Conformity in voting

Abstract

A group of agents has to decide whether to accept or reject a proposal. Agents vote in favor of or against the proposal and, if the number of agents in favor is greater than a certain threshold, the proposal is accepted. Conformist agents vote based on not only their opinion but also the votes of other agents. Independent agents consider only their own opinions. If all agents are conformists and vote simultaneously, there are undominated Nash equilibria in which the decision is different from that obtained if all agents vote for their opinions. Next, we provide the number of independent agents sufficient to obtain, in any equilibrium, the decision obtained when all agents vote for their opinions. This number depends on the total number of agents, the threshold, and the conformity measure. If agents vote sequentially, the voting behavior of conformist agents does not affect the decision.

Appendix

This appendix includes the proofs of Theorem 1 and Lemmas 1, 2, 3, and 4. Before proceeding with the proofs, we need to introduce some notation. Let lower-case letters c, s denote the cardinality of subsets of agents $C,S\subset N$.

Proof of Theorem 1

We distinguish two cases, depending on the q-threshold rule. Case 1 presents the proof for the $1-$ and the $n-$threshold rule and Case 2 presents the proof for the remaining thresholds, $q\in \{2,\ldots ,n-1\}.$

Case 1. Let $I\ne \emptyset$ and the $1-$threshold rule (the $n-$threshold rule is symmetric). If $\#I=\min \{h,n-h\}$, the $1-$truthful social choice function is implemented in equilibrium when $\succeq ^{0}$. In doing so, we show that $m_{i}^{1}$ is weakly dominated by $m_{i}^{0}$ for any $i\in N\backslash I$. By assumption, any $i\in N\backslash I$ knows that there are $\#I$ voting for 0. We distinguish two cases:

Case 1.1. [Case 1.2.] $n-h>h$ [$n-h\le h$]. Let $\#I=h$ [$\#I=n-h$]. Take any agent $i\in N\backslash I$. For any $m=(m_{S}^{1},m_{N\backslash S}^{0})\in UN((M,g^{1}),\succeq ^{0},h)$, $I\subseteq N\backslash S$. Since $n-s\ge \#I$ and by $h-$conformity, $(1;m_{i}^{0},m_{S\backslash \{i\}} ^{1},m_{N\backslash S}^{0})$ $\succeq _{i}^{0}(1;m_{S}^{1},m_{N\backslash S}^{0})$. By selfishness, $(0;m_{i}^{0},m_{-i}^{0})$ $\succ _{i}^{0} (1;m_{i}^{1},m_{-i}^{0})$ and $m_{i}^{1}$ is weakly dominated by $m_{i}^{0}$ for any $i\in N\backslash I.$ Therefore, the result follows. Finally, if $\#I<h$ [$\#I<n-h\le h$] there are $m\in UN((M,q),\succeq ^{0},h)$ such that the $1-$truthful social choice function is not implemented. Let $m=(m_{S} ^{1},m_{N\backslash S}^{0})$ where $S=N\backslash I$. For any $i\in S$, by $h-$conformity and since $\#I+1<h$, $(1;m_{S}^{1},m_{N\backslash S}^{0} )\succ _{i}^{0}\ (1;m_{i}^{0},m_{S\backslash \{i\}}^{1},m_{N\backslash S}^{0} )$.

Case 2. Let $I\ne \emptyset$ and any q-threshold rule such that $q\in \{2,\ldots ,n-1\}$. We show that if $\#I=\min \{n-1,\max \{q-1,n-q\}+\min \{h,n-h\}\}$, the q-truthful social choice function is implemented in equilibrium for any $q\in \left\{ 2,\ldots ,n-1\right\}$ and any $\succeq$. In doing so, we show that for any $\succeq _{i}$, $m_{i}\ne t_{i}$ is weakly dominated by $m_{i}^{t_{i}}$ for any $i\in N\backslash I$. By assumption, any $i\in N\backslash I$ knows there are $\#I$ independent agents voting for their true opinions. For clarification, we divide the proof into three cases depending on the threshold.

