 probability gamma player bids probability gamma player 0.2

s*=10 0.0

0 200 400 600 800 1000

value of item − v

Figure 3: Equilibrium bidding probability for the underdog in the critical state s∗ −1; δ = 1.

two rivals, the surplus-dissipating subgame equilibrium is played from then on, with the underdog dropping out for good and the follower topping the frontrunner with probability 1 − 2δ/v in any round until the game ends (cf. Ødegaard and Zheng [29]). Thus, any such equilibrium exhibits only bilateral rivalry and, once the game has entered the third round, only the top two bidders may remain in the race and only the current frontrunner gets a nonzero payoﬀ, 2δ.

6.2 Pareto Superiority of Trilateral Rivalry

From the viewpoint before the start of the game, there is a continuum of bilateral rivalry equilibria that yield zero expected payoﬀ to all bidders, but there is also a bilateral rivalry one that yields the highest total expected payoﬀ, v −δ, for the bidders, where escalation does not occur and the randomly selected initial frontrunner wins (cf. Ødegaard and Zheng [29]). However, given that players, be they in classroom experiments or in the real world, are

17 so easily trapped in escalating wars of attrition, it makes sense to consider their welfare after escalation has started so that the top two rivals have emerged, that is, when the state of the game is such that s ≥ 2. In any bilateral rivalry equilibrium, if s ≥ 2 then the subgame equilibrium is the surplus-dissipating one, giving an expected payoﬀ 2δ to the current frontrunner and zero to each of the other two (Section 6.1). In any trilateral rivalry equilibrium with dropout state s∗, by contrast, the surplus-dissipating subgame equilibrium does not occur on path at all, whereas the path—when the state s ∈ {2, . . . , s∗ −1}—exhibits trilateral rivalry that yields a total expected payoﬀ larger than 2δ:

Theorem 3 In any trilateral equilibrium with dropout state s∗, in any subgame where the

state s ∈ {2, . . . , s∗ − 1}, we have Vs > 2δ, Ms > 0 and Ls ≥ 0, with Ls > 0 if s < s∗ − 1.

The theorem is based on the law of motion of the value functions (6)–(8), coupled with the strategy proﬁle speciﬁed by Theorem1. The case of V2 > 2δ, however, is obvious: By

V2 = s∗δ (Theorem1.iv) and s∗ ≥ 4 (Theorem1.i), V2 = s∗δ ≥ 4δ. Theorem3 implies that, once the two rounds of the game has elapsed so that two players have emerged as the top two rivals, the two can avoid the detrimental bilateral rivalry if they could engage the third player in a trilateral rivalry equilibrium.

7 Extensions

We extend the analysis to two directions. One is to consider other tie-breaking rules for the event where multiple players try to raise their bids in the same round. The other is to have more than three bidders. Our analysis so far is based purely on the original model introduced by Shubik [33]. In particular, Shubik’s tie-breaking rule, randomly selecting only one of the simultaneous bidders to commit to a higher payment and become the current frontrunner, is conducive to our equilibrium construction. However, the kind of dynamic patterns described in Theorem1 can still obtain even if the tie-breaking rule is replaced by one that requires all players who try raising bids simultaneously to commit to their bid increments. We demonstrate that in Section 7.1 with two such alternative tie-breaking rules. In considering more than three bidders, we ﬁnd that the welfare advantage of trilateral rivalry over bilateral rivalry can be partially extended to that of involving more rivals in

18 a competitive loop. Section 7.2.1 presents a four-player case where a quadrilateral rivalry equilibrium generates larger total surplus than a trilateral rivalry one. Section 7.2.2 presents an m-rivalry equilibrium, with m an integer larger than three, where m players endogenously take turn to bid while the other players stay put, implicitly accommodating his leapfrog eﬀorts. A striking feature of this equilibrium is that the players’ total surplus it generates can go up to just one bid-increment shy of the entire value of the prize.

7.1 Alternative Tie-Breaking Rules

7.1.1 Multiple Frontrunners

Let us consider an alternative tie-breaking rule that allows multiple players to emerge as frontrunners in the same round: Given any current highest bid p, if m players bid in the current round, each of such bidders increases his committed payment to p + δ and becomes a frontrunner, tying with the other m−1 frontrunners; in any round with multiple frontrunners, all players—frontrunners, followers and underdogs—have the option to bid; in any round with a single frontrunners, only the players in the follower- or underdog-role have the option to bid; when no further bid is submitted, the auction ends and the good is allocated if there is a single frontrunner; if there are multiple frontrunners but no further bid is submitted, the auction ends without the good allocated. In short, anyone who bids has to commit to the increment of his bid, but only when a single frontrunner remains unchallenged is a winner declared and the good allocated. Now that multiple frontrunners are permitted, states with multiple frontrunners or multiple followers can occur, and the bidding dynamics becomes more complicated. However, trilateral-rivalry equilibria that dominate the bilateral ones still exist. We demonstrate that with a trilateral equilibrium where the underdog remains active as long as his gap to the frontrunner does not exceed 2δ. A feature similar to the trilateral equilibria in earlier sections is that, when the underdog is indifferent between bidding and dropping out, the follower stays put to accommodate the former’s leapfrog efforts. With multiple players possibly occupying the same position, the set of states is enlarged. On the equilibrium path are five trilateral states where all three players are active: ααα is the state when all three players are tied as frontrunners; ααβ when two are tied as frontrunners and one lags by δ as the follower; ααγ when two are tied as frontrunners and one lags by 2δ

19 as the underdog; αββ when one is the single frontrunner and two lag by δ as tied followers; and αβγ as the consecutive conﬁguration with a frontrunner, a follower lagging by δ, and an underdog lagging by 2δ from the frontrunner. Once the underdog lags from the frontrunner by more than 2δ, only the top two players remain active in this equilibrium, and we denote all such bilateral states by B. There are two terminal states: W means that the game ends with a winner, and Λ without a winner.

Analogous to previous sections, Vs, Ms and Ls represent the continuation values for the α-, β- and γ-players in state s that ranges in the above list. To denote a player’s strategy

given each state, we encircle the player under consideration. For example, π○α αβ denotes the probability with which each of the two typing frontrunners bids when the third player lags

behind by only δ; π○α αγ the probability with which each of the two tying frontrunners bids

when the third lags behind by 2δ; πα○β β the probability with which each of the two tying

followers bids; παα○γ the probability with which the underdog bids when the other two are tying as frontrunners, etc.

Proposition 1 If 9 ≤ v/δ ≤ 108, the following strategy proﬁle constitutes a subgame perfect equilibrium given the alternative tie-breaking rule that allows for multiple frontrunners:

i. In states when all are tied as frontrunners (s = ααα), everyone bids with probability 1/2 π○α αα = 1 − (9δ/v) ;

ii. In states with two frontrunners and one follower (s = ααβ), the frontrunners bid with

probability π○α αβ = 2/3, and the follower does not bid, παα○β = 0;

iii. In states with two frontrunners and one underdog (s = ααγ), the frontrunners bid with

probability π○α αγ = 1/2, and the underdog does not bid, παα○γ = 0;

iv. In states with one frontrunner and two followers (s = αββ), the followers bid with

probability πα○β β = 1/3 (the single frontrunner cannot bid);

v. In the consecutive state consisting of a frontrunner, a follower and an underdog (s =

αβγ), the follower does not bid (πα○β γ = 0), and the underdog bids with probability

παβ○γ = 1 − 3δ/v;

vi. In states where the underdog has dropped out and the two remaining players continue bidding, the bilateral subgame equilibrium is given by: (a) in tied frontrunner states αα,

20 Λ

α, α, β α, β, β W inner

α, α, α α, β, γ

α, α, γ Bilateral w/ties

Figure 4: Multi-frontrunner tie-breaking rule: Thick arrows denote the equilibrium path

each bids with probability 1/2; and (b) in consecutive frontrunner-follower states αβ, the follower bids with probability 1 − 2δ/v.

This equilibrium generates the following continuation values. In states αββ, αβγ, and (bilateral) αβ, the frontrunner’s values are Vαββ = v/9, Vαβγ = 3δ, and Vαβ = 2δ. For any other state or any other player, the continuation value is zero: Vααα = Vααβ = Vααγ = Mααβ =

Mαββ = Mαβγ = Lααγ = Lαβγ = Vαα = Mαβ = 0.

Appendix A.8 proves the above proposition. Figure4 illustrates the equilibrium, with solid arrows representing the transitions due to some player’s bidding, and dashed arrows the transitions when no further bid is submitted. Arrows that are grey and extra thick indicate the equilibrium path, and thin arrows the oﬀ-path transitions. The assumption 9 ≤ v/δ ≤ 108 in Proposition1 plays a similar role to its counterpart in Theorem2. The lower bound 9 ≤ v/δ ensures that players have an incentive to bid in the tied frontrunner state ααα. If v were too low relative to δ, at the start of the auction no one would bid. The upper bound v/δ ≤ 108 ensures that the underdog does not want to bid in the tied frontrunner state ααγ. We conjecture that equilibria where the underdog remains active even when he lags by more than 2δ are possible, in which case the upper bound v/δ ≤ 108 could be relaxed upward. The equilibrium in Proposition1 has three interesting properties. First, in both the

21 tied-follower state αββ and the consecutive state αβγ, the subgame admits the bilateral surplus-dissipating equilibrium, same as the one in Section 3.2. However, it yields smaller total surplus than the trilateral equilibrium constructed here does. Thus the welfare advantage of trilateral rivalry in the main model is extended here. Second, the equilibrium has a similar feature of leapfrog and accommodation as those in the main model. The underdog bids with positive probability in the consecutive state αβγ while the follower stays put. In addition to αβγ, there is another state where the follower stays put without conceding: it is ααβ, where he stays put and, if exactly one of the tying frontrunners bids so that the state becomes αβγ, bids again in the role of γ. Finally, the equilibrium adds two features to the dynamic pattern of trilateral rivalry. First is that on the equilibrium path ties occur (ααβ, ααγ, αββ, etc.), where multiple players raise their bids simultaneously thereby occupying the same position. The second feature is a nontrivial pattern of reentry, whereby we mean that a player stays put for at least a round and, if the auction still goes on after the stay-put period, raises his bid again. In the equilibrium, when the state becomes ααβ, with two players tied as frontrunners followed by a single β-player, it is now the lowest bidder, the β-player, who stays put for the top two (tying) rivals to bid against each other with a probability strictly between zero and one. In the next round, if only one of the previously top two rivals has bid so the state becomes αβγ, the previous β-player—now the γ-player—bids again while the previous α-player who did not bid—now the β bidder—stays put to accommodate the former’s reentry leapfrog.

7.1.2 Single Provisional Winner

Consider another tie-breaking rule: Given any current highest bid p, if m players bid in the current round (m ∈ {1, 2, 3}), each of such bidders increases his committed payment to p + δ, and exactly one of them is randomly selected with probability 1/m to be the provisional winner; in the next round, all but the provisional winner have the option to bid; if no one bids, the provisional winner gets the prize; else the game continues to another round, with the previous provisional winner falling to the second place (and the ﬁrst place occupied by possibly multiple bidders, exactly one of whom being the current provisional winner). In short, anyone who bids has to commit to the increment of his bid, and only one of those who bid becomes the provisional winner with equal probability. This rule is similar to those in online penny auctions.

22 Given this tie-breaking rule, once the game has entered the second round, with exactly one provisional winner among three players, only two players get to move from now on and hence, after the second round, a tie can occur only between two players. Thus any subgame after the second round is either a sequential state in the sense that the currently committed payments of the three players are pairwise distinct, i.e., no two players occupy the same position, or a tie that results from exactly two players bidding in the previous round to both become the current frontrunners, with exactly one of them randomly selected as the provisional winner. Note that a sequential state is fully described by the state variable s that represents the current gap between the underdog and the frontrunner (Section 3.1). Thus, once the game has continued to the third round, the set of possible states thereafter is the disjoint union {2, 3,..., } t {tie}. Conditional on a tie in any round after round two, denote α1 for the current provisional winner, α2 for the other frontrunner, and β for the third player, whose current committed payment is the second highest—and the lowest—among all three players. There is subgame equilibrium that makes ties unpleasant to each tying bidder:

Lemma 4 Given the provisional-winner tie-breaking rule and conditional on a tie in any round after the second round, there is an equilibrium where:

i. in any round with a tie state, player α2 bids for sure, and player β stays put;

ii. in any round at a sequential state, the frontrunner and the follower play the surplus- dissipating subgame equilibrium in Lemma1;

iii. in this equilibrium, the expected payoﬀ for the current provision winner is equal to zero, and that for the other frontrunner is equal to δ.

