Group formation in a dominance-seeking contest

Abstract

We study a group formation game. Players with different strengths form groups before expending effort to win a prize. The prize has the nature of the reward for outdoing in competition such as holding a dominant position among players or being recognized as a dominant status. So, it has the nature of public goods within a winning group (group-specific public goods). In open membership game, we find that a single player stays alone and the others form a group together in equilibrium. The stand-alone player can be anyone except for the first and second strongest players in the contest. However, strong (Nash) equilibrium predicts that the weakest player is isolated. Similarly, we find that in exclusive membership game, every structure can emerge in equilibrium but the weakest player is isolated in the strong equilibrium.

Appendices

Appendix A: Mathematical Proofs

Proof of Remark 1

(a) First, suppose that $K=2$, i.e., a group structure $G=\left\{ G_1 , G_2 \right\}$ is at equilibrium. Then, in the second stage of the game, the two groups will be active by Stein (2002). Stein (2002) showed that at least two players will be active in the contest among any $N \ge 2$ players having different composite strengths. This implies that the two groups $G_1$ and $G_2$, formed in the first stage of our game, will be active in the second stage regardless of the strongest players’ composite-strength levels in each group. Now, suppose that $K \ge 3$, i.e, a group structure $G=\left\{ G_1 , G_2 , \cdots , G_K \right\}$ is formed in the first stage. We claim that all K groups should be active in the second stage if the group structure G is at equilibrium by showing that its transposition is true.

Given the group structure G, suppose that there exists at least one group inactive in the second stage. Then, according to Proposition 1 of Stein (2002), we know that the composite-strength level of the strongest player in the inactive group is lower than those of the strongest players in the active groups. This means that every group member in the inactive group has an incentive to deviate from his group and move to one of the active groups in the first stage, and thus the group structure G cannot be an equilibrium outcome (see Lemma 4 and 5 for details). Therefore, if the group structure $G=\left\{ G_1 , G_2 , \ldots , G_K \right\}$ is an equilibrium outcome, all the K groups should be active in the second stage.

(b) Lemma 1 shows the effort levels chosen by the players of each group in the second stage. The conditions for $e_{1k}^*$ for all $k=1, \ldots , K$ to be strictly positive are given as (7). $\square$

Proof of Lemma 5

Given the group structure $G=\left\{ G_1 , \ldots , G_K\right\}$ where $K > 2$, the payoff for player 1k in $G_k$ is $v_{1k} \left( 1-\frac{K-1}{K} \frac{H}{B_{1k}} \right) ^2$. If player 1k departs from his own group and joins $G_1$, he obtains payoff $v_{1k} \left( 1-\frac{K-2}{K-1} \frac{H^{'}}{B_{11}} \right)$ where $H^{'} = \frac{K-1}{\sum _{l=1, \ne k}^K B_{1l}^{-1}}$. Comparing these two payoffs, we show that player 1k is better off when he deviates from his own group and moves to $G_1$ as follows:

$$\begin{aligned}&v_{1k} \left( 1-\frac{K-2}{K-1} \frac{H^{'}}{B_{11}} \right) - v_{1k} \left( 1-\frac{K-1}{K} \frac{H}{B_{1k}} \right) ^2 \\&\quad> v_{1k} \left( 1-\frac{K-2}{K-1} \frac{H^{'}}{B_{11}} \right) - v_{1k} \left( 1-\frac{K-1}{K} \frac{H}{B_{1k}} \right) \\&\quad =v_{1k} \left( \frac{K-1}{B_{1k} \sum _{l=1}^{K} B_{1l}^{-1}} - \frac{K-2}{B_{11} \sum _{l=1, \ne k}^{K} B_{1l}^{-1}} \right) \\&\quad = \frac{v_{1k} \left\{ (K-1) B_{11} \sum _{l \ne k} B_{1l}^{-1} - (K-2) B_{1k} \sum B_{1l}^{-1} \right\} }{B_{11} B_{1k} \sum B_{1l}^{-1} \sum _{l \ne k} B_{1l}^{-1}} \\&\quad = \frac{v_{1k} \left\{ (K-1) B_{11} \left( \sum B_{1l}^{-1} - B_{1k}^{-1}\right) - (K-2) B_{1k} \sum B_{1l}^{-1} \right\} }{B_{11} B_{1k} \sum B_{1l}^{-1} \sum _{l \ne k} B_{1l}^{-1}} \\&\quad = \frac{v_{1k} \left\{ (K-1) B_{11} \sum B_{1l}^{-1} - (K-2)B_{1k} \sum B_{1l}^{-1} - (K-1) B_{11} B_{1k}^{-1} \right\} }{B_{11} B_{1k} \sum B_{1l}^{-1} \sum _{l \ne k} B_{1l}^{-1}} \\&\quad> \frac{v_{1k} \left\{ (K-1) B_{11} \sum B_{1l}^{-1} - (K-2)B_{11} \sum B_{1l}^{-1} - (K-1) B_{11} B_{1k}^{-1} \right\} }{B_{11} B_{1k} \sum B_{1l}^{-1} \sum _{l \ne k} B_{1l}^{-1}} \\&\quad = \frac{v_{1k} B_{11} \left\{ \sum B_{1l}^{-1} - (K-1) B_{1k}^{-1} \right\} }{B_{11} B_{1k} \sum B_{1l}^{-1} \sum _{l \ne k} B_{1l}^{-1}} > 0 \text { from (7)}. \end{aligned}$$

