A Condorcet Jury Theorem for Large Poisson Elections with Multiple Alternatives

Abstract

Herein, we prove a Condorcet jury theorem (CJT) for large elections with multiple alternatives. Voters have common interests that depend on an unknown state of nature. Each voter receives an imprecise private signal about the state of nature and then submits one vote (simple plurality rule). We also assume that this is a Poisson voting game with population uncertainty. The question is whether the simple plurality rule aggregates information efficiently so that the correct alternative is elected with probability tending to one when the number of voters tends to infinity. The previous literature shows that the CJT holds for large elections with two alternatives, but there is also an example of a large election with three alternatives that has an inefficient equilibrium. We show that there always exists an efficient equilibrium, independent of the number of alternatives. Under certain circumstances (informative types), it is unique in elections with two alternatives. The existence of inefficient equilibria in elections with more than two alternatives is generic.

1. Introduction

Consider voters in a large election. Also assume that there is uncertainty about the state of nature. If voters knew the state of nature, they would all agree on which alternative was best (common interests). Each voter receives an imprecise private signal about the state of nature. As it is quite common in the recent literature on large elections, we assume that the setting is a large Poisson game in which voters are not quite sure about the actual number of voters in the game.1

Myerson [2] proved a generalised Condorcet jury theorem (CJT) for such elections with two alternatives and two states of nature: If voters use the simple plurality rule (one person, one vote) and vote strategically, there exists an (limit) equilibrium (as the number of voters converges to infinity), in which the probability of electing the correct alternative in each state of nature is one.2 On the other hand, Goertz and Maniquet [11] provides an example of a limit equilibrium in the same type of election, but with three alternatives and three states of nature, which is inefficient because the correct alternative is selected with probability zero in one of the states of nature.

We seek to reconcile those two papers and answer the following questions: (1) Can we prove a general Condorcet jury theorem for all elections, independent of the number of alternatives? (2) Are there inefficient equilibria in all such elections, independent of the number of alternatives? (3) What can we say about the co-existence of efficient and inefficient equilibria in the same election?

Myerson [2] presents two examples of elections with two alternatives and hints at the fact that a certain type of inefficient equilibrium exists for one of the examples, but not for the other. Of course, given his CJT for two-alternative elections, we already know that an efficient equilibrium exists for both of his examples. We add the following important qualifications: We provide a condition under which a certain type of inefficient equilibrium (unresponsive equilibrium) exists in each election that satisfies a condition on prior and signal distributions (non-informative signals/types), independent of the number of alternatives. It turns out that one of Myerson’s examples satisfies this condition (the one with the inefficient equilibrium), but the other does not. This condition has a quite natural interpretation and shows that these inefficient equilibria may be less likely than anticipated by reading Myerson’s example. We also show that—besides unresponsive equilibria—no other inefficient equilibrium can exist in two-alternative elections. So, when signals/types are informative, all equilibria in two-alternative elections are efficient.

Our results add to the common wisdom that two-alternative elections are a “quite different animal” with properties that do not extend to more alternatives. Even if signals/types are informative, inefficient equilibria exist in elections with more than two alternatives (Goertz and Maniquet [11] shows an example of such). As we argue below (and as it has been done elsewhere), more general results for elections with more than two alternatives are currently beyond reach. However, given our results, we now know that an efficient equilibrium always exists, even if it may not be the only one.3

2. The Setting

2.1. The Model

We keep our notation similar to the most closely related previous work with more than two alternatives, Goertz and Maniquet [11], but extend it to any number of alternatives beyond three.

Voters vote for one of a finite number of alternatives $\{A_1,A_2...\}=\mathbf{A}$. There is a finite number of states of nature $\{{\omega }_1,{\omega }_2...\}=\mathbf{\Omega }$, with $|\mathbf{A}|=|\mathbf{\Omega }|$. Voters are uncertain about the underlying state of nature; $q_i$ is the prior probability of state ${\omega }_i$, with $q_i\in (0,1)$ for all i. Voters have common, state-dependent, dichotomous preferences. They prefer a particular alternative in each state of nature, and are indifferent between the remaining ones:

$$\begin{array}{ccc}\hfill u(A_i|{\omega }_i)& =& 1\forall i,\hfill \\ \hfill u(A_j|{\omega }_i)& =& 0\forall j\ne i.\hfill \end{array}$$