To extend the trilateral rivalry equilibria in the basic model (Theorem1) to this tie- breaking rule in a general manner, we expand the equilibrium concept to allow for tie-avoiding coordination among the players. Let us make a sunspot state assumption that in each round there is a sunspot (`ala Shell [32] and Cass and Shell [10]) the realized value of which is commonly observed by the players: In the initial round, there are three possible realized values, labeled as 1, 2 and 3, each with probability 1/3; in any other round, there are two realized values, labeled as 1 and 2, each with probability 1/2. Before the game starts, name the players as 1, 2 and 3. Then the trilateral rivalry equilibria in the main model can be

23 replicated through bidders’ coordination in each round to decide who is supposed to bid in the current round according to the realized sunspot state.4 Pick any trilateral equilibrium in the original model (Theorem1), which is of the form ∞ (π1, π2, (πβ,s, πγ,s)s=2) such that, for some dropout state s∗, πβ,s∗−1 = 0, πγ,s∗−1 ∈ (0, 1),

πβ,s = πγ,s = 1 for all s < s∗ − 1, and πβ,s = 1 − 2δ/v and πγ,s = 0 for all s ≥ s∗. In our basic model given the original tie-breaking rule (Section2), the original equilibrium ∞ (π1, π2, (πβ,s, πγ,s)s=2) entails the following transition of roles:

In the initial round, each player gets to be the frontrunner in the second round with probability 1/3. In the second round, each of the two current followers gets to be the round-three frontrunner with probability 1/2. In any subsequent

round with state s between 2 and s∗ − 2, the follower and the underdog each

get to be the next-round frontrunner with probability 1/2. In state s∗ − 1, the follower does not become the next-round frontrunner, and the underdog becomes

the next frontrunner with probability πγ,s∗−1. In any state s ≥ s∗, the follower

becomes the next-round frontrunner with probability πβ,s, and the underdog has zero probability to be the frontrunner.

Now consider the current model given the provisional-winner tie-breaking rule considered in this section. Construct the following strategy proﬁle through modifying the original ∞ equilibrium (π1, π2, (πβ,s, πγ,s)s=2):

i. In the initial round, for any realized value i ∈ {1, 2, 3} of the sunspot state, player i bids for sure and the other two stay put.

ii. In the second round, denote the two followers by their names j and k such that j < k; if the realized sunspot state is 1, then player j bids for sure and k stays put; if the realized state is 2, then k bids and j stays put.

4The recommendation made by the sunspot device is not binding. Once the sunspot picks a designated bidder, it is in the interest of each bidder to bid accordingly based on the expectation that others will follow their assigned roles. See Garratt et al. [17] for a theoretical construct of a bidding ring where bidders use a sunspot state to pick who among them will win an auction without competition, as well as their Footnote 2 for references to a real-world episode of such sunspot coordination (through the phase of the moon) among oligpolists. Outside auction theory, sunspot equilibrium has been a standard notion in the incomplete market theory of ﬁnancial economics (e.g., Bowman and Faust [6]).

24 iii. In any round given a sequential state s for which 2 ≤ s < s∗ −1, if the realized sunspot state is 1, the follower bids for sure and the underdog stays put; else the underdog bids and the follower stays put.

iv. In any round given a sequential state s ≥ s∗ − 1, the bidding probability is the same as

that in the original equilibrium. That is, at the critical state s∗ − 1, the follower stays

put and the underdog bids with probability πγ,s∗−1; at any state s ≥ s∗, the follower

bids with probability πβ,s and the underdog stays put.

v. In the oﬀ-path event where the player who is supposed to stay put deviates to bid thereby tying with the other bidder, the subgame is played according to the equilibrium in Lemma4.

It is obvious that provisions (i)–(iv) together replicate the above-described transition in the original equilibrium, provided that provisions (i)–(v) constitute a subgame perfect equilibrium (SPE) with the sunspot state assumption. They do constitute an SPE. The subgame equilibrium conditional on a tie state, according to Lemma4, gives a continuation value δ/2 to each involved in the tie, which is less than the sunk cost of making the bid. Thus no player wants to deviate to bid when he is not supposed to bid. Hence we have—

Proposition 2 Any trilateral equilibrium in the original model (Theorem1) is also an SPE in the model given the provisional-winner tie-breaking rule and the sunspot state assumption.

7.2 Multilateral Rivalry

7.2.1 Quadrilateral Rivalry

Theorem3 implies that, once two players have emerged as the top two rivals, a trilateral rivalry equilibrium generates larger total expected payoﬀ for the bidders than a bilateral rivalry one does. This implication is extended to a four-player case as follows. Let there be four players, denoted according to their current positions as frontrunner α, follower β, underdog γ1, and “bottomdog” γ2 for the fourth-place bidder. Let t ∈ {0, 1, 2, 3,...} denote the bottomdog’s lag from the frontrunner. We construct an quadrilateral equilibrium where the dropout state for the bottomdog is t = 4. In the subgame once this dropout state is reached, the other three players play the trilateral-rivalry equilibrium with dropout state s∗ = 4. Speciﬁcally—

25 Proposition 3 When there four players, the dollar auction game (Section2) admits an equilibrium with the following strategy proﬁle:

i. In the initial round (t = 0), everyone bids for sure.

ii. In the second (t = 1) and third (t = 2) rounds, every non-frontrunner bids for sure.

iii. In any round where t = 3:

(a) if the current conﬁguration (where the left column denotes the players, and the second their committed payments) is   α p    β p − δ      (18)  γ1 p − 2δ    γ2 p − 3δ

for some p ≥ 3δ, then β, γ1 and γ2 each bid for sure;

(b) else then γ1 and γ2 are in the same position, which by the rule of the game means that it is the fourth round and the conﬁguration is in the form   α 3δ    β 2δ      , (19)  ∅ δ    {γ1, γ2} 0

then the play mimics the trilateral-rivalry equilibrium at the critical state s∗ − 1 = 1/2 3: β stays put, and γ1 and γ2 each bid with probability 1 − (1 − πγ,3) ; where πγ,3

is the one speciﬁed by Theorem1.iv given s∗ = 4.

iv. If t ≥ 4, then γ2 quits from now on, and the other three play the trilateral-rivalry

equilibrium with dropout state s∗ = 4; in the oﬀ-path event where γ2 leapfrogs to the top, he and the previous frontrunner and follower constitute a consecutive 3-player conﬁguration, and the three play the trilateral-rivalry equilibrium (with dropout state

s∗ = 4) from now on and γ1, acting as the new bottomdog, quits for good.

26 Appendix A.10 proves that the above strategy proﬁle constitutes a subgame perfect equilibrium. The appendix further proves that, in any subgame where the players are positioned as (18), this quadrilateral rivalry equilibrium produces larger total expected payoﬀ

for the four players than the trilateral rivalry equilibrium with dropout state s∗ = 4. This quadrilateral equilibrium illustrates how the number of rivals in the dollar auction shrinks in such a discontinuous manner that one does not see in stopping games. The equilibrium starts with every non-frontrunner vying for the frontrunner position in the ﬁrst three rounds. Then the position at the start of the fourth round is either (18) or (19). If it is (18),

quadrilateral rivalry either repeats itself if γ2 leapfrogs into the next frontrunner position, or

reduces to trilateral rivalry if γ2 fails to leapfrog. If it is (19), either the quadrilateral rivalry

shrinks to a trilateral one if γ1 or γ2 manages to leapfrog into the front, or the game ends

if both γ1 and γ2 fail to do so. Thus, the number of active rivals may shrink from four to three and from three to zero, without ever shrinking to two.

7.2.2 Cyclic Patterns among n Players

Let us extend the model to include any ﬁnite number of bidders and construct an equilibrium where these players take turn to make a leapfrog bid while all other players stay put. Let there be n players, for any n ≥ 3. Pick any m ∈ {2, . . . , n} such that m ≤ v/δ. Call any node of the game a consecutive state up to m if and only if there are m among

the n players, listed as i1, . . . , im, whose current committed payments, denoted respectively

by pi1 , . . . , pim , satisfy

k ∈ {1, . . . , m − 1} ⇒ pik+1 = pik − δ,

j 6∈ {i1, . . . , im} ⇒ pim ≥ pj + δ.

That is, the m top bidders are consecutively positioned according to their current committed payments, one following immediately the other, and none of them shares the same position (level of current committed payment) with others. Note that a consecutive state up to m allows the case where players below the top m bidders share the same position.

Proposition 4 Given the original tie-breaking rule (Section2), for any m ∈ {2, . . . , n} such that m ≤ v/δ, conditional on any consecutive state up to m, there exists a subgame equilibrium in each round of which the current mth-place player bids with probability 1−mδ/v

27 and all other players stay put, and this equilibrium generates a total surplus equal to mδ for the players.

The strategy proﬁle for the equilibrium is:

a. Conditional on any consecutive state up to m, the mth-place player bids with probability 1 − mδ/v and all other players stay put. If he does bid, the next round remains to be a consecutive state up to m (with the previous (m − 1)th-place player becoming the mth one now), and the subgame is played by repeating this step; else the game ends with the current frontrunner getting the prize v.

b. If the current node is not a consecutive state up to m but is a consecutive state up to m0 such that m0 ∈ {2, . . . , m − 1} (such m0 exists because the second-place player is always exactly one step behind the ﬁrst-place player), then the subgame thereafter is played according to provision (a) with m0 replacing the m there.

If everyone abides by plan (a), one can show (Appendix A.11) that the continuation value of being the first-place player is equal to mδ, hence it is a best response for the mth- place player to bid thereby incurring a cost mδ to become the frontrunner. It follows that it is the best response for all the players below him to stay put. For any non-frontrunner above him, say in the kth place with k = 2, . . . , m − 1, deviation to bidding is unprofitable: If she deviates and becomes the next frontrunner, this kth-place player pays kδ to get the continuation value equal to kδ, because her leapfrogging to become the next frontrunner leaves a 2δ-distance between the two players adjacent to her currently, hence the next round is a consecutive state up to only k. That triggers plan (b) in the above strategy profile, with m0 = k. Thus the deviation is not profitable to her. While Proposition4 assumes the original tie-breaking rule for expository convenience, the cases with the alternative tie-breaking rules (Section 7.1) are similar. Except the mth- place player, everyone else becomes more unwilling to bid now than given the original tie- breaking rule. We need only to augment the above strategy profile with a provision that in the off-path event where a tie occurs the subgame thereafter is played according to a punishing equilibrium such as the one in Lemma4, or the equilibrium conditional on states ααβ or ααγ in Proposition1. The remaining question is how the game reaches a consecutive state hypothesized in Proposition4. One way is to use the sunspot state formulated in Section 7.1.2 to coordinate

28 among players on which subgame equilibrium to play. For example, suppose m = n = 3, so the consecutive state we want to reach is αβγ. Given the original tie-breaking rule (Section2), if everyone bids in round one and each follower bids in round two, then αβγ is reached in round three. Should players expect at the outset to play the 3-cyclic equilibrium according to Proposition4, however, they would not bid in round one, because one can show that the 3-cyclic equilibrium gives zero continuation value to the frontrunner at the start of round two. To solve this problem, consider the following sunspot recommendation:

In round three, if the state is the consecutive αβγ, then the 3-cyclic equilibrium is played if the realized sunspot value is 1, and the trilateral equilibrium with

dropout state s∗ = 4 is played if the realized sunspot value is 2.

Since both the 3-cyclic equilibrium and the trilateral equilibrium are SPEs conditional on the state αβγ, it is a best response for each player to follow the recommendation in round three. Thus, in round two each follower would bid for sure: If he bids and becomes the round-three frontrunner, the bidder gets either V2 − 2δ if the trilateral equilibrium is to be played, or 3δ − 2δ = δ if the 3-cyclic equilibrium is to be played, each with probability 1/2.

By contrast, if he does not bid, he gets either L2 in the trilateral equilibrium, or zero in the

3-cyclic equilibrium. Since V2−δ > L2 by (23), we have (V2−2δ+δ)/2 > (L2+0)/2 and hence each follower in round two strictly prefers to bid. Thus, the frontrunner in round two will become the round-three follower for sure. Hence his continuation value is equal to either M2 if the trilateral equilibrium is to be played, or zero if it is the 3-cyclic equilibrium. Thus, the continuation value is M2/2, which by Lemma 10 is greater than (V2 + δ/2)/2 = 9δ/4

(V2 = 4δ). It follows that, in round one, each player is willing to bear the sunk cost δ to become the round-two frontrunner. Hence the consecutive state αβγ is reached. If we abstract away the dollar-auction aspect that all players start with equal footing, a consecutive state could be taken as a standalone model for multiplayer escalation situations where the initial positions of the top players are sequentially ordered. For instance, the United States, Russia and China could be ranked in such a manner in terms of their nuclear weapon stockpiles. Then Proposition4 illustrates a drastic diﬀerence that coordination could make to the total surplus. Given a consecutive state up to m, the players can still play the surplus-dissipating equilibrium of bilateral rivalry,5 which generates a total surplus equal to

5The surplus-dissipating equilibrium can be regarded as a trivial cyclic equilibrium in the case where

29 only 2δ. Alternatively, they can play the m-cyclic equilibrium where only the mth-placed player escalates at all, which generates a total surplus equal to mδ. Moreover, since the only condition about m is that m ≤ v/δ, the total surplus can be almost as large as v. In other words, for any y ∈ R let byc denote the largest integer less than or equal to y; then bv/δc ≤ v/δ. Conditional on any consecutive state up to bv/δc, there is a cyclic equilibrium that generates a total surplus equal to bv/δcδ, no less than just a δ short of the full value v.

