The best-shot all-pay (group) auction with complete information

Abstract

We analyze an all-pay group contest in which individual members’ efforts are aggregated via the best-shot technology and the prize is a public good for the winning group. The interplay of within-group free-riding and across-group competition allows for a wide variety of equilibria, according to how well groups overcome internal free-riding. In contrast with the existing literature, we derive equilibria of a symmetric model in which multiple agents per group are active. Our findings differ qualitatively from those of the individualistic all-pay auction: rents are not necessarily dissipated in equilibrium, total expected efforts vary across equilibria, and participation is expected to be greater. Moreover, equilibria with greater symmetry of behavior within a group are shown to have more “wasted” effort but also greater payoffs as overall efforts are lower. In contrast to standard economic intuition, free-riding can be beneficial for players as it reduces competition among groups. Examples of asymmetric group contests are also studied.

1 Introduction

Group conflict is a common experience of daily life. For example, the results of sports team competitions are often front page news, as are the outcomes of elections contested by coalitions of multiple parties. In these and many other instances, individual group members make costly contributions that improve their group’s chance to succeed in the across-group struggle. And all members of the successful group may receive a benefit that is independent of the individual effort actually expended. Thus, a player faces a trade-off: He has an incentive to rely on his teammates’ efforts and lessen his own, but doing so may reduce his team’s chance of winning the contest.

There are many ways to model group conflict; interesting facets include the nature of the prize, the rule used to determine the winner, and the way individual efforts are transformed into the group’s effort (sometimes referred to as group-effort technology). We make three key assumptions. First, we consider a pure public-good structure for the prize, so that all members of the winning coalition benefit from victory regardless of their individual efforts.¹ Second, we identify a group’s effort, as a function of individual members’ efforts, through the best-shot aggregator function (Hirshleifer 1983). This is in contrast with the most common assumption of the summation aggregation function.² Third, to determine the outcome of group competition, we consider an all-pay auction setup: The group with the largest best-shot effort is the winner, and efforts are expended by both victors and vanquished.³

Our paper is most closely related to Chowdhury et al. (2013a), as we share the first two assumptions above. Thus, many of the examples they provide (e.g., Tour de France, Formula 1, competing defense coalitions such as NATO and the Warsaw Pact) apply for our situation as well. Our paper is also closely related to Baik et al. (2001) and Chowdhury et al. (2013b), since we share all assumptions but the form of the aggregator function. There is a close relationship between that contest of Baik et al. (2001) and the individualistic all-pay auction of Baye et al. (1996), since, notably, Baik et al. find that in equilibrium only one player is active in each group. Therefore, our paper is closely related to Baye et al. (1996) as well, the only difference being that we consider group competition rather than individualistic competition.

On the one hand, therefore, a narrow view of our contribution is technical, as we fill in a gap in the literature on group all-pay auctions. On the other hand, and more importantly, with the best-shot aggregator, we believe we offer a framework especially suited to describe competition in which groups bring forth arguments or ideas, rather than aggregate monetary contributions. For example, the institution of amicus curiae (AC) in the US legal system illustrates the kind of group competition we have in mind.

ACs, i.e., “friends of the court,” are parties other than the petitioner or the respondent who add arguments to legal proceedings to bolster the case of one of the two contesting sides. AC briefs are typically employed at the appellate level (the federal Supreme Court and many state Supreme Courts), and cases of extreme social importance attract many ACs.⁴ Clearly, an AC brief that simply repeats the ideas of one of the contesting parties adds very little to the justices’ decision-making process. Therefore, a summation-of-effort technology (as in Baik et al. 2001) may be less appropriate than the best shot to capture the contribution to the group’s success of an individual AC brief. And all interested parties benefit from a favorable decision, regardless of expended effort, thus justifying our assumption that the prize is a public good within a group. While Chowdhury et al. (2013a) also consider a best-shot aggregation technology, in their equilibria, only one agent per group takes action. In contrast, we show equilibria in which multiple agents per group exert effort; therefore, our setup can account for multiple AC briefs being filed on one (or both) sides.⁵

Because of the within-group public-good nature of prize, our paper is related to the literature on the private provision of public goods (PPPG) and the volunteer’s dilemma (VD). Free-riding is often predicted in PPPGs (see, e.g., Olson 1965; Palfrey and Rosenthal 1984; Bergstrom et al. 1986; Vicary 1997; Xu 2001 and Barbieri and Malueg 2008a, b and Barbieri and Malueg 2014a). The VD is a situation where the discrete effort of just one player is enough to benefit all members of the group. The best-shot PPPG can therefore be seen as a VD with continuous effort. It is commonly documented that in symmetric equilibria of VDs and best-shot PPPGs the probability that a public good is produced (or its quantity) decreases as the group becomes larger (Diekmann 1985; Harrington 2001; Barbieri and Malueg 2014a). Thus, the free-riding problem becomes so severe that the expected value of the public good provided decreases despite the fact that there are more contributors. Indeed, we find a similar effect in our contests.

Finally, our work is relevant to the literature on contests with identity-dependent externalities, or IDE (e.g., Funk 1996; Das Varma 2002; Klose and Kovenock 2012, 2013). In such contests, a losing player may care who has won, i.e., the payoff of a loser may depend on the identity of the winner. IDE contests nest the best-shot all-pay auction we study: (1) A player is equally happy if anyone in her group wins the auction; (2) a player is indifferent to the identity of the winning player if not in her group; and (3) the difference in the payoff each player receives if a player in her group wins rather than a player in another group is constant across all players within the same group. Because of the stronger structure we impose, our results considerably sharpen (for our more restrictive setup) the insights gleaned from the examples in Klose and Kovenock (2013), for instance.⁶

We find a variety of equilibria where more than one agent within a group is active (in some equilibria, all players are active). This is an interesting finding, especially because many papers in the contest literature focus on equilibria where only one player in each group is active (Baik et al. 2001; Baik 2008; Chowdhury et al. 2013a). Beyond adding to the realism of equilibrium predictions, the outcomes we find are qualitatively different from those of the standard, individualistic all-pay auction (Baye et al. 1996), which is what arises if only one agent per group is active. Three major differences stand out.

First, rents in symmetric contests might not be completely dissipated, even for active agents in our perfectly discriminating setup.⁷ When at least two players in a group are active, rents are not completely dissipated. Rent dissipation is complete, however, if only one player in each group is active (which resonates with the individualistic all-pay auction and the group all-pay auction setting analyzed by (Baik et al. 2001). Also, regardless of the number of active players per group, as the number of groups approaches infinity, the rent dissipation among active players becomes complete. Second, payoffs may vary across equilibria, and identical players may earn different payoffs.⁸ This observation contrasts with the remarkable results regarding the payoff equivalence of equilibria of individual all-pay auctions (Baye et al. 1996) and generic contests (Siegel 2009). Further, the sum of expected efforts varies across equilibria of a symmetric contest (in contrast to Baik et al. 2001; Baik 2008; Topolyan 2014). Third, in asymmetric contests, a wider participation is possible. While in Baye et al. (1996) some players may be inactive in the symmetric-valuation case, in our analogous settings, all groups must be active, if at least two members are active in some group. And, while Baye et al. (1996) demonstrate all players with value outside the top 2 must be inactive, in our analogous asymmetric setup, some groups can be active even if their value is not in the top 2.

The reason behind these differences is that a wide variety of equilibria exist in our setup, varying according to how well group members cope with the problem of internal free-riding. Indeed, there exist equilibria in which group members are unable to overcome the internal free-riding problem, group efforts are low on average, and, therefore, if free-riding afflicts all groups, group members fare well, since competition across groups is low. And if groups are better able to overcome internal free-riding tendencies, then across-group competition becomes fiercer and equilibrium payoffs decrease, up to the point at which rents are completely dissipated and some groups may remain completely inactive.

We identify two sources for the incentive to free-ride. The first is the number of active agents: Because of the link with VDs and best-shot PPPGs previously described, a broader participation within each active group creates a larger incentive to free-ride among active agents and therefore generates a smaller rent dissipation, a larger payoff for active agents, and, in asymmetric contests, a smaller barrier to the participation of groups with lower values for the public good.

The second reason an agent has to shade his contribution is the possible wasteful duplication of effort within a group. We offer a measure of this waste and provide a family of equilibria where active players within the same group randomize over different intervals; in particular, active fellow group members’ strategies have different supports composed of interlaced intervals.⁹ The finer the interlacing, the larger the probability of wasted contribution, the larger the incentive to shade one’s contribution, again with the consequence that overall efforts are reduced.

In comparison to other group contests, the most striking difference brought about by the best-shot technology is a return of the “group-size paradox”: In semi-symmetric equilibria with asymmetric groups, the effect of group size on a group’s probability of winning is negative (i.e., a larger group is less likely to win, other things equal).¹⁰ Our result is a counterpoint to Katz et al. (1990), Riaz et al. (1995), Esteban and Ray (2001) and Nitzan and Ueda (2011).¹¹ The main reason for this difference is the strong within-group free-riding described above.

The rest of the paper is organized as follows. Section 2 describes the basic characteristic of the model. Section 3 analyzes symmetric contests, exploring both symmetric equilibria and equilibria in which group efforts are symmetric, but individual active agents’ efforts are not. Section 4 analyzes asymmetric contests, with special attention to two- and three-group settings. Section 5 concludes.

2 All-pay auction contests with best-shot group performance

We model a situation in which several groups compete for a prize that is a public good for the winning group. This eliminates concerns about how to share a prize in the event of winning; however, it introduces free-riding among group members as one can benefit from a win even when putting forth relatively little effort. Now, individuals must weigh the benefit of free-riding against the risk of losing out altogether to another group that provides greater effort. When there are differences across groups, we model those as differences in values for the good.

We envision $N$ groups possibly competing for the prize, with group $i$ having $n_i$ members, $i = 1, \ldots , N$. Let $\underline{n}\equiv \min \{n_1, \ldots , n_N\}$ denote the size of the smallest group. We index members of group $i$ by $i_1, i_2, \ldots , i_{n_i}$, $i = 1, \ldots , N$. Throughout the paper, players within the same group are assumed to be identical with respect to their cost of effort and valuation of the prize. The constant marginal cost of effort common to all players is without loss of generality normalized to $1$. We let $v_{i}>0$ denote the common gross benefit of winning to members of group $i$. A member of group $i$ exerting effort $x$ then earns final payoff $v_i - x$ if group $i$ wins and $-x$ if it does not.

The groups compete in an all-pay contest where each group is judged according to its best effort (“best shot”). So the performance of group $i$ is given by $X_i = \max \{x_{i_1}, \ldots , x_{i_{n_i}}\}$. Group $i$ wins if $X_i > X_j$, for all $j \ne i$; in the case that $k$ groups tie for maximum performance, each of the tying groups has probability $1/k$ of being designated the winner.

We say that a player is active if he exerts strictly positive effort with strictly positive probability, and a group is active if it has at least one active player.

3 Symmetric active groups

We begin by analyzing the case where each group has a common number of active players. Such would be typical, for example, of most sporting events (in Sect. 4, we allow active groups to differ in size). In this section, we further assume the value of winning is common across groups, so $v_i = v$ for all $i$.

3.1 Semi-symmetric equilibria

We construct semi-symmetric equilibria in which all players’ equilibrium payoffs are positive (we say semi-symmetric because, while active players all use the same strategy, in equilibrium, some players may remain inactive). In particular, we show that for any $m \in \{1, \ldots , \underline{n}\}$, there is an equilibrium in which exactly $m$ players in each group are active (the others are “extreme” free-riders), and if $m > 1$, then all players’ payoffs are positive. We look for a semi-symmetric mixed-strategy equilibrium, where each active player exerts effort according to a cumulative distribution function (cdf) $F$. As usual, $F$ does not admit any atoms of probability and its support has lower limit 0; the upper limit will be determined as part of the equilibrium.¹² Thus, $F$ is continuous, which implies the equilibrium chance of ties is zero.

Supposing $m \in \{1, \ldots , \underline{n}\}$ players in each group choose effort according to cdf $F$, we now derive the cdf that just makes these active players willing to follow this randomization. We will then show that no inactive player would choose to become active. Consider the problem from the point of view of an active player in group 1, say, player $1_1$. The cdf of the maximum of all efforts other than player $1_1$’s is $(F)^{Nm-1}$. Group 1 can win in either of two ways, given that player $1_1$’s contribution is $x$: Either player $1_1$’s effort is the global best shot or the global best shot is larger than $x$ and is achieved by someone else in group 1. The first event happens with probability $(F(x))^{Nm-1}$, and the second with probability $\left( 1-(F(x))^{Nm-1}\right) \times (m-1)/(Nm-1).$ Therefore, player $1_1$’s expected payoff is

$$\begin{aligned} V(x)&= -x + v \left[ [F(x)]^{Nm-1} + \frac{m-1}{Nm-1}\left( 1-[F(x)]^{Nm-1}\right) \right] \nonumber \\&= -x + v\left[ \frac{m-1}{Nm-1} + \frac{Nm-m}{Nm-1}[F(x)]^{Nm-1} \right] . \end{aligned}$$

(1)

Since $F$ is continuous, $V$ is continuous, which implies that the support of $F$ is an interval $[0, \bar{x}],$ for some $\bar{x}>0$.¹³ The absence of atoms in the equilibrium strategy implies for an active player the payoff is $V(0) = \left( \frac{m-1}{Nm-1} \right) v$, so, from (1), indifference over the randomization interval implies

$$\begin{aligned} F(x) = \left[ \left( \frac{Nm-1}{Nm-m}\right) \frac{x}{v}\right] ^{\frac{1}{Nm-1}}, \qquad \text {with } \qquad \bar{x}= \left( \frac{Nm-m}{Nm-1}\right) v. \end{aligned}$$

(2)

The above analysis shows that indeed each active player can do no better than to use this cdf $F$. To complete the verification of the equilibrium, we must show that no inactive players would wish to become active. We do so in the Appendix and obtain the following result.