Assume that dichotomous preferences simplifies the strategic environment for the voters. With dichotomous preferences, they care only about those election outcomes in which their vote changes the outcome from any of the $|\mathbf{A}|-1$ disliked alternatives to the preferred alternative; they do not need to consider election outcomes in which they are pivotal between any of the disliked alternatives. This assumption has been made in the previous literature (e.g., Goertz and Maniquet [13]). Goertz and Maniquet [11] do not assume dichotomous preferences in their example. We assume dichotomous preferences because it is common in the literature and because it makes the analysis of expected utilities much easier. None of our results depend on the assumption (i.e., it is not used in any of our proofs). They would all hold with a more general utility function as well.

We assume that voters are strategic and vote as a function of their expected impact on the outcome of the election. So, each voter should have a positive probability of being the pivotal voter because his/her vote never matters otherwise. Since we are interested in large electorates, we incorporate this feature by assuming that the actual number of voters is uncertain (population uncertainty).4 This is a fairly common assumption in the literature (e.g., Feddersen and Pesendorfer [5,6,8], Myerson [1,2,14], and Goertz and Maniquet [13,15]). We follow Myerson [1,2,14] and assume that the population size is Poisson-distributed with parameter n. The probability that there are exactly N voters is

$$P(N|n)=\frac{e^{-n}{n}^N}{N!}.$$

(1)

If the actual number of voters is a random variable, then each voter has a strictly positive probability of being decisive in any equilibrium. This is true even in equilibria in which all voters vote for the same alternative because there is a strictly positive probability, albeit small, that only one or no other voter shows up.5

Another feature of a Poisson game is that strategies are defined type-by-type instead of agent-by agent because, in a Poisson game, "players" individual identities are not globally recognised (Myerson [2]). Thus, when we refer to a specific voter, we always mean a specific voter type. This corresponds to an assumption of symmetric equilibria in Bayesian games in which strategies are defined agent-by-agent.

Before the election, each voter receives a private signal about the state of nature which is informative but imprecise. Signals are independent and identically distributed. In this regard, our model is similar to Feddersen and Pesendorfer [7,8], Myerson [2], and Goertz and Maniquet [11,13].6 The signal determines a voter’s type in our Poisson voting game. For simplicity, we assume that there are only $|\mathbf{\Omega }|$ types, so that $\mathbf{T}=\{t_1,...,t_{|\mathbf{\Omega }|}\}$. We denote by $r_i(t_j)$ the probability that a voter is of type $t_j$ in state ${\omega }_i$.

Myerson [2] shows: In a two-alternative election, an efficient equilibrium exists whenever there exist some types t such that $r_1(t)\ne r_2(t)$, which, by some appropriate relabelling of types, immediately implies that $r_i(t_i)>r_j(t_i)$ for all $i,i\ne j$ (i.e., type $t_i$ is always more likely in state ${\omega }_i$ than in any other state, an important feature for information aggregation).The same feature is used in our Theorem 1 (existence of an efficient equilibrium). Unfortunately, simply assuming $r_i(t_i)\ne r_j(t_i)$ is not sufficient to lead to the desired feature in elections with more than two alternatives. We must assume explicitly:

$$r_i(t_i)>r_j(t_i)\forall i\ne j.$$

(2)

Notice, however, that this assumption is less restrictive than assuming that type $t_i$ is the most likely type in state ${\omega }_i$, i.e., $r_i(t_i)>r_i(t_j)$ for all $i,j,i\ne j$ which is, for example, assumed by the original Condorcet jury theorem and in Feddersen and Pesendorfer [7]. It simply suffices to assume that type $t_i$ is more likely in state ${\omega }_i$ than in any other state. For the interested reader: This also implies that informative voting (voting for one’s type) is not automatically efficient because type $t_i$ is not necessarily the most likely type in state ${\omega }_i$.

