Voting ParadoxesA Geometric Analysis of Unexpected Election Behaviors

Damaris AlarconAdvisor: Charles Peltier

2/23/2015

Contents

1 Introduction to problem to be discussed 3

1.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.2 The Problem with Condorcet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.3 The Borda Count and Positional Voting Methods . . . . . . . . . . . . . . . . . . . . 7

1.4 A Break from Traditional Combinatoric Methods . . . . . . . . . . . . . . . . . . . . 12

1.5 Voter types . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2 Case 1: Types in Condorcet Profile 17

2.1 Fundamentals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2.2 Limiting Probabilities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

2.3 Positional Outcomes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

2.4 Consequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

3 Case 2: Types in Beverage Example 40

3.1 The Set-Up . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

3.2 Pairwise Outcomes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

1

3.3 Positional Outcomes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

3.4 Implications of The Tie Point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

3.5 General Probabilities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

4 Conclusion 52

References 54

2

1 Introduction to problem to be discussed

There is no mystery surrounding two-candidate elections. To elect a single winner, all the voter

has to do is cast a single vote for his or her chosen candidate, thus producing one winner and one

loser. However, the story dramatically changes with elections of three or more candidates as these

multicandidate elections admit all sorts of paradoxes (counter-intuitive conclusions) that must be

taken seriously since they can generate doubt about the meaning or even the integrity behind an

election. This paper’s particular interest is on analyzing the voting paradoxes that arise with three-

candidate elections. Any results derived in the following sections were first developed and shown

in the article “Geometry, Voting, and Paradoxes” [4] by Donald G. Saari and Fabrice Valognes and

the book “Basic Geometry of Voting” [3] by Donald G. Saari as well. To see only a small portion

of what can go wrong, let us begin this study by describing the problems a 15-member committee

runs into when trying to select their beverage of choice from among three options.

1.1 Problems

It is common knowledge that to derive the winner of an election, we just need to count how many

people favor each candidate. Voting is so elementary that even nursery school children know how

to vote to select their juice of choice before nap time. However, a lot can go wrong with voting.

Mathematicians have shown that most voting methods fall apart at the mere sight of three or more

candidates as electoral procedures don’t do what we expect them to do: yield an outcome that best

reflects the views of the voters. To illustrate this dilemma, consider the following simple fifteen

voter example extracted from [4].

Suppose that we wish to select a common beverage from among M (Milk), B (Beer), and W

(Wine) that will best reflect the preferences of the voters. If “ � ” means “is preferred to” and we

let the voters’ preferences in this hypothetical 15-member election to be the following:

3

Number Preference

6 M � W � B

5 B � W � M

4 W � B � M

Table 1: List of the voter preferences for common beverage election.

then the plurality outcome (where each person votes for his or her favorite beverage) is M �W � B

with a tally of 6:5:4. Apparently, Milk is the beverage of choice, yet, how true is this? Elections are

useful only if we trust them so how can we trust that milk is truly the voters’ beverage of choice?

Well, one way is to put this outcome to the test. If Milk is truly the voter’s beverage of choice,

then we would expect voters to prefer Milk to Beer. Is this the case? Surprisingly, no. As the next

table shows, these voters actually prefer Beer to Milk when these two options are pitted against

each other.

Number Preference Milk Beer

6 M � W � B 6 0

5 B � W � M 0 5

4 W � B � M 0 4

Total 6 9

Table 2: Pairwise Comparison of Milk versus Beer.

Nine voters prefer Beer to Milk. Tallying up the votes in a similar manner shows that nine voters

prefer Wine to Milk and ten prefer Wine to Beer. These pairwise comparisons suggest that the

voters actually prefer W � B � M, a ranking opposite to the plurality outcome! This contradiction

and potential controversy among the party goers begs the question, what went wrong?

4

1.2 The Problem with Condorcet

The beverage paradox just presented makes it clear that our ultimate goal, then, is to choose a

voting procedure that always honors the voters’ beliefs, yet each voting method promises to provide

its users with an accurate measurement of this; how do we know which “correct one” is correct?

This radical disagreement raises interesting theoretical questions, namely how does a majority vote

ranking of a pair relate to its relative ranking within a plurality outcome? Can anything go wrong

with pairwise rankings? Mathematician, philosopher, and politician Nicolas de Condorcet argued

that a natural way to rank candidates is with pairwise competitions. In 1785, Condorcet wrote

“Essai Sur l’Application De L’Analyse a La Probabilite Des Decisions Rendues a la pluralite des

voix” (Essay on the Application of Analysis to the Probability of Majority Decisions) where he

introduced the concept of a Condorcet winner and Condorcet loser, a criterion for the absolute

winner and absolute loser in an election and an immediate extension of the election phenomenom

developed in Section 1.1.

Definition 1.1. A candidate ck is a Condorcet winner if she wins all pairwise majority vote

elections against all other candidates. Candidate cj is a Condorcet loser if she loses all pairwise

elections.

Example 1.2. In the beverage example, Wine and Milk are, respectively, the Condorcet winner

and loser. Note that there is at most one Condorcet winner and (or) loser per election.

It is easy to embrace the idea of a Condorcet winner since it comes with the comfort of using the

familiar pairwise, majority vote elections. This concept captures a sense of rugged individualism

because ck reigns as the Condorcet winner only if she can beat all other options. But, just as

the Condorcet winner was almost universally accepted as the ultimate choice, its problems became

known. The Condorcet winner need not always exist, in which case we are dealing with the

disturbing results of the Condorcet Paradox.

5

Definition 1.3. The Condorcet Paradox is a situation in which the following is true.

Assume that a society has three individuals n1, n2, n3 who must choose their most preferred

candidate from among three alternatives: c1, c2, c3. Let profile denote the set of voters’ preferences.

Then, if the (strict) preferences of the individuals, known as the Condorcet profile, are as follows:

Individual Preference

n1 c1� c2� c3n2 c2� c3� c1n3 c3� c1� c2

Table 3: The Condorcet Profile.Then:

If c1 is pitted against c2, then c1 defeats c2.

If c2 is pitted against c3, then c2 defeats c3.

If c1 is pitted against c3, then c3 defeats c1.

Hence, there is no Condorcet winner or Condorcet loser [3].

The Condorcet Paradox tells us that when the outcome is determined by pairwise majority

voting as shown above, the majority wishes can be in conflict with each other. That is, the

Condorcet Paradox implies the following three consequences:

1. The social preference relation is not transitive.

2. The outcome is not path independent, i.e., the final outcome depends upon the order in which

the alternatives are chosen to be pitted against one another.

3. Individuals who have the power of deciding the order in which the alternatives are chosen

may exercise that power to obtain an outcome more favorable to themselves. In other words, agenda

control becomes a serious strategic issue.

The division of the voters described by Definition 1.3, which proves that sincere pairwise elec-

tions can create cycles, has been rediscovered many times since by important contributors to this

6

area. In fact, this election paradox plays an important role in motivating Arrow’s Theorem where

in 1951 economist and mathematician Kenneth Arrow proved the impossibility of ever constructing

a method for three or more alternatives that satisfies certain desirable, yet seemingly innocuous

properties. More on this later. Real life scenarios of the Condorcet Paradox can be observed in

trivial elections such as a collegiate department’s selection of a new textbook.

In view of Definition 1.3, a “Condorcet winner”, then, is a “sometimes” concept- sometimes it

is useful; sometimes it is not. This in itself suggests that the Condorcet winner should be critically

re-examined as the above (Condorcet) profile, which defines the cyclic election outcomes

A � B, B � C, C � A,

whereby whichever candidate is voted upon last wins the election- decisively, cannot be trusted

upon to honor the voters’ beliefs. Further, these outcomes, which fail transitivity, are also not path

independent since pairwise comparisons are susceptible to manipulation schemes. For example, if

an individual’s most preferred candidate among A, B, and C were to be A, then he would pit B

against C, {B,C}, first so that B wins the initial pairwise comparison only to be beaten by A in the

second {A,B} pairwise comparison. Hence, this procedure lacks integrity; it generates too many

pairwise winners.

The Condorcet profile makes it clear that pairwise comparisons may lead to intransitive out-

comes, even though the preference ordering of every voter is transitive. Since the purpose of elec-

tions is to decide, competing approaches have been devised to avoid stalemates. As such, related

methods such as the Borda Count possess distinct advantages over pairwise voting.

1.3 The Borda Count and Positional Voting Methods

In 1770, fifteen years before Condorcet introduced the concept of a Condorcet winner, mathe-

matician Jean-Charles Borda questioned whether the French Academy of Science was electing to

membership whom they really wanted. The beverage example illustrates his main concern, namely

7

that the ”winner” of the widely used plurality vote can be the candidate the voters view as “infe-

rior”. Consequently, his attempts to find a method to capture the true views of the voters in 1770

and then again in 1784 led to the development of the Borda Count, providing us with the following

alternate voting procedure for a three candidate election.

Definition 1.4. The Borda Count for a three candidate election is an alternative voting procedure

which assigns 2, 1, and 0 points, respectively, to a voter’s top, middle, and bottom-ranked candidate.

Candidates are then ranked according to the sum of assigned points.

The Borda count for a three candidate election above determines the outcome of a debate or the

winner of an election by giving each candidate, for each ballot, a number of points corresponding to

the number of candidates ranked lower. Borda claimed the superiority of his (2,1,0) point system

for a three candidate election by showing that at least for the beverage example of Section 1.1, the

“correct” candidate is elected. His arguments must have been persuasive since not only does this

point system describe the most commonly used weights for a Borda Count for a three candidate

election, but it was also adopted by the French Academy until the 1800s when Napoleon Bonaparte

exerted his influence to have it overturned. Note that if we wished, we could extend this definition

of the Borda count for an n candidate election.

Definition 1.5. The n candidate Borda Count assigns n−j points to a voter’s jth-ranked candidate,

for j = 1, 2, ..., n. The social ranking is determined by the total number of points obtained by each

candidate where “more is better”. If exactly one candidate gets the maximum total points then it

is the winner.

Example 1.6. To see how this method can change the outcome for a three candidate election,

consider the following Borda Count Tally for the beverage example of Section 1.1.