8 Conclusion

In considering the dollar auction—a dynamic game that albeit specific has deep roots in the conflict literature—this paper demonstrates a novel dynamic pattern where three contestants perpetuate the trilateral rivalry through a cyclic pattern of leapfrog and accommodation. Importantly, the three-sidedness of the contest is not assumed in our model but rather arises as an equilibrium while there is another equilibrium where any player who happens to fall behind the top two immediately drops out. The normative advantage of the former, the trilateral-rivalry equilibrium, is that it generates larger total surplus across all players than the latter, the bilateral-rivalry equilibrium. Given the current polarizing political climate in the United States, our normative result could be taken as a timely—albeit stylistic— suggestion that adding a third political party to the US two-party system might help to mitigate the more and more acute conflict between the two sides. To sustain such collectively advantageous trilateral rivalry, the contestants accommodate one another at critical junctures of the equilibrium: when the third-place player lags behind the frontrunner by a distance beyond which he will drop out forever, the rival near the top stays put to better the chance that the third-place player can catch up through a leapfrog bid. Thus the game either ends, when the latter fails to make the leapfrog bid, or repeats the trilateral rivalry cycle, when he makes the leapfrog. While the paper focuses on trilateral rivalry, we make a partial extension to a quadrilateral rivalry case. Similar to our main result, at the juncture where the top three rivals have emerged, the quadrilateral rivalry equilibrium we construct produces larger total sur-

m = 2. By deﬁnition of consecutive states, any consecutive state up to m is also a consecutive state up to m0 for all m0 ∈ {2, . . . , m − 1}. Thus the hypothesis of Proposition4 implies that the state is also consecutive up to 2 and hence the surplus-dissipating equilibrium is one of the equilibria characterized by the theorem.

30 plus for the players than its trilateral rivalry counterpart. Furthermore, with the idea of sunspot coordination we construct a subgame perfect equilibrium given any n players such that more than three players take turn to bid against and accommodate one another in a cyclic manner. We have also shown that the multilateral rivalry equilibrium is robust to changes of the tie-breaking rule for the case where multiple players bid simultaneously. Our main result is based on the one originally assumed by Shubik [33]: when multiple players bid simultaneously, only one of them is randomly selected to commit to the bid increment. However, we have extended the equilibrium partially to two other tie-breaking rules, one of which similar to those in online penny auctions, whereby all players who bid simultaneously have to commit to their bid increments. To focus on the question who wins the pie in a conﬂict rather than the question how to allocate the pie eﬃciently, we assume complete information and common value of the contested prize. As this paper demonstrates, the complete-information assumption does not assume away the dynamic subtlety of the game. Analysis of the dollar auction with asymmetric information is doable at least in the two-bidder case, and awaits future work in cases with more than two bidders.

A Proofs

A.1 Lemma1

Claim (i) has been proved in Ødegaard and Zheng [29]. Here we prove (ii), the uniqueness claim. Consider any equilibrium where the underdog stays put forever. Then only the follower’s strategy needs to be speciﬁed. By independence condition (1) of nonparticipants, the follower’s strategy is independent of the state variable, which measures merely the distance from the forever inactive underdog. Thus, the follower’s strategy is a constant probability π of bidding. Let V∗ denote the expected payoﬀ for the frontrunner in any such equilibrium. In the current round, he either wins the prize v with probabilityσ ˜2 or becomes the follower next round. In the latter case, by the symmetry condition in our equilibrium concept, he bids with probability π thereby paying 2δ to become the frontrunner again. Thus, the Bellman

31 equation is

2 V∗ = (1 − π) v + π (−2δ + V∗) .

Note that π > 0, otherwise (π = 0), V∗ = v; with v > 2δ by assumption, the follower would bid for sure, contradicting the supposition π = 0. Also note that π < 1, otherwise the

Bellman equation implies V∗ = −2δ + V∗, contradiction. Now that 0 < π < 1, the follower

is indiﬀerent about bidding, hence V∗ = 2δ. Plug this into the Bellman equation to obtain

π = 1 − 2δ/v. That proves Claim (ii).

A.2 Lemma2

Lemma2 By deﬁnition of Ls, the equilibrium expected payoﬀ for an underdog whose lag from the frontrunner is s, we know that Ls = 0 for all s ≥ v/δ. Starting from any such s and use backward induction towards smaller s, together with the law of motion (8) and the

fact V2 − (s + 1)δ < 0 for all s ≥ s∗ due to the deﬁnition of s∗, we observe that Ls = 0 for

all s ≥ s∗. At any state s ≥ s∗, by (8), an underdog gets zero expected payoﬀ if he does not bid; if he bids then by Eq. (2) there is a positive probability with which he gets a negative

payoﬀ V2 − (s + 1)δ; hence his best response is uniquely to not bid at all. Hence

s ≥ s∗ =⇒ Ls = 0 and πγ,s = qγ,s = 0, (20)

which proves Claim (i) of the lemma. Apply backward induction to (8) starting from s = s∗ and we obtain

2 ≤ s ≤ s∗ − 1 =⇒ V2 − (s + 1)δ ≥ Ls ≥ Ls+1 ≥ 0, (21)

with the inequality Ls ≥ Ls+1 being strict whenever s < s∗ − 1. Thus, for any s < s∗ − 1,

Vs − (s + 1)δ > Ls+1 ≥ 0; hence Eqs. (2) and (8) together imply that an underdog’s best response is uniquely to bid for sure:

2 ≤ s < s∗ − 1 =⇒ Ls > 0 and πγ,s = 1, (22)

which proves Claim (ii) of the lemma.

32 A.3 Claims (i) and (iv) of Theorem1

First, we make several observations ﬁrst. By (8) and (21), L2 is a convex combination

between L3 and V2 − 3δ, with V2 − 3δ ≥ L3 when s∗ ≥ 3. Thus,

s∗ ≥ 3 =⇒ L2 ≤ V2 − 3δ. (23)

Eq. (10), combined with (7) and (8), implies

(23) + Ms∗−1 = qγ,s∗−1L2 ≤ (V2 − 3δ) . (24)

Lemma 5 s∗ 6= 3

Proof Suppose, to the contrary, that s∗ = 3. Hence 0 ≤ V2 − 3δ < δ. Thus, by Eq. (24),

M2 < δ. Then (4) requires that π1 < 1, otherwise V1 = M2 < δ, implying a contradiction that no one would bear the sunk cost δ to become the initial α player. Now consider the decision of any non-α player at the state s = 1, as depicted by (5). Since V2 − 2δ > V2 − 3δ ≥ L2, with the second inequality due to (23), each non-α player at s = 1 would maximize the

probability of becoming the α in the next round, i.e., π1 = 1, contradiction.

Lemma 6 If s∗ ≥ 4 then V3 − 2δ ≥ M2 ≥ L2 > 0.

Proof Suppose that V3 − 2δ < L2. Then, by the fact πγ,2 = 1 (Lemma2.ii and s∗ ≥ 4) and

Eq. (2), the β player at state s = 2 would rather stay put than bid, hence πβ,2 = 0. This, combined with (6) in the case s = 2 and the fact πγ,2 = 1, implies that V2 = M2. Since

V3 − 2δ < L2 coupled with (7) implies M2 ≤ L2, we have a contradiction V2 ≤ L2 < V2, with

the last inequality due to (8). Thus we have proved V3 − 2δ ≥ L2. Therefore, with M2 a

convex combination between V3 − 2δ and L2 (since πγ,2 = 1), V3 − 2δ ≥ M2 ≥ L2. Finally,

to show L2 > 0, note from the hypothesis s∗ ≥ 4 and deﬁnition of s∗ that V2 − 3δ > 0. This positive payoﬀ the underdog at state s = 2 can secure with a positive probability through

bidding. Hence L2 > 0 follows from (8).

Lemma 7 If s∗ ≥ 4 then πγ,s∗−1 > 0.

Proof Section 4.1.

33 Proof of Theorem1.i (Impossibility of Odd Dropout States) Suppose, to the con-

trary, that the dropout state s∗ of an equilibrium is an odd number. Since s∗ ≥ 3 by

hypothesis of the lemma, s∗ ≥ 5. By Lemma6, L2 ≤ M2 ≤ V3 − 2δ. Apply (6) to the

case s = s∗ − 2 and use the fact that πγ,s∗−2 = 1 (thereby ruling out Vs∗−2 → v) due to

Lemma2.ii. Then we have Ms∗−1 ≤ L2, due to (24), and hence Ms∗−1 ≤ M2 ≤ V3 − 2δ.

Reason backward along the transition chain (13), starting from Ms∗−1 ≤ V3 − 2δ and using

the fact L2 ≤ M2 ≤ V3 − 2δ (Lemma6), to obtain

Ms∗−1 ≤ V3 − 2δ ⇒ Vs∗−2 ≤ max{V3 − 2δ, M2} ≤ V3 − 2δ

⇒ Ms∗−3 ≤ max{−2δ + Vs∗−2,L2} ≤ max{V3 − 2δ, L2} ≤ V3 − 2δ

⇒ Vs∗−4 ≤ max{Ms∗−3,M2} ≤ V3 − 2δ ⇒ · · ·

Since s∗ is an odd number and s∗ ≥ 5, this chain eventually reaches V3, i.e., 3 = s∗ − 2m for

some positive integer m. Hence we obtain the contradiction V3 ≤ V3 − 2δ.

Proof of Theorem1.iv That the β player stays put for sure at state s∗ − 1 has been

proved in Section 4.2, where the hypothesis s∗ ≥ 4 is true due to Lemma5. For the rest of

the claim, we ﬁrst prove 0 < πγ,s∗−1 < 1. The ﬁrst inequality follows from Lemma7 since

s∗ ≥ 4. To prove πγ,s∗−1 < 1, suppose to the contrary that πγ,s∗−1 = 1. Then by the fact

πβ,s∗−1 = 0 and (6) applied to the case s = s∗ − 1, we have Vs∗−1 = M2 and Ms∗−1 = L2.

Reason backward along the transition chains (12) and (13), starting from Vs∗−1 = M2 and

Ms∗−1 = L2 and using the fact L2 ≤ M2 (Lemma6). Then we have two chains of inequalities:

Vs∗−1 = M2 ⇒ Ms∗−2 ≤ max{−2δ + M2,L2} ≤ M2

⇒ Vs∗−3 ≤ max{Ms∗−2,M2} ≤ M2

⇒ Ms∗−4 ≤ max{−2δ + Vs∗−3,L2} ≤ max{M2,L2} ≤ M2 ⇒ · · ·

Ms∗−1 = L2 ⇒ Vs∗−2 ≤ max{L2,M2} ≤ M2

⇒ Ms∗−3 ≤ max{−2δ + Vs∗−2,L2} ≤ max{M2,L2} ≤ M2

⇒ Vs∗−4 ≤ max{Ms∗−3,M2} ≤ M2 ⇒ · · ·

The two chains combined lead to Vs ≤ M2 for all s ≤ s∗ − 1. Hence V3 ≤ M2, contradicting

Lemma6. Thus we have proved that πγ,s∗−1 < 1.

With πγ,s∗−1 < 1, bidding is not the unique best response for the γ player at state

s∗ − 1, hence V2 ≤ s∗δ (otherwise the bottom branch of (8) in the case s = s∗ − 1 is strictly

34 positive and, by (20), is strictly larger than the middle branch, so the γ player would strictly

prefer to bid). By deﬁnition of s∗, V2 ≥ s∗δ. Thus V2 = s∗δ.

A.4 Claims (ii) and (iii) of Theorem1 (Action of β When s ≤ s∗−2)

The part of Theorem1.iii on the γ player follows from Lemma2.ii. We prove the part on the β player here, through proving Lemmas9 and 11. The former shows that bidding is a follower’s unique best response at even-number states, and the latter, odd-number states. Lemma 11 uses a Lemma 10, which also implies Claim (ii) of the theorem.

Lemma 8 If the dropout state is an even number s∗ ≥ 4, then L2 < V2 ≤ V3 − 2δ.

Proof Since πβ,s∗−1 = 0 (Theorem1.iv), Ms∗−1 ≤ L2. Thus, since πγ,s∗−2 = 1 (Lemma2.ii),

Vs∗−2 is a convex combination between Ms∗−1, which is less than L2, and M2, which is a convex combination between V3 − 2δ and L2, as πγ,2 = 1. Thus Vs∗−2 is between L2 and

V3 − 2δ. Consequently, Ms∗−3, a convex combination between L2 and Vs∗−2 − 2δ (since

πγ,s∗−3 = 1), is between L2 and V3 − 2δ. Repeating this reasoning, with s∗ being an even number, we eventually reach 2 = s∗ − 2m for some integer m ≥ 1, and obtain the fact that

V2 is a number between L2 and V3 − 2δ. Thus, L2 < V3 − 2δ, otherwise the fact L2 < V2 by (8) would be contradicted. Hence L2 < V2 ≤ V3 − 2δ.