Proposition 1

(Semi-symmetric equilibrium) For each $m \in \{1, \ldots , \underline{n}\}$, there exists an equilibrium in the all-pay auction where $m$ players in each group independently contribute effort according to the cdf in (2) and all others exert no effort. Each group wins with probability $1/N$, and expected payoffs are $v/N$ to each inactive player and $\frac{m-1}{Nm-1}v$ to each active player. Given $m$ active players per group, (2) describes the unique semi-symmetric equilibrium strategy.

Proposition 1 reveals a first important contrast with the symmetric individualistic all-pay auction. Baye et al. (1996) found that all players earn a payoff of zero in all equilibria. In contrast, we find in the semi-symmetric equilibrium with $m \ge 2$, all players earn strictly positive payoffs. It is the possibility of winning based on fellow group members’ efforts that ensures even an active player a positive equilibrium payoff, leading to efforts surely less aggressive than $v$.

We now further pursue the effects of having multiple active players within a group by considering the comparative statics with respect to $m$. It is straightforward to see that as $m$ increases, the equilibrium strategy $F$ in (2) shifts leftward in the sense of first-order stochastic dominance (FOSD). Moreover, even as the number of active players in each group increases, the distribution of a group’s best shot shifts leftward in the sense of FOSD.¹⁴ A group’s total expected effort is

$$\begin{aligned} m \int _0^{\bar{x}} y \, \mathrm{d}F(y) = \frac{mv}{N} \times \frac{N-1}{Nm - 1}, \end{aligned}$$

(4)

which decreases monotonically toward $\frac{N-1}{N^2}v$ as $m$ increases. Most interestingly, as active groups symmetrically get larger the total expected benefit to active members becomes arbitrarily large while the group’s total expected cost remains bounded, implying that “rent dissipation” effectively becomes insignificant. Indeed, from (4), we see that the ratio of a group’s expected effort to its expected gross benefit of active members ($mv/N$) is $(N-1)/(Nm-1)$, which monotonically decreases in $m$ with limit 0. Not surprisingly, then, we find that groups prefer equilibria with a greater number of active players. Moving from $m$ to $m+1$ active players per group has no effect on payoffs of players who remain inactive. A player who continues to be active earns a larger payoff if the number of active players per group increases (because $\bar{x}$ decreases with $m$), but the player switching from inactive to active suffers a decrease in payoff. Overall, because an increase in $m$ leaves all groups’ chances of winning at $1/N$ and expected efforts are lower, each group’s collective payoff increases as the number of active players in a semi-symmetric equilibrium increases.

Intuitively, then, the effect of an increase in $m$ is to increase the incentive to free-ride for active members, because agents can count on more fellow group members to take action. Therefore, groups’ efforts decrease, just as in VDs and best-shot PPPGs. In contrast to what happens there, though, in our case, more free-riding is beneficial for players, because it reduces competition among groups.

We further analyze competition among groups by considering the effects of increasing $N$. In the individualistic all-pay auction equilibrium ($m =1$), increases in $N$ cause a FOSD shift to lower efforts. In contrast, when $m \ge 2$, the shift is not fully in accord with FOSD. This follows because, while for lower efforts, increases in $N$ do increase $F$, at the same time $\bar{x}$ increases, meaning that maximum possible efforts are larger as $N$ increases. As the number of groups becomes arbitrarily large, the distribution of a group’s maximum effort, $(F(x))^m$, converges in probability to a unit mass at 0, even as the distribution of the overall best shot converges to the uniform distribution on $[0, v]$. Finally, if $N > 2$, then there are equilibria in which some groups are inactive; however, the possibilities are limited, as shown next.

Proposition 2

(The possibility of inactive groups) Consider a semi-symmetric equilibrium in which each active group has $m$ active players. If $m=1$, then there may be as many as $N-2$ inactive groups. If $m \ge 2$, then all groups are active.

It is trivial to see that any equilibrium has at least two active groups. For $m=1$, Proposition 2 follows from the analysis of Baye et al. (1996). If, however, $m \ge 2$, then all groups are active, for when $m \ge 2$, the largest effort of any active group member will be less than $v$ (cf. (2)), in which case a member of an inactive group could raise her payoff above zero by exerting effort $\bar{x}$.¹⁵

Proposition 2 identifies another difference with the individualistic all-pay auction: Participation is encouraged when groups have multiple active members. While Baye et al. (1996) found that the number of active players in the symmetric individualistic all-pay auction is at least 2 and may be more, we find that in equilibria with at least two active players per group, all groups must be active. This expansion of participation will be reinforced in the analysis of asymmetric groups in Sect. 4.

3.2 Asymmetric equilibria

Here, we explore the possibility that equilibria may be asymmetric even though the active groups are symmetric. For the symmetric individualistic all-pay auction, Baye et al. (1996) showed asymmetric equilibria exist. In particular, they found equilibria where two players randomize over $[0, v]$ and other players randomize over sets of the form $\{0\} \cup [b_i, v]$, where $b_i < v$ is a free player-specific parameter. However, such strategies cannot be part of an equilibrium here, as the following proposition shows. Here, we let $\text {supp }F$ denote the support of an arbitrary cdf $F$.

Lemma 1

(Strategy supports have no interval overlaps for group members using different strategies) Consider $N \ge 2$ competing groups and any Nash equilibrium. Suppose group 1 includes two active players $a$ and $b$ who use strategies $F_a$ and $F_b$, respectively. If $\text {supp }F_a$ and $\text {supp }F_b$ can be expressed as the unions of finitely many intervals and if $F_a \ne F_b$, then $\text {supp }F_a \cap \text {supp }F_b$ contains no nonempty open interval.

Lemma 1 leads us to seek equilibria in which active fellow group members’ strategies have (essentially) nonoverlapping supports. In this section, we focus on the case of two active groups, each with two active players. We designate the groups as 1 and 2, and we suppose each group has a member $a$ and a member $b$.

Example 1

(Symmetry across groups, asymmetry within a group)

We look for an equilibrium in which the $a$ players in each group use strategy $F_a$ with support $[0, x_m]$ and the $b$ players use strategy $F_b$ with support $\{0\} \cup [x_m, \bar{x}]$. Here, $F_b(x_m)=F_b(0)$ denotes the probability with which a $b$ player chooses effort 0. Also, $H(x) = F_a(x) F_b(x)$ is the cdf of a group’s best shot.

First consider group 1’s player $b$. The player’s payoff to zero effort is

$$\begin{aligned} V_{b}(0)&= v \int _0^{x_m} H(y) \, \mathrm{d}F_a(y) = v \int _0^{x_m} F_b(x_m) F_a(y) \, \mathrm{d}F_a(y) \nonumber \\&= \frac{v}{2}F_b(x_m) \int _0^{x_m} \frac{\mathrm{d}}{\mathrm{d}y} \left( F_a^2(y)\right) \, \mathrm{d}y&\nonumber \\&= \frac{v}{2}F_b(x_m), \end{aligned}$$

(5)

where the final equality uses $F_a(0) = 0$ and $F_a(x_m) = 1$. Here, (5) establishes the payoff that player $b$ is also to obtain from efforts in $[x_m, \bar{x}]$. In particular, for effort $x_m$, the player’s payoff then satisfies

$$\begin{aligned} \frac{v}{2}F_b(x_m) = V_{b}(x_m) = -x_m + v H(x_m) = - x_m + v F_b(x_m), \end{aligned}$$

which implies

$$\begin{aligned} x_m = \frac{v}{2}F_b(x_m). \end{aligned}$$

(6)

Now for $x \in (x_m, \bar{x}]$, the player’s payoff must satisfy $\frac{v}{2}F_b(x_m) = V_{b}(x) = -x + v H(x) = - x + v F_b(x),$ from which, by (6), we obtain

$$\begin{aligned} F_b(x) = \frac{x_m + x}{v} \qquad \forall x \in [x_m, \bar{x}]. \end{aligned}$$

(7)

Next consider player $a$ in group 1. Her payoff from $x \in (0, x_m]$ is $V_{a}(x) = -x + v\left( 1- \int _x^{\bar{x}} F_b(y) \, \mathrm{d}H(y)\right) .$ Indifference over $(0, x_m]$ implies $0 = V'_{a}(x) = -1+ v F_b(x_m) F_b(x_m) \,\mathrm{d}F_a(x), \, \forall x \in (0, x_m),$ from which we obtain, after integration,

$$\begin{aligned} F_a(x)&= \frac{x}{v (F_b( x_m))^2 } \qquad \forall x \in [0, x_m]. \end{aligned}$$

(8)

From (6) and (8), we find $v (F_b( x_m))^2 = x_m = \frac{v}{2}F_b(x_m),$ which in turn implies

$$\begin{aligned} F_b(x_m) = \frac{1}{2}\qquad \text {and} \qquad x_m = \frac{v}{4}. \end{aligned}$$

(9)

Using (9) with (7), we find from $F_b(\bar{x}) = 1$ that $\bar{x}= 3v/4$.

From the above analysis, we propose the following as equilibrium strategies:

$$\begin{aligned} F_a(x) = {\left\{ \begin{array}{ll} 1 &{} \text {if}\;x > v/4\\ \dfrac{4x}{v} &{} \text {if}\; x \le v/4 \end{array}\right. } \quad \text {and}\quad F_b(x) = {\left\{ \begin{array}{ll} 1 &{} \text {if}\; x > 3v/4\\ \dfrac{v+4x}{4v} &{} \text {if}\; v/4 < x \le 3v/4\\ \dfrac{1}{2} &{} \text {if}\; x \le v/4 \end{array}\right. }\quad \end{aligned}$$

(10)

(see Fig. 1). Construction of these strategies shows that players’ payoffs are constant over their strategies’ supports. At these strategies, a type-$a$ player earns payoff $u_a^* = 3v/8$ and a type-$b$ player earns payoff $u_b^* = v/4$.

Fig. 1

Equilibrium cumulative distribution functions in Example 1

It only remains to show that no player has a profitable deviation. For player $a$ in group 1 contemplating $x \in (v/4, 3v/4]$, the payoff is

$$\begin{aligned} V_{a}(x)&= -x + v \left( 1- \int _x^{\bar{x}} F_b(y) \, \mathrm{d}H(y)\right) = -x + v \left( 1- \int _x^{3v/4} \dfrac{v+4y}{4v}\times \frac{1}{v} \, \mathrm{d}y \right) \\&= \frac{3v}{8} - \frac{(4x - v)(5v-4x)}{32v}< \frac{3v}{8} \qquad \forall x \in (v/4, 3v/4], \end{aligned}$$

showing the player $a$ has no profitable deviation. A similar calculation shows player $b$ has no profitable deviation (we omit the simple verification that inactive players would not choose to become active). Thus, the foregoing analysis establishes an equilibrium between two groups where each group has one player using $F_a$ and one player using $F_b$, as given in (10). $\square$

To further our understanding of the extent and effects of free-riding in our setup, we refine the structure of the previous example to obtain, in addition to the number of players, a second source of equilibrium variation. Naturally, agents desire to avoid wasting contributions due to a higher effort by a fellow group member. Because contributions other than the maximum contribution of a group member have no effect on a group’s chance of winning, we define a group’s waste as its total contributions minus the group’s best shot:

$$\begin{aligned} W&=\int \int \big (x + y - \max \{x, y\} \big )\, \mathrm{d}F_{a}(x)\, \mathrm{d}F_b(y) =\int x\left( 1-F_{b}\left( x\right) \right) \mathrm{d}F_{a}\left( x\right) \nonumber \\&\quad +\int x\left( 1-F_{a}\left( x\right) \right) \mathrm{d}F_{b}\left( x\right) . \end{aligned}$$

(11)

Waste frequently arises in PPPG problems with nonconvex technology, e.g., nonrefunded insufficient contributions in threshold public-good games. Our notion of waste is thus close to the so-called “assurance” problem discussed in Isaac et al. (1989) or List and Lucking-Reiley (2002), for instance.¹⁶

Clearly, waste is zero if only one member is active per group. The equilibrium described by (10) in Example 1 introduces some waste, but, heuristically, waste remains small, since active group members tend to coordinate on different efforts (indeed, waste is exactly zero for player $b$). In contrast, the probability that an agent wastes his contribution appears large in the semi-symmetric equilibrium in (2), because agents’ contributions often overlap. The following example introduces an additional situation, intermediate between Example 1 and the semi-symmetric equilibrium.