8

Number Preference Milk Beer Wine

6 M � W � B 6x2 0 6x1

5 B � W � M 0 5x2 5x1

4 W � B � M 0 4x1 4x2

Total 12 14 19

Table 4: Borda count for beverage example.

This brings us back to the outcome W � B � M, which agrees with the pairwise election rankings

from Section 1.1, the Condorcet winner!.

Therefore, we arrive at Wine as the winner for the Borda count without having to worry about

the peril of cyclic outcomes that comes with pairwise elections. This advantage presents the Borda

Count as the “correct” voting procedure for this example. But what happens in general? Are

there examples of sets of voter preferences (profiles) for which the Borda Count performs poorly?

Why not use weights such as (6,5,0) or (4,1,0) rather than Borda’s choice of (2,1,0)? Due to

the variability in the candidates’ point-assignments, these tallying methods that assign a specified

number of points to a voter’s first, second and third ranked candidate, called positional voting

methods, are often accused of creating “winners” with the selection of an appropriate positional

voting rule.

Definition 1.7. For a three candidate election, a positional voting method is defined by a voting

vector W3 = (w1, w2, w3) ∈ R3 where

w1 ≥ w2 ≥ w3. (1)

In tallying a ballot, wj points are assigned to the voter’s jth-ranked candidate, where j = 1, 2, 3.

The candidates are ranked according to the number of points assigned to each of them.

9

But, in general we have:

Definition 1.8. For n ≥ 3 candidates, a positional voting method is defined by a voting vector

Wn = (w1, w2, ..., wn) ∈ Rn where

wi ≥wi+1 , i = 1, ..., n− 1. (2)

In tallying a ballot, wj points are assigned to the voter’s jth-ranked candidate, where j = 1, 2, ..., n.

The candidates are ranked according to the number of points assigned to each of them.

Therefore, if we are not careful, an election outcome can more accurately reflect the choice of

an election rule rather than the voters’ wishes. Let us look at three very well-known positional

voting methods.

For three candidate plurality elections,

Definition 1.9. A three candidate plurality election assigns one point to a voter’s top-ranked

candidate and zero to the middle and bottom-ranked candidate so that W3 = (1, 0, 0) ∈ R3.

We observe that:

Example 1.10. With the positional rule of “vote for one” (a plurality election represented by

(1,0,0) meaning a ballot is tallied by giving one point to the candidate positioned first and zero for

the others), Milk wins in our beverage example. This procedure effectively requires a voter to vote

directly for his or her top-ranked candidate and be against the rest.

Note that, if we wished, we could extend the definition for the three candidate plurality election

for an n candidate election, as shown below.

Definition 1.11. A n candidate plurality election assigns one point to a voter’s top-ranked candi-

date and zero to all others so that Wn = (1, 0, ..., 0) ∈ Rn.

10

Likewise, for the antiplurality rule,

Definition 1.12. The antiplurality rule for a three candidate election assigns zero points to a voter’s

bottom-ranked candidate and one to the middle and top-ranked candidate so that W3 = (1, 1, 0)

∈ R3.

it can be shown that:

Example 1.13. With the positional rule of “vote for two” ( the antiplurality rule given by (1,1,0)),

Wine wins. Note that this voting method requires a voter to vote against his or her bottom-ranked

candidate.

The definition for the three candidate antiplurality election for an n candidate election is shown

below.

Definition 1.14. The n candidate antiplurality rule assigns zero points to a voter’s bottom-ranked

candidate and one to all the rest so that Wn = (1, 1, ..., 1, 0) ∈ Rn.

Hence, using the positional voting rule for antiplurality, the outcome agrees with our result

using the Borda Count, namely Wine, but by using the plurality election rule, we conclude that

Milk is the beverage of choice instead. When these weights are normalized to assign a simple point

to a voter’s top ranked candidate, the point assignment defines a normalized voting vector wλ =

(1, λ, 0), where 0 ≤ λ ≤ 1.

Example 1.15. For instance, the normalized forms of (6,5,0) and the Borda Count, (2,1,0), are

represented by the normalized vectors w 56= (66 ,56 ,0) and w 1

2= (1,12 ,0), respectively.

The wλ normalization makes it clear that there is a continuum of tallying methods in between

0 and 1, inclusively, where each is characterized by the weight (the λ - value) placed on a voter’s

second-ranked candidate. In view of all of the different possibilities for the value of λ, it was only

11

natural for Borda’s mathematical collegues and future mathematicians to question which λ method

is optimal in the sense that its outcome best reflects the views of the voters. Which (if any) election

rule faithfully yields outcomes that best capture the views of the voters? In the mid-20th century,

another major advance in voting theory made by the Nobel laureate economist Kenneth Arrow

made a breakthrough in this continuous debate. With contributions from Blau and Marakami,

Arrow used axiomatic methods to prove that there cannot exist an entirely fair voting system

when we choose the following four criteria: Universality, Independence of Irrelevant Alternatives,

Citizen’s Sovereignty, and Non-Dictatorship. He proved that if the first three criteria were true,

then it must be a dictatorship [1]. This, in turn, was a profound message for voting theorists. It

meant that we could not possibly construct a fair voting system! Instead, we can only attempt to

decide which system is “most” fair. In the late 20th century, mathematician Donald Saari began

to tackle this question by developing a systematic way to explore, discover, and prove the existence

of new voting properties using various geometric arguments.

1.4 A Break from Traditional Combinatoric Methods

When considering the paradoxical election behavior that can arise with a three-candidate election

such as the Condorcet paradox or the confusion from the positional voting methods just examined,

very few people have looked for solutions in elementary geometric arguments. The geometry of

voting has long served as a powerful tool to provide a global perspective on possible voting outcomes

for scoring rules while exposing unexpected relationships, yet the development of voting theory

relied heavily on traditional combinatoric methods for its first few years. Since the goal of geometry

is to capture the sense that “a picture is worth a thousand words”, the purpose of geometry within

this paper will be to expose new properties about pairwise and positional voting rules without

having to stress about the complexity of the combinatorics.

For instance, in departing from axiomatic methods, further sections will address voting struc-

12

tures as “vector spaces” since, as discussed in Section 1.3, we can discuss profiles and positional

weightings in terms of vectors. We start out by decomposing profile spaces using geometric ideas in

order to explain how paradoxes arise and with what significance. We will also address the weights

(the λ - value) placed under the positional voting methods of plurality, antiplurality, and the Borda

count, and suggest how to create a profile decomposition that can explain all three-candidate para-

doxes with pairwise and positional voting by using simple geometric arguments. Note that although

these methods do not answer which wλ method is the best, it gives us a break from the otherwise

traditional combinatorial way to compare procedures.

Section 2 and Section 3 demonstrate how the geometry of the equilateral triangle allows us

to see where these voting conflicts occur in pairwise elections and positional voting methods for

two related scenarios by using the geometric arguments found in Donald G. Saari’s and Fabrice

Valogne’s article “Geometry, Voting, and Paradoxes” [4].

1.5 Voter types

If the “beverage paradox” of Section 1.1 is not an unusual setting, then it should be easy to

construct many different illustrating examples. For instance, try creating a different example of

voters’ preferences where the plurality election ranking is c1 � c2 � c3 even though the pairwise

election results are c2 � c1, c3 � c1, c3 � c2. Or, try to create an example involving n1 voters

with the ranking c1 � c2 � c3, n2 voters with the ranking c3 � c1 � c2, and n3 voters with the

ranking c2 � c1 � c3. Or, how about an example using just the three other rankings [3]. These are

some of the challenges Saari offers us, making it clear that analyzing voting procedures by creating

concrete examples can be quite difficult, especially if we want to answer questions such as: can the

Condorcet and Borda Count winners differ? From the beverage example we know that different

positional methods ( e.g. the plurality outcome versus the antiplurality outcome) create different

election outcomes, but is there a general description explaining how election results change with

13

changes in the wλ methods? When using different wλ voting vectors to tally ballots in the profile

of Table 1, either Wine, Milk, or both always emerge as top choice. Are there voting profiles

where each candidate is the “winner” for an appropriate wλ? Are the supporting examples isolated

or robust? Can we characterize all possible examples? What is the minimum number of voters

needed to create each election oddity? Hence, in an attempt to answer these questions, let us use

Saari’s systematic way to explore, discover, and prove the existence of new voting properties by

first formally defining what a voter’s “type” is.

Definition 1.16. Assume that each voter has a strict linear ordering of the c candidates, i.e., each

voter compares each pair of candidates in a transitive manner without being indifferent about any

two candidates. Then, a voter’s type is his (or her) ranking of the candidates {c1, c2, ..., cn} in an

election. If a candidate ci receives more votes than a candidate cj , then the ranking is denoted as

ci � cj . The c! = c(c− 1)...(2)(1) different ways to strictly rank c candidates, define c! voter types.

It follows from Definition 1.16 that a three candidate election between candidates A, B, and

C will result in the following 3!=6 voter types described by Table 5 below. For convenience, we

designate these rankings by “type numbers”, e.g., a “type 4” ranking is C � B � A.

Type Preference Type Preference

1 A � B � C 4 C � B � A

2 A � C � B 5 B � C � A

3 C � A � B 6 B � A � C

Table 5: Voter types.

Note that there are 13 different election rankings possible among candidates A, B, and C: we

can have 6 election rankings without ties between candidates (the strict rankings described by Table

5 above), 6 election rankings with a tie vote between a pair of candidates, and an election ranking

14

with a tie vote among all candidates. Using an approach developed in [3], we will model these 13

different election rankings among candidates A, B, and C through the geometry of the equilateral

triangle, where each point in the equilateral triangle uniquely defines a voter’s particular ranking of

these three candidates. Let us begin by assigning each candidate to a vertex, as depicted in Figure

1.

Figure 1: Representation Triangle and Ranking Regions

Each point in the equilateral triangle above, Figure 1, is assigned an ordinal ranking of the

candidates according to how close this point is to each vertex; the closer a point is to a vertex of

the equilateral triangle, the more preferred the candidate assigned to that vertex is in the election.

