Stable coalition structures in symmetric majority games: a coincidence between myopia and farsightedness

Abstract

The objective of this paper is to study stable coalition structures in symmetric majority games. We assume that players deviate from a coalition structure to another to maximize their power given by the Owen power index. We introduce three myopic core concepts and one farsighted stability concept, the farsighted vNM stable set. Our main result is that the pessimistic core, the largest myopic core, coincides with some farsighted vNM stable set for any number of players. Moreover, we show that a coalition structure belongs to the pessimistic core and the farsighted vNM stable set if and only if it contains an exact majority coalition.

Appendix

We first prove Proposition 2 and, next, Proposition 1.

1.1 Proof of Proposition 2

Proof

For any partition $\mathcal {P}$, consider the following three cases:

• (A) $\mathcal {P}$ contains a majority coalition whose size is strictly greater than k;

• (B) $\mathcal {P}$ contains an exact majority coalition;

• (C) for every coalition in $\mathcal {P}$, its size is strictly less than k.

We show that the partitions satisfying B are in the pessimistic core and the partitions satisfying A or C are not. Let K be an exact majority coalition. $\square$

Case A Let $\mathcal {P}$ be be a partition satisfying A. Let $K^+\in \mathcal {P}$ be a coalition such that $|K^+|=k^+> k$. Let $\mathcal {P}^{*}$ be be a partition satisfying B such that $K\in \mathcal {P}^{*}$ and $K\subseteq K^+$. Since for any $i\in K\cap K^+=K$, $\phi _i(\mathcal {P}^{*})=\frac{1}{k}>\frac{1}{k^+}=\phi _i(\mathcal {P})$, the players in $K\subsetneq K^+$ have an incentive to deviate to $\mathcal {P}^{*}$. Whatever partition the other players, $N\setminus K$, forms, the players in K get $\frac{1}{k}$. Thus, $\mathcal {P}$ is not in the pessimistic core.

Case B Let $\mathcal {P}$ be a partition satisfying B and $K\in \mathcal {P}$. Let $S\subseteq N$ be a deviating coalition. We typically use $\mathcal {P}^{*}$ with $S\in \mathcal {P}^{*}$ to denote a partition after S’s deviation. First, any coalition $S\subseteq N$ with $|S|\ge k$ does not deviate from $\mathcal {P}$, because for some player $i\in K\cap S$, $\phi _i(\mathcal {P})=\frac{1}{k}\ge \frac{1}{|S|}=\phi _i(\mathcal {P}^{*})$ for any partition $\mathcal {P}^{*}$ with $S\in \mathcal {P}^{*}$. Next, consider a coalition S with $|S|< k$. We focus on the case of $S\cap K\ne \emptyset$, because if $S\cap K= \emptyset$, then $S\subseteq N\setminus K$ and any player i in S gets zero both before and after their deviation: $\phi _i(\mathcal {P})=0=\phi _i(\mathcal {P}^{*})$.

In the case of odd n, $|N\setminus S| \ge k$ Hence, in the pessimistic view, $\phi _S(\mathcal {P}^{*})=0$, because the players in $N\setminus S$ form their coalition $N\setminus S$, and the coalition $N\setminus S$ is a majority coalition, K or $K^+$, in $\mathcal {P}^{*}$. In other words, any S with $|S|<k$ has no incentive to deviate from $\mathcal {P}$.

In the case of even n If $|S|\le k-2$, then it is the same as the case of odd n. If $|S|= k-1=\frac{n}{2}$, then the power of coalition S increases as the number of coalitions except S increases.⁷ Therefore, in the pessimistic view, the players in coalition S expect that the players in $N\setminus S$ form their coalition $\{N\setminus S\}$, where $|S|=|N\setminus S| = \frac{n}{2}=k-1$ and $\mathcal {P}^{*}=\{S,N\setminus S\}$. Hence, each player in coalition S expects their minimum $\frac{1}{2}\cdot \frac{1}{|S|}$ per capita after their deviation. We have, for any $i\in K\cap S$,

$$\begin{aligned} \phi _i(\mathcal {P})-\phi _i(\mathcal {P}^{*})= \frac{1}{k}-\frac{1}{2}\frac{1}{|S|} = \frac{2}{n+2} - \frac{1}{n} = \frac{n - 2}{n(n+2)} \ge 0, \end{aligned}$$

where the last inequality holds as $n\ge 3$.

