Voters’ preference diversity, concepts of agreement and Condorcet’s paradox

Abstract

Gehrlein et al. (Math Soc Sci 66:352–365, 2013) have shown that an increase of the voters’ preference diversity, as measured by the number $k$ of preference types in a voting situation, implies a decrease in the probability of having a Condorcet Winner. The results offered in this paper indicate that this relationship is far from being so clear when we consider instead the proximity of voting situations to having $k$ distinct preference types. This measure of agreement is compared to other measures of group mutual coherence previously analyzed in Gehrlein (Condorcet’s paradox, Springer Publishing, Berlin, 2006). It turns out that our results are completely consistent with the theory introduced by List (Good Soc 11:72–79, 2002) that is based on an important distinction between two different concepts of agreement.

Appendix: Volume computation

Given $\text{ J }\subset \{1,2,\ldots ,6\}$ with $\#\text{ J }=6 - \text{ k }$, let $\mathrm{S_\mathrm{J}} ({\alpha },\infty ,\text{ k })$ be the set of all voting situations that satisfy the set of constraints in (1). One way to evaluate the 4-dimensional volume $\mathrm{C_\mathrm{J}} ({\alpha },\infty ,\text{ k })$ of $\mathrm{S_\mathrm{J}} ({\alpha },\infty ,\text{ k })$ proceeds as follows:

Step 1 Rewrite all the constraints that defined $S_J (\alpha ,\infty ,k)$ in terms of a 4-component vector $y$. To achieve this, choose $s\in J$ and $t\notin J$; then rewrite each constraints in (1) taking into account that $x_s =\alpha +x_s -\mathop \sum \nolimits _{j\in J} x_j$ and $x_t =1-\alpha +x_t -\mathop \sum \nolimits _{j\notin J} x_j$. The new set of constraints denoted by $S_J (y,\alpha ,\infty ,k)$ now depends only on the 4-component vector $y=(x_j )_{j\ne s,t}$ and each constraint can be put in the standard form $c_j y\le \alpha _j ,\,1\le j\le p\,$where $c_j$ is a 4-component vector, $\alpha _j$ is an affine function of the parameter $\alpha$ and $p$ is the total number of such constraints.

Step 2 Find the set V(J,y) of all vertices of $S_J ( {y,\alpha ,\infty ,k})$ . To do this, consider any subset $\left\{ {j_1 ,j_2 ,j_3 ,j_4 } \right\}$ of $\left\{ {1,2,\ldots ,p} \right\}$ such that $c_{j_1 } ,\,c_{j_2 } ,\,c_{j_3 } \,\text{ and }\,c_{j_4 }$ are linearly independent and let $v$ be the unique solution to $c_{j_1 } y=\alpha _{j_1 } ,\,c_{j_2 } y=\alpha _{j_2 } ,\,c_{j_3 } y=\alpha _{j_3 } \,\text{ and }\,c_{j_4 } y=\alpha _{j_4 }$. Note that $v$ may not be a vertex of $S_J (y,\alpha ,\infty ,k)$ since there may still exist some other constraints$\,c_j y=\alpha _j$, $j\ne j_1 ,j_2 ,j_3 ,j_4$ that should be satisfied by $v$. Thus $v$ is called a potential vertex and is affected a validity domain that consists in the set

$$\begin{aligned} R_v =\{c_j v\le \alpha _j ,\;\text{ for }\;\text{ all }\;j\ne j_1 ,j_2 ,j_3 ,j_{4\,} \} \end{aligned}$$

Since each $\alpha _j$ is an affine function of $\alpha$, each $R_v$ is either empty or a union of some real ranges on $\alpha \in [0,1]$. Let $I_v$ be the collection of all $r\in [0,1]$ such that $r$ is a bound of an interval in the validity domain of $v$. Collecting all bounds $r$ from all validity domains over potential vertices in a single finite subset $I=\{r_1 ,r_2 ,\ldots ,r_q \}$ of [0,1] with $r_j <r_{j+1}$, the set $V(J,y)$ of all vertices of $S_J (y,\alpha ,\infty ,k)$ for $r_j <\alpha <r_{j+1}$ is the set of all potential vertices the validity domains of which contain the interval $(r_j ,r_{j+1} )$.

Step 3 Find a triangulation of $S_J (y,\alpha ,\infty ,k)$ to derive the volume. Note that each facet $F_j$ of $S_J (y,\alpha ,\infty ,k)$ corresponds to at least one constraint $c_j y\le \alpha _j$ from the definition of $S_J (y,\alpha ,\infty ,k)$. Each vertex can then be attached to the subset of facets it belongs to. Choosing a vertex, said $v^1$, from $V(J,y),$a dissection of $S_J (y,\alpha ,\infty ,k)$ is obtained by considering all pyramids $v^1F_j$ with apex $v^1$ and bases $F_j$ such that $v^1$ is out of $F_{j}$. This is in fact the initial step of the well known Cohen and Hickey algorithm of triangulating a polytope (Cohen and Hickey 1979). This operation is then applied recursively to find a triangulation of $S_J (y,\alpha ,\infty ,k)$ into simplices, each containing five points that are affine independent. Finally the volume of $S_J (y,\alpha ,\infty ,k)$ is the sum of the volumes of each simplex obtained in its triangulation using the following formula of the 4-dimensional volume of a simplex ${\Delta }(a_0 ,a_1 ,a_2 ,a_3 ,a_4 ,a_5 )$ :

$$\begin{aligned} vol( {{\Delta }(a_0 ,a_1 ,a_2 ,a_3 ,a_4 ,a_5 )})=\frac{\left| {\text{ det }(a_1 -a_0 ,a_2 -a_0 ,a_3 -a_0 ,a_4 -a_0 ,a_5 -a_0 )} \right| }{4!}.\qquad \end{aligned}$$