In this manner, each geometric region is identified with one of the 13 different election rankings

among candidates A, B, and C. The six small open triangles represent the 6 strict rankings described

by Table 5, while the seven remaining ranking regions, which involve at least one tie, are portions

of lines. For instance, all points in the triangular sector labeled “1” are closest to A, next closest to

B, and farthest from C, thereby defining the type 1 ranking A � B � C. For example, all points on

the vertical line that is equidistant from A and B represent all voters that are indifferent between

A and B, denoted by A ∼ B. Lastly, the point where all indifference lines intersect represents all

voters who are completely indifferent for all candidates: A ∼ B ∼ C.

In the next section we show how this geometry positions the ranking regions in a manner similar

15

to that of a profile space, the set of all possible profiles showing all possible interactions among

the three candidates, in that adjacent ranking regions differ only by the ranking of an adjacent

pair. An advantage of this geometric profile representation is that it makes it much easier to tally

positional and pairwise ballots as well as identify why, with the same profile, different rules can

have conflicting election outcomes.

16

2 Case 1: Types in Condorcet Profile

This section’s particular interest is in three-candidate elections involving only the three voter types

in the Condorcet profile of Table 3 and how conflict among the pairwise and positional voting out-

comes can occur with only three types of preferences. In view of the question “which voting method

is the best?”, this section will first highlight the features of pairwise outcomes and then compare

these results to the outcomes of different positional voting, wλ, procedures. Special attention will

be given to the plurality vote (λ = 0), the Borda Count (λ = 12), and the antiplurality method

(λ = 1) with regards to what makes each of these wλ procedures desirable in a three-candidate

election.

2.1 Fundamentals

Section 1 showed how considerable insight and unexpected conclusions already arise when the

voters’ beliefs are restricted to only three specified preference types, and even more so when these

three preference types are those involved in the Condorcet paradox: voter types 1, 5, and 3 on Table

5. As it is our purpose to understand the mystery behind pairwise voting cycles, we will now restrict

our analysis of voting paradoxes to these three types (Figure 2a captures this setting) and answer

the questions (1) what are the probabilities of strict transitive rankings with cyclic outcomes?

And (2) how does this probability differ for an odd as opposed to even number of voters? In this

process, our first task is to associate normalized profiles with their election outcomes. As we will

soon see, when voters are restricted to types 1, 3, and 5, the four possible strict pairwise outcomes

include these three rankings, A � B � C , B � C � A , C � A � B, and the cyclic rankings

A � B � C � A.

Definition 2.1. Let pj denote the fraction of all voters that are of the jth type, j = 1, ..., c!. A

normalized profile is the vector p = (p1, ..., pc!)

17

Example 2.2. Consider the vector (16 , 0,13 , 0,

12 , 0). The vector (16 , 0,

13 , 0,

12 , 0) is a normalized

three-candidate profile where 16 of all voters are of type 1, 1

3 are of type 3, 12 are of type 5, and

there are no voters of the remaining three types. As the smallest common denominator is six, the

total number of voters for an associated integer profile is a multiple of six.

Remark 2.3. By Definition 2.1, the normalized profile for a three-candidate election is p =

(p1, p2, p3, p4, p5, p6). However, since we are restricting our election to include only voters of types

1, 5, and 3, there are no voters that are of types 2,4, and 6. Therefore, p2, p4, p6 = 0 so they add

no weight to our election. Hence, this restriction on the voter types converts the normalized profile

space of all three-candidate profiles from a six-dimensional geometric object to a three-dimensional

simplex where p = (p1, p3, p5) [3].

A standard trick in geometry is to invent convenient coordinate systems to simplify the prop-

erties being studied. As Definition 2.1 requires pj ≥ 0 and∑n!

j=1 pj = 1, when expressed as an

algebraic system - two equations and one unknown- this assertion converts the space of normal-

ized profiles to a 2-dimensional geometric object capable of being graphed on our typical xy-plane.

Hence, we may denote the vector p = (p1, p3, p5) in terms of (x, y) coordinates so that we may de-

scribe this normalized profile in terms of its distance in the two coordinate directions. The following

paragraph describes this process.

Let n be equal to the total number of voters in an election between candidates A, B, and C.

Since we are only considering voter types 1,3, and 5, then nj , the number of voters of type j,

satisfy the equation n1 +n3 +n5 = n. Therefore, if we divide n so that x = n1n , y = n5

n , and z = n3n

represent the fraction of voters of types 1,5, and 3, respectively, then x = p1, y = p5, z = p3; this

creates a convenient scaling property where, rather than needing to know the number of voters with

each preference ranking, an election outcome can be determined just by knowing the fraction of all

voters with each preference. Hence, p = (x, y, z). Note that in this trick, we switch the original

18

order of the fractions p3 and p5 so that the fraction of all voters of type 5 rather than type 3 will

be represented by the y-coordinate instead. Table 6 below summarizes these statements.

Type PreferenceFraction of

Voters

1 A � B � C x = n1n

5 B � C � A y = n5n

3 C � A � B z = n3n

Table 6: Description of Admitted Types.

As expected, it follows from this trick that x+ y+ z = 1 or, equivalently, z = 1− (x+ y), where

x, y, z ≥ 0. Therefore, if we plot a profile’s (x, y) values in a Cartesian framework, then the z value

is z = 1 − (x + y), thereby allowing us to represent all possible profiles as the (rational) points of

the triangle

T1 = {(x, y)|x, y ≥ 0, x+ y ≤ 1}

In this manner a profile is identified with point (x,y) in the Fig. 2(b) triangle T1: the closer the

point is to the origin, the larger the z value. Conversely, any (rational) point in T1 defines a profile;

e.g., point (15 ,110) ∈ T1 corresponds to x = 1

5 , y = 110 , z = 1 − (15 + 1

10) = 710 , where the common

denominator 10 identifies one profile for (15 ,110) as n1 = 2, n5 = 1, n3 = 7. (With the scaling, any

positive integer multiple of these nj ’s defines the same T1 point). Thus the set of rational points

in T1 geometrically represents all profiles. Note that only points with rational components, which

are densely positioned in T1, can be identified with actual profiles. But the analysis is simplified

by treating the full triangle T1 -both the rational and irrational points- as the generalized profile

space [2].

To find all possible profiles that support different outcomes, plot the election outcomes on the

generalized profile space T1. The pairwise elections divide the generalized profile space T1 into

19

four major regions - the three corner triangles and the smaller triangle at the center of T1 - where

election outcomes are identified by type numbers.

Figure 2: Condorcet Example Setting

Example 2.4 (Boundary Lines). Triangle T1, Figure 2b above, allows us to easily conclude which

election rankings are associated with 4 distinct regions of profiles. To represent the A,B pairwise

vote, suppose that we want to know under what conditions candidate B will be preferred over

candidate A (B � A). Figure 2a shows that only a type 5 voter votes for B in an election of A

against B so in order to get the pairwise outcome B � A, we must have the proportion of voters

of type 5 be greater than the proportion of voters of types 1 and 3. Hence, candidate B will be

preferred over candidate A whenever y > x+z. However, in order to be able to identify the ranking

region for A � B on the xy-plane, we must get our inequality solely in terms of x and y. Hence, by

using the constraint z = 1−(x+y), we find that B beats A if and only if y > x+z = x+(1−x−y),

or y > 12 . The boundary line for this region is the horizontal dashed line of Figure 2b. Therefore,

the profiles associated with the pairwise outcome B � A are above the y > 12 horizontal dashed

line in T1. Note that it follows from this algebra that there exists a tie between A and B (A ∼

20

B) whenever y = 12 and that A is preferred to B whenever y < 1

2 . In this manner, we can derive

the three types of outcomes possible in a pairwise election of A against B. The analysis of the

remaining two pairs, {A,C} and {B,C}, is similar.

Example 2.5 (Profile Regions). The boundary lines denoting pairwise outcomes divide the profiles

in T1 into 4 different regions of profiles. Of these 4 regions, three produce transitive election

rankings. For example, the profile region to the extreme right, touching the vertex (1, 0) on Triangle

T1, is on the A � B,A � C and B � C sides of the boundary lines, so all of these profiles produce

the type 1 election ranking A � B � C. Similarly, The profile region touching the vertex (0, 1)

produces the election ranking B � C � A , the election ranking for voters of type 5, and the profile

region touching the vertex (0, 0) produces the election ranking C � A � B, the election ranking

for voters of type 3. The last possible strict pairwise outcome belongs to the smaller triangle in

the center, however, unlike the three ranking regions that produce transitive election rankings, this

ranking region identifies all profiles that cause cyclic pairwise outcomes: A � B � C � A; these

profiles produce all cases of the Condorcet Paradox.

This profile coordinate system, then, makes it clear what profiles and how many of them support

different outcomes. For instance, it allows us to easily construct examples for any of the admitted

outcomes possible with our three voter types by giving special attention to the smallest common

denominator for the points (x, y) in each ranking region. We will now show how with no more than

four voters, we can select profiles in T1 that produce one of the following five different outcomes

described by Proposition 2.6.

Proposition 2.6. With no more than four voters, we can create examples of all admitted pairwise

rankings, namely:

1. Unanimity outcomes

2. Non-transitive rankings involving two tie votes

21

3. cyclic outcomes

4. Non-transitive rankings involving one tie vote

5. Strict transitive rankings

Proof. Suppose that our three candidate election includes only one voter so that n = 1. Then,

depending on the type of this one voter, either candidate A, B, or C will be the unanimous winner

of the election. Therefore, with no more than one voter, we can create examples of all unanimity

outcomes. Geometrically, the profiles that will result in unanimity outcomes are the vertices of

triangle T1, that is, the points (1, 0), (0, 1), and (0, 0). Let us consider the profile (0, 0). The profile

(0, 0) describes an election where there are no voters of type 1, no voters of type 5, but there is

one voter of type 3. Since type 3 voters have the preference C � A � B, the outcome C � A � B

implies that C is the unanimous winner of the election. Note that unanimity outcomes are a special

case of strict transitive rankings.