Case 2.1. $q\in \{2,\ldots ,\left\lfloor \frac{n+1}{2}\right\rfloor -1\}$ (Case 2.2. when $q\in \{\left\lceil \frac{n+1}{2}\right\rceil +1,\ldots ,n-1\}$ is symmetric). Then, $q-1<n-q$. We distinguish two cases:

Case 2.1.1. $q-1\le \min \{h,n-h\}$. Then, $q\le h+1$. Let $\#I=n-1$ and $\succeq =(\succeq _{C}^{1},\succeq _{N\backslash C}^{0})\in \times _{i\in N}\mathcal {R}_{i}^{h}$. Let agent i be the only conformist agent. If $c<q-1$, since $c<q-1\le h$, by $h-$conformity, $(0;m_{i}^{0},m_{-i} )\succeq _{i}^{t_{i}}(0;m_{i}^{1},m_{-i})$. If $c=q-1$ and $\succeq _{i}^{1}$ or $c=q$ and $\succeq _{i}^{0}$, since $c=q-1\le h,$ by $h-$conformity, $(0;m_{i}^{0},m_{-i})\succeq _{i}^{t_{i}}$ $(0;m_{i}^{1},m_{-i})$. If $c=q-1$ and $\succeq _{i}^{0}$, by selfishness $(0;m_{i}^{0},m_{C}^{1},m_{N\backslash C\cup \{i\}}^{0})$ $\succ _{i}^{0}(1;m_{i}^{1},m_{C}^{1},m_{N\backslash C\cup \{i\}}^{0})$. If $c=q$ and $\succeq _{i}^{1}$, by selfishness $(1;m_{i}^{1},m_{C\backslash \{i\}}^{1},m_{N\backslash C}^{0})$ $\succ _{i} ^{1}(0;m_{i}^{0},m_{C\backslash \{i\}}^{1},m_{N\backslash C}^{0})$. If $c>q$, when $c<h+1$, by $h-$conformity, $(1;m_{i}^{0},m_{-i})\succeq _{i}^{t_{i}}$ $(1;m_{i}^{1},m_{-i})$; when $c\ge h+1$, by $h-$conformity, $(1;m_{i} ^{1},m_{-i})\sim _{i}^{t_{i}}(1;m_{i}^{0},m_{-i})$. Finally, suppose that $\#I<n-1$. Let $\#I=n-2$ and suppose, wlog, that $1,2\in N\backslash I$. In addition, let $\succeq =(\succeq _{C}^{1},\succeq _{N\backslash C}^{0} )\in \times _{i\in N}\mathcal {R}_{i}^{h}$ be such that $c=q$, and $\succeq _{1}^{1},\succeq _{2}^{1}$. For $m=(m_{1}^{0},m_{2}^{0},m_{-\{1,2\}})$ such that for any $k\in I$, $m_{k}^{t_{k}}$ and $i,j\in \{1,2\}$, $(1;m_{i} ^{0},m_{j}^{0},m_{-\{i,j\}})\succeq _{i}^{1}$ $(1;m_{i}^{1},m_{j} ^{0},m_{-\{i,j\}})$.

Case 2.1.2. $q-1>\min \{h,n-h\}$ . We distinguish two subcases:

Subcase 2.1.2.a. [Subcase 2.1.2.b.] $h<n-h$ [$h\ge n-h$]. Then, $h\le \frac{n}{2}-1$ [$h>\frac{n}{2}-1$] and $q>h+1$ [$q<h+1$]. Let $\#I=n-q+h$ [$\#I=n-q+n-h$] and $\succeq =(\succeq _{C}^{1},\succeq _{N\backslash C}^{0})\in \times _{i\in N}\mathcal {R}_{i}^{h}$. If $c<q$, at most $q-1$ independent agents vote for 1 and at least h independent agents vote for 0. Then, either by selfishness or by $h-$conformity, it is a weakly dominant strategy to vote for 0 for any $i\in N\backslash I$ such that $\succeq _{i}^{0}$. If $c\ge q$, at least h [$n-h$] independent agents vote for 1. Either by selfishness or by $h-$conformity, it is a weakly dominant strategy to vote for 1 for any $i\in N\backslash I$ such that $\succeq _{i}^{1}$. Finally, suppose that $\#I<n-q+h$ [$\#I<n-q+n-h$]. Let $\#I=n-1-(q-h)$ [$\#I=n-q+n-h-1$], and suppose $\succeq =(\succeq _{C}^{1},\succeq _{N\backslash C}^{0})$ is such that for $h-1$ [$n-h-1$] independent agents $\succeq _{i}^{1}$, for $n-q$ independent agents $\succeq _{j}^{0}$ and all $k\in N\backslash I$ are such that $\succeq _{k}^{1}$. Note that the decision implemented by the q-truthful social choice function is 1. Let m be such that all independent agents tell the truth, and all conformist agents report $m_{k}=0$. Since $h\le \frac{n}{2}-1$ [$h>\frac{n}{2}-1$], by $h-$conformity, $(0;m_{k}^{0},m_{-k})\succ _{k} ^{1}(0;m_{k}^{1},m_{-k})$.

