The $q$-majority efficiency of positional rules

Abstract

According to a given quota $q$, a candidate $a$ is beaten by another candidate $b$ if at least a proportion of $q$ individuals prefer $b$ to $a$. The $q$-majority efficiency of a voting rule is the probability that the rule selects a candidate who is never beaten under the $q$-majority, given that such a candidate exits. Closed form representations are obtained for the $q$-majority efficiency of positional rules (simple and sequential) in three-candidate elections. It turns out that the $q$-majority efficiency is: (i) significantly greater for sequential rules than for simple positional rules; and (ii) very close to the $q$-Condorcet efficiency, the conditional probability that a positional rule will elect the candidate who beats all others under the $q$-majority, when one exists.

Appendices

Appendix 1: Infinite case: formulae for $q$-Condorcet efficiency

The $q$-Condorcet efficiency for a given simple positional rule $F_\alpha$, a given $q$ and when a given $n$ tends to infinity, will be denoted $CE(F_{\alpha },n,q)$, while it will be denoted $CE(\overline{F}_{\alpha },n,q)$ for a given sequential positional rule.

We first begin with the simple positional rules.

Proposition 13

Let $F_{_{0}}$ be the PR. Then for large electorates,

$$\begin{aligned} CE\left( F_{0},\infty ,q\right) ={\left\{ \begin{array}{ll} -\frac{1}{135}\frac{27q^{5}+675q^{4}-1890q^{3}+1860q^{2}-775q+112}{(q-1)^{4}q} &{} \mathrm{if}\; \dfrac{{\scriptstyle 1}}{{\scriptstyle 2}}<q\le \dfrac{{\scriptstyle 2}}{{\scriptstyle 3}}\\ 1 &{} \mathrm{if}\; \dfrac{{\scriptstyle 2}}{{\scriptstyle 3}}\le q\le 1 \end{array}\right. } \end{aligned}$$

Proposition 14

Let $F_{_{1}}$ be the PR. Then, for large electorates,

$$\begin{aligned} CE\left( F_{1},\infty ,q\right) ={\left\{ \begin{array}{ll} \frac{1}{135}\frac{27q^{5}-135q^{4}+270q^{3}-240q^{2}+85q-6}{(q^{4}-4q^{3}+6q^{2}-4q+1)q} &{} \mathrm{if}\; \dfrac{{\scriptstyle 1}}{{\scriptstyle 2}}<q\le \dfrac{{\scriptstyle 2}}{{\scriptstyle 3}}\\ \frac{7q-2}{5q} &{} \mathrm{if}\; \dfrac{{\scriptstyle 2}}{{\scriptstyle 3}}\le q\le 1 \end{array}\right. } \end{aligned}$$

Proposition 15

Let $F_{_{2}}$ be the PR. Then, for large electorates,

$$\begin{aligned} CE\left( F_{2},\infty ,q\right) ={\left\{ \begin{array}{ll} \frac{1}{45}\frac{1017q^{5}-3420q^{4}+4590q^{3}-3060q^{2}+1005q-128}{(q^{4}-4q^{3}+6q^{2}-4q+1)q} &{} \mathrm{if}\; \dfrac{{\scriptstyle 1}}{{\scriptstyle 2}}<q\le \dfrac{{\scriptstyle 2}}{{\scriptstyle 3}}\\ 1 &{} \mathrm{if}\; \dfrac{{\scriptstyle 2}}{{\scriptstyle 3}}\le q\le 1 \end{array}\right. } \end{aligned}$$

We now consider the sequential positional rules.

Proposition 16

Let $\overline{F}_{0}$ be the SPR. Then, for large electorates,

$$\begin{aligned} CE\left( \overline{F}_{0},\infty ,q\right) ={\left\{ \begin{array}{ll} -\frac{1}{540}\frac{999q^{5}-2025q^{4}+1080q^{3}+120q^{2}-140q-16}{(q^{4}-4q^{3}+6q^{2}-4q+1)q} &{} \mathrm{if}\; \dfrac{{\scriptstyle 1}}{{\scriptstyle 2}}<q\le \dfrac{{\scriptstyle 2}}{{\scriptstyle 3}}\\ 1 &{} \mathrm{if}\; \dfrac{{\scriptstyle 2}}{{\scriptstyle 3}}\le q\le 1 \end{array}\right. } \end{aligned}$$