If type distributions satisfy Equation (2), there is no aggregate uncertainty in the population. If all private information was public, the state of nature would be known and voters would unanimously vote for the correct alternative (the one that maximises every voter’s utility). We use this as a benchmark and call an equilibrium informationally efficient if the elected alternative is the same as the one that would be selected if all information was public.

The voting rule is the plurality rule with the possibility of abstention. Each voter can vote for one alternative or abstain. So, the action space of a voter type is $\mathbf{C}=\mathbf{A}\cup \varphi$, where $\varphi$ denotes abstention. The alternative with the largest number of votes is elected. We assume a particular tie-breaking rule that is without loss of generality and has been used in the previous literature (e.g., Goertz and Maniquet [11,13]): Any tie involving alternatives $A_i$ and $A_j$ is broken in favor of $A_i$ as long as $i<j$; otherwise, it is broken in favor of $A_j$.7

An economy in our model is defined by a list $(\mathbf{A},\mathbf{\Omega },q,\mathbf{T},r,n,\mathbf{C}$) that satisfies Equation (2). For any expected size of the population n, a strategy is a function ${\sigma }_n:\mathbf{T}\to \Delta (\mathbf{C})$, associating a voter type with a probability distribution over $\mathbf{C}$. Let ${\sigma }_n^C(t)\ge 0$ denote the probability that a voter of type t chooses action $C\in \mathbf{C}$. It has to be true that ${\sum }_{C\in \mathbf{C}}{\sigma }_n^c(t)=1\forall t\in \mathbf{T}$. We simplify the notation slightly by denoting with ${\sigma }_n^i(t)$ the probability that a voter chooses to vote for alternative $A_i$. Suppose that ${\sigma }_n^{*}$ is an equilibrium of the extended Poisson voting game with n expected voters.8 We are interested in limit equilibria ${\sigma }^{*}$ such that ${\sigma }_n^{*}\to {\sigma }^{*}$ as $n\to \infty$. Henceforth, we refer to a limit equilibrium whenever we use the term equilibrium.

We denote $P_{{\sigma }_n}(A_i|{\omega }_j)$ as the probability that alternative $A_i$ is elected in state ${\omega }_j$ given strategy ${\sigma }_n$.

Definition 1.

Informationally Efficient Equilibrium. A limit equilibrium ${\sigma }^{*}$ of an economy $\mathcal{E}$ is informationally efficient if it is true that $P_{{\sigma }^{*}}(A_i|{\omega }_i)=1\forall i$.

Notice that this condition is quite strong: It requires that the correct alternative is elected with probability tending to one in each and every state of nature. However, this is a common assumption in the previous literature and is often possible because of large numbers at least in one equilibrium (see the proof of Theorem 1).

We need a few additional definitions to distinguish between different types of equilibria. Following Feddersen and Pesendorfer [7], we call an equilibrium responsive, if voters “change their vote as a function of their private information with positive probability” (p. 26). So, we call a limit equilibrium responsive if ${\sigma }^{*}(t_i)\ne {\sigma }^{*}(t_j)$ for at least two different $t_i,t_j\in \mathbf{T}$, so that not all voter types vote exactly the same way. We call an equilibrium unresponsive, if all voter types vote the same way, independent of their type. More formally, we call a limit equilibrium unresponsive if ${\sigma }^{*}(t_i)={\sigma }^{*}(t_j)$ for all $t_i\ne t_j\in \mathbf{T}$.

The literature also tends to be interested in informative voting; i.e., voters voting for their type with probability one. If all types engage in informative voting, i.e., ${\sigma }^{i*}(t_i)=1$ for all i, we call the equilibrium informative.

2.2. Voter Behaviour

Before we can prove any results, we need to consider how rational voters vote in a large Poisson voting game. We will only present overarching principles of voter behaviour in this section and relegate the specifics to the respective proofs and the Appendix A. In the Appendix A, we present the two most important tools for the analysis of a Poisson voting game—the magnitude theorem and the offset theorem—from Myerson [1].

Recall that voters derive utility from the outcome of the election alone. A rational voter considers pivotal events in which his or her vote changes the outcome of the election from a less to a more preferred alternative. Pivotal events depend on the particular ballot that a voter wants to submit, and to some extent, on the tie-breaking rule. A voter who considers ballot $A_2$, for example, is pivotal if alternative $A_1$ has the same number of votes as alternative $A_2$ and both have at least as many votes as any other alternative. If a voter considers ballot $A_1$, on the other hand, the voter is pivotal if some alternative has one more vote than alternative $A_1$ and any other alternative is sufficiently behind.