Example 2

(Equilibrium with interlacing interval supports)

As in Example 1, we assume there are two groups, each with two players, player $a$ and player $b$. The following strategies are an equilibrium:

$$\begin{aligned}&F_a(x) = {\left\{ \begin{array}{ll} \frac{8}{27}+\frac{x}{v} &{} \text {if}\quad \frac{10}{27}v \le x \le \frac{19}{27}v\\ \frac{2}{3} &{} \text {if}\quad \frac{2}{27}v \le x \le \frac{10}{27}v\\ \frac{9x}{v} &{} \text {if}\quad 0 \le x\le \frac{2}{27}v \end{array}\right. } \qquad \text {and} \qquad \nonumber \\&F_b(x) = {\left\{ \begin{array}{ll} 1 &{} \text {if}\quad \frac{10}{27}v \le x \le \frac{19}{27}v\\ \frac{1}{6}+\frac{9x}{4v} &{} \text {if}\quad \frac{2}{27}v \le x \le \frac{10}{27}v\\ \frac{1}{3} &{} \text {if}\quad 0 \le x\le \frac{2}{27}v. \end{array}\right. } \end{aligned}$$

(12)

Because the proof of equilibrium follows from the more general case analyzed in the Appendix, here we have simply reported the equilibrium cdfs, which are depicted in Fig. 2. $\square$

Fig. 2

Equilibrium cumulative distribution functions in Example 2

Taken together, Proposition 1 and Examples 1 and 2 reveal a second important contrast with the individualistic all-pay auction where all players have common value. Baye et al. (1996) found the sum of expected efforts to be constant across all equilibria of the symmetric all-pay contest. By contrast, our examples show varying total expected efforts. Even if symmetric groups act symmetrically, and holding fixed the number of active players, a group’s total expected payoff varies over the three equilibria considered: the semi-symmetric equilibrium (see (2) for $N=m=2$) and the equilibria of Examples 1 and 2. A group’s total expected payoff is $135v/216$ in Example 1, $140v/216$ in Example 2, and $144v/216$ in the semi-symmetric equilibrium. Correspondingly, total expected efforts vary across the three equilibria. And so does waste: Using (11), we have waste equal to $v/16 = 0.0625v$ in Example 1, $16v/243 \approx 0.0658v$ in Example 2, and $v/15 \approx 0.0667v$ in the semi-symmetric equilibrium. Surprisingly, while the greater symmetry of strategies within a group is associated with greater waste, it is also associated with higher payoffs as overall efforts are lower.

In the Appendix, we generalize Example 2 by presenting equilibria with strategies taking values over an arbitrary number of interlaced intervals. Thus, as we increase the number of intervals, we cover the gamut from the highly asymmetric (within each group) equilibrium in Example 1 to the semi-symmetric equilibrium described by (2). The results presented in the Appendix formally verify the conjecture one may draw from our examples that the finer the interlacing of contribution strategies, the larger waste becomes. This decreases an agent’s incentive to contribute. However, if all groups coordinate on such an “inefficiently low” level of effort, as the one in the semi-symmetric equilibrium described by (2), then all groups end up benefiting from the reduced competition across groups. And the strategy profile is not vulnerable to unilateral increases in effort because the within-group free-riding prevents any one agent from fully capturing the benefits of additional effort. In summary, therefore, the larger the asymmetry in contributors’ equilibrium strategies within each group (that is to say, the coarser the interlacing), the weaker the incentive to free-ride within each group, the harsher the between-group competition, the smaller the payoff for every group, as exemplified above.

4 Asymmetric active groups

4.1 Two-group contests

We now investigate the effects of asymmetry among active groups. We assume $N=2$ but allow active groups to differ in their sizes and values of winning. We return to the original formulation allowing values $v_1$ and $v_2$ to differ between groups.¹⁷ We derive an equilibrium, which again we call semi-symmetric, where the $m$ active members of group 1 use the same strategy and the $n$ active members of group 2 use the same strategy (obviously, $1 \le m \le n_1$ and $1\le n \le n_2$). Let $F$ be the equilibrium cdf used by each active member of group 1, and let $G$ be the equilibrium cdf used by each active member of group 2.

Now, let $F^M$ be the cdf of the maximum of $m-1$ draws from $F$; let $G^M$ be the cdf of the maximum of $n$ draws from $G$. Then, $F^M(x) = (F(x))^{m-1}$ and $G^M(x) = (G(x))^n$. With player $1_{1}$ exerting effort $x$, we have

$$\begin{aligned} \Pr \!\left( \left. \text {group 1 wins }\right| \,x_{1_1} = x\right) =1-\int _x^\infty F^M(y)\, \mathrm{d}G^M(y). \end{aligned}$$

(13)

To understand (13), note that group 2 wins only if its best shot $y$, say, (having cdf $G^M$) exceeds $x$, and then only if all other group 1 members have efforts less than $y$, which happens with probability $F^M(y)$. Thus, the integral in (13) is just the probability that group 2 wins, given that $x_{1_1} = x$. It now follows that player $1_{1}$’s payoff is

$$\begin{aligned} V_1(x) = -x + v_1 \left[ 1-\int _x^\infty F^M(y)\, \mathrm{d}G^M(y) \right] . \end{aligned}$$

(14)

Lemma 2

The strategies $F$ and $G$ in a semi-symmetric equilibrium (i) are continuous and strictly increasing on their common support $[0, \bar{x}]$, for some $\bar{x}\in \left( 0, \min \{v_1, v_2\}\right]$; (ii) do not both have atoms at 0; and (iii) are continuously differentiable on $(0, \bar{x})$.

Standard arguments, as used in Sect. 3, show $F$ and $G$ have common interval support and cannot have atoms above 0; hence, they are continuous on that support. To complete the proof of Lemma 2, it only remains to establish part (iii) and that is done in the Appendix .

Because $V_1$ is constant over $[0, \bar{x}]$, it is differentiable on $(0, \bar{x})$, with $0 \equiv V_1'(x) = -1 +v_1 F^M(x)\, \mathrm{d}G^M(x),$ which can be rewritten as

$$\begin{aligned} \frac{1}{nv_1}&= \Big (F(x)\Big )^{m-1} \Big (G(x)\Big )^{n-1} G'(x), \end{aligned}$$

(15)

thus providing a differential equation for $G.$ Analogous reasoning for an active member of group 2 yields a second differential equation:

$$\begin{aligned} \frac{1}{mv_2} = \Big (F(x)\Big )^{m-1} \Big (G(x)\Big )^{n-1} F'(x). \end{aligned}$$

(16)

From (15) and (16), we conclude that

$$\begin{aligned} G'(x) = \left( \frac{v_2/n}{v_1/m} \right) F'(x), \end{aligned}$$

(17)

for $x > 0$. Let $\bar{x}$ denote the (common) upper end of the support of the cdfs. We now have

$$\begin{aligned} 1-G(x) =\int _x^{\bar{x}} G'(y)\, \mathrm{d}y = \left( \frac{v_2/n}{v_1/m} \right) \int _x^{\bar{x}} F'(y) \, \mathrm{d}y = \left( \frac{v_2/n}{v_1/m} \right) (1- F(x)), \end{aligned}$$

or

$$\begin{aligned} G(x) = \left( 1- \frac{v_2/n}{v_1/m} \right) + \left( \frac{v_2/n}{v_1/m} \right) F(x). \end{aligned}$$

(18)

In particular, if $v_1/m>v_2/n$, then $G$ puts an atom on 0 while $F$ does not; moreover, the strategy of an active player in group 1 strictly FOSD that of an active player in group 2. Thus, individual active group 1 players exert greater effort than active group 2 players when $\frac{v_1}{m} > \frac{v_2}{n}$, that is, when, for example, group 1 has a relatively larger values or relatively less diffuse responsibility for effort.

Returning to the final determination of $F$ and $G$, with $v_2/n \le v_1/m$, we proceed as follows. Substitute $G$ from (18) into (16) to obtain the following differential equation for $F$:

$$\begin{aligned} \frac{1}{mv_2} = \Big (F(x)\Big )^{m-1} \left[ \left( 1-\frac{v_2/n}{v_1/m}\right) +\left( \frac{v_2/n}{v_1/m}\right) F(x)\right] ^{n-1} F'(x). \end{aligned}$$

(19)

Solve this for $F$, with the boundary condition $F(0) = 0$. Then, determine $\bar{x}$ as the solution to $F(x) = 1$, and, finally, determine $G$ using (18). By construction of the cdfs $F$ and $G$, active players have no profitable deviations. To conclude the verification that the proposed strategies constitute an equilibrium, we must check that no inactive player has a profitable deviation. The following lemma, proven in the Appendix, does so. Moreover, it shows that a unique pair of functions $F$ and $G$ is identified by the previous procedure.

Lemma 3

(Semi-symmetric equilibrium and unique solution) Suppose $v_1/m \ge v_2/n$. There is a unique solution $F$ to the differential equation (19) with initial condition $F(0)=0.$ Therefore, (18) identifies a unique $G$. Moreover, $F$ and $G$ determined through (18) and (19) describe an equilibrium.

While the equilibrium identified in Lemma 3 generally does not yield convenient closed-form solutions for arbitrary $m$ and $n$, $v_1$ and $v_2$, we consider two examples where it does.

Example 3

Suppose $v_1/m = v_2/n$.

From (18), we see active players use a common strategy $F$. The differential equation (19) for the common equilibrium strategy $F$ simplifies to

$$\begin{aligned} \frac{1}{m v_2}&= \left( F(x)\right) ^{m+n-2} f(x) =\left( \frac{1}{m+n-1} \right) \frac{\mathrm{d}}{\mathrm{d}x}\left( F(x)\right) ^{m+n-1}, \end{aligned}$$

from which, with the boundary condition $F(0) = 0$, we obtain $F(x) = \left[ \left( \frac{m+n-1}{m v_2}\right) x \right] ^{\frac{1}{m+n-1}}$, with upper limit of the support equal to $\bar{x}= \frac{nv_1}{m+n-1}= \frac{mv_2}{m+n-1}$.

Because all players use the same strategy, each player is equally likely to be the “best shot.” Therefore,

$$\begin{aligned} \Pr \!\left( \text {group 1 wins}\right) = \frac{m}{m+n} = \frac{\frac{v_1}{v_2}\times n}{\frac{v_1}{v_2}\times n+n} = \frac{v_1}{v_1+v_2}. \end{aligned}$$

Expected payoffs are $\left( \frac{m-1}{m+n-1}\right) v_1$ to active group 1 members, $\left( \frac{n-1}{m+n-1}\right) v_2$ to active group 2 members.$\square$

We next investigate the groups’ performances, asking which group is more likely to win. Giving a general answer turns out to be nontrivial. We proceed incrementally, first analyzing how the equilibrium strategies identified in Lemma 3 vary with parameters. We provide a partial analysis by rewriting (19) as

$$\begin{aligned} 1= v_2\left[ 1-\frac{v_2/n}{v_1/m}\left( 1 - \Big [H(x)\Big ]^{1/m}\right) \right] ^{n-1} H'(x), \end{aligned}$$

(20)

where $H(x) \equiv (F(x))^m$ denotes the cdf of group 1’s best-shot and $H$ satisfies the boundary condition $H(0) = 0$. The differential equation (20) yields the following comparative statics for group 1’s best shot.

Proposition 3

Suppose $v_1/m > v_2/n$. Then, decreasing $m$ or increasing $v_1$ causes the distribution of group 1’s best shot in the semi-symmetric equilibrium derived earlier to shift to the right in the sense of FOSD. Correspondingly, the individual efforts of active group-1 players are also larger in the sense of FOSD.

Thus, increasing the value to group 1 or reducing its size leads the group to improve its performance, as judged by its best-shot cdf. However, without further assumptions, it is unclear how changes in $v_2$ or $n$ will affect group 1’s best-shot distribution.¹⁸ Moreover, knowing only the change in group 1’s best-effort distribution does not suffice to tell us how group 1’s chance of winning changes. Nevertheless, with (18) and (20), we can now calculate the group’s probability of winning. Given $v_1/m > v_2/n$, we have the cdf of group 1’s best shot given by $(F(\cdot ))^m$ and group 2’s by $(G(\cdot ))^n$, so

$$\begin{aligned} \Pr \!\left( \text {group 1 wins}\right)&= \int _0^{\bar{x}} [G(y)]^n\, \frac{\mathrm{d}}{\mathrm{d}y}\left( [F(y)]^m\right) \, \mathrm{d}y \nonumber \\&= \int _0^{\bar{x}}\, \left[ \left( 1- \frac{v_2/n}{v_1/m} \right) + \left( \frac{v_2/n}{v_1/m} \right) F(y)\right] ^n\, \frac{\mathrm{d}}{\mathrm{d}y}\left( [F(y)]^m\right) \, \mathrm{d}y \nonumber \\&= \int _0^{1} \left[ 1- \left( \frac{mv_2}{nv_1} \right) \left( 1- z^{1/m}\right) \right] ^n \, \mathrm{d}z. \end{aligned}$$

(21)

The integrand in (21) is clearly increasing in $v_1$ and decreasing in $v_2$. As shown in the proof of Proposition 3, we also see the integrand is decreasing in $m$ and, as shown in the proof of the following proposition, increasing in $n$. Therefore, we have the following.