A.4.1 Bidding at Even States

Lemma 9 If the dropout state is an even number s∗ ≥ 4, then πβ,s = 1 for any even

number s such that 2 ≤ s ≤ s∗ − 2.

Proof First, by Lemma8, L2 < V3 − 2δ. Thus at state s = 2 the β player strictly prefers

to bid, i.e., πβ,2 = 1. Second, pick any even number s such that 4 ≤ s ≤ s∗ − 2 and suppose,

to the contrary of the lemma, that πβ,s < 1, which means that the β player at state s does

not strictly prefer to bid. Thus Ms ≤ L2 (as the transition Ms → 0 is ruled out by the

fact πγ,s = 1). Consequently, Vs−1, a convex combination between Ms and M2, is weakly

less than M2, as L2 ≤ M2 by Lemma6. Furthermore, Ms−2, a convex combination between

Vs−1 − 2δ and L2, is less than M2, and that in turns implies Vs−3 ≤ M2. Repeating this

reasoning, with s an even number, we eventually obtain the conclusion that V3 ≤ M2, which

contradicts Lemma6. Thus, πβ,s = 1.

35 If the dropout state is an even number s∗ ≥ 4, since πγ,s = 1 for all s ≤ s∗ − 2 (Lemma2.ii), Eq. (2) and the equal-probability tie-breaking rule together imply

∀s ∈ {2, 3, 4, . . . , s∗ − 2} :[πβ,s = 1 =⇒ qβ,s = qγ,s = 1/2] . (25)

By Lemma9,

2 ≤ s ≤ s∗ − 2 and s is even =⇒ qβ,s = qγ,s = 1/2. (26)

A.4.2 Bidding at Odd States

∞ In the following, we extend the summation notation by deﬁning, for any sequence (ak)k=1,

j X i > j =⇒ ak := 0. (27) k=i

P0 In particular, k=1 ak = 0 according to this notation.

Lemma 10 If the dropout state is an even number s∗ ≥ 4, M2 > V2 + δ/2.

Proof Let m := min{k ∈ {0, 1, 2 ...} : V2k+4 − 2δ ≤ L2}. Note that m is well-deﬁned

because s∗/2 − 2 belongs to the set, as Vs∗ − 2δ = 0 ≤ L2 (Eq. (10)). At any odd state

2k + 1 ≤ 2m + 1 (hence k − 1 < m) we have V2k+2 − 2δ = V2(k−1)+4) − 2δ > L2, with the last inequality due to the deﬁnition of m; hence by (6) in the state s = 2k + 1 the β player bids for sure, i.e. πβ,2k+1 = 1. Thus, (25) implies that qβ,s = qγ,s = 1/2 at any such odd state. Coupled with (26), that means the transition at every state s from 2 to 2m + 2 is that the current β and γ players each have probability 1/2 to become the next α player. Thus,

m m ! m X −2k−1 X −2k−2 −2m−2 X −2k V2 = M2 2 + L2 2 + 2 zm − 2δ 2 , (28) k=0 k=0 k=1

where zm := 1 if 2m+2 < s∗ −2, and zm := 2πγ,s∗−1 −1 if 2m+2 = s∗ −2; and the last series Pm k=1 on the right-hand side uses the summation notation deﬁned in (27) when m = 0.

To understand the term for M2 on the right-hand side, note that M2 enters the calculation of V2 at the even states s = 2, 4, 6,..., 2m − 2, and upon entry at state s and in every round transversing from states s to 2, the M2 is discounted by the transition probability 1/2.

The term for L2 is similar, except that L2 enters at the odd states s = 3, 5, 7,..., 2m − 1, and that the transition probability for the L2 at the last state 2m − 1 is equal to one if

36 2m − 1 < s∗ − 1, and equal to πγ,s∗−1 if 2m − 1 = s∗ − 1. That is why the last two terms within the bracket for L2 are  −2m−2 −2m−2 −2m−1 −2m−2 −2m−2  2 + 2 = 2 if zm = 1 2 +2 zm = −2m−2 −2m−2 −2m−1  2 + 2 (2πγ,s∗−1 − 1) = 2 πγ,s∗−1 if zm = 2πγ,s∗−1 − 1.

The term for −2δ is analogous to that for M2.

With s∗ ≥ 4, V2 − 4δ ≥ 0. Thus, by the above-calculated transition probabilities, 1 1 7 L = (L + V − 3δ) ≤ (V − 4δ + V − 3δ) = V − δ. 2 2 3 2 2 2 2 2 2

This, combined with Eq. (28) and the fact zm ≤ 1 due to its deﬁnition, implies that

m m ! m X 7 X X V ≤ M 2−2k−1 + V − δ 2−2k−2 + 2−2m−2 − 2δ 2−2k 2 2 2 2 k=0 k=0 k=1 m m ! X X 7 < M 2−2k−1 + V 2−2k−2 + 2−2m−2 − δ. 2 2 8 k=0 k=0 Thus, the lemma is proved if

m ! m X X 1 − 2−2k−2 + 2−2m−2 = 2−2k−1, (29) k=0 k=0 as the left-hand side of this equation is clearly strictly between zero and one. To prove (29), we use induction on m. When m = 0, (29) becomes 1 − 2−2 − 2−2 = 2−1, which is true. For any m = 0, 1, 2,..., suppose that (29) is true. We shall prove that the equation is true when m is replaced by m + 1, i.e.,

m+1 ! m+1 X X 1 − 2−2k−2 + 2−2(m+1)−2 = 2−2k−1. (30) k=0 k=0 The left-hand side of (30) is equal to

m ! X 1 − 2−2k−2 + 2−2m−2 + 2−2m−2 − 2−2(m+1)−2 − 2−2(m+1)−2 k=0 m X = 2−2k−1 + 2−2m−2 − 2−2(m+1)−1 (the induction hypothesis) k=0 m X = 2−2k−1 + 2−2m−3, k=0 which is equal to the right-hand side of (30). Thus (29) is true in general, as desired.

37 Lemma 11 If the dropout state is an even number s∗ ≥ 4, then πβ,s = 1 at any odd number

state s such that 1 ≤ s ≤ s∗ − 2.

Proof Pick any odd number s such that s ≤ s∗ −2. It suﬃces to prove that Vs+1 −2δ > L2. Since s + 1 is even, it follows from (26) that

1 1 V = (M + M ) ≥ (M + L ) , s+1 2 2 s+2 2 2 2

with the inequality due to the fact Ms+2 ≥ L2, which in turn is due to the fact that the β

player at state s + 2 can always secure the payoﬀ L2 through not bidding at all. Thus, 1 V − 2δ − L ≥ (M + L ) − 2δ − L s+1 2 2 2 2 2 1 1 = M − L − 2δ 2 2 2 2 1 1 1 1 = M − L + (V − 3δ) − 2δ 2 2 2 2 3 2 2 1 1 1 1 ≥ M − (V − 4δ) + (V − 3δ) − 2δ 2 2 2 2 2 2 2 1 1 1 = M − V − δ, 2 2 2 2 4

with the second inequality due to the deﬁnition of Ls and the fact V2 − 4δ ≥ 0 (s∗ ≥ 4). 1 1 1 Since 2 M2 − 2 V2 − 4 δ > 0 by Lemma 10, Vs+1 − 2δ − L2 > 0, as desired.

Proof of Claims (ii) and (iii) of Theorem1 Claim (iii) of the theorem follows directly from combining Lemmas2.ii,9 and 11. With Claim (iii) established, every non- α player in the state s = 1 (i.e., in the second round) bids for sure, hence (4) implies that the transition

V1 → M2 happens for sure. Thus, the payoﬀ from becoming the initial frontrunner is equal to

M2 − δ. By contrast, in failing to become the initial frontrunner a player gets the payoﬀ M1, 1 which by (5) and Claim (iii) of the theorem is equal to 2 (V2 − 2δ + L2). Thus, the net gain from becoming the initial frontrunner is equal to

1 1 1 M − δ − (V − 2δ + L ) > V + δ/2 − δ − (V − 2δ + L ) = (V − L + δ) > 0, 2 2 2 2 2 2 2 2 2 2 2 with the ﬁrst “>” due to Lemma 10, and the last “>” due to (23). Thus, every player at the initial round strictly prefers being the frontrunner. Hence Claim (ii) of the theorem follows.

38 A.5 Lemma3

All lemmas in this subsection assume the hypotheses in Lemma3, that s∗ ≥ 4 is an even ∞ number and a strategy proﬁle π0, π1, (πβ,s, πγ,s)s=2 according to Theorem1 is given, with

the associated value functions (Vs,Ms,Ls)s derived from (6)–(8) and Eq. (14).

Lemma 12 For any positive integer m such that 2m + 1 ≤ s∗ − 1, if V2m+1 − 2δ ≤ L2 then

V3 − 2δ < L2.

Proof Pick any m speciﬁed by the hypothesis such that V2m+1 − 2δ ≤ L2. Suppose, to the

contrary of the lemma, that V3 − 2δ ≥ L2. Thus, the law of motion (6) in the case s = 2,

with πγ,2 = 1, implies that M2 is between L2 and V3 − 2δ, hence V3 − 2δ ≥ M2 ≥ L2. By the law of motion (7) in the case s = 2m, M2m is a convex combination among zero, V2m+1 − 2δ and L2. Thus the hypothesis implies that M2m ≤ L2. Consequently, the law of motion (6) in the case s = 2m − 1, together with πγ,2m−1 = 1 and M2 ≥ L2, implies that V2m−1 ≤ M2 and hence V2m−1 − 2δ ≤ M2 − 2δ. Then (7) in the case s = 2m − 2 implies M2m−2 ≤ L2. Repeating this reasoning backward, with 3 being odd, we eventually reach state s = 3 and

obtain V3 ≤ M2. But since V3 − 2δ ≥ M2, we have a contradiction V3 − 2δ ≥ M2 ≥ V3.

Lemma 13 Denote x := πγ,s∗−1. For any integer m such that 1 ≤ m ≤ s∗/2 − 1,

m−1 m−1 m−1 ! X −2k+2 X −2k X −2k+1 −2(m−1) Ms∗−(2m−1) = −δ 2 + M2 2 + L2 2 + 2 x , (31) k=1 k=1 k=1 m−1 m m−1 ! X −2k+1 X −2k+1 X −2k −2m+1 Vs∗−2m = −δ 2 + M2 2 + L2 2 + 2 x , (32) k=1 k=1 k=1 m−1 m−1 X −2k+1 −2(m−1) X −2k Vs∗−(2m−1) = −δ 2 + 2 (1 − x)v + L2 2 (33) k=1 k=1 m−1 ! X −2k+1 −2(m−1) +M2 2 + 2 x , k=1 m−1 m X −2k −2m+1 X −2k+1 Ms∗−2m = −δ 2 + 2 (1 − x)v + L2 2 (34) k=0 k=1 m−1 ! X −2k −2m+1 +M2 2 + 2 x , k=1 −s∗+3 L2 = δ s∗ − 4 + 2 . (35)

39 Proof First, we prove Eqs. (31) and (32). When m = 1, Eq. (31), coupled with the

summation notation deﬁned in (27), becomes Ms∗−1 = xL2 = πγ,s∗−1L2, which follows

from (7) and the fact that Vs = 2δ and Ms = 0 for all s ≥ s∗, due to Theorem1. This coupled with Eq. (14) implies that

Vs∗−2 = (Ms∗−1 + M2)/2 = M2/2 + xL2/2, which is Eq. (32) when m = 1 (using again the summation notation in (27)). Suppose, for 0 0 0 any integer m with 1 ≤ m ≤ s∗/2 − 2, that Eqs. (31) and (32) are true with m = m . By the induction hypothesis of (32) and Eq. (14), 1 M 0 = (V 0 − 2δ + L ) s∗−(2m +1) 2 s∗−2m 2 m0−1 ! m0 m0−1 ! 1 X M2 X L2 X 0 = −δ 1 + 2−2k+1 + 2−2k+1 + 1 + 2−2k + 2−2m +1x , 2 2 2 k=1 k=1 k=1 0 0 which is Eq. (31) when m = m + 1. By the above calculation of Ms∗−(2m +1) and Eq. (14), 1 V 0 = M 0 + M s∗−(2m +2) 2 s∗−(2m +1) 2 m0−1 ! m0 ! δ 1 X M2 X = − 1 + 2−2k+1 + 1 + 2−2k+1 2 2 2 k=1 k=1 m0−1 ! L2 X 0 + 1 + 2−2k + 2−2m +1x , 4 k=1 which is Eq. (32) in the case m = m0 + 1. Thus Eqs. (31) and (32) are proved. Next we prove Eqs. (33) and (34). When m = 1, Eq. (33), coupled with the notation P0 k=1 ak = 0, becomes Vs∗−1 = (1 − x)v + xM2, which is true by deﬁnition of x and the fact

πβ,s∗−1 = 0 (Theorem1). Then by Eq. (14)