Proposition 17

Let $\overline{F}_{1}$ be the SPR. Then, for large electorates,

$$\begin{aligned} CE\left( \overline{F}_{1},\infty ,q\right) ={\left\{ \begin{array}{ll} \frac{1}{135}\frac{297q^{5}-1215q^{4}+1890q^{3}-1380q^{2}+455q-48}{(q^{4}-4q^{3}+6q^{2}-4q+1)q} &{} \mathrm{if}\; \dfrac{{\scriptstyle 1}}{{\scriptstyle 2}}<q\le \dfrac{{\scriptstyle 2}}{{\scriptstyle 3}}\\ 1 &{} \mathrm{if}\; \dfrac{{\scriptstyle 2}}{{\scriptstyle 3}}\le q\le 1 \end{array}\right. } \end{aligned}$$

Appendix 2: Some technical details

In this section, we give an outline of the techniques used to evaluate the $q$-majority efficiency. Firstly, we explain the computerized evaluation method. Secondly, the systems of inequalities are given both for simple positional rules and sequential positional rules.

1.1 Computerized evaluation method

In order to derive our formulae, we use the technique whose origins are in Gehrlein and Fishburn (1976) and more specifically based on Ehrhart polynomials (see for example Huang and Chua 2000 or Gehrlein 2002). The whole derivation process has been written as a computer program (available from the authors upon simple request) from which we obtain the total number of anonymous profiles that satisfy each set of linear inequalities obtained. Roughly, the total number of anonymous profiles, that satisfy each finite set of linear inequalities with integer coefficients, can be represented by a polynomial in $n$ and $p$ with periodic coefficients. This is in fact a well-established result in the literature on counting the number of integer points in a lattice polytope (see Ehrhart 1962; Huang and Chua 2000 or Lepelley et al. 2008). Lepelley et al. (2008) provide a pedagogical presentation of several computation approaches (or algorithms) while Gehrlein and Lepelley (1999) contains a detailed application of the specific method applied here.

1.2 Positional rules

In an election with three candidates $(a,b,c)$ and $n$ voters, the six possible preference rankings are given by

$$\begin{aligned} \begin{array}{cccccc} n_{1} &{}\quad n_{2} &{}\quad n_{3} &{}\quad n_{4} &{}\quad n_{5} &{}\quad n_{6}\\ a &{}\quad a &{}\quad b &{}\quad b &{}\quad c &{}\quad c\\ b &{}\quad c &{}\quad a &{}\quad c &{}\quad a &{}\quad b\\ c &{}\quad b &{}\quad c &{}\quad a &{}\quad b &{}\quad a \end{array} \end{aligned}$$

A voting situation $\widetilde{n}$ is represented as $\widetilde{n}=\left( n_{1},n_{2},n_{3},n_{4},n_{5},n_{6}\right)$.

Let $VS\left( B,F_{\alpha },q\right)$ be the set of all voting situations $\widetilde{n}$ at which no candidate in the winning set $B$ ($F_{\alpha }\left( \widetilde{n}\right) =B)$ is $q-$majority beaten. Similarly, given a subset $C$ of candidate, $VS(C,q)$ is the set of voting situations with $n$ voters at which no candidate in C is q-majority beaten.

The $q$-majority efficiency of a given positional rules consists to evaluate the following ratio :

$$\begin{aligned} \frac{\left| VS\left( F_{\alpha },q\right) \right| }{\left| VS(q)\right| } \end{aligned}$$

where :

• (i) $VS(q)$ is the set of all voting situations at which there exists at least a candidate who is not q-majority beaten. Note that

  $$\begin{aligned} VS\left( q\right)&= \left| VS\left( \left\{ a\right\} ,q\right) \cup VS\left( \left\{ b\right\} ,q\right) \cup VS\left( \left\{ c\right\} ,q\right) \right| \\&= 3\left| VS\left( \left\{ a\right\} ,q\right) \right| -3\left| VS\left( \left\{ a,b\right\} ,q\right) \right| +\left| VS\left( \left\{ a,b,c\right\} ,q\right) \right| \end{aligned}$$