Generally, $E_k^{ij}$ denotes the pivotal event in which one additional vote for alternative $A_i$ changes the outcome of the election from alternative $A_j$ to alternative $A_i$ in state ${\omega }_k$. And $pi{v}_k^{ij}$ denotes the probability of this pivotal event. The probability of a pivotal event depends on the underlying strategy. To save on notation, we will avoid this additional index, if it is not misleading.

Besides pivotal events, a rational voter also considers the posterior distribution of the states of nature conditional on his/her type. $q_i(t)$ denotes the posterior probability of state ${\omega }_i$ conditional on a voter being of type t.

If voter type t considers voting for alternative $A_i$ rather than abstaining, the expected utility gain can be written as

$$EU(A_i|t)=\sum _{j\ne i}[q_i(t)pi{v}_i^{ij}-q_j(t)pi{v}_j^{ij}].$$

(3)

If $EU(A_i|t)$ is larger than zero, the voter prefers voting for $A_i$ to abstaining. If the expected utility gain of some other alternative $A_j$ is larger, then the voter prefers voting for $A_j$ instead.

To discuss equilibrium strategies, we have to be able to evaluate equations such as Equation (3). In large elections, the probability of a pivotal event converges to zero, and so do entire equations, such as Equation (3). However, Myerson [1] shows that probabilities of pivotal events do not converge to zero at the same speed. Events with probabilities that converge to zero faster than others are infinitely less likely, and can therefore be ignored. The difference in the speeds of convergence of pivotal probabilities makes comparisons between expected utilities from different ballots meaningful.

Myerson [1] proposes the magnitude as a measure for the speed of convergence in a large Poisson game. The magnitude $\mu$ of the probability of a pivotal event is defined as

$$\mu (E_k^{ij})=li{m}_{n\to \infty }\frac{log(pi{v}_k^{ij})}{n}.$$

Events with larger magnitude converge slower than those with smaller magnitude, and are therefore, infinitely more likely. The magnitude of an event is either zero or negative and can be calculated by solving a maximisation problem. In the Appendix A, we discuss in a little more detail, how magnitudes and precise probabilities of pivotal events are calculated (magnitude theorem and offset theorem from Myerson [1]) for the cases arising in our proofs. For a more detailed discussion, we would like to refer the reader to Myerson [1] or to Goertz and Maniquet [13,15].

3. Results

First, we show the existence of an efficient limit equilibrium for any economy, independent of the number of alternatives. So, indeed, the CJT extends to economies with more than two alternatives.

Theorem 1.

There exists an informationally efficient limit equilibrium for any economy $\mathcal{E}$ with multiple alternatives if type distributions satisfy Equation (2).

Proof. 

The proof is divided into two steps. In step 1, we show that for any $\mathcal{E}$ with a finite number of alternatives that satisfies Equation (2), there exists a sequence of strategies ${\sigma }_n$ such that $li{m}_{n\to \infty }{P}_{{\sigma }_n}(A_i|{\omega }_i)=1$ for all i. To guarantee that this is true, it is sufficient (due to the law of large numbers) to verify that the expected fraction of votes for alternative $A_i$ in state ${\omega }_i$ is larger than the expected fraction of votes for each of the other alternatives in that state.9 In step 2, we deduce from step 1 that there exists a limit equilibrium ${\sigma }^{*}$ that aggregates information efficiently.