Case 2.3. n is even and $q\in \{\frac{n}{2},\frac{n}{2}+1\}.$ ¹⁸ Then, $n-q>q-1$. We distinguish two cases:

Case 2.3.1. $q-1\le \min \{h,n-h\}$. Then, $q\le h+1$. Let $\#I=n-1$ and $\succeq =(\succeq _{C}^{1},\succeq _{N\backslash C}^{0})\in \times _{i\in N} \mathcal {R}_{i}^{h}$. Let agent i be the only conformist agent. If $c\le q,$ the proof is analogous to Case 2.1.1. If $c>q$, when $q<c<h+1$, at least q independent agents vote for 1 and at most h agents vote for 1. Since $q\le h+1\le q+1,$by $h-$conformity, $(1;m_{i}^{0},m_{-i})$ $\succeq _{i}^{t_{i}}(1;m_{i}^{1},m_{-i})$. If $q\le h+1\le c$, it could be either $q\le h+1<c$ or $q<h+1\le c.$ If $q\le h+1<c,$ at least $h+1$ independent agents vote for 1. If $q<h+1\le c,$ at least h independent agents vote for 1. Then, by $h-$conformity $(1;m_{i}^{1},m_{-i})\succeq _{i}^{t_{i} }(1;m_{i}^{0},m_{-i})$. Finally, suppose that $\#I<n-1.$ Let $\#I=n-2$ and suppose, wlog, that $1,2\in N\backslash I$. In addition, let $\succeq =(\succeq _{C}^{1},\succeq _{N\backslash C}^{0})$ be such that $c=q-1$, and $\succeq _{1}^{1},\succeq _{2}^{1}$. For $m=(m_{1}^{0},m_{2}^{0} ,m_{-\{1,2\}})$ such that for any $k\in I$, $m_{k}^{t_{k}}$, and $i,j\in \{1,2\}$, since $q-1<n-q$ we have $(0;m_{i}^{0},m_{j}^{0} ,m_{-\{i,j\}})\succeq _{i}^{1}(0;m_{i}^{1},m_{j}^{0},m_{-\{i,j\}})$.

Case 2.3.2. $q-1>\min \{h,n-h\}$. We distinguish two subcases:

Subcase 2.3.2.a. $h<n-h$. Then, $h\le \frac{n}{2}-1$ and $q>h+1.$ Let $\#I=n-(q-h)$ and $\succeq =(\succeq _{C} ^{1},\succeq _{N\backslash C}^{0})\in \times _{i\in N}\mathcal {R}_{i}^{h}$. If $c<q$, for $i\in N\backslash I$ such that $\succeq _{i}^{1}$, they vote for 1 when the independent agents voting for 1 are enough to conform to at least h. Otherwise, by $h-$conformity, they vote for 0. For the particular case in which $q-1$ independent agents vote for 1 and h independent agents vote for 0,  since $q>h+1$ all $i\in N\backslash I$ such that $\succeq _{i}^{0}$ vote for 0 and all $i\in N\backslash I$ such that $\succeq _{i}^{1}$ vote for 1 as $q>h+1.$ The rest of the proof is analogous to Subcase 2.1.2.a.