Ms∗−2 = (Vs∗−1 − 2δ + L2) /2 = ((1 − x)v + xM2 − 2δ + L2) /2, P0 which is Eq. (34) when m = 1 (again using the notation k=1 ak = 0). Suppose, for any 0 0 0 integer m with 1 ≤ m ≤ s∗/2 − 2, that Eqs. (33) and (34) are true with m = m . By the induction hypothesis and Eq. (14), 1 V 0 = (M 0 + M ) s∗−(2m +1) 2 s∗−2m 2 m0−1 m0 δ X 0 L2 X = − 2−2k + 2−12−2m +1(1 − x)v + 2−2k+1 2 2 k=0 k=1 m0−1 ! −1 −1 X −2k −1 −2m0+1 +M2 2 + 2 2 + 2 2 x , k=1

40 which is Eq. (33) in the case m = m0 + 1. By the above calculation and Eq. (14), 1 M 0 = V 0 − 2δ + L s∗−(2m +2) 2 s∗−(2m +1) 2 m0 ! 1 X 0 = −δ 1 + 2−2k+1 + 2−12−2m (1 − x)v 2 k=1 m0 ! m0 ! 1 X M2 X 0 +L + 2−1 2−2k + 2−2k+1 + 2−2m x , 2 2 2 k=1 k=1 which is Eq. (34) in the case m = m0 + 1. Hence Eqs. (33) and (34) are proved. Finally we prove Eq. (35). Applying Eq. (14) to (8) recursively we obtain, for any integer s∗ ≥ 4, that 1 1 1 1 L = V − 3δ + V − 4δ + ··· + (V − (s − 1)δ) 2 2 2 2 2 2 2 2 ∗ δ 1 1 1 = s − 3 + s − 4 + ··· + · 1 2 ∗ 2 ∗ 2 2 1 1 1 1 = δ (s∗ − 3) + (s∗ − 4) + (s∗ − 5) + ··· + , 2 22 23 2s∗−3 which is equal to the right-hand side of (35). In the above multiline calculation, the ﬁrst and second lines are due to V2 = s∗δ (Theorem1.iv).

Lemma 14 Vs∗−2 − 2δ ≥ L2 =⇒ ∀m ∈ {1, . . . , s∗/2 − 1} : Vs∗−2m − 2δ ≥ L2.

Proof By the law of motion and Eq. (14), Eqs. (31), (32), (33), (34) and (35) hold. Denote

µ(m) := 2−2m+1,

−s∗+3 µ∗ := 2 .

Pm−1 −2k −2m+2 With the fact k=1 2 = (1 − 2 )/3, Eq. (32) becomes 2 2 1 V = −δ · (1 − 2µ(m)) + M (1 − 2µ(m)) + µ(m) + L (1 − 2µ(m)) + µ(m)x . s∗−2m 3 2 3 2 3 Hence 2 2 V − 2δ − L = −δ (1 − 2µ(m)) + 2 + M (1 − 2µ(m)) + µ(m) s∗−2m 2 3 2 3 1 −L 1 − (1 − 2µ(m)) − µ(m)x 2 3 4 1 = − (2 − µ(m))δ + (2 − µ(m))M 3 3 2 2 − (s − 4 + µ ) δ (1 + µ(m)) − µ(m)x , ∗ ∗ 3

41 with the second equality due to (35). Thus, Vs∗−2m − 2δ ≥ L2 is equivalent to 1 4 2 (2 − µ(m))M ≥ δ (2 − µ(m)) + (s − 4 + µ ) (1 + µ(m)) − µ(m)x , 3 2 3 ∗ ∗ 3 i.e., M 2(1 + µ(m)) − 3µ(m)x 2 ≥ 4 + (s − 4 + µ ). (36) δ 2 − µ(m) ∗ ∗

Since s∗ − 4 ≥ 0 by hypothesis, and d 2(1 + µ(m)) − 3µ(m)x (2 − µ(m))(2 − 3x) + 2(1 + µ(m)) − 3µ(m)x = dµ(m) 2 − µ(m) (2 − µ(m))2 6(1 − x) = ≥ 0, (2 − µ(m))2 the right-hand side of (36) is weakly increasing in µ(m), which in turn is strictly decreasing in m. Thus the right-hand side of (36) is weakly decreasing in m. Consequently, Vs∗−2m −

2δ − L2 ≥ 0 is satisﬁed for all m if the inequality holds at the minimum m = 1, i.e., if

Vs∗−2 − 2δ − L2 ≥ 0, as claimed.

Proof of Lemma3 Let s ∈ {1, 2, . . . , s∗ − 2}. If s is even and V3 − 2δ ≥ L2, then

Lemma 12 implies Vs+1 − 2δ > L2; thus, by (7) and by the fact that πγ,s = 1 due to the strategy proﬁle speciﬁed in Theorem1, the β player at s gets L2 if he does not bid, and 1 1 2 (Vs+1 − 2δ) + 2 L2 if he does. Hence bidding is the unique best response for β at s. If s is odd and Vs∗−2 − 2δ ≥ L2, then Lemma 14 implies that Vs+1 − 2δ ≥ L2; thus, by the same token as in the previous case, the β player at s weakly prefers to bid.

A.6 Theorem2

Lemma 15 For any even number s∗ ≥ 4, if Eqs. (14) and (35) hold and M2 ≥ V2 = s∗δ, then at the initial and second rounds each player strictly prefers to bid.

Proof First, consider the second round, which means s = 1. For each non-α player, becoming the next α player gives him an expected payoﬀ V2 −2δ = (s∗ −2)δ by the hypothesis

V2 = s∗δ, whereas staying put gives payoff L2, which is less than (s∗ −3)δ by Eq. (35). Thus, each non-α player strictly prefers to bid at state one, hence s = 2 occurs for sure given s = 1. Second, consider the initial state. Based on the analysis of the previous step (from s = 1 to s = 2), becoming the first α yields the expected payoff −δ + M2, whereas staying put yields

42 1 2 (V2 − 2δ + L2). Since M2 ≥ V2 by hypothesis and V2 − 2δ > L2 by the previous analysis, each player strictly prefers to become the ﬁrst α player.

Lemma 16 Any integer s∗ ≥ 3 constitutes an equilibrium if s∗ is an even number and there 3 exists (M2, x, L2) ∈ R+ such that—

a. (M2, x, L2) ∈ R+ × [0, 1] × R+ and it solves simultaneously Eq. (32) in the case m =

s∗/2 − 1, Eq. (34) in the case m = s∗/2 − 1 such that V2 = s∗δ, and Eq. (35);

b. M2 ≥ s∗δ (which is equivalent to V3 − 2δ ≥ L2, cf. (12));

c. Ineq. (36) is satisﬁed in the case m = 1.

Proof Pick any even number s∗ ≥ 4 and assume Conditions (a)–(c). Consider the strategy profile such that everyone bids in the initial round and, in any future round, acts according to the strategy profile specified in Theorem1. This strategy profile implies Eq. (14), which

s∗ allows calculation of the value functions (Vs,Ms,Ls)s=2 via the law of motions. By Condi-

tions (a) and (b), M2 ≥ V2 = s∗δ, hence Lemma 15 implies that bidding at the initial round is a best response for each player, and bidding at second rounds a best response for each non-α player. The incentive for each player to abide by the strategy proﬁle at any state

s ≥ s∗ is the same as in the surplus-dissipating subgame equilibrium. At the state s∗ − 1, bidding with probability x is a best response for the γ player because he is indiﬀerent about

bidding, since V2 −s∗δ = 0 = Ls∗ , and not bidding at all is the best response for the β player

because Vs∗ − 2δ = 0 < L2. At any state s with 2 ≤ s ≤ s∗ − 2, bidding is the best response

for the γ player because V2 − (s + 1)δ > Ls+1 (by Eq. (8)); Condition (c) by Lemma 14 suﬃces the incentive for the β player at every odd state to bid. To incentivize the β player

at every even state s ≤ s∗ − 2 to bid, Lemma 12 says that it suﬃces to have V3 − 2δ ≥ L2, which is equivalent to M2 ≥ L2 since, by the law of motion and Eq. (14), M2 is the midpoint

between V3 − 2δ and L2. Since L2 < s∗δ by Eq. (35), the condition M2 ≥ L2 is guaranteed by Condition (b), M2 ≥ s∗δ.

Lemma 17 For any s∗ ≥ 4, Condition (c) in Lemma 16 implies Condition (b) in Lemma 16.

Proof Condition (c) in Lemma 16 is Ineq. (36) in the case m = 1, i.e., when µ(m) = 2−2m+1 = 1/2. Hence the condition is equivalent to M 2 ≥ 4 + (2 − x)(s − 4 + µ ). (37) δ ∗ ∗ 43 To prove that this inequality implies Condition (b), i.e., M2/δ ≥ s∗, it suﬃces to show

4 + (2 − x)(s∗ − 4 + µ∗) > s∗,

i.e.,

(1 − x)(s∗ − 4) + µ∗(2 − x) > 0,

−s∗+3 which is true because s∗ ≥ 4, µ∗ = 2 > 0 and x ≤ 1.

Lemma 18 Condition (a) in Lemma 16 is equivalent to existence of an x ∈ [0, 1] that solves Eq. (15).

Proof Condition (a) requires existence of (M2, x, L2) ∈ R+ × [0, 1] × R+ that satisﬁes

Eqs. (32), (34) and (35) in the case of m = s∗/2 − 1 and V2 = s∗δ. Combine (32) with (35)

−s∗+3 Pm−1 −2k −2m+2 and use the notation µ∗ := 2 and the fact k=1 2 = (1 − 2 )/3 to obtain 2 2 1 s∗δ = V2 = −δ · (1 − 2µ∗) + M2 (1 − 2µ∗) + µ∗ + δ(s∗ − 4 + µ∗) (1 − 2µ∗) + µ∗x , 3 3 | {z } 3 L2 i.e., M2 1 = (3s∗ + 2(1 − 2µ∗) − (s∗ − 4 + µ∗)(1 − 2µ∗ + 3µ∗x)) . (38) δ 2 − µ∗ By the same token, (34) coupled with (35) is equivalent to

1 1 2 M 1 − (1 − 2µ ) − µ x = −δ 1 + (1 − 2µ ) +(1−x)µ v+δ(s −4+µ ) (1 − 2µ ) + µ , 2 3 ∗ ∗ 3 ∗ ∗ ∗ ∗ 3 ∗ ∗ i.e., M 3µ v 2 (2(1 + µ ) − 3µ x) = ∗ (1 − x) + (2 − µ )(s − 6 + µ ). (39) δ ∗ ∗ δ ∗ ∗ ∗ Plug (38) into (39) and we obtain Eq. (15).

Lemma 19 For any even number s∗ ≥ 4, suppose that Eq. (38) holds. Then Condition (c)

in Lemma 16 is equivalent to Ineq. (16), which is implied by x ≥ 0 if and only if s∗ ≤ 6.

Proof Condition (c) in Lemma 16 has been shown to be equivalent to Ineq. (37). Provided that Eq. (38) is satisﬁed, Ineq. (37) is equivalent to

1 4 + (2 − x)(s∗ − 4 + µ∗) ≤ (3s∗ + 2(1 − 2µ∗) − (s∗ − 4 + µ∗)(1 − 2µ∗ + 3µ∗x)) . 2 − µ∗

44 This inequality, given the fact 1 − 2µ∗ ≥ 0, is equivalent to 1 3(s∗ − 2) x ≥ 5 − 4µ∗ − , 2(1 − 2µ∗) s∗ − 4 + µ∗ i.e., Ineq (16). Given Condition (a) in Lemma 16, which implies x ≥ 0, Ineq. (16) is redundant if and only if the right-hand side of (16) is nonpositive, i.e., 3(2 − µ ) ∗ ≥ 1, 2(1 − 2µ∗)(s∗ − 4 + µ∗) i.e., −s∗+3 −s∗+3 3 (2 − 2 ) s∗ ≤ 4 − 2 + . 2 (1 − 2−s∗+4)

This inequality is satisﬁed when s∗ ∈ {4, 6}, as its right-hand side is equal to ∞ when s∗ = 4, and 61/8 when s∗ = 6. The inequality does not hold, by contrast, when s∗ ≥ 8, as

−s∗+2 −6 −s∗+2 −6 1 − 2 1 − 2 63 s∗ ≥ 8 ⇒ 2 ≤ 2 ⇒ ≤ = 1/4 − 2−s∗+2 1/4 − 2−6 15 −s∗+3 −s∗+3 3 (2 − 2 ) 3 2 63 ⇒ 4 − 2 + < 4 + · · < 8 ≤ s∗. 2 (1 − 2−s∗+4) 2 4 15

Thus, for all even numbers s∗ ≥ 4, Ineq. (37) follows if and only if s∗ ≤ 6.

Lemma 20 Any s∗ ≥ 3 constitutes an equilibrium if s∗ is an even number and—

i. either s∗ ≤ 6 and Eq. (15) admits a solution for x ∈ [0, 1];

ii. or s∗ ≥ 8 and Eq. (15) admits a solution for x ∈ [0, 1] such that Ineq. (16).