• (ii) and $VS\left( F_{\alpha },q\right)$ is the set of all voting situations at which no candidate in the winning set of $F_{\alpha }$ is $q-$majority beaten. Note that

  $$\begin{aligned} VS\left( F_{\alpha },q\right)&= VS\left( \left\{ a\right\} ,F_{\alpha },q\right) \cup VS\left( \left\{ b\right\} ,F_{\alpha },q\right) \cup VS\left( \left\{ c\right\} ,F_{\alpha },q\right) \\&\cup VS\left( \left\{ a,b\right\} ,F_{\alpha },q\right) \cup VS\left( \left\{ a,c\right\} ,F_{\alpha },q\right) \cup VS\left( \left\{ b,c\right\} ,F_{\alpha },q\right) \\&\cup VS\left( \left\{ a,b,c\right\} ,F_{\alpha },q\right) \end{aligned}$$

Since $F_{\alpha }$ is a neutral social choice correspondence, it follows that

$$\begin{aligned} \left| VS\left( \left\{ a\right\} ,F_{\alpha },q\right) \right| =\left| VS\left( \left\{ b\right\} ,F_{\alpha },q\right) \right| =\left| VS\left( \left\{ c\right\} ,F_{\alpha },q\right) \right| \end{aligned}$$

and

$$\begin{aligned} \left| VS\left( \left\{ a,b\right\} ,F_{\alpha },q\right) \right| =\left| VS\left( \left\{ a,c\right\} ,F_{\alpha },q\right) \right| =\left| VS\left( \left\{ b,c\right\} ,F_{\alpha },q\right) \right| \end{aligned}$$

Thus,

$$\begin{aligned} \frac{\left| VS\left( F_{\alpha },q\right) \right| }{\left| VS(q)\right| }\!=\!\frac{3\left| VS\left( \left\{ a\right\} ,F_{\alpha },q\right) \right| \!+\!3\left| VS\left( \left\{ a,b\right\} ,F_{\alpha },q\right) \right| \!+\!\left| VS\left( \left\{ a,b,c\right\} ,F_{\alpha },q\right) \right| }{\left| VS(q)\right| } \end{aligned}$$

We then derive the following sets of constraints used to compute $\dfrac{\left| VS\left( F_{\alpha },q\right) \right| }{\left| VS(q)\right| }$

$$\begin{aligned} \widetilde{n}\in VS\left( \left\{ a\right\} ,F_{\alpha },q\right) \Longleftrightarrow \left\{ \begin{array}{l} n_{1}+n_{2}+\alpha \left( n_{3}+n_{5}\right) >n_{3}+n_{4}+\alpha \left( n_{1}+n_{6}\right) \\ n_{1}+n_{2}+\alpha \left( n_{3}+n_{5}\right) >n_{5}+n_{6}+\alpha \left( n_{2}+n_{4}\right) \\ n_{3}+n_{4}+n_{6}<qn\\ n_{4}+n_{5}+n_{6}<qn \end{array}\right. \\ \widetilde{n}\in VS\left( \left\{ a,b\right\} ,F_{\alpha },q\right) \Longleftrightarrow \left\{ \begin{array}{l} n_{1}+n_{2}+\alpha \left( n_{3}+n_{5}\right) =n_{3}+n_{4}+\alpha \left( n_{1}+n_{6}\right) \\ n_{1}+n_{2}+\alpha \left( n_{3}+n_{5}\right) >n_{5}+n_{6}+\alpha \left( n_{2}+n_{4}\right) \\ n_{3}+n_{4}+n_{6}<qn\\ n_{4}+n_{5}+n_{6}<qn\\ n_{1}+n_{2}+n_{5}<qn\\ n_{2}+n_{5}+n_{6}<qn \end{array}\right. \\ \widetilde{n}\in VS\left( \left\{ a,b,c\right\} ,F_{\alpha },q\right) \Longleftrightarrow \left\{ \begin{array}{l} n_{1}+n_{2}+\alpha \left( n_{3}+n_{5}\right) =n_{3}+n_{4}+\alpha \left( n_{1}+n_{6}\right) \\ n_{1}+n_{2}+\alpha \left( n_{3}+n_{5}\right) =n_{5}+n_{6}+\alpha \left( n_{2}+n_{4}\right) \\ n_{3}+n_{4}+n_{6}<qn\\ n_{4}+n_{5}+n_{6}<qn\\ n_{1}+n_{2}+n_{5}<qn\\ n_{2}+n_{5}+n_{6}<qn\\ n_{1}+n_{2}+n_{3}<qn\\ n_{1}+n_{3}+n_{4}<qn \end{array}\right. \end{aligned}$$