Thus, the group structure $G=\left\{ G_1 , \ldots , G_K\right\}$ with $K > 2$ cannot be an equilibrium outcome. In addition to Lemma 3, this implies that $K=2$ should be held at equilibrium, i.e., only two groups can be formed at equilibrium. $\square$

Proof of Proposition 1

In the body paragraph following right after Lemma 5, we have shown that $G=\left\{ G_1 , G_2 \right\}$ where $G_2 =\left\{ 2 \right\}$ is not an equilibrium structure, for player 1 has an incentive to deviate from $G_1$ and join $G_2$.

Now we consider the group structure $G=\left\{ G_1 , G_2 \right\}$ where $G_2 =\left\{ i \right\}$ for $i=3, \ldots , n$. We first show that player 1 (in $G_1$) has no interests to deviate from his group. Given the group structure, player 1 gets his payoff $v_1 \left( \frac{a_1 v_1}{a_1 v_1 + a_i v_i} \right) ^2$. If he departs from $G_1$ to $G_2$, he gets his payoff $v_1 \left( \frac{a_1 v_1}{a_1 v_1 + a_2 v_2} \right) ^2$, which is less than the payoff obtained from staying at $G_1$. Thus, player 1 has no incentives to deviate from $G_1$ and join $G_2$. What about deviating from $G_1$ and staying alone? If he does so, the group structure changes to $G^{'}=\left\{ G_1 , G_2 , G_3 \right\}$ where $G_1 =\left\{ 2, 3, \ldots , i-1, i+1, \ldots , n\right\}$, $G_2=\left\{ i\right\}$, and $G_3 =\left\{ 1 \right\}$. Given the new group structure, player 1 obtains his payoff $v_1 \left( \frac{(a_1 v_1 ) (a_2 v_2 ) + (a_1 v_1 ) (a_i v_i ) - (a_2 v_2 ) (a_i v_i ) }{(a_1 v_1 ) (a_2 v_2 ) + (a_1 v_1 ) (a_i v_i ) + (a_2 v_2 ) (a_i v_i )} \right) ^2$, which is less than the payoff $v_1 \left( \frac{a_1 v_1}{a_1 v_1 + a_i v_i} \right) ^2$ obtained from staying at $G_1$. So, player 1 doesn’t have any incentive to become a singleton or join $G_2$ by deviating from $G_1$.

Now consider deviation incentives of the other group members in $G_1$. Given the group structure G, player $j (\ne 1)$ in $G_1$ has his payoff $v_j \left( \frac{a_1 v_1}{a_1 v_1 + a_i v_i} \right)$. If he deviates from his group and joins $G_2$, he gets either $v_j \left( \frac{a_i v_i}{a_1 v_1 + a_i v_i} \right)$ in case of $a_j v_j < a_i v_i$ or $v_j \left( \frac{a_j v_j}{a_1 v_1 + a_j v_j} \right) ^2$ in case of $a_j v_j > a_i v_i$. Obviously, player j’s payoff obtained by staying at $G_1$ is greater than these payoffs obtained by joining $G_2$, and he thus has no incentives to depart from $G_1$ to $G_2$. Does he then have an incentive to deviate from $G_1$ and become a singleton? No, he doesn’t, because we have already known that belonging to $G_1$ is better than staying alone in Lemma 5. See the proof of Lemma 5. Hence, any player $j \in G_1$ doesn’t have the deviation incentives.

Last, does player i in $G_2$ have an incentive to join $G_1$? No, because he gets nothing under the grand coalition, while he gets payoff $v_i \left( \frac{a_i v_i}{a_1 v_1 + a_i v_i} \right) ^2 ~(>0)$ from staying alone as a singleton.

In sum, every player has no incentives to deviate from the group structure G in the first stage of the game, given the other players’ decisions unchanged, and the group structure G is an equilibrium group structure appearing at the SPE. $\square$

Proof of Proposition 2

Consider an equilibrium group structure $G=\left\{ G_1 , G_2 \right\}$ where $G_1 =I \backslash \left\{ i\right\}$ and $G_2 =\left\{ i\right\}$ for $i \in \left\{ 3, 4, \ldots , n-1 \right\}$. For this to be the strong equilibrium group structure, no group of players should be able to collectively deviate in a way that they are all better off. Suppose that a group of players, consisting of player 1, player 2, $\ldots$, player $i-1$, jointly deviate from $G_1$ and join $G_2$. Then, the group structure is changed to $G^{'}=\left\{ G_1 , G_2 \right\}$ where $G_1 =\left\{ i+1, i+2, \ldots , n\right\}$ and $G_2 =\left\{ 1, 2, \ldots , i\right\}$. Given the new group structure $G^{'}$, the seceders obtain the following higher payoffs than those obtained under G:

$$\begin{aligned} \pi _1= & {} v_1 \left( \frac{a_1 v_1}{a_1 v_1 + a_{i+1} v_{i+1}} \right) ^2> v_1 \left( \frac{a_1 v_1}{a_1 v_1 + a_{i} v_{i}} \right) ^2, \\ \pi _2= & {} v_2 \left( \frac{a_1 v_1}{a_1 v_1 + a_{i+1} v_{i+1}} \right)> v_2 \left( \frac{a_1 v_1}{a_1 v_1 + a_{i} v_{i}} \right) , \\ \pi _3= & {} v_3 \left( \frac{a_1 v_1}{a_1 v_1 + a_{i+1} v_{i+1}} \right)> v_3 \left( \frac{a_1 v_1}{a_1 v_1 + a_{i} v_{i}} \right) , \\&\vdots&\\ \pi _{i-1}= & {} v_{i-1} \left( \frac{a_1 v_1}{a_1 v_1 + a_{i+1} v_{i+1}} \right) > v_{i-1} \left( \frac{a_1 v_1}{a_1 v_1 + a_{i} v_{i}} \right) . \end{aligned}$$

So, any equilibrium group structure $G=\left\{ G_1 , G_2 \right\}$, where $G_1 =I \backslash \left\{ i\right\}$ and $G_2 =\left\{ i\right\}$ for $i \in \left\{ 3, 4, \ldots , n-1 \right\}$, cannot be the strong equilibrium structure.

Then, what about the equilibrium group structure $G^* =\left\{ G_1 , G_2 \right\}$ where $G_1 =I \backslash \left\{ n\right\}$ and $G_2 =\left\{ n\right\}$? Given the group structure $G^*$, neither any individual player nor any group of players do have the incentives to deviate from the group structure, and thus the group structure $G^*$ is the unique strong equilibrium outcome. $\square$

Proof of Proposition 3

Suppose that $G=\left\{ G_1, \ldots , G_K \right\}$, where $K \ge 2$ and $|G_k |\ge 2$ for all $k=1, \ldots , K$. We show that any player in $G_k$ has no incentives to deviate from his group. For convenience, we denote players in $G_k$ by player 1k, 2k, $\ldots$ , $n_k k$, where $n_k =|G_k|$, i.e., $G_k =\left\{ 1k, 2k, \ldots , n_k k\right\}$. First, consider the deviation incentive for player 1k (the strongest player in $G_k$). From Lemma 1, given the group structure G, player 1k obtains the following payoff:

$$\begin{aligned} \pi _{1k}=v_{1k} p_k^2=v_{1k} \left( 1-\frac{(K-1)B_{1k}^{-1}}{\sum _{l=1}^K B_{1l}^{-1}} \right) ^2 . \end{aligned}$$

If player 1k changes his decision $g^{1k}$ to $g_d^{1k} (\ne g^{1k})$, given the other players’ decisions unchanged, his group $G_k$ collapses and all the players in $G_k$ become singletons. That is, the new group structure $G^{'}=\left\{ G_1 , \ldots , G_{k-1}, \left\{ 1k\right\} , \left\{ 2k\right\} , \ldots , \left\{ n_k k\right\} , G_{k+1}, \ldots , G_{K} \right\}$ appears. Given the new group structure $G^{'}$, player 1k obtains the following payoff:

$$\begin{aligned} \pi _{1k}^d =v_{1k} p_{1k}^2=v_{1k} \left( 1-\frac{(K+n_k -2)B_{1k}^{-1}}{\sum _{l=1}^K B_{1l}^{-1}+\sum _{i=2}^{n_k}(a_{ik}v_{ik})^{-1}}\right) ^2 . \end{aligned}$$

From conditions in (7), we have the following prerequisites for all the groups under the new group structure $G^{'}$ to be active:

$$\begin{aligned} \sum _{l=1}^K B_{1l}^{-1} + \sum _{i=2}^{n_k} (a_{ik} v_{ik})^{-1} > (K+n_k -2)(a_j v_j)^{-1} \quad \text { for all}\quad j=2k, 3k, \ldots , n_k k. \end{aligned}$$

Summing up the conditions for each players $2k, 3k, \ldots , n_k k$, we have

$$\begin{aligned}&(n_k -1) \sum _{l=1}^K B_{1l}^{-1} + (n_k -1) \sum _{i=2}^{n_k} (a_{ik} v_{ik})^{-1}> (K+n_k -2) \sum _{i=2}^{n_k} (a_{ik} v_{ik})^{-1}\\&\quad \Leftrightarrow (n_k -1) \sum _{l=1}^K B_{1l}^{-1} - (K-1) \sum _{i=2}^{n_k} (a_{ik} v_{ik})^{-1} > 0.\quad (*) \end{aligned}$$

We now compare player 1k’s payoffs, $\pi _{1k}$ and $\pi _{1k}^d$:

$$\begin{aligned} \pi _{1k}-\pi _{1k}^d= & {} v_{1k} (p_k + p_{1k}) (p_k - p_{1k}) \\= & {} v_{1k} (p_k + p_{1k}) \left( \frac{(K+n_k -2)B_{1k}^{-1}}{\sum _{l=1}^K B_{1l}^{-1}+\sum _{i=2}^{n_k}(a_{ik}v_{ik})^{-1}} - \frac{(K-1)B_{1k}^{-1}}{\sum _{l=1}^K B_{1l}^{-1}} \right) \\= & {} v_{1k} (p_k +p_{1k}) B_{1k}^{-1} \times \frac{\left\{ (n_k -1) \sum _{l=1}^K B_{1l}^{-1} - (K-1) \sum _{i=2}^{n_k} (a_{ik} v_{ik})^{-1}\right\} }{\left( \sum _{l=1}^K B_{1l}^{-1}+\sum _{i=2}^{n_k}(a_{ik}v_{ik})^{-1}\right) \left( \sum _{l=1}^K B_{1l}^{-1}\right) }. \end{aligned}$$

From $(*)$, we see $\pi _{1k}-\pi _{1k}^d >0$, which implies that player 1k in $G_k$ does not have the deviation incentive.