Case C Let $\mathcal {P}$ be a partition satisfying C. We consider the players who obtain at least $\frac{1}{k}$ in $\mathcal {P}$. Let $\hat{K}=\{j\in N | \phi _j(\mathcal {P})\ge \frac{1}{k}\}$ and $\hat{k}=|\hat{K}|$. We must have $\hat{k} \le k$ because of $\sum _{j\in N}\phi _j(\mathcal {P})=1$. Moreover, we claim that $\hat{k}\ne k$ as follows.

Claim 1

$\hat{k}\le k-1$.

Proof

Assume that $\hat{k}= k$. Let $\mathcal {Z}$ be a partition of $\hat{K}$, namely, $\mathcal {Z}$ is a subpartition of $\mathcal {P}$. Now, $\mathcal {Z}$ must consist of at least two coalitions, because $\mathcal {Z}$ is a partition of exactly k players and for every coalition in $\mathcal {P}$ its size is strictly less than k. We consider the following order of coalitions in $\mathcal {P}$. First, we arrange the coalitions in $\mathcal {Z}$ in order of their size (the biggest coalition in $\mathcal {Z}$ is located at the top) and remove (one of) the smallest coalition(s) in $\mathcal {Z}$ from the order. At least $\frac{k}{2}$ players are now lined up. Next, we arrange the coalitions which are not in $\mathcal {Z}$. The number of the players who are not in $\mathcal {Z}$ is $n-\hat{k}=n-k=\frac{n}{2}-1$. Hence, there exists a coalition which becomes a pivot and is not in $\mathcal {Z}$. In other words, the power index, $\phi$, assigns a positive value to this coalition. This, however, contradicts $\hat{k}= k$, because $\hat{k}= k$ implies that any player not in $\hat{K}$ gets zero.

We assume $\hat{k}\le k-1$ hereafter.

In the case of odd n The fact $\hat{k}\le k-1$ implies that $n-\hat{k}\ge n-(k-1)$: in $\mathcal {P}$, the number of players who get strictly less than $\frac{1}{k}$ is greater than $n-(k-1)= n-(\frac{n+1}{2}-1)=k$. Hence, these players have an incentive to form coalition S such that $|S|=k$ and deviate from $\mathcal {P}$.

In the case of even n If $\hat{k}\le k-2$, then it is the same as the case of odd n. If $\hat{k}= k-1=\frac{n}{2}$, the other $\frac{n}{2}$ players (namely, $N\setminus \hat{K}$) obtain strictly less than $\frac{1}{k}$ in $\mathcal {P}$. We show that, in the pessimistic view, the players in $N\setminus \hat{K}$ have an incentive to form a deviating coalition S such that $S= N\setminus \hat{K}$ with $|S|= n-\hat{k} = \frac{n}{2}$. If $S=N\setminus \hat{K}$ and $|S|=\frac{n}{2}$, then the deviating players in S expect to get $\frac{1}{2}\frac{1}{|S|}=\frac{1}{n}$ per capita after their deviation (see the case of even n in Case B). Therefore, it suffices to show that $\phi _i(\mathcal {P})<\frac{1}{n}$ for any $i\in N\setminus \hat{K}$.

Claim 2

If $\hat{k}=k-1=\frac{n}{2}$, then $\phi _i(\mathcal {P})<\frac{1}{n}$ for any $i\in N\setminus \hat{K}$.

Proof

Let $T=\{i\in N\setminus \hat{K}| \phi _i(\mathcal {P})\ge \frac{1}{n}\}$ and $t=|T|$. We have

$$\begin{aligned}&\hat{k}\cdot \frac{1}{k} + t\cdot \frac{1}{n} +0 \le \sum _{j\in \hat{K}}\phi _j(\mathcal {P}) +\sum _{j\in T}\phi _j(\mathcal {P}) + \sum _{j\in N\setminus (\hat{K}\cup T)}\phi _j(\mathcal {P})=1,\quad \text {and}\\&\hat{k}\cdot \frac{1}{k} + t\cdot \frac{1}{n} = \frac{n}{2}\cdot \frac{2}{n+2}+\frac{t}{n} = \frac{n}{n+2}+\frac{t}{n}. \end{aligned}$$