1.3 Sequential positional rules

Let $\overline{F}_{\alpha }$ be the sequential positional rule associated with the scoring vector $\left( 1,\alpha ,0\right)$. To derive the winning set under $\overline{F}_{\alpha }$ for a voting situation $\widetilde{n}$, the collection of scores recorded by candidates is ranked as a decreasing sequence $s_{1}$ (score of candidate $1$), $s_{2}$ and $s_{3}$ . Three cases arise: (i) if $s_{1}=s_{2}=s_{3}$, then $\overline{F}_{\alpha }\left( \widetilde{n}\right)$ is the set of all candidates; (ii) if $s_{1}>s_{2}=s_{3}$, then $\overline{F}_{\alpha }\left( \widetilde{n}\right)$ consists of the single candidate with the maximal score; (iii) if $s_{1}\ge s_{2}>s_{3}$, then the candidate with the smallest score is ruled out, and the two remaining candidates are opposed in a simple majority comparison and $\overline{F}_{\alpha }\left( \widetilde{n}\right)$ consists of the winner of the comparison or contains both candidates in case of a tie.

Denote by $VS\left( B,\overline{F}_{\alpha },q\right)$ the set of all voting situations $\widetilde{n}$ at which no candidate in the winning set $B$ ($\overline{F}_{\alpha }=B)$ is $q-$majority beaten. Therefore, we are interested in evaluating the frequency

$$\begin{aligned} \frac{\left| VS\left( \overline{F}_{\alpha },q\right) \right| }{\left| VS(q)\right| } \end{aligned}$$

where $VS\left( \overline{F}_{\alpha },q\right)$ is the set of all voting situations at which no candidate in the winning set of $\overline{F}_{\alpha }$ is $q-$majority beaten.

Since $\overline{F}_{\alpha }$ is also a neutral social choice correspondence.

$$\begin{aligned} \frac{\left| VS\left( \overline{F}_{\alpha },q\right) \right| }{\left| VS(q)\right| }\!=\!\frac{3\left| VS\left( \left\{ a\right\} ,\overline{F}_{\alpha },q\right) \right| +3\left| VS\left( \left\{ a,b\right\} ,\overline{F}_{\alpha },q\right) \right| +\left| VS\left( \left\{ a,b,c\right\} ,\overline{F}_{\alpha },q\right) \right| }{\left| VS(q)\right| } \end{aligned}$$

In order to use symmetries in computations, we rewrite $VS\left( \left\{ a\right\} ,\overline{F}_{\alpha },q\right)$ as

$$\begin{aligned} VS\left( \left\{ a\right\} ,\overline{F}_{\alpha },q\right) =VS_{\left\{ b,c\right\} }\left( \left\{ a\right\} ,\overline{F}_{\alpha },q\right) \cup VS_{\left\{ c\right\} }\left( \left\{ a\right\} ,\overline{F}_{\alpha },q\right) \cup VS_{\left\{ b\right\} }\left( \left\{ a\right\} ,\overline{F}_{\alpha },q\right) \end{aligned}$$

where $VS_{E}\left( \left\{ a\right\} ,\overline{F}_{\alpha },q\right)$ is the set of all voting situations at which $E$ is the set of candidates ruled out at the first step and $a$ is the unique winner. By neutrality,

$$\begin{aligned} \left| VS\left( \left\{ a\right\} ,\overline{F}_{\alpha },q\right) \right| =\left| VS_{\left\{ b,c\right\} }\left( \left\{ a\right\} ,\overline{F}_{\alpha },q\right) \right| +\left| VS_{\left\{ b\right\} }\left( \left\{ a\right\} ,\overline{F}_{\alpha },q\right) \right| \end{aligned}$$

and

$$\begin{aligned} \left| VS\left( \left\{ a\right\} ,\overline{F}_{\alpha },q\right) \right| =\left| VS_{\left\{ b,c\right\} }\left( \left\{ a\right\} ,\overline{F}_{\alpha },q\right) \right| +2\left| VS_{\left\{ b\right\} }\left( \left\{ a\right\} ,\overline{F}_{\alpha },q\right) \right| \end{aligned}$$