Next, consider the deviation incentive for player $jk (\ne 1k)$ in $G_k$. Given the group structure G, player jk has his payoff

$$\begin{aligned} \pi _{jk} = v_{jk} p_k = v_{jk} \left( 1-\frac{(K-1)B_{1k}^{-1}}{\sum _{l=1}^K B_{1l}^{-1}} \right) . \end{aligned}$$

If player jk changes his decision $g^{jk}$ to $g_d^{jk} (\ne g^{jk})$, given the other players’ decisions unchanged, the new group structure $G^{'}=\left\{ G_1 , \ldots , G_{k-1}, \left\{ 1k\right\} , \left\{ 2k\right\} , \ldots , \left\{ n_k k\right\} , G_{k+1}, \cdots , G_{K} \right\}$ appears. Under the new group structure $G^{'}$, he has the following payoff:

$$\begin{aligned} \pi _{jk}^d =v_{jk} p_{jk}^2=v_{jk} \left( 1-\frac{(K+n_k -2) (a_{jk} v_{jk})^{-1}}{\sum _{l=1}^K B_{1l}^{-1}+\sum _{i=2}^{n_k}(a_{ik}v_{ik})^{-1}}\right) ^2 . \end{aligned}$$

We now compare player jk’s payoffs, $\pi _{jk}$ and $\pi _{jk}^d$:

$$\begin{aligned} \pi _{jk}-\pi _{jk}^d= & {} v_{jk} (p_k - p_{jk}^2 ) \ge v_{jk} (p_k - p_{jk}) \\= & {} v_{jk} \left( \frac{(K+n_k -2) (a_{jk} v_{jk})^{-1}}{\sum _{l=1}^K B_{1l}^{-1}+\sum _{i=2}^{n_k}(a_{ik}v_{ik})^{-1}} - \frac{(K-1)B_{1k}^{-1}}{\sum _{l=1}^K B_{1l}^{-1}} \right) \\> & {} v_{jk} \left( \frac{(K+n_k -2) (a_{jk} v_{jk})^{-1}}{\sum _{l=1}^K B_{1l}^{-1}+\sum _{i=2}^{n_k}(a_{ik}v_{ik})^{-1}} - \frac{(K-1) (a_{jk} v_{jk})^{-1}}{\sum _{l=1}^K B_{1l}^{-1}} \right) \\= & {} v_{jk} (a_{jk} v_{jk})^{-1} \times \frac{\left\{ (n_k -1) \sum _{l=1}^K B_{1l}^{-1} - (K-1) \sum _{i=2}^{n_k} (a_{ik} v_{ik})^{-1}\right\} }{\left( \sum _{l=1}^K B_{1l}^{-1}+\sum _{i=2}^{n_k}(a_{ik}v_{ik})^{-1}\right) \left( \sum _{l=1}^K B_{1l}^{-1}\right) }\\> & {} 0 \text { from} ~(*). \end{aligned}$$

Thus, we see $\pi _{jk}-\pi _{jk}^d >0$, which implies that player $jk (\ne 1k)$ in $G_k$ does not have the deviation incentive.

To conclude, no players in $G_k$ have an incentive to deviate from the group structure G. $\square$

Proof of Proposition 5

For all possible equilibrium group structures, we keep to eliminate vulnerable ones to collective deviations of any conceivable group of players. First, consider the group structure $G=\left\{ G_1 , \ldots , G_K \right\}$ where $K\ge 3$. As before, we assume that player 1 belongs to $G_1$, i.e., $1\in G_1$. Given the group structure G, payoffs for players i in $G_1$ and those for players j in $G_k$ are as follows, respectively:

$$\begin{aligned} \pi _{11}=v_{11} p_1^2=v_{11} \left( 1-\frac{(K-1) B_{11}^{-1}}{\sum _{l=1}^K B_{1l}^{-1}} \right) ^2 , \pi _i =v_i p_1\text { for} ~i \in G_1 \backslash \left\{ 11\right\} \end{aligned}$$

and

$$\begin{aligned} \pi _{1k}=v_{1k} p_k^2=v_{1k} \left( 1-\frac{(K-1) B_{1k}^{-1}}{\sum _{l=1}^K B_{1l}^{-1}} \right) ^2 , \pi _j =v_j p_k\text { for} ~j \in G_k \backslash \left\{ 1k\right\} . \end{aligned}$$