We obtain $t\le \frac{2n}{n+2}$. As $n\ge 3$, $1\le \frac{2n}{n+2}< 2$. Hence, t should be 0 or 1. We assume $t=1$ and call the player t. The player t forms his singleton coalition in $\mathcal {P}$, because if he is in a coalition consisting of more than two players, symmetry within a coalition implies that his partners also obtain at least $\frac{1}{n}$, which contradicts $t=1$.⁸ Moreover, similarly, symmetry across coalitions implies that the player t’s singleton coalition is the only singleton coalition in $\mathcal {P}$.⁹ In other words, the other coalitions consist of at least two players. We take a coalition $S'\in \mathcal {P}$ such that $|S'|\ge 2$, $S'\subseteq N\setminus \hat{K}$. Each player in $S'$ gets strictly less than $\frac{1}{k}$ in $\mathcal {P}$ because they are not the members of $\hat{K}$. By local monotonicity and inequality $|S'|>|\{t\}|=1$, we obtain $\phi _{S'}(\mathcal {P})\ge \phi _t(\mathcal {P})=\frac{1}{n}$.¹⁰ Hence, we have

$$\begin{aligned}&\hat{k}\cdot \frac{1}{k} + \frac{1}{n} + \frac{1}{n}+0 \le \sum _{j\in \hat{K}}\phi _j(\mathcal {P}) + \phi _t(\mathcal {P}) + \phi _{S'}(\mathcal {P}) +\sum _{j\not \in (\hat{K}\cup \{t\}\cup S')}\phi _j(\mathcal {P})=1,\quad \text {and}\\&\hat{k}\cdot \frac{1}{k} + \frac{1}{n} + \frac{1}{n} = \frac{n}{n+2} + \frac{2}{n} = \frac{n^2+2n+4}{n^2+2n} > 1. \end{aligned}$$

This is a contradiction. Hence, we have $t=0$, i.e., every player in $N\setminus \hat{K}$ gets strictly less than $\frac{1}{n}$ in $\mathcal {P}$. $\square$

Thus, in $\mathcal {P}$, the players in $N\setminus \hat{K}$ have an incentive to from their coalition $N\setminus \hat{K}$ and deviate to get $\frac{1}{n}$ even in the pessimistic view. $\square$

1.2 Proof of Proposition 1

Proof

For any partition $\mathcal {P}$, consider the following three cases:

• (A) $\mathcal {P}$ contains a majority coalition whose size is strictly greater than k;

• (B) $\mathcal {P}$ contains an exact majority coalition;

• (C) for every coalition in $\mathcal {P}$, its size is strictly less than k.

Let V be a set of partitions satisfying B (the set of partitions satisfying B is the pessimistic core. See Proposition 2). $\square$

Internal stability Let $\mathcal {P},\mathcal {P}'$ be any two different partitions in V. Let $K\in \mathcal {P}$ and $K'\in \mathcal {P}'$ ($K\ne K'$ and $k=k'$) be exact majority coalitions. Since K and $K'$ are exact majority coalitions, we have $K\cap K'\ne \emptyset$. Now, assume that there exists a path from $\mathcal {P}$ to $\mathcal {P}'$ satisfying the conditions of Definition 4.

As K is an exact majority coalition in $\mathcal {P}$, we have $\phi _i(\mathcal {P})=0$ for any $i\in N\setminus K$ and $\phi _i(\mathcal {P})=\frac{1}{k}$ for any $i\in K$. Hence, in the path, the first deviating coalition $S^1$ should be a subset of $N\setminus K$. Let the resulting partition be $\mathcal {P}^1$. As $S^1\subseteq N\setminus K$, K is still in $\mathcal {P}^1$. We have $\phi _i(\mathcal {P}^1)=0$ for any $i\in N\setminus K$ and $\phi _i(\mathcal {P}^1)=\frac{1}{k}$ for any $i\in K$. Hence, the second deviating coalition $S^2$ is a subset of $N\setminus K$. In general, for any step j, we have $\phi _i(\mathcal {P}^{j})=\frac{1}{k}$ for any $i\in K$. On the other hand, for the path to reach partition $\mathcal {P}'$, there must be at least one coalition $S^{j^{*}}$ and partition $\mathcal {P}^{j^{*}}$ (at step $j^{*}$) such that $K\cap K'\subseteq S^{j^{*}}$ and $\mathcal {P}'\succ _{S^{j^{*}}}\mathcal {P}^{j^{*}}$; otherwise, $K=K'$. However, we have $\phi _i(\mathcal {P}^{j^{*}})=\frac{1}{k}=\frac{1}{k'}=\phi _i(\mathcal {P}')$ for any $i\in K\cap K'$. This is a contradiction.