When n = 2, it follows from our definition of profile coordinates that we are looking at all profiles

(x, y) with common denominator two. Geometrically, these are either vertices of T1 or vertices of the

small triangle that produces cyclic rankings. As the vertices of T1 represent unanimity outcomes,

let us examine one of the vertices of the cyclic region. For example, consider the profile (12 , 0).

The pairwise election {A,B} reveals that A � B. However, the remaining two pairwise elections

{B,C} and {A,C} reveal that B ∼ C and A ∼ C. Therefore, the outcome of the election is

A � B ∼ C ∼ A, even though transitivity would mandate for A � C. Thus, the profile (12 , 0)

produces a non-transitive ranking involving two tie votes. Similar results follow for the two other

vertices of cyclic region.

We can construct examples of unanimity outcomes and strict transitive rankings with three

voters, but what is in particular special about n = 3 is that it allows us to access the profile

(13 ,13), located in the center of the cyclic region of Triangle T1. The profile (13 ,

13) corresponds to

22

an election where we have one voter of each type, that is, one individual of type 1, one individual

of type 5, and one individual of type 3. Therefore, by Definition 1.3 in Section 1.2, the pairwise

elections {A,B} , {B,C} and {A,C} reveal no Condorcet winner (or loser), thereby describing a

case of the Condorcet Paradox. Due to these results, the profile (13 ,13) produces the cyclic outcome

A � B � C � A, where the minimum number of voters needed to have these cyclic rankings is

three.

Finally, with four-voter elections (n = 4) we can access the outcomes produced by the profiles

lying along the boundary lines of Triangle T1. For instance, consider the profile (14 ,14) so z = 1

2 .

The pairwise elections A � B, {B,C} and {A,C} reveal the outcome A � B ∼ C � A, even

though transitivity would require for A � C. Therefore, the profile (14 ,14) produces a non-transitive

ranking involving one tie vote, namely B ∼ C. Non-transitive rankings involving one tie vote can

also be result from the points (14 ,12) and (12 ,

14).

A similar analysis shows that with any n = 1, 2, 3, 4, we can create examples of Strict transitive

rankings. Geometrically, these are the profiles located strictly inside the four regions of profiles on

T1. For example, the profile (23 ,13) maps the outcome A � B � C. As expected, the profile (23 ,

13)

results in the strict transitive ranking preference, of type 1 voters. Hence, with no more than four

voters, we can create examples of all admitted pairwise rankings, the set of all different preference

relationships (orderings) possible among candidates A, B, and C.

Triangle T1 makes it easy to compute the likelihood of each outcome described in Proposition

2.6, which in turn, will help us identify where the likelihood of cyclic rankings is the greatest for

an odd or an even number of voters. The following subsection discusses how the probability of

observing Condorcets Paradox is obtained by summing the probabilities that the voting situations

in that subset will be observed. The outcome will obviously be completely driven by the specific

23

mechanism that determines the probability with which each specific voting situation is observed.

2.2 Limiting Probabilities

Geometrically, the set of rational points in T1 represents all profiles possible among our three voter

types. As we have just shown, the boundary lines of T1 partition this profile space into four different

profile regions, whose width and length both equal 12 . Therefore, these four regions of profiles that

result in strict pairwise rankings have equal areas. Now, to illustrate how easy it is to obtain

results from this simple geometry, consider the impartial anonymous culture (IAC) assumption

popularized by W. Gehrlein, P. Fishburn, and others [2], which reduces computational difficulties

by treating each profile as being equally likely. With IAC, the probabilities correspond to the areas

of appropriate regions. For instance, by using the areas of the 4 triangles in T1, it follows that

except for finitely many n, 14 of the generalized profiles have either a type 1 outcome, a type 5

outcome, a type 3 outcome, or a cyclic outcome. Although actual profiles are identified with the

rational points in T1, we get an approximation that holds because the ratios of the number of

rational points with the same common denominator (that is, with a specified number of voters)

in different regions are approximated by the ratios of their areas. Moreover, agreement improves

(surprisingly rapidly) as n, the number of voters, increases.

Similar to IAC, if instead we assume that profile points in T1 are centrally distributed, then all

three transitive outcomes are equally likely.

Definition 2.7. Let p denote a three-dimensional vector (p1, p2, p3) for the three-candidate case,

where pj denotes the probability that a randomly selected voter from the population of potential

voters will have the corresponding preference ranking on candidates that are shown in Table 6.

Then a profile probability is centrally distributed if the likelihood of profile (p1, p2, p3) is the same

as (p2, p1, p3) or any of the four other ways these pj values can be permuted.

This probability model is formed on the basis that once a profile has been established, it is easy

24

to accumulate the voters preferences according to the possible preference rankings to obtain the

associated voting situation for that profile.

Lastly, note that by appealing to the central limit theorem, it follows from either of these

assumptions that these 14 probability values represent limits as the number of voters becomes very

large. To illustrate, we will now identify a wide class of settings where the likelihood of cyclic

rankings is the greatest for an odd number of voters and then for an even number of voters. To

derive this probability, it is sufficient to (1) enumerate all possible voting situations for a specified

n, and (2) identify the subset of all possible voting situations for which a cycle exists. Let us begin

by (1), enumerating all possible voting situations for a specified n.

To calculate the total number of profiles for n voters among our three voter types, we apply the

standard identity:

k∑j=1

j =

(k + 1

2

)=k(k + 1)

2(3)

Proposition 2.8. There are(n+22

)rational points in T1 with common denominator n. Therefore,

n-voters create(n+22

)different profiles among voter types 1,5, and 3 .

Proof. Suppose that j is equal to the total number of voters of both type 1 and type 5. Then n− j

denotes the number of voters of type 3 so z = n−jn , where 0 ≤ j ≤ n. Then for each value of j ,

we add j + 1 profiles to our total number of profiles for n voters. Therefore, the total number of

different profiles n voters create is equal to∑n

j=0 j + 1, which, by using the properties of sums and

Equation 4, can be rewritten as:

n+1∑j=1

j =

(n+ 2

2

)=

(n+ 1)(n+ 2)

2(4)

25

To compute the number of profiles in each profile region, a similar analysis follows. That is,

suppose j is equal to the total number of voters of type 3. Note that a profile (x, y) gives strict

rankings with cyclic outcomes if and only if it lies inside the cyclic region of T1, i.e., satisfies the

constraints: x < 12 , y <

12 , and x + y > 1

2 . Then, in order to (2) identify the subset of all possible

voting situations for which a cycle exists, we must calculate the total number of profiles n voters

create inside the cyclic region of T1. However, the implications of these results for the case when

n is odd differs from the case when n is even as the profiles in T1 need not be equally distributed

among the four profile regions.

Remark 2.9. If only one voter in a large population has type 3 preferences so that z = 1n , then

cycles occur if and only if n is odd and x = y = n−12n .

It follows from Remark 2.9 that in the case when n is odd, j, the total number of type 3 voters,

ranges from 1 to n−12 . Hence, if n is odd, then using Equation 5, there exist

∑n−12

j=1 j = (n−1)(n+1)8

different (x, y) points in the cyclic region that satisfy:

x+ y = 1− j

n= 1− z > 1

2, where j = 1, 2, ...,

n− 1

2. (5)

Likewise, a similar argument shows that if n is even, then there exist∑n−2

2j=2 (j − 1) different

(x, y) points in the cyclic region that satisfy

x+ y = 1− j

n= 1− z > 1

2,where j = 2, 3, ...,

n− 2

2. (6)

Note that by using the properties of sums and Equation 5,∑n−2

2j=2 j − 1 can be rewritten as∑n−2

2−1

j=2 j = (n−2)(n−4)8 . These results imply a smaller number of profiles being located inside

the cyclic region when n is even than when n is odd.

26

Due to the manner in which rational points are distributed in a region, there exist more profiles

in the cyclic region when n is odd than when n is even. However, the probabilities for these two

cases converge to the same number, 14 , as n approaches infinity. The fraction of the T1 points that

are in the cyclic region when n is odd is equal to

(n−1)(n+1)8

(n+1)(n+2)2

=1

4(1− 3

n+ 2). (7)

As n →∞, this quantity tends to 14 . Similarly, for even values of n, we have

(n−2)(n−4)8

(n+1)(n+2)2

=1

4(1− 9n− 6

(n+ 1)(n+ 2)) (8)

which also tends to 14 as n →∞.

Thus, Equation 7 and Equation 8 tell us that regardless of whether the total number of voters

is odd or even, the probability of getting strict rankings with cyclic outcomes is approximately 14 ,

so long as the total number of voters, n, is a relatively large number. For small values of n, these

14 probabilities may not hold.For example, with n = 3, instead of approximately 1

4 of the points

being in the cyclic region, there are only 110 of them. For n = 4 , this probability drops to zero,

then rebounds to 17 for n = 5 only to drop to 1

26 for n = 6. The following theorem reiterates all

of these results, including the likelihood of strict transitive rankings when n is odd and when n is

even.

Theorem 2.10. When voters are restricted to types 1,3, and 5, the four possible strict pairwise

outcomes include these three types and the cyclic rankings A � B � C � A. If profile points in

T1 are assumed to be centrally distributed, then the three transitive rankings are equally likely. In

the case of n voters, and we assume that all points in T1 are equally likely, the probability of strict

27

rankings with cyclic outcomes is 14(1 − 3

n+2) if n is odd and 14(1 − 9n−6

(n+1)(n+2)) if n is even. The

likelihood of a strict transitive ranking is 14(1 + 1

n+2) if n is odd and 14(1− 1

n+1) if n is even. [4]

The boundary lines and the resulting division of profiles identified with these pairwise outcomes

help us associate profiles with their election outcomes, allowing us to calculate the probabilities

stated in Theorem 2.10 above. The same can be said about the boundary lines derived from the

plurality (λ = 0), Borda (λ = 12), and the antiplurality (λ = 1) voting systems. However, conflicts

among the pairwise and positional outcomes emerge from the start, λ = 0, and continue as the λ

in our wλ procedure varies.