$\square$

Now suppose that players in $G_1$ and players in $G_k$ jointly change their decisions to combine the two groups, $G_1$ and $G_k$, given the other players’ decisions unchanged. Namely, players i in $G_1$ change their decisions $g^i =G_1$ to $g_d^i =G_1 \cup G_k$, and player j in $G_k$ change their decisions $g^j =G_k$ to $g_d^j =G_k \cup G_1$. Then, the new group structure $G^{'}=\left\{ G_1^{'}, G_2, \ldots , G_{k-1}, G_{k+1}, \ldots , G_K \right\}$ is chosen, where $G_1^{'} =G_1 (\in G) \cup G_k (\in G)$. Note that $G_1$ and $G_k$ in G are merged and the other groups in G remain the same. Given the new group structure $G^{'}$, the payoffs for the deviators (players i in $G_1$ and players j in $G_k$ under the group structure G) are as follows:

$$\begin{aligned} \pi _{11}^d=v_{11} p_{1d}^2=v_{11} \left( 1-\frac{(K-2) B_{11}^{-1}}{\sum _{l=1, \ne k}^K B_{1l}^{-1}} \right) ^2 , \pi _i^d =v_i p_{1d}\quad \text { for}\quad i \in G_1 {\backslash }\left\{ 11\right\} \end{aligned}$$

and

$$\begin{aligned} \pi _{1k}^d=v_{1k} p_{1d}=v_{1k} \left( 1-\frac{(K-2) B_{11}^{-1}}{\sum _{l=1, \ne k}^K B_{1l}^{-1}} \right) , \pi _j^d =v_j p_{1d}\quad \text { for}\quad j \in G_k \backslash \left\{ 1k\right\} . \end{aligned}$$

Comparing the deviating players’ payoffs above, we have:

• $\pi _{11}^d - \pi _{11}=v_{11} (p_{1d}+p_1 )(p_{1d}-p_1 )=v_{11} (p_{1d}+p_1 )B_{11}^{-1} \times \frac{\sum _{l=1}^K B_{1l}^{-1}-(K-1) B_{1k}^{-1}}{\sum _{l=1}^K B_{1l}^{-1} \sum _{l=1, \ne k}^K B_{1l}^{-1}}>0$ by conditions in (7)

• $\pi _i^d - \pi _i=v_i (p_{1d}-p_1)>0$ for $i\in G_1 \backslash \left\{ 11\right\}$

• $\pi _{1k}^d -\pi _{1k}=v_{1k} (p_{1d}-p_{k}^2)>v_{1k} (p_{1d}-p_k)=v_{1k} \times \frac{(K-1)B_{1k}^{-1} \sum _{l=1, \ne k}^K B_{1l}^{-1}-(K-2) B_{11}^{-1} \sum _{l=1}^K B_{1l}^{-1}}{\sum _{l=1}^K B_{1l}^{-1} \sum _{l=1, \ne k}^K B_{1l}^{-1}}$ $> v_{1k} \times \frac{(K-1)B_{1k}^{-1} \sum _{l=1, \ne k}^K B_{1l}^{-1}-(K-2) B_{1k}^{-1} \sum _{l=1}^K B_{1l}^{-1}}{\sum _{l=1}^K B_{1l}^{-1} \sum _{l=1, \ne k}^K B_{1l}^{-1}}=v_{1k} B_{1k}^{-1} \times \frac{(K-1) \sum _{l=1, \ne k}^K B_{1l}^{-1}-(K-2) \sum _{l=1}^K B_{1l}^{-1}}{\sum _{l=1}^K B_{1l}^{-1} \sum _{l=1, \ne k}^K B_{1l}^{-1}}$ $=v_{1k} B_{1k}^{-1} \times \frac{\sum _{l=1}^K B_{1l}^{-1}-(K-1) B_{1k}^{-1}}{\sum _{l=1}^K B_{1l}^{-1} \sum _{l=1, \ne k}^K B_{1l}^{-1}}>0$ by conditions in (7)

• $\pi _j^d -\pi _j = v_j (p_{1d}-p_k ) >0$ $j\in G_k \backslash \left\{ 1k\right\} .$

All the deviators are better off under the new group structure $G^{'}$, which means that, under the group structure G, the players in $G_1$ and the players in $G_k$ have an incentive to collectively deviate from the group structure G. So, every group structure $G=\left\{ G_1 , \ldots , G_K \right\}$ with $K\ge 3$ is defenceless against the collective deviation of some players.

Second, consider the group structure G with $K=2$, i.e., $G=\left\{ G_1 , G_2 \right\}$. Let $G_2 =\left\{ 12, 22, \ldots , n_{2}2\right\}$, where $n_2 =|G_2 |\ge 2$. Given the group structure G, payoffs for the players in $G_1$ and those in $G_2$ are as follows:

$$\begin{aligned} \pi _{11}=v_{11} p_1^2=v_{11} \left( 1-\frac{B_{11}^{-1}}{B_{11}^{-1}+B_{12}^{-1}} \right) ^2 , \pi _i =v_i p_1\text { for} ~i \in G_1 \backslash \left\{ 11\right\} \end{aligned}$$

and

$$\begin{aligned} \pi _{12}=v_{12} p_2^2=v_{12} \left( 1-\frac{B_{12}^{-1}}{B_{11}^{-1}+B_{12}^{-1}} \right) ^2 , \pi _j =v_j p_2\text { for} ~j \in G_2 \backslash \left\{ 12\right\} . \end{aligned}$$