For any K and any two different partitions $\mathcal {P},\mathcal {P}'$ such that $K\in \mathcal {P}$ and $K\in \mathcal {P}'$, we have $\phi _i(\mathcal {P})=\phi _i(\mathcal {P}')$ for any player $i\in N$. Thus, there is no indirect dominance within V.

External stability We show that for any partition $\mathcal {P}$ satisfying A or C, there exists a partition $\mathcal {P}^{*}\in V$ such that $\mathcal {P}^{*}$ indirectly dominates $\mathcal {P}$.

Case A For any $\mathcal {P}$ satisfying A, let $K^+\in \mathcal {P}$ denote a majority coalition whose size $k^+$ is strictly greater than k. There exists a partition $\mathcal {P}^{*}\in V$ and an exact majority coalition $K\in \mathcal {P}$ such that $K\subsetneq K^+$ and $\phi _i(\mathcal {P}^{*})=\frac{1}{k}>\frac{1}{k^+}= \phi _i(\mathcal {P})$ for any $i\in K$. Thus, $\mathcal {P}^{*}$ (in)directly dominates $\mathcal {P}$.

Case C Let $\mathcal {P}$ be a partition satisfying C. We consider the set of players who obtain at least $\frac{1}{k}$ in $\mathcal {P}$. Let $\hat{K}=\{j\in N | \phi _j(\mathcal {P})\ge \frac{1}{k}\}$ and $\hat{k}=|\hat{K}|$. As Claim 1 in the proof of Proposition 2, we have $\hat{k} \le k-1$.

In the case of odd n $\hat{k} \le k-1$ means that at least k players get strictly less than $\frac{1}{k}$ in $\mathcal {P}$. They form coalition S consisting of k players and directly deviate to partition $\{S\}\cup (\mathcal {P}|_{N\setminus S})\in V$.

In the case of even n If $\hat{k}\le k-2$, then it is the same as the case of odd n. If $\hat{k}= k-1=\frac{n}{2}$, then the other $\frac{n}{2}$ players (namely, $N\setminus \hat{K}$) obtain strictly less than $\frac{1}{k}$. Let $\mathcal {Q}$ be a partition of $N\setminus \hat{K}$, namely $\mathcal {Q}$ is a subpartition of $\mathcal {P}$. We first show the following claim.

Claim 3

$|\mathcal {Q}|\ge 2$.

Proof

Assume that $|\mathcal {Q}|=1$ (i.e., $\mathcal {Q}=\{N\setminus \hat{K}\}$, which contains $k-1$ players). We have

$$\begin{aligned} \phi _{N\setminus \hat{K}}(\mathcal {P})=1-(|\mathcal {P}|-1)\frac{(|\mathcal {P}|-2)!}{|\mathcal {P}|!}=1-(|\mathcal {P}|-1)\frac{1}{|\mathcal {P}|(|\mathcal {P}|-1)} =\frac{|\mathcal {P}|-1}{|\mathcal {P}|}, \end{aligned}$$

(5)

for $2\le |\mathcal {P}| \le k$.¹¹ This attains its minimum at $|\mathcal {P}|=2$. Hence, $\phi _{N\setminus \hat{K}}(\mathcal {P})\ge \frac{1}{2}$. However, since $\phi _i(\mathcal {P})\ge \frac{1}{k}$ for any $i\in \hat{K}$, we have $\phi _{N\setminus \hat{K}}(\mathcal {P})=1-\phi _{\hat{K}}(\mathcal {P})\le 1-(\frac{1}{k}\cdot \hat{k})=1-(\frac{k-1}{k})=\frac{1}{k}$, which contradicts $\phi _{N\setminus \hat{K}}(\mathcal {P})\ge \frac{1}{2}$ as $n\ge 4$ ($k\ge 3$ when $n\ge 4$, and 4 is the smallest even n). $\square$

Next, we construct a path from $\mathcal {P}$ to $\mathcal {P}^{*}\in V$ via a partition $\bar{\mathcal {P}}$ given by

$$\begin{aligned} \bar{\mathcal {P}}= & {} (\mathcal {P}|_{\hat{K}}) \cup \{N\setminus \hat{K}\}, \\ \mathcal {P}^{*}= & {} (\mathcal {P}|_{\hat{K}\setminus \{i^{*}\}}) \cup \{(N\setminus \hat{K})\cup \{i^{*}\}\}, \end{aligned}$$

where $i^{*}$ is any player in $\hat{K}$. Note that $\mathcal {P}^{*} \in V$ because coalition $(N\setminus \hat{K})\cup \{i^{*}\}$ is an exact majority.