Subcase 2.3.2.b. $h\ge n-h$ . Then, $h>\frac{n}{2}-1$. Let $\#I=n-q+n-h$, and $\succeq =(\succeq _{C}^{1},\succeq _{N\backslash C}^{0} )\in \times _{i\in N}\mathcal {R}_{i}^{h}$. If $c\ge q$, at least $n-h$ independent agents vote for 1, and there are strictly less than $n-q$ independent agents voting for 0. Given that $n-q>q-1$ and $q<h$, either by selfishness or by $h-$conformity, it is a weakly dominant strategy to vote for 0 for any $i\in N\backslash I$ such that $\succeq _{i}^{0}$. The rest of the proof is analogous to Subcase 2.1.2.b.

Proof of Lemma 1

We distinguish two cases, depending on the q-threshold rule. Case 1 presents the proof for the $1-$ and the $n-$threshold rule and Case 2 presents the proof for the remaining thresholds, $q\in \{2,\ldots ,n-1\}.$

Case 1. Let the $1-$threshold rule (the $n-$threshold rule is symmetric). Then, the $1-$truthful social choice function is not implemented when $\succeq ^{0}$. Let $\#I=h$ and $h=\frac{n}{2}$ ($h=\frac{n+1}{2}$ if n is odd). By assumption, any conformist agent knows there are h independent agents voting for 0. Take any agent $i\in N\backslash I$. Since $\#I+i>\frac{n}{2}-i$, by $h-$conformity of agent $i\in N\backslash I$, $(0;m_{i}^{0},m_{S\backslash \{i\}}^{1},m_{N\backslash S}^{0})$ $\succeq _{i}^{0}(1;m_{S}^{1},m_{N\backslash S}^{0})$. Finally, if $\#I=h-1$, by $h-$conformity and since $\#I+i\le h$, $(1;m_{S}^{1},m_{N\backslash S} ^{0})\succ _{i}^{0}(0;m_{i}^{0},m_{S\backslash \{i\}}^{1},m_{N\backslash S} ^{0})$.

Case 2. Let the q-threshold rule where $q\in \{2,\ldots ,n-1\}.$ We show that if $\#I=n-1$, the q-truthful social choice function is implemented for any $q\in \left\{ 2,\ldots ,n-1\right\}$ and any $\succeq =(\succeq _{C}^{1},\succeq _{N\backslash C}^{0})\in \times _{i\in N}\mathcal {R}_{i}^{h}$. By assumption, the conformist agent knows there are $n-1$ independent agents voting for their true opinions. Let agent i be the unique conformist agent. If $\succeq _{i}^{0}$, by selfishness $(0;m_{i} ^{0},m_{C}^{1},m_{N\backslash C\cup \{i\}}^{0})$ $\succ _{i}^{0}(1;m_{i} ^{1},m_{C}^{1},m_{N\backslash C\cup \{i\}}^{0})$ and if $\succeq _{i}^{1}$, by selfishness $(1;m_{i}^{1},m_{C\backslash \{i\}}^{1},m_{N\backslash C}^{0})$ $\succ _{i}^{1}$ $(0;m_{i}^{0},m_{C\backslash \{i\}}^{1},m_{N\backslash C}^{0} )$. Therefore, in both cases, the q-truthful social choice function is implemented in equilibrium.