Step 1: Consider an economy $\mathcal{E}$ that satisfies Equation (2). Let ${ϵ}_1$, ${ϵ}_2$, ${ϵ}_3$, ...,${ϵ}_{|\mathbf{A}|}$ be small positive numbers such that

$${ϵ}_1{r}_1(t_1)={ϵ}_2{r}_2(t_2)=...={ϵ}_K{r}_{|\mathbf{A}|}(t_{|\mathbf{A}|}).$$

Consider a sequence of strategies in which for each $t_i\in \mathbf{T}$, ${\sigma }_n^i(t_i)={ϵ}_i$ and ${\sigma }_n^{\varphi }(t_i)=1-{ϵ}_i$; i.e., a voter of type $t_i$ mixes between voting for alternative $A_i$ and abstention. Denote by ${\lambda }_j^i$, the expected fraction of votes for alternative $A_i$ in state ${\omega }_j$. Given the strategy, the expected fractions of votes for the different alternatives in state ${\omega }_i$ are

$$\begin{array}{ccc}\hfill {\lambda }_i^i& =& r_i(t_i){ϵ}_i\hfill \\ \hfill {\lambda }_i^j& =& r_i(t_j){ϵ}_j\forall j\ne i.\hfill \end{array}$$

With Equation (2), it is true for each ${\omega }_i$ that ${\lambda }_i^i>{\lambda }_i^j$$\forall j\ne i$. So, by the law of large numbers, $li{m}_{n\to \infty }{P}_{{\sigma }_n}(A_i|{\omega }_i)=1$ for all i as $n\to \infty$. Notice that ${\sigma }_n$ does not depend on n.

Step 2: Let ${\sigma }_n^{*}$ be a sequence of strategies that maximises the ex-ante utility of the voters. Such strategies exist, as they maximise a continuous function on a compact set. We claim that they are equilibrium strategies. Indeed, the existence of a profitable deviation would contradict the fact that ${\sigma }_n^{*}(t)$ maximises expected utilities. Additionally, it is impossible that the expected utility from ${\sigma }_n^{*}(t)$ is lower than the expected utility from ${\sigma }_n(t)$ as defined above, because the expected utility of ${\sigma }_n(t)$ tends to 1. So, it has to be true that $P_{{\sigma }_n^{*}}(A_i|{\omega }_i)\to 1$ for all i as $n\to \infty$, and therefore, $li{m}_{n\to \infty }{\sigma }_n^{*}$ is a limit equilibrium that aggregates information efficiently.10 ☐

The CJT extends to all elections, independent of the number of alternatives. So, an efficient equilibrium also exists for the example in Goertz and Maniquet [11]. Our Theorem 1 and their example, together imply that efficient and inefficient equilibria may co-exist for the same economy/election. In Goertz and Maniquet [11], voters vote informatively in the inefficient equilibrium. Since it is only an example, we do not know whether this type of equilibrium exists for other economies as well. Our Theorem 2, however, establishes that a certain type of inefficient equilibrium exists for every economy, independent of the number of alternatives, if a certain condition on prior and posterior probabilities is satisfied. It is an unresponsive equilibrium in which all voters vote for the same alternative.

The existence of unresponsive equilibria in voting games is often thought of as trivially true. After all, who has an incentive to change their vote if—by construction—nobody is ever pivotal? Notice that this argument does not actually apply to Poisson voting games with population uncertainty. There is always a positive probability—however small—with which every single voter is the only one who shows up for the election. So, voting for the same alternative is not a weakly dominated strategy in a Poisson voting game. Theorem 2 shows that unresponsive equilibria exist only for certain elections that satisfy a condition on prior and posterior probabilities. Since voters do not play weakly dominated strategies, there is no reason to assume that they are less likely to coordinate on these types of equilibria than on other ones, especially other inefficient ones.

Theorem 2.

There exists an unresponsive limit equilibrium in which all voter types vote for the same alternative $A_i$ for economies that satisfy Equation (2) if and only if $\mathcal{E}$ is such that $\frac{r_i(t)}{r_j(t)}\ge \frac{q_j}{q_i}foralli\ne jandallt\in \mathbf{T}$.

Proof. 

Recall that each voter has a positive probability of being pivotal. So, we have to check whether voting for $A_i$ is the best response for a voter when pivotal. According to the statement of the theorem, ${\sigma }^i(t_j)=1\forall j$, so that in all states of nature ${\omega }_j$, ${\lambda }_j^i=1$, and ${\lambda }_j^k=0$$\forall j$ and $\forall k\ne i$. This means that each respective pivotal event has the same probability in each state of nature. The pivotal events depend on which alternative is voted for, so let us consider the two possible cases one after the other.