The first deviating coalition (from $\mathcal {P}$ to $\bar{\mathcal {P}}$) is $N\setminus \hat{K}$. For every player $i\in N\setminus \hat{K}$, $\phi _i(\mathcal {P}^{*}) > \phi _i(\mathcal {P})$, because $\phi _i(\mathcal {P})<\frac{1}{k}$ and $\phi _i(\mathcal {P}^{*})=\frac{1}{k}$. Moreover, by $|\mathcal {Q}|\ge 2$, all players in $\mathcal {Q}$ can form one coalition $\{N\setminus \hat{K}\}$ and move to $\bar{\mathcal {P}}$.

The second deviating coalition (from $\bar{\mathcal {P}}$ to $\mathcal {P}^{*}$) is $(N\setminus \hat{K})\cup \{i^{*}\}$. We show that for any $i\in N\setminus \hat{K}$, $\phi _i(\bar{\mathcal {P}})<\frac{1}{k}$ (for $i^{*}$; see Claim 5).

Claim 4

In $\bar{\mathcal {P}}$, for any $i\in N\setminus \hat{K}$, $\phi _i(\bar{\mathcal {P}})<\frac{1}{k}=\phi _i(\mathcal {P}^{*})$.

Proof

We first show that $\phi _i(\bar{\mathcal {P}})\le \frac{1}{k}$. We have

$$\begin{aligned} \frac{1}{k}-\phi _i(\bar{\mathcal {P}}) =\frac{1}{k}-\frac{|\bar{\mathcal {P}}|-1}{|\bar{\mathcal {P}}|}\cdot \frac{1}{k-1} =\frac{1}{k(k-1)}\left( \frac{k}{|\bar{\mathcal {P}}|}-1 \right) \ge 0, \end{aligned}$$

where the first equality holds in the same manner with (5) and the final inequality follows from $k\ge |\bar{\mathcal {P}}|$. Hence, if $k= |\bar{\mathcal {P}}|$, then $\phi _i(\bar{\mathcal {P}})=\frac{1}{k}$. Below, we show that this case does not occur. Assume that $|\bar{\mathcal {P}}|=k$. Note that $\hat{k}= k-1$. From $\bar{\mathcal {P}}= (\mathcal {P}|_{\hat{K}}) \cup \{N\setminus \hat{K}\}$, it follows that $\mathcal {P}|_{\hat{K}}$ is a partition of $\hat{K}$ into $\hat{k}$ singletons, which means that the players in $\hat{K}$ are singletons in the first partition $\mathcal {P}$, because the first deviating coalition is $\{N\setminus \hat{K}\}$. The local monotonicity of the Owen power index implies that for any $i\in \hat{K}$ and any $j\in N\setminus \hat{K}$, $\phi _i(\mathcal {P}) \le \phi _j(\mathcal {P})$, because $i\in \hat{K}$ forms a singleton coalition in $\mathcal {P}$ and $j\in N\setminus \hat{K}$ belongs to a coalition consisting of at least one player in $\mathcal {P}$. This contradicts $\hat{K}$, namely, $\phi _i(\mathcal {P})\ge \frac{1}{k}$ for $i\in \hat{K}$ and $\phi _j(\mathcal {P})< \frac{1}{k}$ for $j\in N\setminus \hat{K}$.

We next show that in $\bar{\mathcal {P}}$, there exists $i^{*}\in \hat{K}$ which agrees with the second deviation.

Claim 5

In $\bar{\mathcal {P}}$, there exists $i^{*}\in \hat{K}$ such that $\phi _{i^{*}}(\bar{\mathcal {P}}) < \frac{1}{k}=\phi _{i^{*}}(\mathcal {P}^{*})$.