Finally, suppose that $\#I<n-1.$ Let $\#I=n-2$ and suppose wlog that $1,2\in N\backslash I$. For clarification, we divide the proof into four cases depending on the threshold. Take $q<\frac{n}{2}$. Let $\succeq =(\succeq _{C}^{1},\succeq _{N\backslash C}^{0})$ be such that $c=q$, and $\succeq _{1}^{1},\succeq _{2}^{1}$. For $m=(m_{1}^{0},m_{2}^{0},m_{-\{1,2\}})$ such that for any $k\in I$, $m_{k}^{t_{k}}$, and $i,j\in \{1,2\}$, $(0;m_{i}^{0},m_{j}^{0},m_{-\{i,j\}} )\succeq _{i}^{1}(0;m_{i}^{1},m_{j}^{0},m_{-\{i,j\}})$. Take $q=\frac{n}{2}$ and n is even. Let $\succeq =(\succeq _{C}^{1},\succeq _{N\backslash C}^{0})$ be such that $c=q-1$, and $\succeq _{1}^{0},\succeq _{2}^{0}$. For $m=(m_{1} ^{1},m_{2}^{1},m_{-\{1,2\}})$ such that for any $k\in I$, $m_{k}^{t_{k}}$, and $i,j\in \{1,2\}$, $(1;m_{i}^{1},m_{j}^{1},m_{-\{i,j\}})\succeq _{i}^{0}$ $(1;m_{i}^{1},m_{j}^{0},m_{-\{i,j\}})$. Take $q=\frac{n+1}{2}$ and n is odd. Let $\succeq =(\succeq _{C}^{1},\succeq _{N\backslash C}^{0})$ be such that $c=q-1$, and $\succeq _{1}^{0},\succeq _{2}^{0}$. For $m=(m_{1}^{1},m_{2} ^{1},m_{-\{1,2\}})$ such that for any $k\in I$, $m_{k}^{t_{k}}$, and $i,j\in \{1,2\}$, $(1;m_{i}^{1},m_{j}^{1},m_{-\{i,j\}})\succeq _{i}^{0}$ $(1;m_{i}^{1},m_{j}^{0},m_{-\{i,j\}})$. Take $q>\frac{n}{2}$. Let $\succeq =(\succeq _{C}^{1},\succeq _{N\backslash C}^{0})$ be such that $c=q-1$, and $\succeq _{1}^{0},\succeq _{2}^{0}$. For $m=(m_{1}^{1},m_{2}^{1} ,m_{-\{1,2\}})$ such that for any $k\in I$, $m_{k}^{t_{k}}$, and $i,j\in \{1,2\}$, $(1;m_{i}^{1},m_{j}^{1},m_{-\{i,j\}})\succeq _{i}^{0}$ $(1;m_{i}^{0},m_{j}^{1},m_{-\{i,j\}})$.

The proof of Proposition 2 makes use of three lemmas. Lemma 2 applies for the $1-$truthful social choice function, Lemma 3 for the $n-$truthful social choice function, and Lemma 4 for the remaining q-truthful social choice functions, $q\in \{2,\ldots ,n-1\}.$

Lemma 2

For any number of agents n, the $1-$truthful social choice function is implemented in any subgame perfect Nash Equilibrium on $R^{h_{i}}$.

Proof

Let n and the $1-$threshold rule. Let $\succeq ^{0}$. Then, $x=0$. We have to show that $SPN(G^{\{1,\ldots ,n\},1})\ne \emptyset$ and that for any $m\in SPN(G^{\{1,\ldots ,n\},1})$, $g^{1}(m)=0$. We solve the game starting at stage n where agent n votes (remember that $o=(1,\ldots ,n)$). Since $\succeq _{n}^{0}$, in equilibrium $m_{n}$ is such that for any $\widetilde{m}_{i}\in M_{i}$ where $i<n$, by selfishness $\alpha _{n}\left( \widetilde{m}_{1},\ldots ,\widetilde{m}_{n-1},m_{n}\right) =0$ if agent n is pivotal relative to $\left( \widetilde{m}_{1},\ldots ,\widetilde{m}_{n-1}\right)$, and by $h-$conformity, $\alpha _{n}\left( \widetilde{m}_{1},\ldots ,\widetilde{m}_{n-1},m_{n}\right)$ would be 0 or 1 depending on h if agent n is not pivotal. Out of the path of play, agent n is indifferent. We proceed to stage $n-1$, that is, to agent $n-1$. Applying the same reasoning iteratively, we reach stage 1. Again, since $\succeq _{1}^{0}$, in equilibrium, $m_{1}$ is such that $(m_{2},\ldots ,m_{n})$ as described above, by selfishness $\alpha _{1} (m_{1},\ldots ,m_{n})=0$, since agent 1 is pivotal relative to $\left( m_{2},\ldots ,m_{n}\right)$. Note that $g^{1}(m_{1},\ldots ,m_{n})=0$. Then, $m\in SPN(G^{\{1,\ldots ,n\},1})$ and $g^{1}(m)=0$.