Proposition 4

Suppose $v_1/m > v_2/n$. At the semi-symmetric equilibrium earlier derived, group 1’s probability of winning increases as $v_1$ increases, $m$ decreases, $v_2$ decreases, or $n$ increases.

These comparative statics are intuitive: Increasing $v_1$ makes winning more desirable for active group-1 players, and reducing $m$ lessens the incentives to free-ride. Both effects increase the probability that group 1 wins. Similarly, reducing $v_2$ or increasing $n$ weakens group 2 and makes it less likely to win. And while the integral in (21) has no convenient closed form, for any particular values of $m$ and $n$, it is easily calculated. Table 1 provides such calculations, which reflect conclusions of Proposition 4.

Table 1 The probability that group 1 wins: $v_1 = v_2$, $m$ active group-1 members and $n$ active group-2 members

Propositions 3 and 4 give us insight into how group 1’s performance varies with parameters, but they are silent on the two groups’ relative performance for given parameters. We next investigate group 1’s probability of winning, first establishing a FOSD comparison of the distributions of the groups’ best shots. When $v_1/m > v_2/n$ it is immediate from (18) that the distribution of an individual active group-1 player FOSD that of an active group-2 player. However, this does not yet establish whether the best shot of group 1 FOSD that of group 2 because group 2 may have more active players. As shown in Example 3, when $v_1/m = v_2/n$, the cdf of the best shot of the group with the larger value (strictly) FOSD the cdf of the other group’s best shoot. The following proposition extends this stochastic dominance comparison to the case in which $v_1/m>v_2/n$.¹⁹

Proposition 5

Suppose $(m,n) \in \{(i,j)\,|\, 1 \le i \le n_1, 1 \le j \le n_2\}$ and suppose $v_1/m > v_2/n$. Let $F$ and $G$ be determined through Lemma 3. Then, there exists an equilibrium in which group 1 has $m$ active members using cdf $F$, group 2 has $n$ active members using cdf $G$, and all other players are inactive. Let $F^{\mathrm{max}}$ and $G^{\mathrm{max}}$ denote the cdfs of the maximum efforts of groups 1 and 2, respectively.

• (i) If $v_1 \ge v_2$, then $F^{\mathrm{max}}$ strictly dominates $G^{\mathrm{max}}$ in the sense of FOSD.

• (ii) If $v_1 < v_2$, then $F^{\mathrm{max}}$ and $G^{\mathrm{max}}$ cannot be ranked by FOSD.

Proposition 5 implies that if $v_1/m > v_2/n$ and $v_1 \ge v_2$, then group 1 wins with probability exceeding 1/2. As part (ii) suggests, however, either group may be more likely to win if $v_1/m > v_2/n$ but $v_1 < v_2.$ ²⁰

As illustrated in Table 1, for fixed $v_1$, $v_2$, and $m$, as $n$ becomes arbitrarily large, group 1’s probability of winning is bounded away from 1. This occurs because the distribution of group 2’s best-shot effort approaches a nondegenerate distribution. We next determine which group is more likely to win when both groups are large. Suppose both groups grow large, say, in constant proportion: $m = tn$, where $t$ satisfies $1 \ge mv_2/nv_1 = t v_2/v_1$, that is, $t \le v_1/v_2$. Then, we have by the Bounded Convergence Theorem that

$$\begin{aligned}&\lim _{n \rightarrow \infty } \Pr \!\left( \text {group 1 wins}\, | \, m = tn \right) = \int _0^{1} \lim _{n \rightarrow \infty } \left[ \left( 1- \frac{tv_2}{v_1}\right) + \left( \frac{tv_2}{v_1} \right) z^{\frac{1}{tn}} \right] ^n \, \mathrm{d}z\nonumber \\&\quad = \int _0^{1} z^{\frac{v_2}{v_1}} \, \mathrm{d}z = \frac{v_1}{v_1 + v_2}. \end{aligned}$$

(22)

Remarkably, (22) is independent of $t$ (given $t \le v_1/v_2$); so as the two groups’ active members become arbitrarily numerous, the probabilities of winning become independent of the relative sizes of the active membership of the groups, and only relative values matter.²¹

We conclude Sect. 4.1 with an example. Beyond Example 3, a second case in which equilibrium strategies can be exactly derived is where one player faces many. We fix $m=1$ but allow arbitrary $v_1$, $n$, and $v_2$. The example again highlights the importance of the relative size of $v_1/m$ and $v_2/n$.

Example 4

(One vs. many). $m=1$

Part i): $v_1>v_2/n$. By (18), active group-2 players will place an atom on 0. With $m=1$, (15) becomes

$$\begin{aligned} \frac{1}{nv_1}&= \Big (G(x)\Big )^{n-1} G'(x) = \frac{\mathrm{d}}{\mathrm{d}x} \Big (G(x)\Big )^{n}; \end{aligned}$$

integrate the previous equation to obtain

$$\begin{aligned} \frac{x}{v_1} = \Big (G(x)\Big )^{n} - \Big (G(0)\Big )^{n} = \Big (G(x)\Big )^{n} - \left( 1-\frac{v_2}{nv_1}\right) ^{n} , \end{aligned}$$

where the second equality uses (18), with $F(0) = 0$. Therefore, $G(x) = \left[ \frac{x}{v_1} + \left( 1-\frac{v_2}{nv_1}\right) ^{n} \right] ^{\frac{1}{n}}.$ Solving $G(\bar{x}) = 1$, we find

$$\begin{aligned} \bar{x}= \left( 1 - \left( 1-\frac{v_2}{nv_1}\right) ^{n} \right) v_1, \end{aligned}$$

(23)

which is strictly less than $v_1$ and, if $n \ge 2$, is less than $v_2$.²² Consequently, when $n \ge 2$, the one active player on group 1 actually achieves a strictly positive expected payoff, even if $v_1 < v_2$—the free riding among active players on group 2 opens the door to this possibility. Indeed, as $n$ increases, each active group-2 player is increasingly likely to exert 0 effort; moreover, the equilibrium $G$ becomes larger, corresponding to stochastically lower equilibrium individual efforts. The equilibrium $F$ is found from (18) to be

$$\begin{aligned} F(x)&= \frac{nv_1}{v_2}\left( G(x) - \left( 1-\frac{v_2}{n v_1}\right) \right) \nonumber \\&= \frac{nv_1}{v_2}\left( \left[ \left( 1-\frac{v_2}{n v_1}\right) ^n + \frac{x}{v_1}\right] ^{1/n} - \left( 1-\frac{v_2}{n v_1}\right) \right) . \end{aligned}$$

(25)

Next, using (21), we explicitly calculate the probability that group 1 wins as

$$\begin{aligned}&\Pr \!\left( \text {group 1 wins}\right) = \int _0^{1} \left[ \left( 1-\frac{v_2}{nv_1}\right) +\frac{v_2}{nv_1}z\right] ^n \, \mathrm{d}z\\&\quad = \frac{n}{n+1}\times \frac{v_1}{v_2}\left( 1-\left( 1-\frac{v_2}{nv_1}\right) ^{n+1}\right) . \end{aligned}$$

Here, in part (i), $v_1/v_2$ ranges from $1/n$ to $\infty .$ By Proposition 4, therefore, we see that group 1’s probability of winning increases with its relative value ($v_1/v_2$) in the contest of one active player against many, and this probability can be anywhere from $1/(n+1)$ to $1$. In contrast, if $v_1 = v_2$, then group 1’s probability of winning increases in $n$, but even as the number of active group-2 players grows arbitrarily large, group 1’s probability of winning remains bounded above by $1-e^{-1}\approx 0.6321$, well below 1.

Part ii): $v_1 < v_2/n$. Rearranging (18), we obtain

$$\begin{aligned} F(x) = \left( 1-\frac{nv_1}{v_2}\right) + \frac{nv_1}{v_2} G(x). \end{aligned}$$

(26)

Now, we see player 1 places an atom on 0; hence, player 1 will earn payoff 0 in equilibrium. Therefore, $0 = -x + v_1 (G(x))^n,$ which implies $G(x) = (x/v_1)^{1/n}$ and $\bar{x}= v_1$ for the active group-2 players. And now the active group-1 player uses the following strategy on $[0, v_1]$: $F(x) = \left( 1-\frac{nv_1}{v_2}\right) + \left( \frac{x}{v_1}\right) ^{\frac{1}{n}}.$ Here, in part (ii), group 1’s value is very low relative to group 2’s, even accounting for the attendant free-riding among active group-2 players. The best the active player in group 1 can do is to earn a zero payoff.²³ $\square$

Section 3 considered symmetric groups in which members act symmetrically or asymmetrically within their group. In this section, we have considered asymmetric groups in which members act symmetrically within their group. One can easily construct examples in which groups act asymmetrically and active players within a group act asymmetrically. Numerous examples are possible—most simply, one group could have one active player and the other have two, with those two players using strategies similar to those in Example 1.

4.2 Three-group contests

We conclude with an example involving three groups, not all identical, to illustrate the consequences of having more than one active member per group on the incentives of low-value groups to participate.

Example 5

(An example with three active groups, $v_1=v_2 =v \ge v_3$)

Suppose groups 1 and 2 have 2 active members, and group 3 has none. From Proposition 1, we see that the active players would randomize over $[0,2v/3]$. Therefore, if $v_3< 2v/3$, group 3 players are effectively “blockaded” from active participation.²⁴ But what happens if $v_3 > 2v/3$? It turns out that two cases are to be considered, $v_3 \in [2v/3, 3v/4)$ and $v_3 \in [3v/4, v]$.

Part i): $v_3 \in [3v/4, v]$. So now suppose groups 1 and 2 have 2 active members, group 3 has 1. We look for an equilibrium that is symmetric across groups 1 and 2, with agents in these groups exerting effort using cdf $F_1 = F_2 = F,$ while the active agent in group 3 uses $F_3.$ In the Appendix, we show the following is an equilibrium: $\bar{x}= v_3- 2^{1/3}\Big (4v_3-3v\Big )^{4/3} / (8v_3^{1/3} ),$ and

$$\begin{aligned} F(x)&= \left( \frac{v_3-\bar{x}+x}{v_3}\right) ^{\frac{1}{4}}, \quad F_3(x)\\&= \frac{4(v_3)^{\frac{3}{4}}}{3v} (v_3-\bar{x}+x)^{\frac{1}{4}}\left[ 1 -\left( \frac{v_3-\bar{x}}{v_3-\bar{x}+x}\right) ^{\frac{3}{4}}\right] \quad \forall x \in [0, \bar{x}]. \end{aligned}$$

If $v_3 = v$, one can show that $F_3$ stochastically dominates $(F)^2$. Therefore, group 3 has a larger probability of winning than does either group 1 or group 2. This reinforces the earlier finding in the two-group contest that when values are equal, the smaller active group is more likely to win. And, if one takes $v_3 < v$ sufficiently close to $v$, continuity implies that the low-value group 3 remains the most likely to win.

Part ii): $v_3 \in [2v/3, 3v/4)$. From part (i), one sees player 3’s payoff is zero if $v_3 = 3v/4$. Hence, for smaller $v_3$ player 3’s expected payoff will also be zero. Therefore, for $v_3 \in (2v/3, 3v/4)$, player 3 participates in the contest, but only earns zero. Correspondingly, in equilibrium, $\bar{x}= v_3$. Surprisingly, it turns out that in equilibrium player 3 will not randomize over the entire interval $[0, v_3]$. Rather, groups 1 and 2 randomize over $[0, v_3]$; but player 3 continuously randomizes only over $[b, v_3]$ and puts all remaining mass on 0.²⁵

In the Appendix, we show the following is an equilibrium: $b\!=\!\left[ \frac{6v}{(v_3)^{1/4}}\big (1\!-\!\frac{4v_3}{3v}\big ) \right] ^{4/3}$, and

$$\begin{aligned}&F(x) = {\left\{ \begin{array}{ll} \left( \dfrac{x}{v_3}\right) ^{1/4} &{} \text {if }x \in [b, v_3]\\ \dfrac{x^{1/3}}{v_3^{1/4} b^{1/12} } &{} \text {if }x \in [0, b), \end{array}\right. }\\&F_3(x) = {\left\{ \begin{array}{ll} \sqrt{\dfrac{v_3}{x}} \left( 1- \dfrac{4v_3}{3v} + \dfrac{4v_3^{1/4}}{3v} x^{3/4}\right) &{} \text {if }x \in [b, v_3]\\ \dfrac{3}{v}\left( \dfrac{v_3}{2}\right) ^{2/3}(3v-4v_3)^{1/3} &{} \text {if }x \in [0, b). \end{array}\right. } \end{aligned}$$

Consequently, $b$ takes value 0 when $v_3 = 3v/4$ (consistent with part (i), where player 3 uses no atoms) and equals $2v/3$ when $v_3 = 2v/3$; in the limit as $v_3$ approaches $2v/3$ from above, group three effectively drops out—$F(0)$ goes to 1 as $v_3$ goes to $2v/3$. $\square$

Example 5 reveals a final important contrast with the individualistic all-pay auction contest. There, with players having values $v_1=v_2>v_3$, Baye et al. (1996) show the unique equilibrium has players 1 and 2 earn zero payoffs and player 3 remain inactive. Example 5 shows neither of these conclusions necessarily applies to group contests. Because groups 1 and 2 have more than one active player, their players earn positive payoffs, which means the maximum effort of players in groups 1 and 2 is less than $v_1$. This raises the possibility that if $v_3$ is not too much smaller than $v_2$, then a player from group 3 could actively participate, even earning a positive payoff, as in part (i) of Example 5.