Case 1: Suppose that $i=1$. If all voters vote for alternative $A_1$, there is no pivotal event in which a single voter can change the outcome of the election from another alternative to $A_1$. So, $EU(A_1|t)=0$. For any other alternative $A_j$, a voter is pivotal if no other voter shows up for voting. The probability of that event (avoiding unnecessary notation) is $pi{v}^0=e^{-n}$ (with Equation (1) and $N=0$). With $EU(A_j|t)=q_j(t)pi{v}^0-q_1(t)pi{v}^0$, voting $A_1$ is a best response for all t if and only if $q_j(t)\le q_1(t)$, or if $\frac{r_i(t)}{r_j(t)}\ge \frac{q_j}{q_i}$, for all $j\ne i$.

Case 2: Suppose that $i>1$. A voter is pivotal for $A_i$ if no other voter shows up and $A_1$ wins otherwise. So, $EU(A_i|t)=q_i(t)pi{v}^0-q_1(t)pi{v}^0$. A voter may deviate by voting for $A_j$ with either $j<i$ or $j>i$. If $j>i$, then $EU(A_j|t)=q_j(t)pi{v}^0-q_1(t)pi{v}^0$. So, voting for $A_i$ is a best response if and only if $q_i(t)\ge q_j(t)$ for all t and all $j>i$.

If $j<i$, a voter is pivotal if either no voter shows up (in which case $A_1$ wins otherwise), or if one other voter shows up (in which case $A_i$ wins otherwise). The probability of one other voter showing up is $pi{v}^1=e^{-n}n$ (again, with Equation (1) and $N=1$). So, $EU(A_j|t)=q_j(t)pi{v}^0-q_1(t)pi{v}^0+q_j(t)pi{v}^1-q_i(t)pi{v}^1$. Voting for $A_i$ i a best response for all t if $pi{v}^0(q_i(t)-q_1(t))\ge q_j(t)pi{v}^0-q_1(t)pi{v}^0+q_j(t)pi{v}^1-q_i(t)pi{v}^1$. Notice that $li{m}_{n\to \infty }\frac{pi{v}^0}{pi{v}^1}=0$. 11 So, in the limit, the comparison of utility gains reduces to $0\ge q_j(t)-q_i(t)$. Voting for $A_i$ is a best response if and only if $q_i(t)\ge q_j(t)$ for all t and all $j<i$.

All unresponsive equilibria are, of course, inefficient because the same alternative is elected in all states of nature.

One of the examples in Myerson [2] satisfies the condition in Theorem 2; the other one does not. This is the reason that an unresponsive equilibrium exists for one of the examples, but not for the other. Myerson moves from one example to the other by changing the prior probabilities. However, Theorem 2 shows that the existence of an unresponsive equilibrium actually depends on the interplay between prior and type distributions.

In fact, the condition in Theorem 2 has quite an intuitive interpretation under which the voting behaviour appears to be natural: It implies that all voters consider the same state of nature the most likely, conditional on their type. The private information is not particularly informative, relative to the prior distribution of states.

We conclude that inefficiencies can arise in any election with two or more alternatives if types are not particularly informative. In such an equilibrium, the inefficiency is caused by what we call ‘information failures’ because the voters’ private information is not particularly informative.

In what follows, we assume away such information failures by focusing on economies that do not satisfy the condition in Theorem 2. The “informative types” condition assures that not all voter types consider the same state of nature as the most likely, conditional on their type.

Definition 2.

Informative Types. For all ${\omega }_i$ for which there exists one $t^{\prime }$ such that $q_i(t)\ge q_j(t)\forall j,j\ne i,$ and all $t\in \mathbf{T}\setminus t^{\prime }$, there exists at least one ${\omega }_k$ such that $q_i(t^{\prime })<q_k(t^{\prime })$.

So far, we know that an efficient equilibrium exists for each economy, independent of the number of alternatives. If types are un-informative, an unresponsive inefficient equilibrium exists as well. So, for economies with un-informative types, efficient and inefficient equilibria co-exist.

Theorem 3 shows that there is an important difference between economies with two and with more than two alternatives. If types are informative, all equilibria in two-alternative elections are efficient.