Let $\succeq \in \mathcal {R}^{h_{i}}$ be such that for some $\succeq _{i}$, $\succeq _{i}^{1}$. Then, $x=1$. We have to show that $SPN(G^{\{1,\ldots ,n\},1})$ $\ne \emptyset$ and that for any $m\in SPN(G^{\{1,\ldots ,n\},1})$, $g^{1}(m)=1$. In doing so, we show the proof by contradiction, that is, for any $m\in SPN(G^{\{1,\ldots ,n\},1} )$, $g^{1}(m)=0$. We solve the game starting at stage n where agent n votes. Let wlog, agent n in order o for whom $\succeq _{n} ^{1}$. Since $\succeq _{n}^{1}$, in equilibrium $m_{n}$ is such that for any $\widetilde{m}_{i}\in M_{i}$ where $i<n$, by selfishness $\alpha _{n}\left( \widetilde{m}_{1},\ldots ,\widetilde{m}_{n-1},m_{n}\right) =1$ if agent n is pivotal relative to $\left( \widetilde{m}_{1},\ldots ,\widetilde{m}_{n-1}\right)$, and by $h-$conformity, $\alpha _{n}\left( \widetilde{m}_{1},\ldots ,\widetilde{m}_{n-1},m_{n}\right)$ would be 0 or 1, depending on h whether agent n is not pivotal. Out of the path of play, agent n is indifferent. Note that for any $\widetilde{m}_{i}\in M_{i}$ where $i<n$, $g^{1}(\widetilde{m}_{1},\ldots ,\widetilde{m}_{n-1},m_{n})=$ 1. Therefore, any $i<n$ is not pivotal relative to any $(\widetilde{m}_{1},\ldots ,\widetilde{m}_{n-1},m_{n})$ for any $\widetilde{m}_{j}\in M_{j}$ for $j\ne i$, $j<n$. In equilibrium, for any $i<n$, agent i announces $m_{i}$ according to $h-$conformity at any node in which she plays. Then, m $\in SPN(G^{\{1,\ldots ,n\},1})$ and $g^{1}(m)=1,$ and we get a contradiction. $\square$

Lemma 3

For any number of agents n, the $n-$truthful social choice function is implemented in any subgame perfect Nash equilibrium on $R^{h_{i}}$.

Proof

The proof is symmetric to Lemma 2. $\square$

The proof of Lemma 4 makes extensive use of Lemmas 2 and 3.

Lemma 4

For any number of agents n, the q-truthful social choice function such that $q\in \{2,\ldots ,n-1\}$ is implemented in any subgame perfect Nash Equilibrium on $R^{h_{i}}$.

Proof

Let n and any q-threshold rule such that $q\in \{2,\ldots ,n-1\}$. We show that for any $G^{\{1,\ldots ,n\},q}$, if agents play their equilibrium strategies, the reduced game obtained can be either $G^{\{1,\ldots ,k\},q}$ where $q=k$ or $G^{\{1,\ldots ,k\},q}$ where $q=1,$ which would depend on the preferences of the agents. We solve the game starting at stage n where agent n votes (remember that $o=(1,\ldots ,n)$). Let $\succeq \in \mathcal {R}^{h_{i}}$. We distinguish two claims, depending on $\succeq _{n}$:

Claim A. If $\succeq _{n}^{0}$, in equilibrium $m_{n}$ is such that for any $\widetilde{m}_{i}\in M_{i}$ where $i<n$, by selfishness $\alpha _{n}\left( \widetilde{m}_{1},\ldots ,\widetilde{m}_{n-1},m_{n}\right) =0$ if agent n is pivotal relative to $\left( \widetilde{m}_{1},\ldots ,\widetilde{m}_{n-1}\right)$, and by $h-$conformity, $\alpha _{n}\left( \widetilde{m}_{1},\ldots ,\widetilde{m}_{n-1},m_{n}\right)$ would be 0 or 1, depending on h if agent n is not pivotal. Out of the path of play, agent n is indifferent. Then, the reduced game obtained after agent n plays her equilibrium strategy is a game with agents 1 to $n-1$ and q-threshold, that is, $G^{\{1,\ldots ,n-1\},q}$.

