How frequently do different voting rules encounter voting paradoxes in three-candidate elections?

Abstract

We estimate the frequencies with which ten voting anomalies (ties and nine voting paradoxes) occur under 14 voting rules, using a statistical model that simulates voting situations that follow the same distribution as voting situations in actual elections. Thus the frequencies that we estimate from our simulated data are likely to be very close to the frequencies that would be observed in actual three-candidate elections. We find that two Condorcet-consistent voting rules do, the Black rule and the Nanson rule, encounter most paradoxes and ties less frequently than the other rules do, especially in elections with few voters. The Bucklin rule, the Plurality rule, and the Anti-plurality rule tend to perform worse than the other eleven rules, especially when the number of voters becomes large.

Appendices

Appendix 1: Simulating elections with the spatial model

1.1 A.1.1 A spatial model of vote-share probabilities

The spatial model of voting represents a conceptualization of the distribution of voter ideal points and the locations of candidates in candidate space. Good and Tideman (1976) describe the philosophy and the technical details of the spatial model that we use; here we summarize its essential elements. Consider the case of $m\,=\,3$ candidates. There are two sub-cases to be considered. The first is that the three candidates are collinear, that is, all three lie in a single line. In this case, the candidate that is in the middle will not be the last choice of any voter. Since every candidate is the last choice of some voters in the vast preponderance of the results for elections with three candidates that we have seen, we neglect this case and consider the more general case in which the positions of three candidates are not collinear. In this case the locations of the three candidates define a plane, namely the plane that we call candidate space. Label the candidates $A,\,B$, and $C$, so that the 3! = 6 different strict rankings are ABC, ACB, CAB, CBA, BCA, and BAC. The six strict rankings form a voting situation that describes how many ballots voters have cast for each of the six rankings.

We assume that the locations of voter ideal points in candidate space follow a circularly symmetric bivariate normal distribution. The probability that a voter chosen at random will place the three candidates in a particular order (e.g., ABC) is the integral of the density function of voter ideal points over the portion of the plane where the order of the distances to the three candidates is the order prescribed by the ranking whose probability is being calculated. The portions of the plane assigned to the six rankings of the three candidates are delineated by the perpendicular bisectors of the line segments connecting pairs of candidates. Figure 3 shows, for specific locations of the three candidates, the division of the candidate plane into the six sectors that define the six rankings. The figure also shows the volume of the triangular wedge (the integral under the spherical multivariate normal distribution centered at $O)$ whose integral describes the probability that a randomly chosen voter ranks the three candidates in the order ABC.³⁴

While there are six degrees of freedom in the locations of the three candidates, there are only four degrees of freedom in the probabilities determined by the integrals. One degree of freedom is lost because, if the positions of the three candidates are rotated around the mode of the distribution of voter ideal points, the pattern of perpendicular bisectors rotates by the same amount, and the integrals are unchanged. A second degree of freedom is lost because, if the locations of the three candidates are moved the same distance toward or away from the point that is equidistant from them, the perpendicular bisectors and hence the integrals are unchanged.

There are different ways to use the remaining four degrees of freedom to describe the structure of the spatial model. We found that the following parameterization makes it easiest to simulate rankings that are very similar to the voting situations that we observe in actual elections. We first place the intersection of the three perpendicular bisectors of the triangle formed by the three candidates (that is, the triangle’s circumcenter $T$) at the origin of a coordinate system, and we rotate the coordinate system so that the center of voter ideal points $O$ is on its horizontal axis. We then use the first degree of freedom for the horizontal distance OT, and the remaining three degrees of freedom for the angles formed by the line $\overline{OT}$ and the three perpendicular bisectors that describe the boundaries between pairs of the six strict rankings. To simulate data with the spatial model, we adopt distributions for the parameters based on the analysis of real voting data in Plassmann and Tideman (2011). In particular, we assume that $\overline{OT}$ follows a Weibull distribution with scale parameter $\alpha \,=\,0.6858$ and shape parameter $\beta \,=\,2.4608$, and that the three angles follow a Dirichlet distribution with parameters $\delta _{1}\,=\,26.47,\,\delta _{2}\, =\,23.37$, and $\delta _{3}~=~23.65$.³⁵