5 Conclusion

We have analyzed a group contest problem in which individual members’ efforts are aggregated into a group effort via the best-shot technology. The interplay of within-group free-riding and across-group competition turns out to be especially interesting in this framework because it makes possible a wide variety of equilibria.

Indeed, the group contest has almost a coordination-game flavor: Equilibria with large group efforts coexist along with ones having small group effort (and with many intermediate cases). The best-shot technology is key to this result, along with the fact that the prize is a public good within a group. Indeed, as known from the theory of best-shot public goods (e.g., Barbieri and Malueg 2014a), the largest group effort occurs when only one agent within a group is active, while the smallest group effort arises when all agents within the group are active. In the former case, the lone active agent in a group does not find it profitable to lower his effort from this high (average) level because his group’s effort would be lowered one-to-one, and so he would bear the whole impact of the decreased probability of winning. In the latter case, one of the many active agents within a group does not find it profitable to increase his effort from his low (average) level because the probability of his being the group’s best-shot is small, since all other members of the same group are contributing as well; thus, the marginal positive effect on the probability of his group’s winning is small.

This wide variety of equilibria makes our results qualitatively different from those of the individualistic all-pay auction of Baye et al. (1996): Rents are not necessarily dissipated in equilibrium, total expected efforts vary across equilibria, and participation is expected to be larger. An intriguing finding is that, in contrast to much of the literature, free-riding can be beneficial for players as it reduces competition among groups. Less free-riding, in contrast, results in a cut-throat competition where the total value of the prize is the same but the total amount of effort expended is higher. Some equilibria derived here had never been, to the best of our knowledge, documented in the contest literature. In particular, equilibrium strategies with interlacing supports are entirely new, having been identified in neither group nor individualistic contests. We use these equilibria to pinpoint another source of free-riding: the wasteful duplication of efforts within groups.

Since our setup can be reinterpreted as a contest with IDE, our results can be thought of as complementing and expanding some of those in Klose and Kovenock (2012, 2013), who also find that identical players may earn different payoffs and the expected payoffs may vary across equilibria, in contrast to Baye et al. (1996). In particular, Proposition 6 in Klose and Kovenock (2012) and Example 3 in Klose and Kovenock (2013) deal with a situation that can be mapped into our framework by the competition between a 2-member group and an individual. Most importantly, we have spotlighted two sources of equilibrium payoff variability, namely the number of active agents and the wasteful duplication of efforts within groups. But we also show identical active members of the same group may receive different payoffs, extending the result of Example 3 in Klose and Kovenock (2013). And our Proposition 5 establishes a group-level first-order stochastic dominance ordering stronger than the member-by-member comparison of Proposition 6 in Klose and Kovenock (2012).

Many aspects of best-shot group contest problem deserve further attention. For instance, we have explored neither the consequences of preference asymmetries within groups nor of the possibility of sequential contributions within a group.²⁶ Both features are likely to be relevant for applied analysis, for instance, in determining the effects of the different state-by-state rules concerning amicus curiae briefs. In this paper, we have also abstracted from private information. These issues are the subjects of future research (see Barbieri and Malueg 2014b).

Appendix

Proof of Proposition 1

To complete the discussion in the main body of the paper and prove the proposition, we must show that no inactive players would wish to become active. Because groups are acting symmetrically, each of the inactive players earns payoff $v/N$. An inactive player considering a deviation would only consider efforts in $(0, \bar{x}]$. The payoff to such a deviator would be

$$\begin{aligned} V_D(x)&= -x + v \left[ [F(x)]^{Nm} + \frac{m}{Nm}\left( 1-[F(x)]^{Nm}\right) \right] \nonumber \\&= -x + v\left[ \frac{1}{N} + \frac{N-1}{N}[F(x)]^{Nm} \right] . \end{aligned}$$

Because $[F(x)]^{Nm}$ is proportional to $x^{\frac{Nm}{Nm-1}}$, we see that $V_D$ is strictly convex on $[0, \bar{x}]$, implying that an inactive player’s best-effort level will be either 0 or $\bar{x}$. And his payoff from effort level $\bar{x}$ is

$$\begin{aligned} v - \bar{x}= v - \left( \frac{Nm-m}{Nm-1} \right) v = \frac{m-1}{Nm-1}v < \frac{v}{N}. \end{aligned}$$

Consequently, becoming active is unprofitable—an inactive player has no profitable deviation. $\square$

Proof of Lemma 1

Let $G$ denote the cdf of the best shot of all group-1 players other than $a$ and $b$. Let $H$ denote the cdf of the best shot of all players not in group 1. Then, $F_a$, $F_b$, $G$, and $H$ are continuous with the minimum of their supports equal to 0, and $H$ is strictly increasing where it has value in $(0, 1)$ (the reasoning follows that in Baye et al. 1996).

The payoffs of players $a$ and $b$, for effort $x$, are

$$\begin{aligned} U_a(x)&= -x +v\left( 1-\int _x^{\infty } G(y) F_b(y) \, \mathrm{d}H(y) \right) \quad \text {and} \quad U_b(x)\\&= -x + v\left( 1-\int _x^{\infty } G(y) F_a(y) \, \mathrm{d}H(y) \right) . \end{aligned}$$

Part I. We first show that if the supports of $F_a$ and $F_b$ overlap on an interval, then $F_a = F_b$ on that interval. Suppose some $(x_1, x_2)$ lies in the supports of both $F_a$ and $F_b$. Indifference for players $a$ and $b$ implies that, for almost all $x \in (x_1, x_2)$,

$$\begin{aligned} 0&= U'_a(x) = -1 + v G(x) F_b(x) \, \mathrm{d}H(x) ; \end{aligned}$$

(27)

$$\begin{aligned} 0&= U'_b(x) = -1 + vG(x) F_a(x) \, \mathrm{d}H(x) . \end{aligned}$$

(28)

Consequently, since $\mathrm{d}H$ is positive on $(x_1, x_2)$, (27) and (28) imply $F_b(x) = F_a(x)$ for almost all $x \in (x_1, x_2)$. And because strategies are continuous on $[0, \infty )$, it must be that $F_b(x) = F_a(x)$ for all $x \in (x_1, x_2)$.

Part II. We now show that if $F_a \ne F_b$ then there does not exist an interval $(x_1, x_2)$ with $x_1 < x_2$ and $(x_1, x_2) \subset \text {supp }F_a \cap \text {supp }F_b$; that is, two active group members using different strategies have no (nondegenerate) interval overlap in their strategies’ supports. The proof is by contradiction, so suppose $F_a \ne F_b$, and there exists a nonempty nondegenerate interval $(x_1, x_2) \subset \text {supp }F_a \cap \text {supp }F_b$. Then, without loss of generality, we may suppose $\exists y_a \in \text {supp }F_a$ such that $y_a>0$, $y_a \notin \text {supp }F_b$, and $F_a(y_a) \ne F_b(y_a)$. Either $y_a > x_2$ or $y_a < x_1$; both possibilities yield contradictions. We conduct the proof only for $y_a > x_2$; the other case is similar.

Suppose $y_a > x_2$. Define $x_3 \equiv \inf \{y \ge x_2 \,|\, F_a(y) \ne F_b(y)\}$. Then, $x_3$ is well defined and at least as large as $x_2$ (it may be that $x_3 = x_2$). Intuitively, $x_3$ is the first point at least as large as $x_2$ above which $F_a$ and $F_b$ begin to diverge. Because above zero $F_a$ and $F_b$ have no atoms and their supports consist of finitely many intervals, it must be that, as one cdf begins to increase from $x_3$, it does so for a nondegenerate interval while the other cdf is constant over that interval. Thus, there exists $\varepsilon > 0$ such that for any $y \in (x_3, x_3+\varepsilon )$, either $F_a(y)$ is strictly increasing and $F_b(y)$ is constant or $F_a(y)$ is constant and $F_b(y)$ is strictly increasing. Without loss of generality, suppose the first case applies.

Then, $F_a(y) = F_b(y)$ for all $y \in [x_2, x_3]$ and

$$\begin{aligned} F_a(y) > F_a(x_3) = F_b(x_3) = F_b(y) \qquad \forall y \in (x_3, x_3 + \varepsilon ], \end{aligned}$$

(29)

where the inequality follows because $(x_3, x_3+\varepsilon ]$ lies in the support of $F_a$; the first equality follows because $F_a(x_2) = F_b(x_2)$ (from part I) and because the definition of $x_3$ implies $F_a$ and $F_b$ coincide on $[x_2, x_3]$; and the second equality follows because $(x_3, x_3+\varepsilon )$ is not in the support of $F_b$.

In equilibrium, player $a$ is indifferent between efforts $x_2$ and $x_3+\varepsilon$, so

$$\begin{aligned} 0&= U_a(x_3\!+\!\varepsilon ) \!-\! U_a(x_2)\nonumber \\&= v\int _{x_2}^{x_3+\varepsilon } G(y) F_b(y) \, \mathrm{d}H(y) \!-\! (x_3+\varepsilon - x_2)\nonumber \\&=\!v\int _{x_2}^{x_3} G(y) F_b(y) \, \mathrm{d}H(y)\nonumber \\&\ \qquad \qquad + v\int _{x_3}^{x_3\!+\!\varepsilon } G(y) F_b(x_3) \, \mathrm{d}H(y) \!-\! (x_3\!+\!\varepsilon \!-\! x_2).\quad \text {(by (29))} \end{aligned}$$

(30)

And, in equilibrium, effort $x_2$ is at least as good for player $b$ as is effort $x_3+\varepsilon$, so

$$\begin{aligned} 0 \ge U_b(x_3+\varepsilon ) - U_b(x_2) = v \int _{x_2}^{x_3+\varepsilon } G(y) F_a(y)\, \mathrm{d}H(y) - (x_3+\varepsilon - x_2). \end{aligned}$$

(31)

Therefore,

$$\begin{aligned}&v\int _{x_2}^{x_3} G(y) F_b(y) \, \mathrm{d}H(y) + v\int _{x_3}^{x_3+\varepsilon } G(y) F_b(x_3) \, \mathrm{d}H(y) \nonumber \\&\quad = x_3+\varepsilon - x_2&\text {(by (30))} \nonumber \\&\quad \ge v\int _{x_2}^{x_3+\varepsilon } G(y) F_a(y)\, \mathrm{d}H(y)&\text {(by (31))} \nonumber \\&\quad > v\int _{x_2}^{x_3} G(y) F_b(y)\, \mathrm{d}H(y)\nonumber \\&\quad \quad + v\int _{x_3}^{x_3+\varepsilon } G(y) F_b(x_3)\, \mathrm{d}H(y).&\text {(by (29))} \end{aligned}$$

(32)

The extremes of (32) establish a contradiction, from which we conclude that a nondegenerate interval such as $(x_1, x_2)$ does not exist. $\square$

Generalization of Example 2: equilibria with interlacing interval supports

Here, we characterize all equilibria with interlacing interval supports with two symmetric groups and two players in each group acting asymmetrically. We are interested in the evolution of expected payoffs, efforts, and waste as the interlacing of players’ strategies gets finer. We use the notation and terminology of Example 2, and we consider an arbitrary odd number $k$ of disjoint intervals covering $(0, \bar{x})$ (a similar construction can be carried out for even values of $k$). As in Example 1, we assume there are two groups, each with two players, one type-$a$ and one type-$b$. The support of $a$’s equilibrium strategy is $\left[ 0,x_{1}\right] \cup \left[ x_{2},x_{3}\right] \cup \cdots \cup \left[ x_{k-3},x_{k-2}\right] \cup \left[ x_{k-1},\bar{x}\right] ,$ while the support of $b$’s equilibrium strategy is $\{0\} \cup \left[ x_{1},x_{2}\right] \cup \left[ x_{3},x_{4} \right] \cup \cdots \cup \left[ x_{k-2},x_{k-1}\right] .$

We first determine the values of these interval extremes, along with equilibrium strategies $F_a,$ $F_b$ and the groups’ best shot cdf $H = F_a \cdot F_b.$ Then, we perform a comparative statics exercise with respect to $k$ that formally demonstrates how, as we decrease $k$, the asymmetry in contributors’ equilibrium strategies within each group increases, the incentive to free-ride within each group diminishes and that the payoff for every group becomes smaller.