Proof The lemma follows from Lemma 16, where Condition (a) has been characterized by Lemma 18, Condition (b) by Lemmas 17 can be dispensed with, and Condition (c), by

Lemma 19, can be dispensed with when s∗ ≤ 6 (hence Claim (i) of the lemma) and is equivalent to Ineq (16) when s∗ > 6 (hence Claim (ii) of the lemma).

Lemma 21 If x = 1, the left-hand side of (15) is less than the right-hand side of (15).

2 Proof When x = 1, the left-hand side of (15) is equal to (2 − µ∗) (s∗ − 6 + µ∗), and the right-hand side equal to

(2(1 + µ∗) − 3µ∗) (3s∗ + 2(1 − 2µ∗) − (s∗ − 4 + µ∗)(1 − 2µ∗ + 3µ∗)) 2 = (2 − µ∗) 2s∗ + 6 − µ∗ − µ∗s∗ − µ∗ .

45 Thus, the lemma follows if

2 (2 − µ∗)(s∗ − 6 + µ∗) < 2s∗ + 6 − µ∗ − µ∗s∗ − µ∗,

−s∗+3 i.e., 9µ∗ < 18, which is true because µ∗ = 2 .

Lemma 22 s∗ = 4 constitutes an equilibrium if and only if v/δ > 35/2, and s∗ = 6 constitutes an equilibrium if and only if v/δ > 6801/120 (= 56.675).

Proof By Lemma 20, with s∗ ≤ 6 the necessary and suﬃcient condition for equilibrium is that Eq. (15) admits a solution for x ∈ [0, 1]. By Lemma 21, the left-hand side of that equation is less than its right-hand side when x = 1. Thus, it suﬃces to show that the left-hand side is greater than the right-hand side when x = 0, i.e., 3µ v ∗ (2−µ )+(2−µ )2(s −6+µ ) > 2(1+µ ) (3s + 2(1 − 2µ ) − (s − 4 + µ )(1 − 2µ )) , δ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ which is equivalent to v (2 − µ ) > s (4 + µ ) + (6 − µ )(2/µ − 2 − µ ). δ ∗ ∗ ∗ ∗ ∗ ∗

Since µ∗ is equal to 1/2 when s∗ = 4, and equal to 1/8 when s∗ = 6, the above inequality is equivalent to v/δ > 35/2 when s∗ = 4, and v/δ > 6801/120 when s∗ = 6.

Proof of Theorem2 Claims (i) and (ii) of the theorem are just Lemma 22. To prove

Claim (iii), pick any even number s∗ ∈ {8, 10, 12,...}. By Lemma 20.ii, s∗ constitutes an equilibrium if Eq. (15) admits a solution for x ∈ [0, 1] that satisﬁes Ineq. (16). By Lemma 21, the left-hand side of (15) is less than its right-hand side when x = 1. Thus, it suﬃces to show that the left-hand side is greater than the right-hand side when x is equal to some number greater than or equal to the right-hand side of Ineq. (16). To that end, note from s∗ ≥ 8

−s∗+3 that µ∗ = 2 ≤ 1/32, hence 2 − µ∗ ≥ 63/32 and 1 + µ∗ < 33/32. Thus, the left-hand side of (15) is greater than

3µ v 63 632 ∗ (1 − x) + (s − 6), δ 32 32 ∗ and the right-hand side of Ineq. (16) 3(2 − µ ) 3 × 63/32 1 − ∗ < 1 − . 2(1 − 2µ∗)(s∗ − 4 + µ∗) 2 × 1 × (s∗ − 3)

46 Therefore, it suﬃces, for s∗ to constitute an equilibrium, to have 3µ v 63 632 ∗ (1 − x) + (s − 6) δ 32 32 ∗ greater than or equal to the right-hand side of (15) when 3 × 63 x = x∗ := 1 − . 64(s∗ − 3)

To that end, denote φ(s∗, x) for the right-hand side of (15), i.e.,

φ(s∗, x) := (2(1 + µ∗) − 3µ∗x) (3s∗ + 2(1 − 2µ∗) − (s∗ − 4 + µ∗)(1 − 2µ∗ + 3µ∗x))

−s∗+3 (recall that µ∗ = 2 ). Note, from 0 < x < 1, that −1/32 < µ∗(2 − 3x) < 1/16. Hence 63 1 1 33 = 2 − < 2(1 + µ ) − 3µ x < 2 + = , 32 32 ∗ ∗ 16 16 15 1 1 33 = 1 − < 1 − 2µ + 3µ x < 1 + = . 16 16 ∗ ∗ 32 32

Thus, the ﬁrst factor 2(1 + µ∗) − 3µ∗x of φ(s∗, x) is positive for all x ∈ (0, 1). If the second

factor of φ(s∗, x) is nonpositive when x = x∗ then φ(s∗, x∗) ≤ 0 and we are done, as the left-hand side of (15) is positive. Hence we may assume, without loss of generality, that

3s∗ + 2(1 − 2µ∗) − (s∗ − 4 + µ∗)(1 − 2µ∗ + 3µ∗x∗) > 0.

Consequently, φ(s∗, x∗) can only get bigger if we replace its ﬁrst factor by the upper bound

33/16, and the term 1−2µ∗ +3µ∗x in the second factor by its lower bound 15/16 (note that,

in the second factor, s∗ − 4 + µ∗ > 0 because s∗ ≥ 8). I.e., φ(s∗, x∗) is less than 33 15 33 33 23 79 3s + 2(1 − 2µ ) − (s − 4 + µ ) = s + − µ 16 ∗ ∗ 16 ∗ ∗ 16 16 ∗ 4 16 ∗ 33 33 23 < s + 16 16 ∗ 4

< 5s∗ + 12.

Therefore, the above observations put together, we are done if

3µ v 63 632 ∗ (1 − x ) + (s − 6) ≥ 5s + 12 δ ∗ 32 32 ∗ ∗ In other words, it suﬃces to have 2 3µ∗v 3 × 63 63 63 · · + (s∗ − 6) ≥ 5s∗ + 12, δ 64(s∗ − 3) 32 32

47 i.e., 32µ v 32 × 64 ∗ ≥ −2(s − 6)(s − 3) + (5s + 12) (s − 3). δ ∗ ∗ 632 ∗ ∗ 32×64 With 632 ≈ 0.516, the above inequality holds if 32µ v ∗ ≥ −2(s − 6)(s − 3) + (5s + 12)(s − 3), δ ∗ ∗ ∗ ∗ i.e., 9µ v ∗ ≥ 3s2 + 15s − 72, δ ∗ ∗

−s∗+3 which, coupled with µ∗ = 2 , is equivalent to the hypothesis (17) of Claim (iii).

A.7 Theorems3

First, when s = s∗ = 1, clearly Ms = πβ,s∗−1L2 = πβ,s∗−1δ/2 > 0 and Ls = Ls∗−1 = 0.

Second, we show that if s < s∗ − 1 then Ms > 0 and Ls > 0. By Eq. (8) and Theorem1, 1 1 Ls ≥ 2 (V2 − (s + 1)δ) = 2 (s∗δ − (s + 1)δ) > 0, with the last inequality due to s ≤ s∗ − 2.

By (7), the follower can secure an expected payoﬀ no less than L2, through staying put at s

(while the underdog bids for sure), hence M2 ≥ L2 > 0. Third, for any state s ≤ s∗ − 1, we show that Vs > 2δ. With s ≤ s∗ − 1, the frontrunner’s surplus, by (6), is

i. either Vs = (M2 + Ms+1)/2 (if s < s∗ − 1)

ii. or Vs = (1 − πγ,s∗−1)v + πγ,s∗−1M2 (if s = s∗ − 1).

In Case (i), his surplus is

(M2 + Ms+1)/2 > M2/2 > (V2 + δ/2) /2 ≥ (4δ + δ/2) /2 > 2δ,

with the second inequality due to Lemma 10, and the third inequality due to V2 = s∗δ and s∗ ≥ 4. In Case (ii), his surplus is

(1 − πγ,s∗−1)v + πγ,s∗−1M2 > (1 − πγ,s∗−1)2δ + πγ,s∗−14δ > 2δ, with the ﬁrst inequality due to v > 2δ by assumption, the fact M2 > V2 by Lemma 10, and the fact V2 ≥ 4δ as in Case (i).

48 A.8 Details of Section 7.1.1 (Multiple Frontrunners)

Recall the notations in Section 7.1.1. The configuration from a player’s perspective is denoted by the state with the player’s position encircled: trilateral-rivalry configurations: ○α αα, ○α αβ, ○α αγ, ○α ββ, ○α βγ, α○β β, α○β γ, αα○β , αα○γ , and αβ○γ ; bilateral-rivalry configurations: ○α α, ○α β, and α○β ; and the terminal states W and Λ. Given any nonterminal configuration s, πs denotes the bidding probability of the encircled player, and Vs, Ms and Ls denote the continuation values. Recall the description of the hypothesized equilibrium:

Assumption 9 ≤ v/δ ≤ 108.

1/2 Strategy proﬁle π○α αα = 1 − (9δ/v) , π○α αβ = 2/3, π○α αγ = 1/2, πα○β β = 1/3, πα○β γ = 0,

παα○β = 0, παα○γ = 0, παβ○γ = 1 − 3δ/v, πα○β = 1 − 2δ/v,, π○α α = 1/2

Continuation values V○α βγ = 3δ, V○α ββ = v/9, V○α β = 2δ, VW = v, V○α αα = V○α αβ =

V○α αγ = Mαα○β = Mα○β β = Mα○β γ = Lαα○γ = Lαβ○γ = V○α α = Mα○β = LΛ = 0.

1/2 Note that the bidding probability π○α αα = 1 − (9δ/v) is well-deﬁned because of the lower bound assumption 9 ≤ v/δ.

Given each conﬁguration s, we shall verify that the above-listed bidding probability πs is a best response, and the continuation values according to the Bellman equation are equal to the ones stated above.

Bilateral configurations ○α α, ○α β and α○β According to [29, Prop.5.2], the continuation value for the α- and β-players in the bilateral-rivalry configurations αβ are 2δ and 0, and the continuation value for the bilateral tie-configuration αα is also 0; the equilibrium bidding probability is equal to 1 − 2δ/v for the β-player in αβ, and equal to 1/2 for each of the typing α-players in αα.

(Trilateral) conﬁguration ○α αα To bid at this position, a player commits to an additional δ payment thereby getting an expected payoﬀ equal to

2 2 −δ + π○α ααV○α αα + 2π○α αα(1 − π○α αα)V○α αβ + (1 − π○α αα) V○α ββ 2 = δ + 0 + 0 + (9δ/v)1/2 (v/9) = 0.

49 If the player does not bid, his expected payoﬀ is equal to

2 2 π○α ααMαα○β + 2π○α αα(1 − π○α αα)Mα○β β + (1 − π○α αα) · 0 = 0.

1/2 Thus, the bidding probability π○α αα = 1 − (9δ/v) is a best response, and V○α αα = 0.

Conﬁguration ○α αβ For an α-player, bidding means incurring a sunk cost δ thereby getting an expected payoﬀ equal to

−δ + π○α αβπαα○β V○α αα + 1 − π○α αβ παα○β V○α αβ + 1 − π○α αβ 1 − παα○β V○α βγ = −δ + 0 + 0 + (1 − 2/3)(1 − 0)3δ = 0.

And staying put gives an α-player an expected payoﬀ is equal to

π○α αβπαα○β Mαα○β + (1 − π○α αβ)παα○β Mα○β β + π○α αβ(1 − παα○β )Mα○β γ = 0 + 0 + 0 = 0.

Thus, the bidding probability π○α αβ = 2/3 is a best response, and V○α αβ = 0.

Conﬁguration ○α αγ In bidding, an α-player commits to the δ additional payment thereby getting an expected payoﬀ equal to

−δ + π○α αγπαα○γ V○α αα + (1 − π○α αγ)παα○γ V○α αβ + π○α αγ(1 − παα○γ )V○α α + (1 − π○α αγ)(1 − παα○γ )V○α β = −δ + 0 + 0 + 0 + (1/2)(1 − 0)2δ = 0.

And staying put gives an α-player an expected payoﬀ is equal to

π○α αγπαα○γ Mαα○β +(1−π○α αγ)παα○γ Mα○β β +π○α αγ(1−παα○γ )Mα○β +(1−π○α αγ)(1−παα○γ )·0 = 0.

Thus, the bidding probability π○α αγ = 1/2 is a best response, and V○α αγ = 0.

Conﬁguration ○α ββ At this conﬁguration, the α-player has no decision to make. Given

the proposed bidding probabilities πα○β β = 2/3,

2 2 2 V○α ββ = πα○β βMαα○β + 2πα○β β(1 − πα○β β)Mα○β γ + (1 − πα○β β) v = 0 + 0 + (1 − 2/3) v = v/9.