1.2 A.1.2 A model of vote-casting outcomes

The spatial model yields a vote share probability $p_{r}$ for each of the six strict rankings $r$, with $\Sigma p_{r}~=~1$. To complete the model of vote-casting, it is necessary to describe the relationship between the vote share probabilities (six real numbers) and the number of ballots that $n$ voters submit for each of the six rankings (six integers). We assume that the $p_{r}$s represent the expected shares of the ballots. We also assume that the distribution of the vector of $p_{r}$s does not vary with the size of the electorate. To formalize these assumptions, let $n_{r}$ be the number of ballots submitted for ranking $r$, with $\Sigma n_{r}=n$ and $r=1,\ldots ,6$, and let $N_{r}$ be a random variable that describes the distribution of $n_{r}$. If voters cast their ballots independently of each other, then in a given election in which the expected vote shares are given by a vector of $p_{r}$s, the six $N_{r}$s follow a multinomial distribution with $E[N_{r}]= np_{r}$ (and thus $E[{N_r /n}]=p_r),\,Var[N_{r}]=np_{r}\,(1 - p_{r})$, and Cov [$N_{r},\,N_{s}]=-np_{r}p_{s}$. One way of accommodating dependent ballots is to assume that the six $n_{r}$s follow a multinomial-Dirichlet distribution with $E[N_{r}]=np_{r},\,Var[N_{r}]=np_{r}(1 - p_{r})\Psi$, and $Cov[N_{r},\,N_{s}]=- np_{r}p_{s}\Psi$, where $\Psi = (\psi +n)/(\psi + 1)$. The parameter $\psi$ describes the additional variance that is introduced through the Dirichlet distribution; in the limit as $\psi$ approaches infinity, the multinomial-Dirichlet distribution approaches the multinomial distribution. Plassmann and Tideman (2011) show that the multinomial distribution leads to less variation among simulated ballots than what is observed in ballots from actual elections, while the variation among ballots simulated with a multinomial-Dirichlet distribution with $\psi ~=~330$ is very close to the variation among observed ballots. Thus a specified distribution for the $p_{r}$s leads to a distribution of election outcomes, for any size of electorate, that can be parameterized and simulated.

1.3 A.1.3 A model of voter behavior versus a model of vote-casting outcomes

The spatial model can be interpreted either as a model of voter behavior or as part of a model of vote-casting outcomes. It is important to distinguish the two interpretations. When viewed as a model of voter behavior, the task would be to identify the positions of the ideal points in actual elections. Knowledge of these positions would explain why one observes a particular voting situation. The key question would be whether observed voting situation can be interpreted as revealed preferences that provide information about the positions of the ideal points. This question has received considerable attention in the literature; Henry and Mourifié (2011) assess and reject the validity of the spatial model of voting when it is interpreted as a model of voter behavior.

In contrast, when the spatial model serves as a model of vote share probabilities and thus as part of a model of vote-casting outcomes, the focus is on the distribution of voting situations rather than the positions of ideal points. The task is to parameterize the spatial model so that the model of vote-casting outcomes yields a distribution of voting situations that corresponds to the distribution of observed voting situations. Knowledge of the positions of ideal points is not required for this task. Thus our work is neither related to nor affected by the literature on revealed preferences. We adopt the spatial model as a model of vote-casting outcomes solely because, among all contenders for a model of vote share probabilities of which we are aware, it comes closest to describing the distribution of observed voting situations, not because we want to defend it as a model of voter behavior.

Appendix 2: Estimating the frequencies of five voting paradoxes

Let “voting rule $X$” be the voting rule for which the frequency of occurrence of a paradox is to be determined. Let $P$ be the number of occurrences of the paradox, where $P\,=\,0$ before the first voting situation is drawn. Let $N$ be the number of voting situations to be drawn.

1.1 A.2.1 The strong lack of monotonicity paradox

• 1. Use the calibrated spatial model to draw a voting situation $R$.

• 2. Use voting rule $X$ to determine the winner of $R$.

• 3. There are four pairs of rankings across which the winner is ranked in consecutive positions.³⁶ Consider each of these four pairs in turn, starting with the first pair. If the ranking in which the winner is ranked lower has at least one vote, then decrease the number of votes for that ranking by one and increase the number of votes for the other ranking by 1. Denote the new voting situation by $R^\prime$. If the ranking in which the winner is ranked lower has no votes, then continue with the next pair of rankings. If this was the fourth pair of rankings, then continue with Step 1.

• 4. Use voting rule $X$ to determine the winner of $R^\prime$.

• 5a. (Ignoring ties) If voting rule $X$ chooses different winners for $R$ and $R^\prime$, then increment $P$ by 1. Continue with Step 1.

• 5b. (Accounting for ties) If voting rule $X$ chooses either (1) different winners for $R$ and $R^\prime$, or (2) the same winner for $R$ and $R^\prime$ and the winner is unique for $R$ but not unique for $R^\prime$, then increment $P$ by 1. Continue with Step 1.

• 6. If $P$ was not incremented in Step 5, then the following must be true: voting rule $X$ chooses the same winner for the two voting situations. Continue with Step 3 with a different pair of rankings. After the fourth pair, continue with Step 1.