Now suppose that the players in $G_1$ and the players, except for player $n_{2}2$, in $G_2$ jointly change their decisions to form a new group, given player $n_2 2$’s decision unchanged. Namely, players i in $G_1$ change their decisions $g^i =G_1$ to $g_d^i =G_1 \cup G_2 \backslash \left\{ n_2 2\right\}$, and players j, other than player $n_2 2$, in $G_2$ change their decisions $g^j =G_2$ to $g_d^j =G_2 \backslash \left\{ n_2 2\right\} \cup G_1$. Then, the new group structure $G^{'}=\left\{ G_1^{'}, G_2^{'}\right\}$ appears, where $G_1^{'}=I\backslash \left\{ n_2 2\right\}$ and $G_2^{'}=\left\{ n_2 2\right\}$. Given the new group structure $G^{'}$, the payoffs of the deviators are as follows:

$$\begin{aligned} \pi _{11}^d =v_{11} p_{1d}^2=v_{11} \left( 1-\frac{B_{11}^{-1}}{B_{11}^{-1}+(a_{n_2 2}v_{n_2 2})^{-1}} \right) ^2 , \pi _i^d =v_i p_{1d}\text { for} ~i \in G_1 \backslash \left\{ 11\right\} \end{aligned}$$

and

$$\begin{aligned} \pi _{12}^d =v_{12} p_{1d}=v_{12} \left( 1-\frac{B_{11}^{-1}}{B_{11}^{-1}+(a_{n_2 2}v_{n_2 2})^{-1}} \right) , \pi _j^d =v_j p_{1d}\text { for} ~j \in G_2 \backslash \left\{ 12, n_2 2\right\} . \end{aligned}$$

Comparing the deviating players’ payoffs above, we have:

• $\pi _{11}^d - \pi _{11} >0$

• $\pi _i^d - \pi _i >0$ for $i\in G_1 \backslash \left\{ 11\right\}$

• $\pi _{12}^d -\pi _{12}>0$

• $\pi _j^d -\pi _j >0$ $j\in G_2 \backslash \left\{ 12, n_2 2\right\} .$

All the deviators are better off under the new group structure $G^{'}$, which means that, under the group structure G, the players in $G_1$ and the players in $G_2$, other than the weakest player in $G_2$, have an incentive to collectively deviate from the group structure G. So, every group structure $G=\left\{ G_1 , G_2 \right\}$ with $|G_2|\ge 2$ is vulnerable to the collective deviation of the players.

Finally, consider the group structure $G=\left\{ G_1 , G_2 \right\}$ with $|G_2 |=1$. Specifically, suppose that $G_1 =I \backslash \left\{ j\right\}$ and $G_2 =\left\{ j\right\}$ where $j\ne n$. Given the group structure G, players have the following payoffs:

$$\begin{aligned} \pi _{11}=v_{11} p_1^2=v_{11} \left( 1-\frac{B_{11}^{-1}}{B_{11}^{-1}+(a_j v_j)^{-1}} \right) ^2 , \pi _i =v_i p_1\text { for} ~i \in G_1 \backslash \left\{ 11\right\} \end{aligned}$$

and

$$\begin{aligned} \pi _{j}=v_{j} p_j^2=v_{j} \left( 1-\frac{(a_j v_j)^{-1}}{B_{11}^{-1}+(a_j v_j)^{-1}} \right) ^2 . \end{aligned}$$

Now suppose that the players, other than player n, in $G_1$ and player j jointly change their decisions to organize a new group, given player n’s decisions unchanged. In other words, players i, other than player n, in $G_1$ change their decisions $g^i =I\backslash \left\{ j\right\}$ to $g_d^i=I\backslash \left\{ n\right\}$, and player j changes his decision $g^j =\left\{ j\right\}$ to $g_d^j =I\backslash \left\{ n\right\}$. Then, the new group structure $G^{'}=\left\{ G_1^{'}, G_2^{'}\right\}$ appears, where $G_1^{'}=I\backslash \left\{ n\right\}$ and $G_2^{'}=\left\{ n\right\}$. Given the new group structure $G^{'}$, the payoffs of the deviators are as follows:

$$\begin{aligned} \pi _{11}^d=v_{11} p_{1d}^2=v_{11} \left( 1-\frac{B_{11}^{-1}}{B_{11}^{-1}+(a_n v_n)^{-1}} \right) ^2 , \pi _i^d =v_i p_{1d}\text { for} ~i \in G_1 \backslash \left\{ 11, n\right\} \end{aligned}$$

and

$$\begin{aligned} \pi _{j}^d=v_{j} p_{1d}. \end{aligned}$$

Comparing the deviating players’ payoffs above, we have:

• $\pi _{11}^d - \pi _{11} >0$

• $\pi _i^d - \pi _i >0$ for $i\in G_1 \backslash \left\{ 11, n\right\}$

• $\pi _j^d -\pi _j >0$.