We begin with the determination of $F_a$ and $F_b$, and thereby $H$. We start with the FOC for type-$a$’s utility maximization relative to a point $x$ belonging to one of the intervals in which $F_{a}$ is strictly increasing and $F_b>0$, say $\left[ x_{k^{\prime }-1}, x_{k^{\prime }} \right]$ (in other words, take $k^{\prime }$ odd and $k^{\prime }<k-1$). This is nothing but equation (27), which in our context yields $1=vF_{b}\left( x_{k^{\prime }}\right) h(x)$, where we denote $h\left( x\right) =\mathrm{d}H\left( x\right)$. Therefore, by continuity of $H,$ integration yields

$$\begin{aligned} H(x_{k^{\prime }})-H(x_{k^{\prime }-1})=\frac{x_{k^{\prime }}-x_{k^{\prime }-1}}{vF_{b}\left( x_{k^{\prime }}\right) }. \end{aligned}$$

(33)

At the same time, $b$ must be indifferent between contributing $x_{k^{\prime }}$ and $x_{k^{\prime }-1}$, leading to

$$\begin{aligned} x_{k^{\prime }}-x_{k^{\prime }-1}&=v\int _{x_{k^{\prime }-1}}^{x_{k^{\prime }}}F_{a}\left( y\right) \mathrm{d}H\left( y\right) = v F_{b}\left( x_{k^{\prime }-1}\right) \int _{x_{k'-1}}^{x_{k'}} F_a(y) \, \mathrm{d}F_a(y) \nonumber \\&= \frac{v}{2} F_{b}\left( x_{k^{\prime }-1}\right) \left[ (F_a(x_{k'}))^2 - (F_a(x_{k'-1}))^2 \right] \nonumber \\&=\frac{1}{2}\frac{v}{F_{b}\left( x_{k^{\prime }-1}\right) }\left[ H^{2}\left( x_{k^{\prime }}\right) -H^{2}\left( x_{k^{\prime }-1}\right) \right] , \end{aligned}$$

(34)

where the last equality uses $F_b(x_{k'}) = F_b(x_{k'-1})$. Combining (33) and (34), and again noting that $F_{b}\left( x_{k^{\prime }-1}\right) =F_{b}\left( x_{k^{\prime }}\right) ,$ we derive

$$\begin{aligned} 2F_{b}\left( x_{k^{\prime }-1}\right) =F_{a}\left( x_{k^{\prime }}\right) +F_{a}\left( x_{k^{\prime }-1}\right) . \end{aligned}$$

(35)

Similarly, when considering intervals $[x_{k'-2}, x_{k'-1}]$ in which $F_{b}$ is strictly increasing, one obtains

$$\begin{aligned} 2F_{a}\left( x_{k^{\prime }-2}\right) =F_{b}\left( x_{k^{\prime }-1}\right) +F_{b}\left( x_{k^{\prime }-2}\right) . \end{aligned}$$

(36)

There are $(k-1)/2$ equalities described by (35) and $(k-1)/2 - 1$ equalities described by (36). Moreover, we have the two boundary conditions $F_{b}(x_{k-1})=1,$ and $F_{a}(0)=0.$ Finally, note that, in addition to $F_{a}(0),$ only $\frac{ k-1}{2}$ other values of $F_{a}$ need to be determined, since $F_{a}(x_{k^{\prime }+1})=F_{a}(x_{k^{\prime }})$ for an odd $k^{\prime }$, and that similar considerations hold for $b$. Therefore, maintaining $k^{\prime }$ odd and $k^{\prime }<k-1$, the solution of this linear system of equations is

$$\begin{aligned} F_{a}(x_{k^{\prime }+1})=\frac{k^{\prime }+1}{k}=F_{a}(x_{k^{\prime }}), \quad F_{b}(x_{k^{\prime }})=\frac{k^{\prime }}{k}=F_{b}(x_{k^{\prime }-1}). \end{aligned}$$

(37)

Indeed, to verify (35) is satisfied, note that,

$$\begin{aligned} 2F_{b}\left( x_{k^{\prime }-1}\right) = 2\left( \frac{k^{\prime }}{k}\right) =\frac{k^{\prime }+1}{k}+\frac{k^{\prime }-1}{k} =F_{a}\left( x_{k^{\prime }}\right) +F_{a}\left( x_{k^{\prime }-1}\right) , \end{aligned}$$

while to verify (36) we see that

$$\begin{aligned} 2F_{a}\left( x_{k^{\prime }-2}\right) =2\left( \frac{k^{\prime }-1}{k}\right) =\frac{ k^{\prime }}{k}+\frac{k^{\prime }-2}{k} = F_{b}\left( x_{k^{\prime }-1}\right) +F_{b}\left( x_{k^{\prime }-2}\right) . \end{aligned}$$

Conditions (37) and (27) further imply that

$$\begin{aligned} f_{a} (x)\! =\! \frac{1}{v} \left( \frac{k}{k'}\right) ^2 \, \text {for } x \in [x_{k^{\prime }-1},x_{k^{\prime }}], \, f_{b} (x) \!=\! \frac{1}{v} \left( \frac{k}{k'-1}\right) ^2 \, \text {for } x \!\in \![x_{k^{\prime }-2},x_{k^{\prime }-1}], \end{aligned}$$

and that, for all $\tilde{k} \le k-1,$ we have

$$\begin{aligned} H(x_{\tilde{k}}) =\frac{\tilde{k}(\tilde{k}+1)}{k^2}. \end{aligned}$$

(38)

The above ensures that $a$ is indifferent between contributing any amount $x \in \left[ 0,x_{1}\right] \cup \left[ x_{2},x_{3}\right] \cup \cdots \cup \left[ x_{k-3},x_{k-2}\right] \cup \left[ x_{k-1},\bar{x}\right] .$ To see that $a$ does not benefit from a contribution that $b$ also makes, note that for any $x$ in an interval in which $F_b$ is strictly increasing, the first derivative of $a$’s utility, using (27), is $-1+v F_b (x) h(x),$ and $h(x)$ is constant, since both $f_b (x)$ and $F_a (x)$ are constant in this interval. Therefore, $a$’s utility is strictly convex in an interval in which $F_b$ is strictly increasing, so its maximum is at the extremes of the interval. But the above reasoning already ensures that $a$ does not strictly prefer any contribution belonging to $\{x_1,x_2,\ldots ,x_{k-1},\bar{x}\}$ to the equilibrium strategy. Similar considerations apply to $b.$

We now proceed to determining the values of $x_1,x_2,\ldots ,x_{k-1},\bar{x}.$ By (33) and (38), we obtain that, for $k^{\prime }$ odd, $v\cdot \frac{2\left( k^{\prime }\right) ^{2}}{k^{3}} =x_{k^{\prime }}-x_{k^{\prime }-1}.$ For $\tilde{k}$ even, the necessary and sufficient condition to maximize type $b$’s utility leads to the following analog of (33): $vF_{a}\left( x_{\tilde{k}}\right) \left( H(x_{\tilde{k}})-H(x_{\tilde{k} -1})\right) = x_{\tilde{k}}-x_{\tilde{k}-1}.$ Equation (38) then yields

$$\begin{aligned} v\frac{\tilde{k}}{k}\frac{2\tilde{k}}{k^{2}} = x_{\tilde{k}}-x_{\tilde{k}-1}; \end{aligned}$$

therefore, the size of each interval, except for $\bar{x}- x_{k-1},$ follows the same formula (but intervals are all of different length). Starting from $x_0=0,$ then, it is therefore possible to determine all intervals endpoints up to $x_{k-1}.$ Finally, $\bar{x}$ is determined from the indifference condition of type-$a$ when contributing $x \in [x_{k-1},\bar{x}],$ i.e., $vH(x) - x = v - \bar{x}$, which at $x = x_{k-1}$ implies

$$\begin{aligned} \bar{x}=x_{k-1}+v\left( 1-H\left( x_{k-1}\right) \right) = x_{k-1}+v\left( 1-\frac{\left( k-1\right) k}{k^{2}}\right) = x_{k-1}+\frac{v}{k}. \end{aligned}$$

Because the support intervals have union $[0, \bar{x}]$ and they are disjoint except possibly at endpoints, it follows that the sum of their lengths is $\bar{x}$; therefore,

$$\begin{aligned} \bar{x}=v\frac{2}{k^{3}} \sum _{j=1}^{j=k-1}\left( j\right) ^{2}+\frac{v}{k } =\frac{2v}{k^{3}}\frac{\left( k-1\right) \left( k\right) \left( 2k-1\right) }{6}+\frac{v}{k} = v\frac{2k^{2}+1}{3k^{2}}. \end{aligned}$$

We now perform a comparative statics exercise by varying the number of intervals, i.e., $k$. Note that, as a function of $k$, $\bar{x}$ is decreasing and that it converges to $\frac{2}{3}v$ as $k$ goes to infinity. Moreover, the payoff of type $a$ is $v-\bar{x}=v(k^{2}-1)/(3k^{2}),$ while the payoff of $b$ is

$$\begin{aligned} v\left( 1-\int _{x_{k-1}}^{\bar{x}}F_{a}\left( y\right) \mathrm{d}H\left( y\right) \right) -x_{k-1}&=v\left( 1-\frac{1}{2}\left( 1-\Big (H\left( x_{k-1}\right) \Big ) ^{2}\right) \right) -x_{k-1} \\&=v\left( 1\!-\!\frac{1}{2}\left( 1\!-\!\left( \frac{k(k\!-\!1)}{k^{2}}\right) ^{2}\right) \right) \!-\!\left( \bar{x}\!-\!\frac{v}{k}\right) \\&=v\left( \frac{2k^{2}+1}{6k^{2}}\right) . \end{aligned}$$

Therefore, the difference in payoffs within the group is $v/2k^2$ and the total payoff of the group is

$$\begin{aligned} 2\left( v-\bar{x}\right) +\frac{v}{2k^{2}} =v\left( 2\frac{k^{2}-1}{3k^{2}}+ \frac{1}{2k^{2}}\right) = v\frac{4k^{2}-1}{6k^{2}}=v\left( \frac{2}{3}-\frac{1}{6k^{2}}\right) . \end{aligned}$$

Clearly, the difference in payoffs within the group is decreasing in $k$, while the sum is increasing. Also, as $k \rightarrow \infty ,$ the difference in payoffs converges to zero, while the sum converges to $2v/3$, which is the group’s payoff in the fully symmetric equilibrium described in Eq. (2).²⁷ Also, note that the equilibrium is symmetric across groups, so the expected probability of winning for each group remains $1/2$ for any $k$. Since the expected payoff is increasing in $k$, the expected total group effort must therefore be decreasing in $k$ (an explicit calculation follows). And since the difference in payoffs within the group is decreasing in $k$ and converges to zero, expected individual efforts converge to the same limit.