2.3 Positional Outcomes

In Section 2.1 we learned that paradoxes must be anticipated 14 of the time with procedures based on

pairwise rankings as the profiles in the cyclic region of T1 do not abide by the central assumption

of transitivity. This suggests using other methods such as positional voting, which admit only

rational voters. However, the outcomes of the {ci, cj} pairwise and positional rankings need not

have anything to do with one another. To see this, in this section we construct another geometric

representation of the profile space T1 with the purpose to show all possible interactions among

all three-candidate plurality, Borda Count, and antiplurality positional outcomes. The geometry

of this profile space will not only identify all inconsistencies among the outcomes for different

positional rules, but it will also highlight where the positional rankings differ from the pairwise

rankings described in Section 2.1.

Recall the normalized voting vector wλ = (1, λ, 0), where 0 ≤ λ ≤ 1. Its normalization makes

it clear that there is a continuum of tallying methods where each is characterized by the weight λ

placed on a voter’s second-ranked candidate. Let wλ outcomes define the different rankings possible

with voter types 1,3, and 5 given a fixed λ-value. We will now show that the choice of a positional

voting method (λ-value) matters in selecting a winner by constructing the geometric representation

28

of these wλ outcomes.

Although much of our construction of the profile space for positional voting methods mimics the

one described for pairwise outcomes, one crucial difference lies in the manner in which candidates

are ranked. That is, much like we did with pairwise outcomes, we create the profile space for

positional voting methods by using the profile coordinates x, y, and z to derive the set of rankings

possible among candidates A,B and C when we restrict voters to the types 1, 5, and 3. Since these

rankings are dependent on the value of λ, we denote the rankings of positional voting methods as

the wλ rankings. However, unlike pairwise outcomes, wλ rankings are derived by calculating the

tally for each candidate, i.e., the portion of the total number of points in the election gained by

candidates A, B, and C. By employing the following technique developed by Saari [3], we are able

to use these tallies to derive the boundary lines for different positional voting methods.

The normalized voting vector wλ = (1, λ, 0) is a rule that assigns 1 point to a voter’s top-

ranked candidate, λ points to their middle-ranked candidate, and no points to their bottom-ranked

candidate. In view of this rule, the point totals for candidates A, B, and C are n1 + λn3, n5 + λn1,

and n3 + λn5, respectively, so there are a total of n+ λn points in the election.

Example 2.11. We observe from Table 6 that candidate A is top-ranked by voters of type 1,

middle-ranked by voters of type 3, and bottom-ranked by voters of type 5. Therefore, candidate A

receives n1(1) points from voters of type 1 and n3(λ) points from voters of type 3. Hence, the total

number of points gained by A is n1(1) +n3(λ). Note that, as opposed to ranking two candidates at

a time in order to get the general outcome for a given set of voters, positional voting methods derive

the election outcome by ranking all candidates side-to-side such that voters consider all candidates

at once. In this manner, transitivity in the outcomes is guaranteed.

Note that the election ranking that follows from these point totals agrees with the ranking of

the normalized election vector :

29

q = (n1 + λn3n(1 + λ)

,n5 + λn1n(1 + λ)

,n3 + λn5n(1 + λ)

) (9)

Here, qi specifies the portion of the total tally received by candidate i = A, B, C. This technique

allows us to describe the tallies for candidates A, B, and C in terms of our profile coordinates x, y,

and z.

Example 2.12. Consider the portion of the total tally received by candidate A, qA = n1+λn3n(1+λ) . This

quantity is equivalent to

n1n(1 + λ)

+λn3

n(1 + λ)=

x

1 + λ+

zλ

1 + λ(10)

Hence, qA = x+λz1+λ , where the quantity x + λz is the tally, i.e., the total fraction of points for

candidate A. Similarly, we find that the tally for candidate B is y + λx and the tally for candidate

C is z + λy.

As Example 2.11 shows, we can use the profile coordinates x, y, and z to calculate the point

totals for each candidate when voters are restricted to the types 1,5, and 3. In this manner, we

see that the components of the normalized election vector, rather than the actual tallies, dominate

the ranking of the candidates. Hence, much as we did with pairwise outcomes, we use these profile

coordinates to draw the boundary lines for different positional voting methods. However, note that

in order to derive the boundary lines for which A ∼ B, B ∼ C, and A ∼ C, we must have the

tallies of candidates A, B, and C in only x and y terms. Therefore, by substituting 1− (x+ y) in

for z, the tallies described in Example 2.12 reduce to the tallies described in Table 7 below.

30

Candidate Tally

A (1-λ)x− λy + λ

B y + λx

C 1-x+ (λ− 1)y

Table 7. wλ Tallies

Example 2.13. In Example 2.12, we found the tally for candidate A to be x+ λz. Therefore, by

substituting 1− (x+ y) for z, the tally for candidate A is x+ λ(1− (x+ y)) = x+ λ− λx− λy =

(1− λ)x− λy + λ, where λ is a constant.

By equating two of the tallies at a time, we derive the parametrized family of equations for each

candidate pair, i.e., the set of related boundary lines on which each candidate pair is tied. As we

will soon see, these equations shown in the second column of Table 8 below help us classify which

profiles define the relative pairwise rankings for each candidate pair by partitioning the normalized

profile space into multiple profile regions.

Pair Equation Rotation Pt x-intercept Pt

A ∼ B (1-2λ)x− (1 + λ)y = −λ (13 ,13) ( −λ

1−2λ , 0)

A ∼ C (2-λ)x+ (1− 2λ)y = 1− λ (13 ,13) (1−λ2−λ , 0)

B ∼ C (1+λ)x+ (2− λ)y = 1 (13 ,13) ( 1

1+λ , 0)

Table 8. Parametrized Family of Equations

Example 2.14. Suppose that we want to classify which profiles define the relative A � B or

B � A rankings. Then, we must find the A ∼ B boundary line defined by equating the A and B

tallies, i.e., the parametrized family of equations for the candidate pair A ∼ B. Hence, we must

have (1− λ)x− λy + λ = y + λx. The result for this pair is as follows:

31

(1− λ)x− λy + λ = y + λx

x− λx− λy + λ = y + λx

(1− 2λ)x− (1 + λ)y = −λ

So, for example, if we let λ = 13 , then the boundary line denoting all profiles for which A ∼ B

is defined by the equation

(1− 2(1

3))x− (1 +

1

3)y = −1

3⇐⇒ y =

1

4(x+ 1)

All profiles where y > 14(x+ 1) define the pairwise ranking B � A, so, consequently, the profiles

below the line y = 14(x+1) define the pairwise ranking A � B. Note that the Table 8 tallies always

define a linear equation, or a straight line, given a fixed λ value.

Note that a slight change in the λ-value will cause a movement in the A ∼ B, A ∼ C, and

B ∼ C boundary lines, thereby changing the point of intersection of these three lines. This change

in the point of intersection, i.e., the profile which defines a completely tied wλ election outcome,

may, consequently, also change the election outcome associated with a fixed profile in T1 since the

boundary lines’ division of the profile space will no longer be the same. Therefore, a convenient

way to discover conflicts is to trace how the location of this “tie point” changes with λ. By tracing

how the tie point changes with λ, we can discover how all other wλ election rankings change with λ

. However, Saari notes that for the types (rankings) involved in the Condorcet profile, there is no

need to trace the location of the tie point as the parametrized family of equations for all candidate

pairs are satisfied when x = y = 13 for all λ values. What does this mean? Within the scope of this

section, this merely means that for any given λ, the three boundary lines for the candidate pairs

will “rotate” about the point (13 ,13), thereby intersecting at the same point for any λ value and

32

conserving the same set of election outcomes in the normalized profile space as λ changes in value.

To summarize, for any given λ, we do not have to worry about new kinds of election outcomes

coming into the picture, i.e., occurring in the normalized profile space other than the initial 6 (soon

to be explained). As will be shown in the next section, this common rotation point among our

three pairs of candidates is the exception, rather than the rule.

Example 2.15. Let x = y = 13 and λ ∈ [0, 1]. When we evaluate the parametrized family of

equations for the candidate pair A ∼ B at this point, we have

(1− 2λ)1

3− (1 + λ)

1

3= −λ ⇐⇒ −λ = −λ

So irrespective of what the value for λ is, when we evaluate the parametrized family of equations

for the candidate pair A ∼ B at x = y = 13 we always have the tie A ∼ B.

As a consequence of this common rotation point, all of the boundary lines for three-candidate

election with only types 1,5, and 3 rankings pass through - intersect at- the point (13 ,13) so by

focusing our attention to the plurality (λ = 0), Borda Count (λ = 12), and antiplurality (λ =

1) positional voting methods we may draw enough conclusions about this continuum of tallying

methods with regards to which election rule (λ) best captures the views of the voters. Figure 3

below shows which election rankings are associated with each profile region defined by the plurality,

Borda Count, and antiplurality boundary lines. By evaluating the equations from Table 8 at each

of these λ values, we observe that these boundary lines partition the normalized profile space for

the Condorcet example setting into 6 profile regions, where each result in one of the six initial voter

preferences, types, described by Table 5.

Remark 2.16. Figure 3 below immediately discloses one very crucial property about the boundary

lines for this setting: each boundary line rotates in a clockwise direction from its λ = 0 setting to

33

Figure 3: Positional Outcomes

reach the adjacent boundary line position when λ = 1. While it is obvious from Figure 3 that each

boundary lines rotates in a clockwise direction, let us consider the boundary line for the candidate

pair A ∼ B at λ = 0 to see how it reaches the adjacent boundary line position when λ = 1. Table 9

below shows the corresponding equation for the A ∼ B boundary line, defined by the parametrized

family of equations (1− 2λ)x− (1 +λ)y = −λ, for the the plurality (λ = 0) , Borda Count (λ = 1),

and antiplurality (λ = 1) election rules.

Rule (λ) Equation

0 y = x12 y = 1

3

1 y = −12x+ 1

2

Table 9. Boundary line for pair A ∼ B

As we can see from Table 9, the equation for the A ∼ B boundary line at λ = 1 is y = −12x+ 1

2 ,

which is the equation that defines the B ∼ C boundary line at λ = 0. Similarly, what was formerly

the boundary line for the pair A ∼ B at λ = 0, y = x, is now the A ∼ C boundary line at λ = 1.