All the deviators are better off under the new group structure $G^{'}$, which means that, under the group structure G, all the players other than player n (the lowest-composite-strength player in the contest) have an incentive to collectively deviate from the group structure G. So, every group structure $G=\left\{ G_1 , G_2 \right\}$ with $G_2 =\left\{ j (\ne n)\right\}$ is not immune to the collective deviation of the players.

Therefore, the unique group structure $G=\left\{ G_1 , G_2 \right\}$, where $G_1 =I\backslash \left\{ n\right\}$ and $G_2 =\left\{ n\right\}$, is unsusceptible to every conceivable collective deviation by players. So, it satisfies the strong equilibrium property.

Appendix B: The share function approach to get the main results

Given a group structure $G=\left\{ G_1, \ldots , G_K \right\}$ where $K\ge 3$, let $X=\sum _{k=1}^{K} X_{k}$. From (8), we then have the winning probability of a group $G_k \in G$ or its share function, $p_k$: $p_k = 1-\frac{X}{B_{1k}}$. By summing up all K groups’ share functions, i.e., $\sum _{k=1}^K p_k =1$, we have (10):

$$\begin{aligned} X^* =\frac{K-1}{\sum _{k=1}^K 1/B_{1k}}. \end{aligned}$$

We denote $p_k$ and X in equilibrium in the second stage by $p_k (K)$ and $X^* (K)$. Since $p_k (K)=1-\frac{X^* (K)}{B_{1k}}$, we have the following lemma.

Lemma A

$p_k (K)>p_l (K)$ for $B_{1k}>B_{1l}$ and $k\ne l$.

Denoting $\pi _{i}$ for $i\in G_k$ in the second-stage equilibrium by $\pi _{i} (K)$, we have:

$$\begin{aligned} \pi _{1k} (K)=v_{1k} \left( p_k (K)\right) ^2 \text {and} ~\pi _{i} (K)=v_i p_k (K) \text { for} ~i\in G_k\backslash \left\{ 1k\right\} . \end{aligned}$$

Suppose that ${\widetilde{K}}<K$. Given a group structure $G=\left\{ G_1, \ldots , G_{{\widetilde{K}}}\right\}$, we then have equilibrium payoffs for players in a group $G_k \in G$:

$$\begin{aligned} \pi _{1k} ({\widetilde{K}})=v_{1k} \left( p_k ({\widetilde{K}})\right) ^2 \text { and} ~\pi _{i} ({\widetilde{K}})=v_i p_k ({\widetilde{K}}) \text { for} ~i\in G_k\backslash \left\{ 1k\right\} . \end{aligned}$$

The payoff differences for players in $G_k$ between the K-group structure ($\left\{ G_1, \ldots , G_K \right\}$) and the ${\widetilde{K}}$-group structure ($\left\{ G_1, \ldots , G_{{\widetilde{K}}}\right\}$) are as follows:

$$\begin{aligned} \pi _{1k} ({\widetilde{K}})-\pi _{1k} (K)= & {} v_{1k}\left( p_k ({\widetilde{K}})+p_k(K) \right) \left( p_k ({\widetilde{K}})-p_k(K) \right) \\= & {} \frac{1}{a_{1k}} \left( p_k ({\widetilde{K}})+p_k(K) \right) \left( X^* (K) - X^* ({\widetilde{K}})\right) \end{aligned}$$

and

$$\begin{aligned} \pi _i ({\widetilde{K}})-\pi _i (K)=v_i \left( p_k ({\widetilde{K}})-p_k(K) \right) =\frac{v_i}{B_{1k}}\left( X^* (K) - X^* ({\widetilde{K}})\right) \text { for} ~i\in G_k\backslash 1k. \end{aligned}$$

We thus have the following lemma.

Lemma B

For ${\widetilde{K}}<K$ and $i \in G_k$, $\pi _i ({\widetilde{K}})>\pi _i (K) \Leftrightarrow p_k ({\widetilde{K}})>p_k (K) \Leftrightarrow X^* ({\widetilde{K}})<X^* (K)$.

Lemma B means, if $X^* ({\widetilde{K}})<X^* (K)$, players in a group $G_k$ are better off when the number of groups changes from K to ${\widetilde{K}}$, i.e., it decreases.

Now suppose that $K\ge 3$ groups are formed in the first stage, i.e., $G=\left\{ G_1 , G_2 , \ldots , G_K \right\}$, and that groups $1, 2, \ldots , K$ are listed in descending order by the composite strengths of their strongest (active) players, thus $1\in G_1$. Note that player 1 is the strongest among our players.

Lemma C

When $K\ge 3$, the active player in $G_2$, player 12, can be better off by joining in $G_1$.

Proof

• (a) Suppose that $|G_2|=1$ and player 12 moves to $G_1$. The number of groups then changes from K to $K-1$, and we have:

  $$\begin{aligned} X^* (K-1) - X^* (K)=\frac{K-2}{\sum _{k=1, \ne 2}^K 1/B_k} - \frac{K-1}{\sum _{k=1}^K 1/B_k}<0. \end{aligned}$$

  From Lemmas A and B, $p_1 (K-1)>p_1 (K)>p_2 (K)$. Thus, $\pi _{12} (K) = v_{12} \left( p_2 (K)\right) ^2 < v_{12} \left( p_1 (K-1)\right) =\text {the payoff of player 12 when joining in } G_1$.