Proof

Assume that for every $i\in \hat{K}$, $\phi _{i}(\bar{\mathcal {P}}) \ge \frac{1}{k}$. We have $\phi _{N\setminus \hat{K}}(\bar{\mathcal {P}})=\frac{|\bar{\mathcal {P}}|-1}{|\bar{\mathcal {P}}|}$ ($2\le |\bar{\mathcal {P}}| \le k$) in the same manner as (5). This attains its minimum $\frac{1}{2}$ at $|\bar{\mathcal {P}}|=2$ (namely, $\phi _{N\setminus \hat{K}}(\bar{\mathcal {P}})=\sum _{j\in N\setminus \hat{K}}\phi _j(\bar{\mathcal {P}})\ge \frac{1}{2}$). Hence,

$$\begin{aligned} \sum _{j\in N}\phi _j(\bar{\mathcal {P}})= \sum _{j\in \hat{K}}\phi _j(\bar{\mathcal {P}})+\sum _{j\in N\setminus \hat{K}}\phi _j(\bar{\mathcal {P}}) \ge (k-1)\frac{1}{k}+\frac{1}{2} =\frac{3}{2}-\frac{1}{k}. \end{aligned}$$

This, however, contradicts $\sum _{j\in N}\phi _j(\bar{\mathcal {P}})=1$ as $k\ge 3$ (i.e., $n\ge 4$). $\square$

Thus, for every player, $i\in (N\setminus \hat{K})\cup \{i^{*}\}$, $\phi _i(\mathcal {P}^{*}) > \phi _i(\bar{\mathcal {P}})$. $\square$

1.3 Proof of Proposition 3

Proof

We show that $\mathcal {P}^{*}=\{K\}\cup [N\setminus K]$ is in the projective core. We consider the following two cases: odd n and even n. Let S be a deviating coalition and $\mathcal {P}$ be the partition after S’s deviation; formally,

$$\begin{aligned} \mathcal {P}:=\{K\setminus S\} \cup \{S\} \cup [(N\setminus K)\setminus S]. \end{aligned}$$

(6)

Odd, condition (i) We show that for any $S\subseteq N$ such that $S\cap K\ne \emptyset$ and $|S|\le k-1$, we have $\phi _i(\mathcal {P}^{*}) \ge \phi _i(\mathcal {P})$.

For any player $i\in S\cap K$.¹² To compute $\phi _S(\mathcal {P})$, let $l=|S\cap K|$ and $r=|(N\setminus K)\setminus S|$. Note that

$$\begin{aligned} 1\le & {} |S|\le k-1, \end{aligned}$$

(7)

$$\begin{aligned} 1\le & {} l \le k-1,\nonumber \\ 0\le & {} r \le n-k. \end{aligned}$$

(8)

The number of orders that S becomes a pivot as a coalition in $\mathcal {P}$ (formally, a pivot in its corresponding weighted majority game) is

$$\begin{aligned} \sum _{j=0}^{l-1} \left[ (1+j)! (r-j)! \left( \begin{array}{cc} r \\ j \\ \end{array}\right) \times 2 \right]= & {} 2\sum _{j=0}^{l-1} [ (j+1) r! ]\nonumber \\= & {} 2r!\sum _{j=1}^{l}j\nonumber \\= & {} r! l(1+l). \end{aligned}$$

(9)

The number of all orders (of coalitions in $\mathcal {P}$) is $(r+2)!$. Hence, by (9), we have

$$\begin{aligned} \phi _S(\mathcal {P})=\frac{r! l(1+l)}{(r+2)!} = \frac{l(1+l)}{(r+2)(r+1)}. \end{aligned}$$

Since $|S|=l+n-k-r$, we have, for any $i\in S$,

$$\begin{aligned} \phi _i(\mathcal {P}) =\frac{l(1+l)}{(r+2)(r+1)}\cdot \frac{1}{l+n-k-r}. \end{aligned}$$

(10)

This increases with respect to l. By $|S|=l+n-k-r$, the right hand side of (7) implies $l\le 2k+r-1-n$. By $k=\frac{n+1}{2}$, we have $l\le r$. Hence, with respect to l, (10) attains its maximum at $l=r$. Assuming $l=r$, (10) is

$$\begin{aligned} \frac{r(1+r)}{(r+2)(r+1)}\cdot \frac{1}{r+n-k-r}=\frac{r}{r+2}\cdot \frac{1}{n-k}. \end{aligned}$$

(11)

This increases with respect to r. By (8), substituting $r=n-k$ for (11), we have

$$\begin{aligned} \frac{1}{n-k+2}= & {} \frac{2}{n+3} < \frac{2}{n+1} = \frac{1}{k}. \end{aligned}$$

(12)

Thus, (10)–(12) imply that for any $i\in S\cap K$, $\phi _i(\mathcal {P})<\frac{1}{k}=\phi _i(\mathcal {P}^{*})$.