So, although the three boundary lines for the λ = 0 and the λ = 1 triangles in Figure 3 partition

34

the normalized profile space into 6 geometrically identical profile regions, each boundary line is

identified with a different pair of candidates.

Note that the six types in Table 5 are always represented in the geometry of the profile space,

the Figure 3 equilateral triangles, as the λ in our wλ procedure varies in value. This is a feature

that follows from the common rotation point (13 ,13). Lastly, note that for each of these three wλ

procedures, the effects are drawn over triangle T1 from Section 2 so that we may easily see where

the pairwise and positional outcomes agree, or (most likely) disagree. To illustrate how easy it is

to obtain results from this simple geometry, let us assume, once more, that each of the profiles in

the normalized profile space is equally likely so that agreement improves (surprisingly rapidly) as

n, the number of voters, increases. Hence, the probabilities for these outcomes correspond to the

areas of appropriate regions so it is worth taking a look into what these areas are and how they

change as λ varies in value.

2.4 Consequences

The rotation of the boundary lines as λ progresses from λ = 0 to λ = 1 creates conflicts among

the pairwise and positional outcomes because the clockwise rotation of the boundary lines causes

the areas of some regions to monotonically decrease and others to increase as λ → 12

−, which, in

turn, will cause the limiting probabilities (areas) for the three outcomes of voter types 1, 3, and 5

to differ. Figure 3 immediately discloses a difference in areas between the 6 voter preference types

described in Table 5 as λ varies through its admissible values. For instance, the three regions for

the outcomes of types 1, 3, and 5 increase in area as λ → 12

−, resulting in a limiting probability

of 29 at λ = 1

2 as opposed to 19 , the limiting probability of the remaining three types. Therefore,

it appears that the Borda Count (λ = 12) favors the outcomes of types 1, 3, and 5. However,

as λ → 1, the areas of these regions change to monotonically approach the common value 16 , the

original value at λ = 0. Hence, the limiting probability of any strict election ranking (in either

35

the set of all profiles or the cyclic region) is 16 for λ = 0, 1. This implies that the likelihood for

the three strict transitive rankings studied in the previous subsection do not remain constant as

λ varies through its admissible values! Moreover, what connects these different values is that the

areas of some regions monotonically decrease, while others increase, as → 12

−. Furthermore, as

λ → 1, they then change to monotonically approach the common value 16 , the original value at

λ = 0. This clockwise rotation of the boundary lines immediately poses us with three consequences.

The following proposition describes the first consequence.

Proposition 2.17. With exception of the point (13 ,13), each profile in T1 experiences three different

wλ election rankings as λ varies through its admissible values. There are two cases to consider,

namely:

1. If a point is on a boundary line when λ = 0, then two of the rankings have ties and one is

strict.

2. Otherwise (for points not on a boundary line when λ = 0), two of the rankings are strict and

one is a pairwise tie.

Example 2.18. Consider the point (16 ,16). This point lies on the line y = x of the λ = 0 triangle;

the boundary line that produces the tie A ∼ B. Therefore, when using the plurality vote ( λ = 0),

the profile (16 ,16) produces a tie between A and B, giving us the outcome C � A ∼ B. However, the

line y = x denoting the tie A ∼ B when λ = 0 rotates in a clockwise position to become the line

y = 13 at λ = 1

2 . Therefore, at λ = 12 , the profile (16 ,

16) now lies in the non-cyclic region denoting

the preference of type 3 voters. Hence, when using the Borda Count (λ = 12), the profile (16 ,

16)

gives the strict election ranking C � A � B. Lastly, at λ = 1, we see that clockwise rotation of

the boundary lines once again places the profile (16 ,16) in the y = x line, except that in this case,

the y = x line produces the tie A ∼ C. Therefore, the antiplurality election outcome for the profile

(16 ,16) is A ∼ C � B.

36

Hence, the profile (16 ,16) lying on the boundary line for A ∼ B when λ = 0 experiences three

different election outcomes as λ varies through the values ≤ λ ≤ 1, namely the two ties A ∼ B and

A ∼ C, and the strict ranking C � A � B.

The next example shows that, due to the clockwise rotation of the boundary lines of the Figure 3

triangles, some profiles in triangle T1 will experience pairwise ties for λ-values other than λ = 0, 12 , 1.

Example 2.19. Consider the point ( 312 ,

712). This point lies on the region producing the strict

ranking B � A � C, as it is located on the non-cyclic region producing the preference ranking of

voters of type 6. However, as the line y = 1 − 2x (producing the tie A ∼ C when λ = 0) rotates

in a clockwise position to become the line x = 13 at λ = 1

2 , the profile ( 312 ,

712) experiences two

different wλ election rankings rather than one. That is, note that for for λ = 0, the profile ( 312 ,

712)

is on the right side of the boundary line denoting the pairwise tie A ∼ C, yet for λ = 12 , the profile

( 312 ,

712) is now on the left side of the boundary line denoting the pairwise tie A ∼ C. Therefore,

the profile ( 312 ,

712) must experience the pairwise tie A ∼ C for some λ-value where 0 < λ < 1

2 and,

as expected, the strict ranking B � C � A for λ = 12 . As λ→ 1 , we see that clockwise rotation of

the boundary lines does not change the election ranking for the profile ( 312 ,

712); the election ranking

remains B � C � A at λ = 1.

Hence, the profile ( 312 ,

712) experiences three different election outcomes as λ varies through the

values 0 ≤ λ ≤ 1, namely the two strict rankings B � A � C and B � C � A and the pairwise tie

B � A ∼ C .

As a second consequence of this clockwise rotation, consider the set of profiles with pairwise

votes that define a particular strict transitive outcome. These profiles belong to the triangle T1

underlying each of the triangles in Figure 3. The following two propositions outline how many strict

election rankings from either the plurality vote, the Borda Count, or the antiplurality method are

associated with each of the three transitive outcomes modeled by T1.

37

Proposition 2.20. The set of profiles that define a particular strict transitive pairwise outcome

allow only two strict election rankings with the plurality (λ = 0) and with the antiplurality vote

(λ = 1). In each case, one of these outcomes agrees with the pairwise ranking.

Example 2.21. Take for instance the region labeled “1” in T1, to the right of the vertical dotted

line in Figure 3. By examining this region in the λ = 0 and λ = 1 triangles, we see that these

profiles allow two different strict plurality and antiplurality election outcomes. That is, the pairwise

A � B � C ranking denoted by region 1 is accompanied by a plurality ranking of either A � B � C

(type 1) or A � C � B (type 2) so only one of these outcomes agrees with the pairwise rankings.

While this difference in outcomes creates a conflict, note that at least the plurality and pairwise

procedures agree on which candidate should be top-ranked; this can be fashioned into a strong

argument in favor of the plurality vote.

Similarly, the pairwise A � B � C ranking denoted by region 1 is accompanied by an antiplu-

rality ranking of either A � B � C (type 1) or B � A � C (type 6). In this case, however, the

conflicting ranking is B � A � C (type 6). This causes the pairwise and antiplurality methods to

agree only on who should be bottom-ranked; they can disagree on the rest of the ranking and who

should win. Under this light, the antiplurality method might be considered as weak.

The clockwise rotation of the boundary lines of Figure 3 emits even more conflicting election

rankings for other wλ procedures, 0 < λ < 1 .

Proposition 2.22. For each λ ∈ (0, 1), wλ admits three different strict election rankings for each

of the three sets of profiles, one of which agrees with the pairwise ranking. This is not only due the

rotation of the boundary (indifference) lines, but also the monotonicity of the x coordinate.

Example 2.23. Recall the region labeled “1” in T1, to the right of the vertical dotted line in Figure

2b. By examining this region in the λ = 12 setting, i.e., the Borda Count, we see that the pairwise

38

A � B � C ranking denoted by region 1 is accompanied by a w 12

ranking of either A � B � C

(type 1), A � C � B (type 2), or B � A � C (type 6). Therefore, the Borda Count allows not only

two, but three strict rankings for profiles from the strict pairwise ranking region 1. Furthermore,

the w 12

ranking of A � B � C agrees with the pairwise A � B � C ranking denoted by region 1 so

at least one of the strict election rankings agrees with the pairwise ranking.

Lastly, before stating Theorem 2, we must address the profile set causing cyclic pairwise out-

comes. As expected, accompanying a pairwise cycle, we can have any strict wλ ranking, any wλ

ranking with one pair tied, or a completely tied outcome. Basically, anything can happen with any

wλ method. Hence, our conclusions for this setting are limited. These statements (and others) are

collected in the following theorem:

Theorem 2.24. If three voter types 1,3, and 5 are allowed, then each profile that is not a Condorcet

profile admits three different wλ election outcomes as λ varies.

The set of profiles with pairwise votes that define a particular strict transitive outcome allows

only two strict election rankings with the plurality and with the antiplurality vote. In each case, one

of these outcomes agrees with the pairwise rankings. All other wλ outcomes admit three different

strict rankings, one of which agrees with the pairwise ranking. The profile set causing cyclic pairwise

outcomes admits all wλ rankings.

If all T1 points are equally likely, then the limiting probability of any strict election ranking (in

either the set of all profiles or the cyclic region) is 16 for λ = 0, 1. For the Borda Count the limiting

probability for either setting is 29 for outcomes of types 1,3, and 5, and 1

9 for the remaining three

types.

The likelihood of an election outcome being of a particular type either strictly increases or strictly

decreases as λ→ 12 .

These results show that even with only three types of voter preferences, conflict can arise

39

among the pairwise and positional election outcomes. As just shown, it is not clear from this

information which wλ procedure is the best. For instance, the plurality and pairwise outcomes

identify the same candidate as being top-ranked, an advantage, yet the rankings of the remaining

two candidates differ. Because of these peculiarities, we conclude that a similar argument can

probably be fashioned to support any other wλ procedure.