50 Conﬁguration α○β β In bidding, a β-player commits to an additional payment 2δ thereby getting an expected payoﬀ equal to

−2δ + πα○β βV○α αβ + (1 − πα○β β)V○α βγ = −2δ + 0 + (1 − 1/3)3δ = 0.

And by not bidding the player gets an expected payoﬀ equal to

πα○β βLαβ○γ + (1 − πα○β β) · 0 = 0.

Thus, the bidding probability πα○β β = 1/3 is a best response, and Mα○β β = 0.

Conﬁguration ○α βγ The α-player here has no decision to make. Given the proposed bidding probabilities πα○β γ = 0 and παβ○γ = 1 − 3δ/v,

V○α βγ = πα○β γπαβ○γ Mαα○β + (1 − πα○β γ)παβ○γ Mα○β γ + πα○β γ(1 − παβ○γ )Mα○β + (1 − πα○β γ)(1 − παβ○γ )v = 0 + (1 − 0)(1 − 3δ/v) · 0 + (1 − 0)(3δ/v)v = 3δ.

Conﬁguration α○β γ In bidding, the β-player commits to an additional 2δ payment thereby getting an expected payoﬀ equal to

−2δ + παβ○γ V○α αβ + (1 − παβ○γ )V○α β = −2δ + 0 + (3δ/v)2δ = −2δ (1 − 3δ/v) , which is negative because 1 − 3δ/v > 0 due to the lower bound assumption v/δ > 9. By contrast, through not bidding, the player gets an expected payoﬀ equal to

παβ○γ Lαβ○γ + 1 − παβ○γ · 0 = 0.

Thus πα○β γ = 0 is the best response and Mα○β γ = 0.

Conﬁguration αβ○γ In bidding, the γ-player commits to an additional payment 3δ thereby getting an expected payoﬀ equal to

−3δ + πα○β γV○α αβ + (1 − πα○β γ)V○α βγ = −3δ + 0 + (1 − 0)3δ = 0; and the player’s expected payoﬀ from not bidding is equal to zero since the β-player does not bid. Thus, παβ○γ = 1 − 3δ/v is a best response, and Lαβ○γ = 0.

51 Conﬁguration αα○β In bidding, the β-player commits to an additional payment 2δ thereby getting an expected payoﬀ equal to

2 2 2δ + π○α αβV○α αα + 2π○α αβ(1 − π○α αβ)V○α αβ + (1 − π○α αβ) V○α ββ = −2δ + 0 + 0 + (1 − 2/3)2v/9 = −2δ + v/81, which is negative since v/δ < 162 by the upper bound assumption v/δ < 108. By contrast, through not bidding the β-player gets an expected payoﬀ equal to

2 2 π○α αβLαα○γ + 2π○α αβ(1 − π○α αβ)Lαβ○γ + (1 − π○α αβ) · 0 = 0.

Thus, the bidding probability παα○β = 0 is the best response, and Mαα○β = 0.

Conﬁguration αα○γ In bidding, the γ-player commits to an additional payment 3δ thereby getting an expected payoﬀ equal to

2 2 −3δ + π○α αγV○α αα + 2π○α αγ(1 − π○α αγ)V○α αβ + (1 − π○α αγ) V○α ββ = −3δ + 0 + 0 + (1/2)2v/9 = −3δ + v/36,

which is negative because of the upper bound assumption v/δ < 108. By contrast, the player

can secure a zero payoﬀ through staying put. Thus, the bidding probability παα○γ = 0 is the

best response, and Lαα○γ = 0. We have exhausted all the nonterminal configuration that the proposed bidding strategy π can reach. For each configuration s, we have verified that π conditional on s is a best response given the continuation values of the successive configurations, and that the

continuation value of s is equal to the asserted amount. Thus Proposition1 is proved.

A.9 Details of Section 7.1.2 (Provisional Winner)

Lemma4 First, consider player α2 in a tie state. Since player β is supposed to stay put

according to (i), if α2 does not bid then the game ends with α1 being the winner and α2

getting zero payoﬀ. If α2 bids, he bears a sunk cost δ and becomes the only frontrunner in the next round, which according to provision (ii) gives him a continuation value equal to 2δ, and gives the provisional winner (who becomes the follower in the next round) zero continuation

value (Lemma1). Thus, bidding is the best response for α2, as prescribed by (i). Note also

that the on-path expected payoﬀ for an α2 player is equal to 2δ − δ = δ.

52 Second, consider player β at a tie state. Since player α2 is supposed to bid for sure,

if β bids then he ties with α2 and is selected the provisional winner with probability 1/2. As noted above, his continuation value is equal to zero if he is selected the provisional winner, and is equal to δ if otherwise. Thus, if β bids then his net expected payoff is equal to 1 1 2 · 0 + 2 δ − δ = −δ/2. By contrast, if he stays put, then according to provision (ii) he gets zero net payoff. Thus staying put, as (i) prescribes, is the best response for β. Third, consider the underdog γ in a sequential state s. If he stays put according to provision (ii), the player gets zero expected payoff. If he deviates to bid, then the player bears a sunk cost equal to (s + 1)δ and either (i) becomes the unique frontrunner (when the follower stays put) or (ii) ties with the follower (when the latter bids). In Case (i), according to provision (ii) the continuation value for this deviating γ is equal to 2δ. In Case (ii), by the calculation in the previous paragraph, the continuation value is equal to −δ/2. Thus, in either case, his net expected payoff is negative, because the sunk cost for this deviation is (s + 1)δ ≥ 3δ. Thus, staying put, as (ii) prescribes for γ, is the best response for γ. Finally, at any sequential state, it is a best response for the current follower to abide by provision (ii). That is because the γ player is supposed to stay put. Hence the play prescribed by (ii) is just the surplus-dissipating subgame equilibrium constructed in Lemma1.

Proposition2 Pick any trilateral equilibrium in Theorem1. If all players abide by Pro- visions (i)–(v) in Section 7.1.2, the transition function among the roles (α, β and γ) is the same as that in the original equilibrium, and so the continuation values of these roles in ∞ various states are equal to those in the original equilibrium, denoted as (Vs,Ms,Ls)s=1. To verify players’ incentive to abide by (i)–(iv), ﬁrst consider any tied state (deﬁned in Section 7.1.2). By Lemma4, provision (v) constitutes a subgame equilibrium conditional on 1 1 that state and each tying bidder gets a continuation value equal to 2 δ + 2 ·0 = δ/2 according to Lemma4.iii and the tie-breaking rule that picks a provisional winner from the two with equal probabilities.

Second, consider any sequential state s for which 2 ≤ s < s∗ − 1. Suppose that the realized sunspot value in that round is for the follower β to bid. The follower gets

Vs+1 − 2δ if he bids, and zero if he stays put (since γ is now supposed to stay put). In the original equilibrium (Theorem1), β is willing to bid and he gets L2 if he does not bid, hence

Vs+1 − 2δ ≥ L2 ≥ 0. Thus, in the current model of Section 7.1.2, β is willing to bid as

53 well. Now consider the underdog γ. If γ stays put, he gets Ls+1, which is nonnegative. If he deviates to bid, he gets δ/2 − (s + 1)δ < 0. Thus γ would stay put. The other case, where the sunspot stipulates the underdog to bid and the follower to stays put, is analogous. Third, consider the second round. Suppose that the realized sunspot value is for follower k to bid and the other follower j to stay put. Player k gets V2 − 2δ (> 0) if he bids, and zero if he does not (as j is supposed to stay put). Player j gets L2 if she stays put, and δ/2 − 2δ < 0 if she bids. Clearly each is willing to abide by the proposed strategy.

Finally, consider the initial round. The player who is supposed to bid gets V1 − δ if he abides by the proposed strategy, and zero if otherwise (since the other two are expected to stay put). Note V1 = M2 by the proposed strategy and V2 + δ/2 > δ by Lemma 10. Each of the other two players gets M1 from staying put, and δ/2 − δ < 0 from deviation to bid.

Thus, each player is willing to abide by the proposed strategy.

A.10 The Quadrilateral Equilibrium in Section 7.2.1

First, from our construction of trilateral-rivalry equilibria, one can obtain the associated value function for the trilateral-rivalry equilibrium with dropout state s∗ = 4:

V2 = 4δ, (40)

V3 = (16 + 3/2 − πγ,3) δ, (41)

M2 = V3/2 − 3δ/4, (42)

L2 = δ/2, (43)

M3 = πγ,3L2 = πγ,3δ/2, (44)

L3 = 0. (45)

Second, in any subgame where the four players are positioned as in (18), let A denote the continuation value for α, B the continuation value for β, C for γ1, and D for γ2. By

Provision (c.i) of the strategy proﬁle proposed in Section 7.2.1, players β, γ1 and γ2 each bid for sure given this positioning. Thus the conﬁguration in the next round is one of the following three (with denoting an empty slot), each with probability 1/3:

[β, α, , γ1, γ2], t = 4; (46)

[γ1, α, β, , γ2], t = 4; (47)

[γ2, α, β, γ1], t = 3. (48)

54 If it is (46) or (47), player γ2 quits and the other three play the trilateral-rivalry equilibrium with s∗ = 4; if it is (48) then each non-frontrunner bids for sure, as in (18). Note that 1 D = (A − 4δ) (49) 3

because the γ2 in (18) quits, thereby getting zero payoﬀ, unless (48) happens. Since (46), (47) and (48) each happen with probability 1/3,

1 A = (M + M + B), (50) 3 3 2 1 (43) 1 3 B = ((V − 2δ) + L + C) = V − δ + C , (51) 3 3 2 3 3 2 and

1 C = (L + (V − 3δ) + D) 3 3 2 1 1 = δ + (A − 4δ) by (40), (45) and (49) 3 3 1 1 = −δ + (M + M + B) by (50) 9 3 3 2 1 π 1 3 1 3 = −3δ + γ,3 δ + V − δ + V − δ + C by (44), (42) and (51). 27 2 2 3 4 3 3 2

Thus,

1 5 1 C = V + −9 + (6π − 15) δ 80 2 3 4 γ,3 (41) 1 5 3 1 = 16 + − π δ + −9 + (6π − 15) δ 80 2 2 γ,3 4 γ,3 1 = (31 − π ) δ. (52) 80 γ,3

This plugged into Eq. (51) gives

1 3 1 B = V − δ + (31 − π ) δ 3 3 2 80 γ,3 (41) 1 3 3 1 = 16 + − π δ − δ + (31 − π ) δ 3 2 γ,3 2 80 γ,3 1 31 1 = 16 + − 1 + π δ. (53) 3 80 80 γ,3

55 Plugging this into Eq. (50), we have 1 1 31 1 A = M + M + 16 + − 1 + π δ 3 3 2 3 80 80 γ,3 (44),(42) 1 π 1 3 1 31 1 = γ,3 δ + V − δ + 16 + − 1 + π δ 3 2 2 3 4 3 80 80 γ,3 (41) 1 π 1 3 3 1 31 1 = γ,3 δ + 16 + − π δ − δ + 16 + − 1 + π δ 3 2 2 2 γ,3 4 3 80 80 γ,3 1 31 1 = 40 + − 1 + π δ (54) 9 80 80 γ,3 Then Eq. (49) implies 1 31 1 D = 4 + − 1 + π δ. (55) 27 80 80 γ,3

Recall that 0 < πγ,3 < 1, which plugged into Eqs. (52), (53), (54) and (55) implies

3 32 2 8 δ < C < 80 δ = 5 δ, (56) 1 1 5δ = 3 · 15δ < B < 3 · 17δ < 6δ, (57) 1 1 4δ < 9 · 39δ < A < 9 · 41δ < 5δ, (58) 1 3 5 9 δ = 27 δ < D < 27 δ. (59)

We verify the equilibrium conditions through backward induction. Let us start with any subgame with state t ≥ 4. Expecting the trilateral-rivalry equilibrium to be played in the subgame (Provision (e)), the current frontrunner, follower and third-place bidder

each ﬁnds it a best response to abide by it. The lowest-place bidder γ2 cannot proﬁt from the deviation of leapfrogging to the top: being on the top gives him a continuation value

equal to V2 = 4δ because he and the previous frontrunner and follower form a consecutive trilateral conﬁguration, while the leapfrog costs him a payment at least as large as 5δ. Thus, the proposed strategy proﬁle is an equilibrium in this subgame.