Odd, condition (ii) For any $S\subseteq N\setminus K$, by (6), we have $\mathcal {P}=\{K\}\cup \{S\}\cup [(N\setminus K)\setminus S]$. Since K is an exact majority coalition, we have $\phi _i(\mathcal {P}^{*})=0=\phi _i(\mathcal {P})$ for any i with $i\in S\subseteq N\setminus K$.

Even, condition (i) The approach is the same as the case of odd n. We have (7)–(8). We change the computation of (9) as follows:

$$\begin{aligned}&\sum _{j=0}^{l-2} \left[ (1+j)! (r-j)! \left( \begin{array}{cc} r \\ j \\ \end{array}\right) \times 2 \right] + l!(r-l+1)!\left( \begin{array}{cc} r \\ l-1\\ \end{array}\right) \nonumber \\&\quad = 2\sum _{j=0}^{l-2}\left[ (j+1) r! \right] + l\cdot r!\nonumber \\&\quad = r! \cdot l^2. \end{aligned}$$

(13)

The number of all orders is $(r+2)!$. Hence, we obtain

$$\begin{aligned} \phi _S(\mathcal {P})=\frac{r! \cdot l^2.}{(r+2)!} = \frac{l^2}{(r+2)(r+1)}. \end{aligned}$$

Since $|S|=(l+n-k-r)$, for any $i\in S$,

$$\begin{aligned} \phi _i(\mathcal {P}) =\frac{l^2}{(r+2)(r+1)}\cdot \frac{1}{l+n-k-r}. \end{aligned}$$

(14)

This increases with respect to l. From $|S|=l+n-k-r$ and (7), we have $l\le 2k+r-1-n$. Moreover, because of $k=\frac{n}{2}+1$, we obtain $l\le r+1$. Hence, with respect to l, (14) attains its maximum at $l=r+1$. Assuming $l=r+1$, (14) is

$$\begin{aligned} \frac{(r+1)^2}{(r+2)(r+1)}\cdot \frac{1}{r+1+n-k-r}=\frac{r+1}{r+2}\cdot \frac{1}{n-k+1}. \end{aligned}$$

(15)

This increases with respect to r. By (8), substituting $r=n-k$ for (15), we have

$$\begin{aligned} \frac{1}{n-k+2}= & {} \frac{2}{n+2} < \frac{2}{n+1} = \frac{1}{k}. \end{aligned}$$

(16)

Thus, from (14)–(16), we have the same conclusion as the case of odd n.

Even, condition (ii) This part is exactly the same as the case of odd n. $\square$

1.4 Proof of Proposition 6

Proof

We first consider the partition $\{K\} \cup [N\setminus K]$. Although every player in $[N\setminus K]$ has an incentive to join coalition K, all players in K refuse it. Moreover, every player in $[N\setminus K]$ has no incentive to merge with any other player in $[N\setminus K]$, because he gets zero even after the merge. Every player in K has no incentive to deviate alone from K, because he gets $\frac{n-2k+3}{(n-k+1)(n-k+2)}\ (<\frac{1}{k})$ after his deviation.¹³ Every player in K has no incentive to merge with any player in $[N\setminus K]$, because he gets $\frac{1}{(n-k+1)(n-k)}$ ($\le \frac{1}{k}$ for $n\ge 3$, odd) or $\frac{1}{2(n-k+1)(n-k)}$ ($<\frac{1}{k}$ for $n\ge 4$, even) after the merge.¹⁴ Next, consider a partition which contains a winning coalition whose size is strictly greater than k. The partition $\{N\}$ is individually stable because it is Nash stable. For a partition which contains a winning coalition $K^+$ with $k^+>k$, every player i in $K^+$ has no incentive to deviate from $K^+$ because $K^+\setminus \{i\}$ is still winning. Although every singleton player in $N\setminus K^+$ has an incentive to join coalition $K^+$, all players in $K^+$ refuse it. $\square$