3 Case 2: Types in Beverage Example

In the first section we reveal why paradoxical cycles occur with pairs of candidates and why election

outcomes vary with the choice of the positional voting method. In Section 2, these ideas were

combined to compare positional with pairwise vote outcomes when we restrict three-candidate

elections to voter types 1, 5, and 3, the Condorcet profile. This material, then, is a step toward

understanding the kinds of difficulties posed by all three-candidate settings with only three rankings.

In this section, we take a look at the voting inconsistencies for another setting: three-candidate

elections restricted to type 2, 4, and 5 rankings.

3.1 The Set-Up

Note that three-candidate elections restricted to type 2, 4, and 5 rankings describe the general case

for the beverage dilemma presented in Section 1.1, which raised interesting theoretical questions

such as how does as how does a majority vote ranking of a pair relate to its relative ranking within

a plurality outcome? Can anything go wrong with pairwise rankings? Whereas before we were not

able to answer any of these questions, we now can using the same geometric techniques developed to

analyze voting and discover sources of election paradoxes in the Condorcet example setting (types

1, 5, and 3) of Section 2.1. As much of the following analysis mimics that of Section 2, we denote

(emphasize) the key differences in the set-up with “remarks”.

Although three-candidate elections define a six-dimensional profile space, we now know that it

40

is possible to envision enough of this space to appreciate why and where different voting rules have

different outcomes by restricting our election to only three voter types. Furthermore, in Section 2.1

we have seen how profile coordinates allow us to easily “see” which profiles define which outcomes by

using algebraic techniques to convert this three-dimensional simplex to a two-dimensional triangle,

capable of being graphed in our familiar xy-plane. These results justify addressing the three-

candidate electoral problems following the voter restriction to types 2, 5, and 4, i.e., the types

involved in the beverage example of Section 1.1, through the same geometric techniques used in

Section 2. Note that the results from our analysis of the beverage example setting will differ as the

ordinal ranking of the candidates is different from those of the Condorcet example setting. These

rankings [Refer to Table 10 below] tell us that two out of the three pertaining rankings regions are

ranking regions adjacent to one another in our Figure 1 equilateral triangle from Section 1.5.

Type PreferenceFraction of

Voters

2 A � C � B x = n2n

5 B � C � A y = n5n

4 C � B � A z = n4n

Table 10. Admitted Types for Beverage Example

3.2 Pairwise Outcomes

Remark 3.1 (Ranking Regions). Table 10 shows that a portion of the voters, voter types 5 and 4 ,

strongly oppose (bottom-rank) candidate A, but split their opinions about the other two candidates,

B and C. Meanwhile, the remaining portion of the voters, voters of type 2, strongly favor (top-rank)

candidate A. This ordinal ranking of the candidates is captured by Figure 4a below, where two of

the preferences (types 4 and 5) share an edge of the Figure 4a triangle, and the third ranking (type

41

2) is from a ranking region touching the remaining vertex A. Hence, we are no longer dealing with

the Figure 2a symmetric “pinwheel” configuration that causes the pairwise cycles, a fact that, as

we will soon explain, is reflected in the composition of the normalized profile space.

Figure 4: The Beverage Example Setting

As before, if we define x = n2n , y = n5

n , and z = n4n for this case, it follows from this notation

that x, y, z ≥ 0 and x + y + z = 1, or, equivalently, z = 1 − (x + y). This restriction allows all

possible profiles to be represented as (rational) points of the Figure 4b triangle

T2 = {(x, y)|x, y ≥ 0, x+ y ≤ 1}.

Therefore, we may again identify a profile with a point (x, y) in triangle T2. For example, the

beverage profile of Section 1.1 corresponds to the point ( 615 ,

415) in T2, implying that z = 5

15 .

To represent the {A,B} pairwise vote, notice that A beats B if and only if x > 12 : in Fig. 4b,

these generalized profiles are to the right of the x = 12 vertical dashed line in T2. Similarly, B beats

C if and only if y > 12 : these generalized profiles are above the horizontal dashed line in T2. The

final A � C relationship also has x > 12 . Consequently, the pairwise elections divide the generalized

profile space T2 into three major regions -the square and the two smaller triangles- where election

outcomes are, as in Section 2.1, identified by type numbers.

42

Remark 3.2 (Profile Regions). As can be observed from Figure 4b, when voters are restricted to

types 2, 5, and 4, the three possible pairwise outcomes include these three types, namely the rankings

A � C � B, B � C � A, and C � B � A, respectively. All election rankings involving tie votes

are transitive, with the exception of the rankings that result from the profiles lying along the A ∼ B

and A ∼ C boundary lines for y < 12 ; these rankings produce non-transitive rankings involving two

tie votes, namely A ∼ C � B ∼ A. As triangle T2, has no cyclic region, this is the only paradoxical

outcome we can encounter with the beverage example setting .

Hence, the types from the profile are all possible pairwise outcomes we could have for the

election. Triangle T1 has a cyclic region because the three types analyzed in that setting are

capable of giving us an outcome with cyclic rankings, i.e., non-transitive strict pairwise outcomes,

which is not the case with the types involved in the beverage example setting: all profiles in T2 will

always produce transitive rankings.

By treating each profile as being equally likely, we find again that the probabilities of outcomes

correspond to the areas of appropriate regions. For instance, by using the areas of the square and

the two triangles in T2, it follows that half of the generalized profiles have a type 4 outcome while

a quarter each have type 5 and type 2 outcomes.

Remark 3.3 (Probabilities). For a large n expect about 12 of the profiles in T2 to have a type 4

outcome while about 14 each have type 2 and 5 outcomes. It is obvious from these computations

that the types involved in the beverage example favor the type 5 outcome, B � C � A. We had

only seen this type of “bias” happen for positional voting methods, namely the Borda Count in

Section 2.4, but as can be observed from T2, pairwise elections are also capable of producing certain

outcomes, and therefore certain winners, more often than others.

Note that the likelihood of these outcomes only gets more accurate as the number of voters

increases. For our purposes, further analysis of these pairwise rankings will not be pursued as this

43

section’s particular interest is in the conflict among the pairwise and wλ outcomes.

3.3 Positional Outcomes

Conflict between the pairwise and positional outcomes are surprisingly easy to find and perhaps

even trace by examining this setting’s wλ tally for each candidate.

Candidate Tally

A x

B y + λz = (1− λ)y − λx+ λ

C z + λ(x+ y) = 1− (1− λ)(x+ y)

Table 11. wλ Tallies for Beverage Example

Positional voting methods are graphed in a similar manner to Section 2.3, i.e., to find all out-

comes, plot the following equations defining A ∼ B, A ∼ C, and B ∼ C ties for a specified Wλ.

According to Table 11, they are the equations listed on the second column of Table 12 below.

Pair Equation Rotation Pt x-intercept Pt

A ∼ B (1 + λ)x− (1− λ)y = λ (12 ,12) ( λ

1+λ , 0)

A ∼ C (2-λ)x+ (1− λ)y = 1 (1,−1) ( 12−λ , 0)

B ∼ C (1- 2λ)x+ 2(1− λ)y = 1− λ (0, 12) ( 1−λ1−2λ , 0)

Table 12. Parametrized Family of Equations for Beverage Example

As noted in Section 2.3, a slight change in the λ-value causes a change in the location of the

boundary lines, predetermined by each boundary lines’ rotation point. We briefly discussed the

properties of the rotation point by describing it as the point about which the boundary lines will

“rotate”, but, by definition, what makes a point a rotation point is that the (x, y) coordinates of

the point satisfy the parametrized family of equations for each of the candidate pairs, irrespective

of what the value for λ is. So, no matter what λ is, the rotation point of each boundary line will

remain fixed, i.e. be the same throughout. The parametrized family of equations for each of the

44

candidate pairs of Section 2.3 were all satisfied by the same point, (13 ,13), thereby resulting in a

“common” rotation point of the Condorcet example setting. Example 2.14 confirmed this for the

candidate pair A ∼ B. We emphasize this occurrence of a common rotation point now because,

more often than not, the rotation point will not be the same one for each candidate pair. As Table

12 shows, the rotation point for the A ∼ B boundary line is (12 ,12), the rotation point for the A ∼ C

boundary line is (1,−1), and the rotation point for the B ∼ C boundary line is (0, 12). These three

rotation points are indicated by the solid dots in Figure 5 for the plurality (λ = 0), the Borda

Count (λ = 12), and the antiplurality(λ = 1) election rules.

Remark 3.4 (Rotation). The rotation of the boundary lines for the beverage example setting is

not always clockwise, as was the case for the boundary lines for the Condorcet example setting of

Section 2.3. In fact, the monotonicity of the x-intercept point in Table 12 shows that the rotation

point of each boundary line causes only the A ∼ C boundary line to have a clockwise rotation, and

the remaining two to have counterclockwise rotations.

The x-intercept point for the A ∼ B boundary line, ( λ1+λ , 0), approaches the point (12 , 0) as

λ→ 1, causing a counterclockwise rotation. Similarly, the x-intercept point for the C ∼ B boundary

line approaches the point (0, 0) as λ→ 1, causing a counterclockwise rotation. Note that since the

x-intercept point ( 12−λ , 0) goes to infinity (is undefined) for λ = 1

2 , the C ∼ B boundary line has no

x-intercept point at λ = 12 . Lastly, the x-intercept point for the A ∼ C boundary line approaches

the point (1,0) , causing the only clockwise rotation as λ→ 1.

Note that it follows from Remark 3.4 that by the time λ reaches λ = 1, they no longer even

interact as they are parallel to one another and therefore, never intersect, causing grave implications

for the tie point of the beverage example, i.e., the profile at which all three boundary lines intersect.

Unlike the Condorcet example setting, it is crucial to make a distinction between the rotation point

for each boundary line and the point of intersection, tie point, for the beverage example setting

45

as none of these are defined by the same point as λ changes. Since by tracing how the tie point

changes with λ we can discover how all other wλ election rankings change with λ and since the

tie point is determined by the rotation point from which the boundary lines rotate about, we have

four different points to keep in mind as λ changes.

Example 3.5. As just explained, the rotation points do not change in value as λ changes. Let us

see how, in comparison, as λ changes, the tie point changes for these three election rules.