Again, we measure waste as in (11). For instance, in Example 2, we have that waste is

$$\begin{aligned} \int _{0}^{\frac{2}{27}v}x \cdot \frac{2}{3} \cdot \frac{9}{v} \, \mathrm{d}x+\int _{\frac{2}{27}v}^{ \frac{10}{27}v}x \cdot \frac{1}{3} \cdot \frac{9}{4v} \, \mathrm{d}x= \frac{16}{243}v. \end{aligned}$$

To calculate how $W$ depends on $k$ in general, note that $W=\int x \,d\left( F_{a}\left( x\right) +F_{b}\left( x\right) \right) -\int x \,\mathrm{d}H\left( x\right) .$ The first term is expected total effort of the group. It can be calculated indirectly from the payoff calculations in the paper. Since a group’s expected winnings are $v$ while payoff is calculated above as $v\left( \frac{2}{3}- \frac{1}{6k^{2}}\right)$, it follows that the expected total effort is $\left( 2k^{2}+1\right) v/6k^{2}.$ The term $\int x \,\mathrm{d}H\left( x\right)$ is usefully decomposed as $\sum \nolimits _{j=1}^{k-1}\int _{x_{j-1}}^{x_{j}}x \,\mathrm{d}H\left( x\right) +\int _{x_{k-1}}^{\bar{x}}x \,\mathrm{d}H\left( x\right) ,$ and exploiting the fact that $H$ is conditionally uniform over $\left( x_{j-1},x_{j}\right)$, we have

$$\begin{aligned} \int x \,\mathrm{d}H\left( x\right)&= \sum \limits _{j=1}^{k-1}\frac{x_{j}+x_{j-1}}{2} \left( H\left( x_{j}\right) -H\left( x_{j-1}\right) \right) +\frac{ \bar{x}+x_{k-1}}{2}\left( 1-H\left( x_{k-1}\right) \right) \\&= \sum \limits _{j=1}^{k-1}\frac{1}{3}\frac{v}{k^{3}}j\left( 2j^{2}+1\right) \left( 2\frac{j}{k^{2}}\right) +\frac{v}{6k^{3}}\left( 4k^{2}-3k+2\right) \\&= \frac{5k^{2}+8k^{4}+2}{30k^{4}}v, \end{aligned}$$

where the second equality uses the formulas for $x_{j}$ and $H\left( x_{j}\right)$ developed earlier. Therefore, waste is calculated as $W=\frac{\left( k^{2}+1\right) \left( k+1\right) \left( k-1\right) }{15k^{4}} v,$ which is easily seen to be a strictly increasing function of $k$. Also, as $k\rightarrow \infty$, $W\rightarrow \frac{1}{15}v$, which one can show is the waste in the semi-symmetric equilibrium. $\square$

Proof of Lemma 2, part (iii)

Expression (14) and parts (i) and (ii) of Lemma 2 show $V_1(x)$ is continuous. If $v_1^*$ denotes the equilibrium payoff to one of the $m$ active players using cdf $F$, then it must be that $V_1(x) = v_1^*$ on a dense subset of $[0, \bar{x}]$; and because $V_1$ is continuous, it follows that $V_1$ is constant on $[0, \bar{x}]$. Hence, $V_1$ is continuously differentiable on $(0, \bar{x})$. Rearranging the expression for $V_1$, we obtain

$$\begin{aligned} 1- \frac{V_1(x) + x}{v_1} = \int _x^\infty F^M(y)\, \mathrm{d}G^M(y). \end{aligned}$$

Since the above left-hand side is continuously differentiable, so will be the right-hand side. And this implies, too, that $G^M \in C'$ for $x \in (0, \bar{x})$. To see this, using the definition of derivative and $V_1' = 0$, we have

$$\begin{aligned} -\frac{1}{v_1}&= \lim _{h \rightarrow 0} \frac{\int _{x+h}^\infty F^M(y)\, \mathrm{d}G^M(y)-\int _x^\infty F^M(y)\, \mathrm{d}G^M(y)}{h}\\&= -\lim _{h \rightarrow 0} \frac{\int _{x}^{x+h} F^M(y)\, \mathrm{d}G^M(y)}{h}, \end{aligned}$$

and, since $F^M$ is continuous, we obtain²⁸

$$\begin{aligned} \frac{1}{v_1} = \lim _{h \rightarrow 0} F^M(x_h') \, \frac{\int _{x}^{x+h} \, \mathrm{d}G^M(y)}{h}=\lim _{h \rightarrow 0} F^M(x_h') \, \frac{G^M (x+h) -G^M (x)}{h}, \end{aligned}$$

where $x_h' \in [x,x+h],$ which, by the usual properties of limits, leads to

$$\begin{aligned} \frac{1}{v_1} \, \frac{1}{F^M(x)} =\lim _{h \rightarrow 0} \frac{G^M (x+h) -G^M (x)}{h}. \end{aligned}$$

(39)

The above demonstrates that $G^M$, and therefore also $G$, is continuously differentiable on $(0, \bar{x})$. Analogous considerations establish that $F^M$ and $F$ are continuously differentiable on $(0, \bar{x})$. $\square$

Proof of Lemma 3

While the Lipschitz condition fails at $0$, it is still possible to show that a unique solution exists by showing that there exists only one possible $\bar{x}$ where $F\left( \bar{x}\right) =1$ and then using this as initial condition, completing the solution at zero by continuity.

By contradiction, suppose there are two distinct upper bounds $\bar{x}_{1}$ and $\bar{x}_{2}$, with $\bar{x}_{2}>\bar{x}_{1}$. Let $F_{2}$ and $F_{1}$ be the appropriate unique solutions, respectively. Then, since both $F_{2}$ and $F_{1}$ must be strictly increasing, we have $F_{1}\left( \bar{x}_{1}\right) >F_{2}\left( \bar{x}_{1}\right) .$ At any $x\le \bar{x}_{1}$ for which $F_{1}\left( x\right) >F_{2}\left( x\right)$, the differential equation (19) implies $F'_1(x)<F'_2(x)$. Therefore, $\left( F_{1}\left( x\right) -F_{2}\left( x\right) \right)$ decreases in $x$, and it is strictly positive at $\bar{x}_{1}$, ruling out the possibility that both $F_{1}$ and $F_{2}$ satisfy the same initial condition at zero, thus establishing a contradiction.

To conclude the verification that the proposed strategies constitute an equilibrium, in addition to the argument in the main text we must check that no inactive player has a profitable deviation. Let $H^M(x) \equiv (F(x))^m$ denote the cdf of the maximum effort of the $m$ active players on group 1; again let $G^M(x)) \equiv (G(x))^n$ denote cdf of the maximum effort of the $n$ active players on group 2 (and let $g^M$ be the associated density). In this situation, an inactive player on group 1 choosing effort $x \in (0, \bar{x}]$ would earn payoff

$$\begin{aligned} V_1(x) = - x + v_1 \left[ H^M(x)G^M(x) + \int _x^\infty G^M(y)\, \mathrm{d}H^M(y) \right] , \end{aligned}$$

yielding

$$\begin{aligned} V_1'(x)&= -1 + v_1 H^M(x) g^M(x) = -1 + v_1 (F(x))^m n (G(x))^{n-1} G'(x)\\&= - 1 + v_1 F(x)\Big [n(F(x))^{m-1} (G(x))^{n-1} G'(x)\Big ]\\&= -1 + v_1 F(x) (1/v_1)&\text {(by (15))}\\&= -(1-F(x))< 0 \quad \forall x \in (0, \bar{x}). \end{aligned}$$

Thus, an inactive player on group 1 would not choose to become active. Similarly, group 2 inactive players would not become active. It follows that the $F$ and $G$ derived above constitute an equilibrium with $m$ active group-1 players, $n$ active group-2 players, and all other players inactive. $\square$

Proof of Proposition 3

We begin by recalling the differential equation for the cdf of group 1’s best-shot effort:

$$\begin{aligned} 1= v_2\left[ 1-\frac{m v_2}{n v_1}\left( 1 - \Big [H(x)\Big ]^{1/m}\right) \right] ^{n-1} H'(x),\quad \text {with} \quad H(0) = 0. \end{aligned}$$

(20)

It is simpler to work with the inverse function of the best-shot cdf, rather than $H$ itself. To this end, we define $\eta \equiv H^{-1}$: $\eta (H(x)) = x$ for all $x \ge 0$. Then, $\eta '(H(x))H'(x) \equiv 1$, so, letting $y=H(x)$, we may rewrite (20) as

$$\begin{aligned} \eta '(y) = v_2\left[ 1-\frac{m v_2}{n v_1}\left( 1 - y^{1/m}\right) \right] ^{n-1} \quad \text {with} \quad \eta (0) = 0. \end{aligned}$$

(40)

First consider increasing $v_1$. In (40), the term in the square brackets is clearly increasing in $v_1$. Therefore, if $\tilde{v}_1 > v_1$, with corresponding inverse best-shot cdfs $\tilde{\eta }$ and $\eta$, then $\tilde{\eta }'(y) > \eta '(y)$ for all $y \in [0, 1)$. Together with $\tilde{\eta }(0) = \eta (0) = 0$, it follows that $\tilde{\eta }(y) > \eta (y)$ for all $y \in [0, 1]$. The upper end of the support of $\tilde{H}$ is higher than that for $H$ because $\tilde{\bar{x}} = \tilde{\eta }(1)>\eta (1) = \bar{x}$. Also, suppose $x = \eta (t)$. Because $\tilde{\eta }(t)> x$, there exists $\tilde{t}< t$ such that $\tilde{\eta }(\tilde{t}))= x$. Now, we see $\tilde{H}(x) = \tilde{H}(\tilde{\eta }(\tilde{t})) = \tilde{t} < t = H(\eta (t)) = H(x)$, that is, $\tilde{H}(x) < H(x)$ for any $x$ such that $H(x) \in (0, 1)$; thus, the cdf $\tilde{H}$ FOSD $H$. Since the number of players in group 1 does not change, it must be that individual group-1 efforts are also higher in the sense of FOSD: $\tilde{F}(x) = (\tilde{H}(x) )^{1/m}< (H(x))^{1/m} = F(x)$ for any $x$ such that $\tilde{H}(x) \in (0, 1)$.

Next consider reducing $m$. We show that the term in square brackets of (20) decreases with increases in $m$. To see this most simply, define $\varphi (m) = m (1-t^{1/m})$, where $t \in (0, 1)$ takes the place of $H(x)$. Then $\varphi '(m) = 1-t^{1/m}+t^{1/m} \log \left( t^{1/m}\right) > 0$ for all $t^{1/m} \in (0, 1)$.²⁹ Therefore, an increase in $m$ reduces the full expression in square brackets. Conversely, reducing $m$ increases this term in square brackets, so if $\tilde{m}< m$ (and, starting with $v_1/m \ge v_2/n$), then arguing as in the previous paragraph, we conclude that $\tilde{\bar{x}} > \bar{x}$ and the cdf $\tilde{H}$ FOSD $H$. Since reducing the number of players stochastically increases efforts, it must be that the fewer active players, acting symmetrically, individually increase their efforts in the sense of FOSD: $\tilde{F}(x) = (\tilde{H}(x) )^{1/\tilde{m}}< (\tilde{H}(x))^{1/m}< (H(x))^{1/m} = F(x)$ for any $x$ such that $\tilde{H}(x) \in (0, 1)$. $\square$

Proof of Proposition 4

It only remains to establish the effect of increasing $n$. In (21), denote $\frac{mv_2}{v_1} \left( 1- z^{1/m}\right)$ by $a$. Notice that since by assumption $\frac{mv_2}{nv_1} \le 1$ and $1-z^{1/m}<1$ for all $0<z<1$, it follows that $0<a<n$. We show that the integrand of (21), now written as $A_n \equiv \left( 1-\frac{a}{n} \right) ^n$, is increasing in $n$. Let $\tilde{A}_n$ be the extension of $A_n$ to $n$ continuous; then, $\tilde{A}_n$ is differentiable in $n$ and

$$\begin{aligned} \frac{\partial \tilde{A}_n}{\partial n} = \left( 1-\frac{a}{n} \right) ^n \left[ \log \left( 1-\frac{a}{n} \right) + \frac{a}{n-a} \right] . \end{aligned}$$

Denoting $\frac{n-a}{n}$ by $t$, we have

$$\begin{aligned} \frac{\partial \tilde{A}_n}{\partial n} = t^n \left[ \log t - 1 +\frac{1}{t} \right] . \end{aligned}$$

Since $\log t - 1 +\frac{1}{t}>0$ for all $0<t<1$, it follows that $\frac{\partial \tilde{A}_n}{\partial n}>0$. Therefore, the integrand in (21) is increasing in $n$, implying that the probability that group 1 wins is increasing in $n$. $\square$

Proof of Proposition 5

Part (i). Because $\left( \frac{v_2}{v_1}\right) \left( \frac{m}{n}\right) < 1$, (18) yields $G(0)>0 = F(0)$, implying $G^\text {max}(0) > F^\text {max}(0)$. It only remains to show that $(G(x))^n > (F(x))^m$ for all $x \in (0, \bar{x})$. Using (18), this is equivalent to showing

$$\begin{aligned} \left( \left[ 1-\left( \frac{v_2}{v_1}\right) \frac{m}{n}\right] +\left( \frac{v_2}{v_1}\right) \frac{m}{n}F(x)\right) ^n > (F(x))^m \qquad \forall x \in (0, \bar{x}), \end{aligned}$$

which, after taking logs, is equivalent to showing

$$\begin{aligned} {\mathrm {log }}\!\left( \, \left[ 1-\left( \frac{v_2}{v_1}\right) \frac{m}{n}\right] +\left( \frac{v_2}{v_1}\right) \frac{m}{n}F(x)\, \right) >\frac{m}{n} {\mathrm {log }}\!\left( \, F(x)\, \right) . \end{aligned}$$

And this inequality follows from the strict concavity of ${\mathrm {log }}\!\left( \, \cdot \, \right)$: for any $x \in (0, \bar{x})$,

$$\begin{aligned} {\mathrm {log }}\!\left( \, \left[ 1-\left( \frac{v_2}{v_1}\right) \frac{m}{n}\right] +\left( \frac{v_2}{v_1}\right) \frac{m}{n}F(x)\, \right)&> \left[ 1-\left( \frac{v_2}{v_1}\right) \frac{m}{n}\right] {\mathrm {log }}\!\left( \, 1\, \right) \\&\quad + \left( \frac{v_2}{v_1}\right) \frac{m}{n} {\mathrm {log }}\!\left( \, F(x)\, \right) \\&= \left( \frac{v_2}{v_1}\right) \frac{m}{n} {\mathrm {log }}\!\left( \, F(x)\, \right) \ge \frac{m}{n} {\mathrm {log }}\!\left( \, F(x)\, \right) , \end{aligned}$$

where the last inequality follows because ${\mathrm {log }}\!\left( \, F(x)\, \right) <0$ and in part (i) we assume $v_1 \ge v_2$.