(27)

For illustration, we now apply this method to evaluate the 4-dimensional volume of $S_J (\alpha ,\infty ,4)$ with $J=\{6,5\}$. By setting $x_6 =\alpha -x_5$, $x_1 =1-\alpha -x_2 -x_3 -x_4$ and $y=(x_2 ,x_3 ,x_4 ,x_5 )$, $S_J (\alpha ,\infty ,4)$ is represented by set $S_J (y,\alpha ,\infty ,4)$ of all 4-component vectors $y$ such that:

$$\begin{aligned} \left( {{\begin{array}{*{20}c} 1 &{}\quad 1 &{}\quad 1 &{}\quad 1 \\ 1 &{}\quad 1 &{}\quad 1 &{}\quad {-1} \\ {-1} &{}\quad 0 &{}\quad 0 &{}\quad {-1} \\ 0 &{}\quad {-1} &{}\quad 0 &{}\quad {-1} \\ 0 &{}\quad 0 &{}\quad {-1} &{}\quad {-1} \\ {-1} &{}\quad 0 &{}\quad 0 &{}\quad 1 \\ 0 &{}\quad {-1} &{}\quad 0 &{}\quad 1 \\ 0 &{}\quad 0 &{}\quad {-1} &{}\quad 1 \\ 0 &{}\quad 0 &{}\quad 0 &{}\quad 1 \\ 0 &{}\quad 0 &{}\quad 0 &{}\quad {-1} \\ \end{array} }}\right) y\le \left( {{\begin{array}{*{20}c} {1-\alpha } \\ {1-2\alpha } \\ {-\alpha } \\ {-\alpha } \\ {-\alpha } \\ 0 \\ 0 \\ 0 \\ \alpha \\ 0 \\ \end{array} }}\right) \end{aligned}$$

(28)

By solving all possible combinations of four equations extracted from the list of constraints in (28), the collection of all validity domains shows that the set of vertices is stable for $\alpha$ in [0,1/5] and [1/5,1/3] respectively. Moreover, the set of constraints in (28) is not feasible for $\alpha \quad >$1/3 and is of dimension lower than 4 for $\alpha$ =0 or $\alpha$ =1/3. For $\alpha$ in [0,1/5], the list of vertices is provided in Table 6.

Table 6 Vertices of $S_J (y,\alpha ,\infty ,4)$ and their repartition on facets

A facet $F_j$ of $S_J (y,\alpha ,\infty ,4)$ corresponds to the constraint $c_j y=\alpha _j$ obtained by saturating the$j{th }$constraint in (28). In Table 6, the presence of $j$ in the column “Facets for the vertex” means that the corresponding vertex lays on $F_j$. Starting with $v^{12}$ as the initial vertex, we obtain a triangulation of $S_J (y,\alpha ,\infty ,4)$ into eight simplexes as shown in Fig. 5.

Fig. 5

A triangulation of $S_J (y,\alpha ,\infty ,4)$

In Fig. 5, each simplex in the triangulation corresponds to a terminal node and consists in the set of the five vertices linked to that node. The 4-dimensional volume of $\mathrm{S_\mathrm{J}} (\text{ y },{\alpha },\infty ,4)$ is then derived by performing (27) for each of the eight simplexes obtained:

$$\begin{aligned} vol( {S_J ( {y,\alpha ,\infty ,4})})=\frac{1}{6}\alpha ( {1-4\alpha })( {17\alpha ^2-8\alpha +1})\;\text{ for }\;{0<}\alpha <\frac{1}{5}. \end{aligned}$$

Since there are fifteen possible $\text{ J }\subset \{1,2,\ldots ,6\}$ with $\# \text{ J }=2$, then for $\,0<\alpha <\frac{1}{5}$

$$\begin{aligned} C( {\alpha ,\infty ,4})&= 15vol( {S_J ( {y,\alpha ,\infty ,4})}) \\&= -170\alpha ^4+\frac{245}{2}\alpha ^3-30\alpha ^2+\frac{5}{2}\alpha . \\ \end{aligned}$$

The same technique applies for ${\alpha }$ in [1/5,1/3], and subsequently for all the other sets of voting situations considered in the paper. To overcome the difficulty due to a huge number of potential vertices and multiple validity domains for distinct value $k$ = 5,4,3,2,1, we have built a program using MAPLE codes for computerized evaluations.²

It is worth noticing that all the results we have obtained have been checked by using an alternative technique based on Barvinok algorithm (see e.g. Lepelley et al. 2008). In this case, we rewrite each set of constraints in terms of $m$ and $n_j ,j=1,2,\ldots ,6$ instead of ${\alpha }=m/n$ and $x_j =n_j /n$, where $m$ denotes the minimum number of voters we should remove from a voting situation in order for the reduced voting situation to have $k$remaining preference ranking types. We then derive, via Barvinok algorithm, the quasi polynomial representing the total number of integer points that satisfy the set of constraints as a function of $m$ and $n$. The corresponding volume is the coefficient of the leading term in $n$ when $m$ is replaced by ${\alpha }n$ in that representation.