Finally, $\frac{\left| VS\left( \overline{F}_{\alpha },q\right) \right| }{\left| VS(q)\right| }$ equals

$$\begin{aligned} \frac{6\left| VS_{\left\{ c\right\} }\left( \left\{ a\right\} ,\overline{F}_{\alpha },q\right) \right| +3\left| VS_{\left\{ b,c\right\} }\left( \left\{ a\right\} ,\overline{F}_{\alpha },q\right) \right| +3\left| VS\left( \left\{ a,b\right\} ,\overline{F}_{\alpha },q\right) \right| +\left| VS\left( \left\{ a,b,c\right\} ,\overline{F}_{\alpha },q\right) \right| }{\left| VS(q)\right| } \end{aligned}$$

The sets of constraints used to compute $\dfrac{\left| VS\left( \overline{F}_{\alpha },q\right) \right| }{\left| VS(q)\right| }$ are obtained as follows

$$\begin{aligned} \widetilde{n}\in VS_{\left\{ b,c\right\} }\left( \left\{ a\right\} ,\overline{F}_{\alpha },q\right) \Longleftrightarrow \left\{ \begin{array}{l} n_{1}+n_{2}+\alpha \left( n_{3}+n_{5}\right) >n_{3}+n_{4}+\alpha \left( n_{1}+n_{6}\right) \\ n_{3}+n_{4}+\alpha \left( n_{1}+n_{6}\right) =n_{5}+n_{6}+\alpha \left( n_{2}+n_{4}\right) \\ n_{3}+n_{4}+n_{6}<qn\\ n_{4}+n_{5}+n_{6}<qn \end{array}\right. \\ \widetilde{n}\in VS_{\left\{ c\right\} }\left( \left\{ a\right\} ,\overline{F}_{\alpha },q\right) \Longleftrightarrow \left\{ \begin{array}{l} n_{1}+n_{2}+\alpha \left( n_{3}+n_{5}\right) >n_{5}+n_{6}+\alpha \left( n_{2}+n_{4}\right) \\ n_{3}+n_{4}+\alpha \left( n_{1}+n_{6}\right) >n_{5}+n_{6}+v\left( n_{2}+n_{4}\right) \\ n_{1}+n_{2}+n_{5}>n_{3}+n_{4}+n_{6}\\ n_{3}+n_{4}+n_{6}<qn\\ n_{4}+n_{5}+n_{6}<qn \end{array}\right. \\ \widetilde{n}\in VS\left( \left\{ a,b\right\} ,\overline{F}_{\alpha },q\right) \Longleftrightarrow \left\{ \begin{array}{l} n_{1}+n_{2}+\alpha \left( n_{3}+n_{5}\right) >n_{5}+n_{6}+\alpha \left( n_{2}+n_{4}\right) \\ n_{3}+n_{4}+\alpha \left( n_{1}+n_{6}\right) >n_{5}+n_{6}+\alpha \left( n_{2}+n_{4}\right) \\ n_{1}+n_{2}+n_{5}=n_{3}+n_{4}+n_{6}\\ n_{3}+n_{4}+n_{6}<qn\\ n_{4}+n_{5}+n_{6}<qn\\ n_{1}+n_{2}+n_{5}<qn\\ n_{2}+n_{5}+n_{6}<qn \end{array}\right. \\ \widetilde{n}\in VS\left( \left\{ a,b,c\right\} ,\overline{F}_{\alpha },q\right) \Longleftrightarrow \left\{ \begin{array}{l} n_{1}+n_{2}+\alpha \left( n_{3}+n_{5}\right) =n_{3}+n_{4}+\alpha \left( n_{1}+n_{6}\right) \\ n_{1}+n_{2}+\alpha \left( n_{3}+n_{5}\right) =n_{5}+n_{6}+\alpha \left( n_{2}+n_{4}\right) \\ n_{3}+n_{4}+n_{6}<qn\\ n_{4}+n_{5}+n_{6}<qn\\ n_{1}+n_{2}+n_{5}<qn\\ n_{2}+n_{5}+n_{6}<qn\\ n_{1}+n_{2}+n_{3}<qn\\ n_{1}+n_{3}+n_{4}<qn \end{array}\right. \end{aligned}$$