Theorem 3.

For any $\mathcal{E}$ with $|\mathbf{A}|=2$ that satisfies Equation (2) and has informative types, all limit equilibria are efficient.

Proof. 

Consider any economy with $|\mathbf{A}|=2$ that has informative types. For alternative $A_i$ and type t, we can write $EU(A_i|t)=q_i(t)pi{v}_i^{ij}-q_j(t)pi{v}_j^{ij}$. Because of Equation (2) and Bayes’ Rule, $q_i(t_i)>q_i(t_j)$ and $q_j(t_i)<q_j(t_j)$. So, we can immediately conclude that type $t_i$ is at least as likely to vote for $A_i$ as type $t_j$ and that type $t_j$ is at least as likely to vote for $A_j$ as type $t_i$ (this is true in any two-alternative election). In addition, at most one of the two types mixes. Because types are informative, it cannot be the case that both types vote for the same alternative with probability one. So, with informative types and in a limit equilibrium, type $t_i$ is, in fact, more likely to vote for alternative $A_i$ than type $t_j$ (${\sigma }^{*i}(t_i)>{\sigma }^{*i}(t_j)$) and type $t_j$ is more likely to vote for alternative $A_j$ than type $t_i$ (${\sigma }^{*j}(t_j)>{\sigma }^{*j}(t_i)$). This does not, however, immediately imply that the correct alternative necessarily receives more votes in each state of nature because type $t_i$ is not necessarily the most likely type in ${\omega }_i$ (see Equation (2)).

Let us consider a limit equilibrium that is inefficient because alternative $A_j$ does not win in ${\omega }_j$. Because of our special tie-breaking rule, this can mean either that $A_i$ and $A_j$ receive the same expected vote share in ${\omega }_j$ (if $j>1$) or that $A_i$ receives a larger expected vote share than $A_j$ in ${\omega }_j$ (if $j=1$). We do not need to make a distinction between those two cases because the proof can be one in one step.

The difference in expected vote shares in the two different states of nature can be written as:

$$\begin{array}{ccc}\hfill {\lambda }_i^i-{\lambda }_i^j& =& r_i(t_i)({\sigma }^{*i}(t_i)-{\sigma }^{*j}(t_i))+r_i(t_j)({\sigma }^{*i}(t_j)-{\sigma }^{*j}(t_j))\hfill \\ \hfill {\lambda }_j^i-{\lambda }_j^j& =& r_j(t_i)({\sigma }^{*i}(t_i)-{\sigma }^{*j}(t_i))+r_j(t_j)({\sigma }^{*i}(t_j)-{\sigma }^{*j}(t_j)).\hfill \end{array}$$

So, if the equilibrium is inefficient because alternative $A_j$ does not win (in expected terms), in state ${\omega }_j$, it must be true that ${\lambda }_j^i-{\lambda }_j^j\ge 0$ (equal to or larger than 0 depending on our tie-breaking rule and the respective alternatives). From above, we know that ${\sigma }^{*i}(t_i)>{\sigma }^{*i}(t_j)$ and that ${\sigma }^{*j}(t_j)>{\sigma }^{*j}(t_i)$. Combining these, we also have $({\sigma }^{*i}(t_i)-{\sigma }^{*j}(t_i))>({\sigma }^{*i}(t_j)-{\sigma }^{*j}(t_j))$. Now, we can compare the expected vote share difference in the two states of nature and must come to the conclusion that if ${\lambda }_j^i-{\lambda }_j^j\ge 0$, then it must also be true that ${\lambda }_i^i-{\lambda }_i^j>{\lambda }_j^i-{\lambda }_j^j$. So, $\mu (E_j^{ji})>\mu (E_i^{ji})$ (see the magnitude theorem in the Appendix A). Then, however, both types should only vote for alternative $A_j$, a contradiction to above. So, an inefficient equilibrium does not exist if types are informative. ☐

In two-alternative elections, inefficiencies only arise because of ‘information failures’; i.e., uninformative types. If types are informative, then there exists a mechanism that leads to an efficient equilibrium: if the expected vote shares are inefficient (and not all voters vote for the same alternative), the pivotal event in the ’inefficient’ state always has the larger magnitude, so that all voter types would prefer to vote more for the correct alternative that is now losing. Inefficient vote shares are not stable as an equilibrium.