• (b) Suppose that $|G_2|\ge 2$ and player 12 moves to $G_1$. Then, although there are still K groups, $p_1 (K)$ increases because player 12’s deviation results in the decrease of $B_{12}$ and hence the decrease of $X^* (K)$. Remind that $X^* (K) =\frac{K-1}{\sum _{k=1}^K 1/B_{1k}}$ and $p_k (K)=1-\frac{X^* (K)}{B_{1k}}$. Thus, $\pi _{12} (K)=v_{12} \left( p_2 (K)\right) ^2< v_{12} \left( p_1 (K)\right) <v_{12}\cdot \text {increased }p_1 (K)=\text {the payoff of player 12 when joining in }G_1$. $\square$

By Lemma C, we see that $K=2$ in equilibrium, that is, $G=\left\{ G_1 , G_2\right\}$. By the same argument as in (b) in the proof of Lemma C, we also know that the active player in $G_2$ has an incentive to join in $G_1$ when $|G_2|\ge 2$. Therefore, $G_1 =I\backslash \left\{ i\right\}$ and $G_2 =\left\{ i\right\}$ in equilibrium. By the way, if $i=2$, player 1 can be better off by moving to $G_2$. So, an equilibrium group structure should be as follows.

Proposition A

A SPE group structure is $G=\left\{ G_1 , G_2 \right\}$ where $G_1 =I\backslash \left\{ i\right\}$, $G_2 =\left\{ i\right\}$, and $i\in I\backslash \left\{ 1, 2\right\}$.

Under a SPE group structure described above, the payoffs of players in $G_1$ are

$$\begin{aligned} \pi _1 (2)= & {} v_1 \left( p_1 (2)\right) ^2 \text { and} ~\\ \pi _j (2)= & {} v_j p_1 (2) \text { for} ~j\in G_1\backslash \left\{ 1\right\} , \end{aligned}$$

where $p_1 (2)=1-\frac{X^* (2)}{a_1 v_1}$ and $X^* (2)=\frac{1}{1/{a_1 v_1}+1/{a_i v_i}}$. When $i=n$, $X^* (2)$ is minimized and thus $p_1 (2)$ is maximized. This implies the players’ payoffs in $G_1$ are maximized when $G_2=\left\{ n\right\}$, which means that players $1, 2, \ldots , n-1$ have a collective incentive to stick together and isolate player n.

Proposition B

The strong equilibrium group structure is $G=\left\{ G_1 , G_2 \right\}$ where $G_1 =I\backslash \left\{ n\right\}$, $G_2 =\left\{ n\right\}$.

In exclusive membership game $\Gamma$, the unanimity condition is required for organizing a group, and it remove a player’s unilateral deviation incentive from his group. Thus, other than the grand coalition, every group structure $G=\left\{ G_1, \ldots , G_K \right\}$ with $K\ge 2$ can be the SPE group structure. However, there exists a unique strong equilibrium group structure.

Proposition C

In exclusive membership game $\Gamma$, the strong equilibrium group structure is $G=\left\{ G_1 , G_2 \right\}$ where $G_1 =I\backslash \left\{ n\right\}$, $G_2 =\left\{ n\right\}$.

Proof

• (a) Consider a SPE group structure $G=\left\{ G_1 , G_2 , \ldots , G_K \right\}$ with $K\ge 3$ and check a collective incentive of all the players in $G_2$ to join in $G_1$. From the same argument as in (a) in the proof of Lemma C, we can see that the players in $G_2$ obtain higher payoffs by collectively joining in $G_1$ because $X^* (K-1)<X^* (K) \Rightarrow p_1 (K-1)>p_1 (K)>p_2 (K)$. So, they have such a collective deviation incentive. In addition, since $p_1 (K-1)>p_1 (K)$, every player in $G_1$ can be better off by allowing $G_2$ to join in his group. In sum, all the players in both $G_1$ and $G_2$ get higher payoffs when $G_1$ and $G_2$ merge. We hence find $K=2$ in strong equilibrium.

• (b) Consider a SPE group structure $G=\left\{ G_1 , G_2 \right\}$ with $|G_2|\ge 2$. By the same argument as in (b) in the proof of Lemma C, we can see that $|G_2|-1$ strongest players in $G_2$ (the players except for the weakest one in $G_2$) have a collective incentive to join in $G_1$. Also, by allowing them to join in $G_1$, all the players in $G_1$ can be better off due to the increase of $p_1 (2)$. In sum, all the players in $G_1$ and the strongest $|G_2|-1$ players in $G_2$ have a collective incentive to merge. That is, we find that $|G_2|=1$ in strong equilibrium.

• (c) Consider a SPE group structure $G=\left\{ G_1 , G_2 \right\}$ with $G_2 =\left\{ i \right\}$ and $i \in I\backslash \left\{ 1\right\}$. By the same argument as in Proposition B, payoffs of players in $G_1$ are maximized when $i=n$, which implies that players $1, 2, \ldots , n-1$ have a collective incentive to move together and isolate player n. Finally, we have $G_2 =\left\{ n\right\}$ in strong equilibrium. $\square$