Next consider any subgame with t = 3, i.e., the conﬁguration (18). For player γ2: if he does not bid then the state becomes t = 4 next round, in which he will quit and get zero; whereas if he bids now and becomes the next frontrunner, he gets a payoﬀ A − 4δ,

which is positive by (58); thus bidding is his best response. For player γ1: if he becomes the next frontrunner (through bidding), then the next conﬁguration becomes (47), giving him a

payoﬀ equal to V2 − 3δ = δ; whereas, if γ1 does not bid, the next round is either (46), giving

him a payoﬀ L3 = 0, or (48), giving him a payoﬀ D < 5δ/27 by (59). Thus, bidding is the

56 best response for player γ1. For player β: if he bids and becomes the next frontrunner, the next configuration is (46), giving him a payoff equal to V3 − 2δ, which is larger than 14δ by Eq. (41); whereas, if β does not bid, then the next configuration is either (47), giving him a payoff L2 = δ/2, or (48), giving him a payoff C < 2δ/5 by (56). Thus bidding is the best response for player β. Therefore, the proposed strategy profile constitutes an equilibrium in any subgame that starts with the consecutive configuration (18). We next consider any subgame with t = 2, which means the third round, with configuration in the form   α    β  . (60)   {γ1, γ2} For player β: if he becomes the next frontrunner, then according to Provision (c.ii) he gets

(41) 3 3 V − 2δ = 16 + − π δ − 2δ = 14 + − π δ; 3 2 γ,3 2 γ,3 by contrast, if he stays put then, since both of the γ players are bidding according to Provision (b), β in the next round will become the third-place bidder in the consecutive configuration and hence gets payoff C < 2δ/5 by (56). Thus bidding is the best response for β. For each of the γ players: if he does not bid then he becomes the fourth-place player in the next round thereby getting only D < 5δ/27 by (59); whereas bidding and becoming the next frontrunner gives him A − 3δ > δ by (58); thus bidding is the best response for each of γ1 and γ2. Hence the proposed strategy profile constitutes an equilibrium in any subgame with t = 2. Next consider any subgame with t = 1, which means the second round, at which the configuration is in the form   α   . (61) {γ1, γ2, γ3} If a γ player does not bid, then in the third round he will become one of the two γ players in the configuration (60) and hence his payoff will be equal to

1 1 1 (58),(59) 1 5 59 · 0 + (A − 3δ) + D < (5δ − 3δ) + δ = δ < δ, 3 3 3 3 81 81 where the zero term on the left-hand side is his payoﬀ in the event where the β player in the third round gets to become the next frontrunner thereby starting the subgame equilibrium

57 according to Provision (c.ii), rendering zero expected payoﬀ for both bottom-row players, this γ one of them. By contrast, if this γ player bids and becomes the next frontrunner, then in the third round he will be the α in conﬁguration (60), which in the fourth round will become one of the following, each with probability 1/3:       β γ1 γ2        α   α   α          ,   ,   ; (62)  ∅   β   β        {γ1, γ2} γ2 γ1

thus, when t = 1, his expected payoﬀ from bidding and becoming the next frontrunner is

1 2 (57) 2 4 M + B − 2δ > · 5δ − 2δ = δ > δ. 3 3 3 3 3

Hence bidding is the best response for the player. Thus, the proposed strategy constitutes an equilibrium in any subgame with t = 1. Finally, consider t = 0, i.e., the initial round. Consider any bidder i. If i bids and becomes the frontrunner (in the second round), then he will become the β in the configuration (60) in the third round and then, in the fourth round, his position is the same as the β’s position in one of the three configurations in (62). Thus his payoff from bidding in the initial round and becoming the first frontrunner is equal to

1 2 (41) 1 3 2 23 2 3 49 −δ + (V − 2δ) + C = −δ + 16 + − π δ − 2δ + C > δ + · δ = δ. 3 3 3 3 2 γ,3 3 6 3 8 12

If i does not bid in the initial round, he becomes in the second round one of the γ players in the conﬁguration (61); then he either (i) pays 2δ to become the frontrunner in the third round (and become the second-place player in the fourth round to get either M3 or B), or (ii) becomes one of the two γ players in the conﬁguration (60) in the third round. In Case (ii), as shown in the previous step on t = 2, the best outcome for the bidder is to get A − 3δ.

Since M3 < δ/2 by (44) and B − 2δ < 4δ by (57), Case (i) renders less than 4δ for him; since A − 3δ < 2δ by (58), Case (ii) gives him less than 2δ. Thus, either case in the alternative of not bidding produces less than 4δ, while bidding and becoming the initial frontrunner yields more than 4δ. Thus, bidding is the best response for each player i in the initial round. Therefore, the quadrilateral-rivalry strategy proﬁle is a subgame perfect equilibrium.

58 A.10.1 Larger total Surplus for the Bidders

First, we calculate

(42) 1 3 (41) 1 3 3 1 M = V − δ = 16 + − π δ − δ = 8 − π δ 2 2 3 4 2 2 γ,3 4 2 γ,3 and hence

M − B (53) 1 1 31 1 2 = 8 − π − 16 + − 1 + π δ 2 γ,3 3 80 80 γ,3 1 31 1 1 1 = 8 − − + 1 + π . (63) 3 80 2 3 80 γ,3

By the fact L2 = δ/2 and V2 = 4δ, we also have

L − C (52) 1 1 9 π 2 = − (31 − π ) = − γ,3 , δ 2 80 γ,3 80 80 A − V (54) 1 31 1 22 1 31 1 1 2 = 40 + − 1 + π − 2δ = + · − 1 + π . δ 9 80 80 γ,3 9 9 80 9 80 γ,3 Combine these two equations with Eqs. (55) and (63) to obtain 1 ((A + B + C + D) − (V + M + L )) δ 2 2 2 1 1 1 1 = (A − V ) − (M − B) − (L − C) + D δ 2 δ 2 δ 2 δ 22 1 31 1 1 1 31 1 1 1 = + · − 1 + π − 8 − + + 1 + π 9 9 80 9 80 γ,3 3 80 2 3 80 γ,3 9 π 1 31 1 − + γ,3 + 4 + − 1 + π 80 80 27 80 80 γ,3 22 8 4 1 31 1 31 1 31 9 = − + + · + · + · − 9 3 27 9 80 3 80 27 80 80 1 1 1 1 1 1 + + + − − 1 + π 2 80 3 9 27 80 γ,3 2 322 1 1 5 1 = − + + + + 1 + π 27 80 · 27 2 80 27 80 γ,3 2 320 > − + 27 80 · 27 2 = . 27 Thus, when the consecutive configuration (18) occurs for the first time, right after the third round (with current price 3δ), the quadrilateral rivalry equilibrium yields larger total expected payoff for all four players than the trilateral rivalry one does.6 6Furthermore, at this node, for the fourth-place bidder (whose committed payment is still zero), staying put (thereby sticking to the trilateral rivalry equilibrium) is weakly outperformed by bidding (thereby playing

59 A.11 Cyclic Equilibria with m Bidders (Proposition4)

Given the explanation immediately below the statement of Proposition4, we need only to verify that the on-path continuation value for the frontrunner is equal to mδ. Let V∗ denote the expected payoff for the current frontrunner when everyone abides by the strategy profile. The frontrunner either wins the prize v now, with probability mδ/v, or becomes the follower next round. In the latter case, either the game ends in one of the next m − 1 rounds, with this player getting zero payoff, or the game continues up to the round where he becomes the mth-ranked player and hence gets to pay mδ to become the frontrunner again. Thus,

m V∗ = (mδ/v)v + (1 − mδ/v) (−mδ + V∗) ,

which is true if and only if V∗ = mδ.

References

[1] P. Amengual and R. Toral. Truels, or survival of the weakest. Computing in Science & Engineering, 8(5):88–95, 2006.1

[2] Reiko Aoki. R&D competition for product innovation: An endless race. American Economic Review: Papers and Proceedings, 81(2):252–256, 1991.1

[3] Ned Augenblick. The sunk-cost fallacy in penny auctions. Review of Economic Studies, 83(1):58–86, 2016.1

[4] Michael R. Baye, Dan Kovenock, and Casper G. de Vries. The all-pay auction with complete information. Economic Theory, 8:291–305, 1996.1

[5] Walter Bossert, Steven J. Brams, and D. Marc Kilgour. Cooperative vs non-cooperative truels: Little agreement, but does that matter? Games and Economic Behavior, 40:185– 202, 2002.1

[6] David Bowman and Jon Faust. Options, sunspots, and the creation of uncertainty. Journal of Political Economy, 105:957–975, 1997.4

the quadrilateral rivalry one): staying put means getting zero; whereas, bidding either ends with V2 − 4δ = 0 (if the leapfrogging does not result in the switch of equilibrium) or A − 4δ > 0 (if the leapfrogging leads to the switch of equilibrium).

60 [7] Steven J. Brams, D. Marc Kilgour, and Bryan Dawson. Truels and the future. Math Horizons, 10(4):5–8, 2003.1

[8] Christopher Budd, Christopher Harris, and John Vickers. A model of the evolution of duopoly: Does the asymmetry between ﬁrms tend to increase or decrease? Review of Economic Studies, 60:543–573, 1993.1

[9] Jeremy Bulow and Paul Klemperer. The generalized war of attrition. American Eco- nomic Review, 89(1):175–189, 1999.1

[10] D. Cass and K. Shell. Do sunspots matter? Journal of Political Economy, 91:193–227, 1983.1, 7.1.2

[11] Eddie Dekel, Matthew O. Jackson, and Asher Wolinsky. Jump bidding and budget con- straints in all-pay auctions and wars of attrition. Discussion Papers 1454, Northwestern University, Center for Mathematical Studies in Economics and Management Science, August 10, 2006.1,2

[12] Eddie Dekel, Matthew O. Jackson, and Asher Wolinsky. Vote buying: Legislatures and lobbying. Quarterly Journal of Political Science, 4:103–128, 2009.1

[13] Gabrielle Demange. Rational escalation. Ann. Econ. Stat., 25(26):227–249, 1992.1

[14] Ulrich Doraszelski. An R&D race with knowledge accumulation. RAND Journal of Economics, 34(1):20–42, 2003.1

[15] Marcus Frean and Edward Abraham. Rock-scissors-paper and the survival of the weakest. Proceedings of the Royal Society B: Biological Sciences, 268:1323–1327, 2001.1

[16] Drew Fudenberg, Richard Gilbert, Joseph Stiglitz, and Jean Tirole. Preemption, leapfrogging and competition in patent races. European Economic Review, 22:3–31, 1983.1

[17] Rod Garratt, Thomas Tr¨oger,and Charles Z. Zheng. Collusion via resale. Econometrica, 77(4):1095–1136, 2009.4

[18] Christopher Harris and John Vickers. Racing with uncertainty. Review of Economic Studies, 54:1–21, 1987.1

61 [19] Toomas Hinnosaar. Penny auctions. International Journal of Industrial Organization, 48:59–87, 2016.1

[20] Johannes H¨orner.A perpetual race to stay ahead. Review of Economic Studies, 71:1065– 1088, 2004.1

[21] Johannes H¨ornerand Nicolas Sahuguet. A war of attrition with endogenous eﬀort levels. Economic Theory, 47:1–27, 2011.1

[22] John E. Hunter. Mathematical models of a three-nation arms race. Journal of Conﬂict Resolution, 24(2):241–252, 1980.1

[23] Ali Kakhbod. Pay-to-bid auctions: To bid or not to bid. Operations Research Letters, 41(5):462–467, 2013.1

[24] D. Marc Kilgour and Steven J. Brams. The truel. Mathematics Magazin, 70(5):315–326, 1997.1,2

[25] Moshe Kress, Jonathan P. Caulkins, Gustav Feichtinger, Dieter Grass, and Andrea Seidl. Lancaster model for three-way combat. European Journal of Operational Research, 264:46–54, 2018.1

[26] Vijay Krishna and John Morgan. An analysis of the war of attrition and the all-pay auction. Journal of Economic Theory, 72:343–362, 1997.1

[27] Wolfgang Leininger. Escalation and cooperation in conﬂict situations: The dollar auction revisited. Journal of Conﬂict Resolution, 33(2):231–254, 1989.1

[28] Fredrik Ødegaard and Chris K. Anderson. All-pay auctions with pre- and post-bidding options. European Journal of Operational Research, 239:579–592, 2014.1

[29] Fredrik Ødegaard and Charles Z. Zheng. Surplus dissipating equilibria in the Dol- lar Auction. INFOR: Information Systems and Operational Research, 2019. DOI: 10.1080/03155986.2019.1629771. 6.1, 6.2, A.1, A.8

[30] Barry O’Neill. International escalation and the dollar auction. Journal of Conﬂict Resolution, 30(1):33–50, 1986.1

62 [31] Christian Seel and Philipp Strack. Continuous time contests with private information. Mathematics of Operations Research, 41:1093–1107, 2016.1

[32] K. Shell. Monnaie et allocation intertemporelle. Mimeo, Seminaire Roy-Malinvaud, Centre National de la Recherche Scientiﬁque, Paris, 1977.1, 7.1.2

[33] Martin Shubik. The dollar auction game: A paradox in noncooperative behavior and escalation. Journal of Conﬂict Resolution, 15(1):109–111, 1971.1,2,7,8

[34] Ron Siegel. All-pay contests. Econometrica, 77:71–92, 2009.1

[35] B. Sinervo and C. M. Lively. The rock-paper-scissors game and the evolution of alternative male strategies. Nature, 380:240–243, March 1996.1

[36] R. Toral and P. Amengual. Distribution of winners in truel games. In P. L. Garrido, J. Marro, and M. A. Munoz, editors, AIP Conference Proceedings, volume 779, pages 128–141. Melville, New York, American Institute of Physics, 2005.1