Unfortunately, this mechanism does not work in elections with more alternatives. It is possible, as the example in Goertz and Maniquet [11] shows, that pivotal events in the inefficient state do not have the largest magnitude, so they are infinitely less likely than pivotal events in the other state(s). As a result, voters do not have an incentive to vote more for the losing alternative and it is not elected when it should be.

In elections with more than two alternatives, inefficient equilibria can arise due to ’information failures’ (when types are uninformative), but also due to ’coordination failures’ as in the example of Goertz and Maniquet [11].

One “unfortunate” feature of the example in Goertz and Maniquet [11] is that voters vote informatively—a particularly intuitive voting strategy that is maybe more likely to occur than other types of more complicated strategies—but that the resulting equilibrium is inefficient. The interested reader may wonder whether this is typical in elections with more than two alternatives. This, however, is not the case. If one changes the example slightly—the type distribution in state ${\omega }_1$ to $r_1(t_1)=0.4$, $r_1(t_2)=r_1(t_3)=0.3$—then informative voting is an equilibrium and is efficient. While informative voting may nor may not be efficient in elections with more than two alternatives, the existence of responsive inefficient equilibria is generic (see Goertz and Maniquet [11] for a more detailed discussion).

4. Conclusions

We have proven a Condorcet jury theorem (CJT) for elections with common interests and uncertainty about the state of nature and about the number of players in the election (Poisson voting game): There always exists an efficient equilibrium, independent of the number of alternatives. We have also established the “informative-types” condition: When types are not informative, each of these elections also has an inefficient equilibrium in which all voters vote for the same alternative. Recall that voters in a Poisson voting game do not play weakly dominated strategies in these inefficient equilibria, so that they cannot be eliminated with the typical equilibrium-selection arguments.

If types are informative, we can also show that all equilibria in elections with two alternatives are efficient. As such, two-alternative elections remain “quite a different animal” than elections with more than two alternatives.12 The mechanism that drives expected vote shares towards efficiency when types are informative does not work in elections with more than two alternatives. While we can show that these elections always have an efficient equilibrium, we know from previous work that they may also have inefficient ones even if types are informative.

Of course, general results about the co-existence of efficient and inefficient equilibria in elections with more than two alternatives (and constructive proofs that show the equilibrium strategies explicitly) would be very desirable. We have to agree with the previous literature that this seems, currently, beyond reach. We will just shortly outline the argument and would like to refer the interested reader to Goertz and Maniquet [13] for a more thorough discussion of this (although their model includes partisans and is more concerned with approval voting than the simple plurality rule).

Myerson [2], among others, shows that strategies in two-alternative elections are step strategies (i.e., in some sense monotonous in the voter’s type). In two-alternative elections, a voter of type $t_j$ is always at least as likely to vote for alternative $A_j$ as a voter of type $t_i$.13 This also implies that only a small set of strategies qualify as possible equilibria: (1) both voter types vote for the same alternative, or (2) one voter type mixes between the two alternatives and the other voter type votes informatively. So, it is not very hard to narrow down the set of equilibria precisely for a two-alternative election.

Unfortunately, this monotonicity argument no longer holds in elections with more than two alternatives.14 So, we have nothing other than trial-and-error to guide us when finding possible equilibria for a particular election. This is the reason that some of the recent important findings were presented only as examples (e.g., Goertz and Maniquet [13]’s Theorem 2 and the example in Goertz and Maniquet [11]). The example in Goertz and Maniquet [11], in which voters vote informatively, shows that inefficient equilibria in elections with more than two alternatives are generic and can also be quite stable (i.e., cannot easily be ruled out with the typical equilibrium selection arguments).

There are other previous works that have investigated other rules than the simple plurality rule for common-interest elections with more than two alternatives and found that there are other rules that may perform better than the simple plurality rule (e.g., Ahn and Oliveros [16] and Goertz and Maniquet [11] found that approval voting has some desirable properties). However, the simple plurality rule remains the most common voting rule in reality. Thus, our focus on that particular rule.