Claim B. If $\succeq _{n}^{1}$, in equilibrium $m_{n}$ is such that for any $\widetilde{m}_{i}\in M_{i}$ where $i<n$, by selfishness $\alpha _{n}\left( \widetilde{m}_{1},\ldots ,\widetilde{m}_{n-1},m_{n}\right) =1$ if agent n is pivotal relative to $\left( \widetilde{m}_{1},\ldots ,\widetilde{m}_{n-1}\right)$, and by $h-$conformity, $\alpha _{n}\left( \widetilde{m}_{1},\ldots ,\widetilde{m}_{n-1},m_{n}\right)$ would be 0 or 1, depending on h whether agent n is not pivotal. Out of the path of play, agent n is indifferent. Then, the reduced game is $G^{\{1,\ldots ,n-1\},q-1}$.

We now distinguish several cases, depending on the number of agents such that $\succeq _{i}^{1}$.

Case 1. $\succeq$ such that $\#\{i\in N:\succeq _{i}^{1}\}<q-1$. We distinguish several cases, depending on the preferences of agents 2 to n. If $\succeq _{k}^{0}$, for $k=\{q+1,\ldots ,n\}$, Claim A applies at stages n to $q+1$ and the reduced game is $G^{\{1,\ldots ,q\},q}$. If $\succeq _{q}^{0}$ and $\succeq _{j}^{1}$ for exactly one agent $j\in \{q+1,\ldots ,n\}$, Claim B applies at stage j, and Claim A otherwise. Then, the reduced game is $G^{\{1,\ldots ,q\},q-1}$. To solve stage q, we apply Claim A and the reduced game is $G^{\{1,\ldots ,q-1\},q-1}$. If $\succeq _{q-1}^{0}$, and $\succeq _{j}^{1}$, $\succeq _{k}^{1}$ for exactly two agents, $j,k\in \{q,\ldots ,n\}$. Claim B applies when we reach stages j and k, and Claim A otherwise. Then, the reduced game is $G^{\{1,\ldots ,q-1\},q-2}$. To solve stage $q-1$, we apply Claim A and the reduced game is $G^{\{1,\ldots ,q-2\},q-2}$. By a similar reasoning, we end up in a case in which $\succeq _{3}^{0}$ and $\succeq _{j}^{1}$ for exactly $q-2$ agents $j\in \{4,\ldots ,n\}$. Claim B applies when each of the $q-2$ agents play, and Claim A otherwise. Then, the reduced game is $G^{\{1,2,3\},2}$. To solve stage 3, we apply Claim A and the reduced game is $G^{\{1,2\},2}$. In all of the abovementioned cases, the q-threshold equals the number of agents remaining in the game. Applying Lemma 3 , the result follows.

Case 2. $\succeq$ such that $q<\#\{i\in N:\succeq _{i}^{1}\}$. We distinguish several cases depending on the preferences of agents 2 to n. If $\succeq _{k}^{1}$ for $k=\{n+2-q,\ldots ,n\}$, Claim B applies at stages n to $n+2-q$. Then, the reduced game is $G^{\{1,\ldots ,n+1-q\},1}$. If $\succeq _{n+1-q}^{1}$ and $\succeq _{j}^{0}$ for exactly one agent $j\in \{n+2-q,\ldots ,n\}$. Claim A applies when we reach stage j, and Claim B otherwise. Then, the reduced game is $G^{\{1,\ldots ,n+1-q\},2}$. To solve stage $n+1-q$, we apply Claim B and the reduced game is $G^{\{1,\ldots ,n-q\},1}$. If $\succeq _{n-q}^{1}$ and $\succeq _{j}^{0}$, $\succeq _{k}^{0}$ for exactly two agents, $j,k\in \{n+1-q,\ldots ,n\}$. Claim A applies when we reach stages j and k, and Claim B otherwise. Then, the reduced game is $G^{\{1,\ldots ,n+1-q\},2}$. To solve stage $n-q$, we apply Claim B and the reduced game is $G^{\{1,\ldots ,n-q-1\},1}$. By a similar reasoning, we end up in a case in which $\succeq _{3}^{1}$ and $\succeq _{j}^{0}$ for exactly $n-q-1$ agents $j\in \{4,\ldots ,n\}$. Claim A applies when each of the $n-q-1$ agents play, and Claim B otherwise. Then, the reduced game is $G^{\{1,2,3\},2}$. To solve stage 3, we apply Claim B and the reduced game is $G^{\{1,2\},1}$. In all of the abovementioned cases, for the obtained reduced games, the threshold is 1. Applying Lemma 2, the result follows.