Part (ii). As in part (i), $G^{\mathrm{max}}(0) > F^{\mathrm{max}}(0)$. Next, consider $x$ in a left neighborhood of $\bar{x}$. Letting $f$ denote the density corresponding to $F$, we see, by taking limits in (16), that $f(\bar{x}) =1/(mv_2)$. Moreover,

$$\begin{aligned} \lim _{x\rightarrow \bar{x}}\frac{\mathrm{d}G^\mathrm{{max}}(x)}{\mathrm{d}x}&= \lim _{x\rightarrow \bar{x}} n \left( G(x)\right) ^{n-1} G'(x)\\&= \lim _{x\rightarrow \bar{x}} n G(x)^{n-1} \left( \frac{v_2}{v_1}\right) \left( \frac{m}{n}\right) f(x)&\text {(by (17))}\\&= \frac{1}{v_1}> \frac{1}{v_2}&(\text {because } v_1<v_2)\\&= \lim _{x\rightarrow \bar{x}} m \left( F(x)\right) ^{m-1} f(x) \\&= \lim _{x\rightarrow \bar{x}} \frac{\mathrm{d}F^{\mathrm{max}}(x)}{\mathrm{d}x}, \end{aligned}$$

showing that at $\bar{x}$, $G^{\mathrm{max}}$ increases strictly faster than does $F^{\text {max}}$. Therefore, for some $\varepsilon > 0$, $G(x) < F(x)$ for all $x \in (\bar{x}- \varepsilon , \bar{x})$. So near 0, $G>F$; and near $\bar{x}$, $G<F$. Thus, $F$ and $G$ cannot be ranked by FOSD. $\square$

Construction of Example 5

Part i): $v_3 \in [3v/4, v]$. Suppose groups 1 and 2 have 2 active members, group 3 has 1. We have the following probabilities of winning when a player in the named group exerts effort $x$:

$$\begin{aligned} \Pr \!\left( \text {group 1 wins} \,|\, x\right)&= F_1(x) (F_2(x))^2 F_3(x) + \int _x^{\bar{x}} (F_2(y))^2 F_3(y) f_1(y)\, \mathrm{d}y, \nonumber \\ \Pr \!\left( \text {group 2 wins} \,|\, x\right)&= (F_1(x))^2 F_2(x) F_3(x) + \int _x^{\bar{x}} (F_1(y))^2 F_3(y) f_2(y)\, \mathrm{d}y, \text { and} \nonumber \\ \Pr \!\left( \text {group 3 wins} \,|\, x\right)&= (F_1(x))^2 (F_2(x))^2 . \end{aligned}$$

(41)

We look for an equilibrium that is symmetric across groups 1 and 2, with $F_1 = F_2 = F$ and associated density $f$. The group-3 player earns payoff $v_3 - \bar{x}$, so, for any $x \in (0, \bar{x}]$, the group-3 player’s payoff is

$$\begin{aligned} v_3 -\bar{x}= V_3(x) = -x + v_3 (F(x))^4, \end{aligned}$$

implying

$$\begin{aligned} F(x) = \left( \frac{v_3-\bar{x}+x}{v_3}\right) ^{1/4} \quad \forall x \in (0, \bar{x}]. \end{aligned}$$

(42)

By right-continuity of cdfs, formula (42) also holds at $x=0$. With symmetry and (41), we see an active group-1 player choosing effort $x$ obtains payoff

$$\begin{aligned} V_1(x) = -x + v \left[ (F(x))^3 F_3(x) + \int _x^{\bar{x}} (F(y))^2 F_3(y) f(y)\, \mathrm{d}y \right] . \end{aligned}$$

Constancy of player 1’s payoff over the randomization interval yields

$$\begin{aligned} 0 = V_1'(x) = -1+v \Big [2(F(x))^2 F_3(x) f(x) + (F(x))^3 f_3(x)\Big ], \end{aligned}$$

which can be rewritten as

$$\begin{aligned} \frac{1}{v F(x)} = \frac{\mathrm{d}}{\mathrm{d}x}\left[ (F(x))^2 F_3(x)\right] . \end{aligned}$$

(43)

If the group-3 player’s payoff is positive, then it must be that $F$ puts an atom at 0 and $F_3$ does not. Therefore, upon integration, the previous differential equation yields

$$\begin{aligned} (F(x))^2 F_3(x)&= \int _0^x \frac{(v_3)^{1/4}}{v} (v_3 - \bar{x}+ y)^{-1/4}\, \mathrm{d}y \\&= \frac{4(v_3)^{1/4}}{3v} \left[ (v_3 - \bar{x}+ x)^{3/4} - (v_3 - \bar{x})^{3/4} \right] . \end{aligned}$$

Substituting for $F(x)$ from (42) and solving for $F_3$, we obtain

$$\begin{aligned} F_3(x) = \frac{4(v_3)^{3/4}}{3v} (v_3-\bar{x}+x)^{1/4}\left[ 1 -\left( \frac{v_3-\bar{x}}{v_3-\bar{x}+x}\right) ^{3/4}\right] \quad \forall x \in [0, \bar{x}]. \end{aligned}$$

Now to find $\bar{x}$, we solve $F_3(\bar{x}) = 1$, a condition that we rewrite as

$$\begin{aligned} \frac{1}{v_3}\left( v_3 - \frac{3v}{4}\right) =\left( \frac{v_3-\bar{x}}{v_3}\right) ^{3/4}. \end{aligned}$$

(44)

Equilibrium requires $\bar{x}\le v_3$, so by (44) the above analysis will be valid if and only if $v_3 \in [3v/4, v]$. So for $v_3 \in [3v/4, v]$, we calculate

$$\begin{aligned} \bar{x}= v_3- \frac{2^{1/3}}{8v_3^{1/3}} \Big (4v_3-3v\Big )^{4/3}. \end{aligned}$$

So we see the foregoing is valid (i.e., yields $\bar{x}\le v_3$) if and only if $v_3 \in \left[ \frac{3}{4}v, v\right]$.

Part ii): $v_3 \in [2v/3, 3v/4)$. We saw from part (i) that player 3’s payoff is zero when $v_3 = 3v/4$. Hence, for smaller $v_3$ player 3’s expected payoff will also be zero. Therefore, for $v_3 \in (2v/3, 3v/4)$, player 3 will choose to participate in the contest but only earns a payoff of zero. Correspondingly, equilibrium will require $\bar{x}= v_3$. Surprisingly, it turns out that in equilibrium player 3 will not randomize over the entire interval $[0, v_3]$. Rather, we look for a strategy where groups 1 and 2 randomize over $[0, v_3]$, but group 3 continuously randomizes only over $[b, v_3]$ and puts all remaining mass on 0. The value for $b$ is not arbitrary, but will be (uniquely) determined as part of the equilibrium.

On $[b, v_3]$ the condition that group 3 player’s payoff is 0 implies groups 1 and 2 use strategy

$$\begin{aligned} F(x) = \left( \frac{x}{v_3}\right) ^{1/4} \qquad \forall x \in [b, v_3]. \end{aligned}$$

(45)

To find $F_3$ on $[b, v_3]$, we integrate the differential equation (43) down from $\bar{x}=v_3$, using the boundary condition $F(v_3)=F_3(v_3)=1$:

$$\begin{aligned} 1-(F(x))^2 F_3(x)&= \int _x^{v_3} \frac{\mathrm{d}}{\mathrm{d}y}\left[ (F(y))^2 F_3(y)\right] \mathrm{d}y \end{aligned}$$

(46)

$$\begin{aligned}&=\int _x^{v_3} \frac{1}{v} (v_3)^{1/4} y^{-1/4}\, \mathrm{d}y&\text {(by (43) and (45))} \nonumber \\&= \frac{4}{3v} v_3^{1/4} \left( v_3^{3/4} - x^{3/4}\right) . \end{aligned}$$

(47)

And the formula for $F_3$ will be derived from (46)–(47) and (45):

$$\begin{aligned} F_3(x) = \sqrt{\frac{v_3}{x}} \left( 1- \frac{4v_3}{3v} + \frac{4v_3^{1/4}}{3v} x^{3/4}\right) \quad \forall x \in [b, v_3]. \end{aligned}$$

Now, consider strategies over $(0, b)$. Here, group 3 places no mass on any $x$ in this range, so $F_3(x) = F_3(b)$ (this is the size of the atom on 0). An active group-1 player choosing $x \in (0, b)$ has payoff

$$\begin{aligned}&V_1(x)= -x+ v \\&\quad \times \left[ (F(x))^3 F_3(b) + \int _x^{b} (F(y))^2 F_3(b) f(y)\, \mathrm{d}y + \int _b^{v_3} (F(y))^2 F_3(y) f(y)\, \mathrm{d}y \right] . \end{aligned}$$

Constancy of $V_1$ over $(0, b)$ yields

$$\begin{aligned} 0 = V_1'(x) = - 1 + 2v F_3(b) (F(x))^2 f(x) = - 1 + \frac{2v F_3(b) }{3}\frac{\mathrm{d}}{\mathrm{d}x}(F(x))^3, \end{aligned}$$

which we integrate to find, using the boundary condition $F(0) = 0$,³⁰

$$\begin{aligned} F(x) = \left( \frac{3x}{2vF_3(b)}\right) ^{1/3}. \end{aligned}$$

(48)

Next, we use continuity of $F$ at $b$: $\left( \frac{b}{v_3}\right) ^{1/4} = F(b) = \left( \frac{3b}{2vF_3(b)}\right) ^{1/3}.$ Cubing the extremes of the previous equation, we find

$$\begin{aligned} F_3(b) = \frac{3}{2} \times \frac{b^{1/4} (v_3)^{3/4}}{v}. \end{aligned}$$

(49)

And continuity requires this $F_3(b)$ must agree with what we found for $F_3$ on $[b, v_3]$:

$$\begin{aligned} \frac{4}{3v} v_3^{1/4} \left( v_3^{3/4} - b^{3/4}\right)&= 1-(F(b))^2 F_3(b)&(\text {by (46) and (47)})\\&= 1 - \left( \frac{b}{v_3}\right) ^{1/2} \times \frac{3}{2} \, \frac{b^{1/4} (v_3)^{3/4}}{v}&(\text {by (48) and (49)})\\&= 1 - \frac{3 (v_3)^{1/4}}{2v} b^{3/4}, \end{aligned}$$

yielding $b^{3/4} = \frac{6v}{(v_3)^{1/4}}\left( 1- \frac{4v_3}{3v}\right) ,$ which is strictly positive if and only if $v_3 < 3v/4$. Thus, the candidate $b$, given $v_3$, is uniquely determined as

$$\begin{aligned} b= \left[ \frac{6v}{(v_3)^{1/4}}\left( 1- \frac{4v_3}{3v}\right) \right] ^{4/3}. \end{aligned}$$

(50)

Our candidate equilibrium is as follows:

$$\begin{aligned}&F(x) = {\left\{ \begin{array}{ll} \left( \dfrac{x}{v_3}\right) ^{1/4} &{} x \in [b, v_3]\\ \dfrac{x^{1/3}}{v_3^{1/4} b^{1/12} } &{} x \in [0, b), \end{array}\right. }\nonumber \\&F_3(x) = {\left\{ \begin{array}{ll} \sqrt{\dfrac{v_3}{x}} \left( 1- \dfrac{4v_3}{3v} + \dfrac{4v_3^{1/4}}{3v} x^{3/4}\right) &{} x \in [b, v_3]\\ \dfrac{3}{v}\left( \dfrac{v_3}{2}\right) ^{2/3}(3v-4v_3)^{1/3} &{} x \in [0, b), \end{array}\right. } \end{aligned}$$

(51)

where $b$ is given by (50). The formula for $b$ equals 0 when $v_3 = 3v/4$ (consistent with part (i), where player 3 uses no atoms) and equals $2v/3$ when $v_3 = 2v/3$; in the limit as $v_3$ approaches $2v/3$ from above, group three effectively drops out—$F(b)$ goes to 1 as $v_3$ goes to $2v/3$.

It only remains to verify that player 3 does not strictly want to choose an $x \in (0, b)$. The payoff to player 3 of effort $x\in [ 0,b]$ is $v_{3}(F( x))^{4} -x$, where $F$ is described in (51). This payoff is strictly convex on $[0, b]$, so on $[0, b]$ it is maximized at $0$ or $b.$ At $x=0$, the payoff is zero; at $x\!=\!b,$ this payoff is also zero; so, contributing $x\in [0,b]$ is not a profitable deviation. Thus, (51) describes equilibrium strategies in part (ii). $\square$