• 7. Determine the frequency of occurrence as $P/N$.

1.2 A.2.2 The strong truncation paradox

Steps 1–2 and 4–7 are the same as in A.2.1

• 3. Consider each of the six possible strict rankings in turn, starting with the first ranking. Denote this ranking by $r$. If ranking $r$ has at least one vote, then decrease the number of votes for ranking $r$ by one-half. Identify the ranking $r^\prime$ that has the same highest-ranked candidate as ranking $r$, and increase the number of votes for ranking $r^\prime$ by one-half.³⁷ Denote the new voting situation by $R^\prime$. If ranking $r$ has no votes, then continue with the next ranking. If ranking $r$ is the sixth ranking, then continue with Step 1.

1.3 A.2.3 The strong no-show paradox

Steps 1–2 and 4–7 are the same as in A.2.1

• 3. Consider each of the six possible strict rankings in turn, starting with the first ranking. Denote this ranking by $r$. If ranking $r$ has at least one vote, then decrease the number of votes for ranking $r$ by one. Denote the new voting situation by $R^\prime$. If ranking $r$ has no votes, then continue with the next ranking. If ranking $r$ is the sixth ranking, then continue with Step 1.

1.4 A.2.4 Violations of the Subset Choice Consistency (SCC) Condition

Steps 1–2 are the same as in A.2.1

• 3. If the winner of voting situation $R$ is not unique, then continue with Step 1.³⁸

• 4. Consider, in turn, each of the two candidates who did not win, starting with the first of these candidates. Eliminate this candidate. Use $R$ to assemble the voting situation $R^{\mathrm{short}}$ for the ensuing two-candidate election.

• 5. Use voting rule $X$ to determine the winner for $R^{\mathrm{short}}$ (this is the majority rule winner for all 14 voting rules that we consider).

• 6a. (Ignoring ties) IF the winner for $R^{\mathrm{short}}$ is not the same as the winner for $R$, THEN increment $P$ by 1. Continue with Step 1.

• 6b. (Accounting for ties) IF (1) the winner for $R^{\mathrm{short}}$ is not the same as the winner for $R$, or (2) the winner for $R^{\mathrm{short}}$ is not unique (there is a two way tie that involves the winner of $R)$, THEN increment $P$ by 1. Continue with Step 1.

• 7. If $P$ was not incremented in Step 6, then the following must be true: the winner for $R^{\mathrm{short}}$ is the same as the winner for $R$. Continue with Step 4 with the other candidate. After the second candidate, continue with Step 1.

• 8. Determine the frequency of occurrence as $P/N$.

1.5 A.2.5 The reinforcement paradox

We examine only the version of the reinforcement paradox for which the two elections under consideration either have the same number of voters or have a number of voters that differs by exactly 1 across the elections. When sampling the elections for the results reported in Table 12, we assumed that the number of voters in each of the two individual elections sum to the number of voters listed in the header of each column.

• 1. Use the spatial model to draw two voting situations, $R$ and $R^\prime$.

• 2. Use voting rule $X$ to determine the winners for $R$ and $R^\prime$.

• 3. If the winners of $R$ and $R^\prime$ are not the same, then continue with Step 1.

• 4. Combine $R$ and $R^\prime$ to form a new voting situation $R^{\prime \prime }$.

• 5. Use voting rule $X$ to determine the winner for $R^{\prime \prime }$.

• 6. EITHER the winners for $R,\,R^\prime$, and $R^{\prime \prime }$ are unique, OR some/all winners are not unique but one decides to ignore ties. One of the following must be true:

  • 6a. The winner for $R^{\prime \prime }$ IS NOT the same as the winner for $R$ and $R^\prime$. If true, then increment $P$ by 1 and continue with Step 1.

  • 6b. The winner for $R^{\prime \prime }$ IS the same as the winner for $R$ and $R^\prime$. If true, then continue with Step 1.

• 7. Reaching Step 7 implies that the winners for $R,\,R^\prime$, and $R^{\prime \prime }$ are not unique and that one decides not to ignore ties. One of the following must be true:

  • 7a. The winner of $R^{\prime \prime }$ is tied with another candidate and the winners of $R$ and/or $R^\prime$ are tied with the same candidate. If true, then continue with Step 1.

  • 7b. The winner of $R^{\prime \prime }$ is tied with another candidate and the winners of $R$ and/or $R^\prime$ are either not tied or tied with a different candidate. If true, then increment $P$ by 1 and continue with Step 1.

  • 7c. The winner of $R^{\prime \prime }$ is not tied and the winners of $R$ and/or $R^\prime$ are tied with another candidate. If true, then increment $P$ by 1 and continue with Step 1.

• 8. Determine the frequency of occurrence as $P/N$.