Case 3. $\succeq$ such that $q-1\le \#\{i\in N:\succeq _{i}^{1}\}\le q.$ We distinguish two subcases.

Subcase 3.1. $n-q<q-1$. We distinguish several cases depending on the preferences of agents 2 to n. If $\succeq _{k}^{0}$ for $k=\{q+1,\ldots ,n\}$, Claim A applies at stages n to $q+1$. Then, the reduced game is $G^{\{1,\ldots ,q\},q}$. If $\succeq _{r+1}^{0}$ and $\succeq _{j}^{1}$ for exactly $q-r$ agents, $q>r>1$, $j\in \{r+2,\ldots ,n\}$. Claim B applies when each of the $q-r$ agents play, and Claim A otherwise, and the reduced game is $G^{\{1,\ldots ,r+1\},r}$. To solve stage $r+1$, we apply Claim A and the reduced game is $G^{\{1,\ldots ,r\},r}$. Lemma 3 applies. If $\succeq _{2}^{0}$ and $\succeq _{j}^{1}$ for exactly $q-1$ agents, $j\in \{3,\ldots ,n\}$. Claim B applies in the stages in which each of the above $q-1$ agents play, and Claim A otherwise, and the reduced game is $G^{\{1,2\},1}$. Lemma 2 applies.

Subcase 3.2. $n-q>q-1.$ We distinguish several cases, depending on the preferences of agents 2 to n. If $\succeq _{k}^{1}$ for $k=\{n+2-q,\ldots ,n\}$, Claim A applies at stages n to $n+2-q$ and the reduced game is $G^{\{1,\ldots ,n+1-q\},1}$. If $\succeq _{r+1}^{1}$ and $\succeq _{j}^{0}$ for exactly $n-q-r$ agents, $n-q>r>0$, $j\in \{r+2,\ldots ,n\}$. Claim A applies when each of the $n-q-r$ agents play, and Claim B otherwise, and the reduced game is $G^{\{1,\ldots ,r+1\},2}$. To solve stage $r+1$, we apply Claim B and the reduced game is $G^{\{1,\ldots ,r\},1}$. Lemma 2 applies. If $\succeq _{3}^{1}$ and $\succeq _{j}^{0}$ for exactly $n-q-r$ agents, $j\in \{3,\ldots ,n\}$. Claim A applies when each of the $n-q-r$ agents play, and Claim B otherwise, and the reduced game is $G^{\{1,2\},1}$. Lemma 2 applies.

Subcase 3.3. $n-q=q-1$. We distinguish several cases, depending on the preferences of agents 2 to n. If $\succeq _{r+1}^{0}$ and $\succeq _{j}^{1}$ for exactly $q-r$ agents, $q>r>1$, $j\in \{r+2,\ldots ,n\}$. Claim B applies when each of the $q-r$ agents play, and Claim A otherwise, and the reduced game is $G^{\{1,\ldots ,r+1\},r}$. To solve stage $r+1$, we apply Claim A and the reduced game is $G^{\{1,\ldots ,r\},r}$. Lemma 3 applies. If $\succeq _{r+1}^{1}$ and $\succeq _{j}^{0}$ for exactly $n-q-r$ agents, $n-q>r>0$, $j\in \{r+2,\ldots ,n\}$. Claim A applies when each of the $n-q-r$ agents play, and Claim B otherwise, and the reduced game is $G^{\{1,\ldots ,r+1\},2}$. To solve stage $r+1$, we apply Claim B and the reduced game is $G^{\{1,\ldots ,r\},1}$. Lemma 2 applies. $\square$

Proof of Proposition 2

This follows from Lemmas 2, 3, and 4.