Observe that at λ = 0, these three boundary lines intersect at the point (13 ,13) so the point (13 ,

13)

is the tie point for this election rule. However, once λ reaches the value 12 , the three boundary

lines rotate so that now they intersect at the point (0, 12). Lastly, note that at λ = 1, the boundary

lines’ rotation results in parallel boundary lines, thus there is no point of intersection for the

antiplurality method, λ = 1. As we will explain next, this discovery of no tie point shows how the

set of admissible wλ election rankings can drastically change with different values of λ.

As opposed to having all 6 types be possible election outcomes for all wλ election rules, we now

have to consider different sets of admitted election outcomes associated with each wλ election rule.

To illustrate, I will show how the set of admitted election outcomes can vary with λ by first taking

a look at the profile space for the plurality (λ = 0), the Borda Count (λ = 12), and the antiplurality

(λ = 1) voting methods.

Note that, once more, the profile regions producing the election outcomes associated with the

plurality, Borda Count, and antiplurality voting methods in are drawn over the profile space for

the pairwise elections of the related profile types, triangle T2.

Remark 3.6 (wλ Profile Regions). Changes in the set of admitted election outcomes must occur

as λ changes in value because the λ = 0 completely tied point is at (13 ,13), it moves to the T2

boundary at (12 ,12) when λ = 1

2 , and it vanishes at infinity (there is no three-way tie point!) when

46

Figure 5: Positional Outcomes for Beverage Example Setting

λ = 1, meaning that the rotation of the boundary lines results in a different division of profiles

with each λ value. As we can observe from above, this rotation results in 6 strict outcomes for

wo, 4 strict outcomes for w 12, and only 2 strict outcomes for w1, producing all sorts of conflicting

election outcomes in view of the T2 pairwise outcomes.

Let us extract some results from this diagram by, e.g., considering the profile region that

produces a type 4 ranking with pairwise outcomes (the square defined by the dotted lines) under

the λ = 0 Figure 5 triangle.

Example 3.7. All 6 strict rankings from Table 5 are possible outcomes for the plurality vote in

the beverage example and they are all equally likely. Furthermore, note that the tie point, i.e., the

point of intersection of all three lines and what causes these 6 profile region to be possible outcomes,

lies inside the type 4 profile region of T2. This means that under the plurality vote, some of the

profiles located inside profile region 4 of T2 may produce a type 4 plurality outcome, as expected

under pairwise elections, but some other portion of profiles may produce any of the other 5 types

of plurality outcomes instead, depending on what the proportion of voters, (x, y) coordinate, is.

For instance, recall Table 1 from the beverage example in Section 1.1. If we identify M, B, and W

47

from this example with candidates A, B, and C from Table 1, then the plurality outcome stated in

Section 1.1 is A � B � C, a type 1 ranking. But, as can be seen from the λ = 0 Figure 5 triangle,

this is only one of the six possible plurality outcomes we could have by restricting our election to

the types involved in the beverage example.

Note that the plurality vote from Section 2.3 (the Condorcet example setting) had profile regions

of T1 that were accompanied by at most three different plurality outcomes and not 6. From Figure

5, we observe that this same disagreement in the pairwise and positional rankings hold for the

Borda Count, and antiplurality voting methods. That is, the λ = 12 Figure 5 triangle shows four

possible outcomes we could have with the Borda Count and the λ = 1 Figure 5 triangle shows

2 possible outcomes we could have with the antiplurality method. Note that the set of admitted

outcomes for a given wλ election becomes smaller as λ increases in value. Hence, due to these

observations, the next natural question is to find whether the same serious conflict holds for all wλ

elections rules.

3.4 Implications of The Tie Point

The set of admitted election outcomes can vary with λ and, consequently, allow multiple rankings

to be associated with each of the profile regions of T2. The different locations of the rotation point

for each pair cause the boundary lines to divide T2 in different ways. By tracing the location of

the tie point, we will now show how these different rotation points are what cause serious conflict

between the pairwise and positional outcomes. The following analysis describes only what happens

to the profiles in the square defined by the dotted lines with a C � B � A pairwise ranking, i.e.,

the type 4 profile region of T2.

As the completely tied election point is the intersection of the A ∼ B and B ∼ C boundary

surfaces, solving these algebraic equations shows that these tied points satisfy

48

(x, y) = (1 + λ

3,1− λ+ λ2

3(1− λ)), 0 ≤ λ ≤ 1, (11)

or, because λ = 3x− 1,

y =1− 3x+ 3x2

2− 3x= −x+

1

3− 1

3(3x− 2). (12)

This curve is plotted in Figure 6a along with the (λ = 0) plurality rankings. The accompanying

Figure 6b magnified version displays the location of the tied point for λ = 14 while the dashed lines

indicate the λ = 14 ranking regions.

Figure 6: Location of the Tie Point

To extract results from Figure 6b, notice how the dashed lines are positioned within the plurality

ranking region one. As these lines identify which regions of profiles cause each of the thirteen w 14

election rankings, it follows, for instance, that a profile can be found with the plurality A � B � C

(type one) outcome together with any selected w 14

ranking; that is, we can select a point that is

strictly inside one of the w 14

ranking regions 2, 3, 4, 5, or 6 that make-up plurality ranking region

one. This same behavior holds for all positional methods where 0 < λ < 12 . In contrast, we learn

49

from Figure 6a that because the y-value approaches infinity as λ → 1 and x → 23 , the boundary

lines representing profiles with tied w1 outcomes become vertical. Hence, by definition, it follows

from the structure of this curve that parallel boundary lines do not hold for any other λ-value other

than 1 so we conclude that for all λ < 1, at least two different wλ strict rankings accompany the

C � B � A pairwise outcomes. However, note that the tie point leaves T2 only after the Borda

Count, so for wλ < 12 , profiles exist where any of the thirteen wλ rankings can accompany the

C � B � A pairwise outcome.

Additionally, this curve also determines how wλ election rankings change with a fixed profile.

Notice that a profile p located between the curve and the A ∼ B plurality line has the plurality

ranking A � B � C. Increasing the λ value moves the tie point along the curve, forcing different wλ

ranking regions to cross p . Thus, p’s election outcome must vary with the procedure. For instance,

the magnified version of T2 in Figure 6b displays the w 14

profile regions. If p has a type 4 election

outcome for w 14, then p has already experienced type 1, 6, and 5 election outcomes for earlier λ

values. In other words, with this fixed profile p, each candidate “can win” by using an appropriate

wλ election rule. That is, when p crosses the profile region for a type 1 election outcome, we know

from Table 5 that the winner will be candidate A, but when p crosses the profile region for a type 5

or 6 outcome, then candidate C will be the winner instead. Considering that the winner is C for w 14,

then based on the λ number of points assigned to the middle-ranked candidate, where 0 ≤ λ ≤ 14 ,

the proportion of voters making up profile p each have the opportunity to have their top-ranked

candidate win the election. Furthermore, by including tied election outcomes, the geometry shows

that each profile in the region between the curve and the A ∼ B plurality boundary line admits

seven different election rankings for different wλ procedures, four of which are strict (types 1, 4, 5,

and 6). A similar argument shows that any profile with a type one plurality outcome that is below

this curve also has seven different positional rankings but now each candidate is bottom-ranked

with some wλ election procedure.

50

Lastly, with regards to the likelihood of these outcomes, note that since the area (derived by

using integration) between the curve and the λ = 0 boundary line for A ∼ B is 112−

19 ln(2) ≈ .0063,

the limiting probability of this peculiar behavior (where any candidate can “win”) is twice this

value because the area of T2 is 12 . Hence, with limiting probability 1

6 −29 ln(2) ≈ .0126, it is possible

for a profile to elect all three candidates when the ballots are tallied with different wλ methods.

Furthermore, considering only profiles in the square (with area 14), the limiting probability for this

behavior is 13 −

49 ln(2) ≈ .0253.

3.5 General Probabilities

Note that a type 4 pairwise ranking C � B � A (given by any generalized profile in the Figure

6a square) can be accompanied by any of the thirteen possible plurality rankings, but what are

some other “likelihood statements” that follow from the triangles? This small subsection addresses

a few.

For instance, the areas of the triangle with a type 3 plurality outcome and the square are,

respectively, 112 and 1

4 . This elementary computation shows that with a C � B � A (type 4)

pairwise outcome, the (conditional) likelihood of a C � A � B (type 3) plurality outcome is the

relative area of these plurality outcomes in the square, or11214

[2]. Again, using a similar simple

area computation we find that if C is the Condorcet (pairwise) winner, then with probability 13 ,

the plurality winner is someone else. However, if either A or B is the Condorcet winner, then, we

know with certainty, that C is the plurality winner.

The Borda Count (λ = 12) and antiplurality (λ = 1) triangles disclose a dramatic decrease in

conflict between the pairwise and positional outcomes; e.g., no longer can all possible positional

outcomes occur; the set of admitted outcomes for a given wλ procedure where 12 ≤ λ < 1 has a

maximum of 4 possible strict outcomes and a minimum of 2 possible strict outcomes. Also, in this

setting, if B or C is the Condorcet winner, then that candidate is the Borda (λ = 12) winner. But

51

notice, when A is the Condorcet winner, the Borda winner could be A or C. The geometry tells

us what to expect: if A is the Condorcet winner, the likelihood C is the Borda winner is 13 . The

antiplurality vote always conflicts with the Condorcet winner if it is A or B.

A small description of these results, along with the probabilities described at the end of Section

3.4, are described by Theorem 3.6 below:

Theorem 3.8. Suppose that profiles are restricted to preferences from the beverage example of

Section 1.1. With limiting probability 16−

29 ln2, it is possible for a profile to elect all three candidates

when the ballots are tallied with different wλ methods. When restricted to where the pairwise

votes define the C � B � A ranking, the probability of this behavior is 13 −

49 ln2. The election

phenomenom where each candidate is bottom-ranked with some wλ procedure has limiting probability

16 − [16 −

29 ln2] = 2

9 ln2.

For λ = 0 the limiting probability of all six possible strict outcomes are equal. If all profiles in

T2 are equally likely, then, given that the pairwise rankings define C � B � A, we have that C is

plurality bottom ranked with (conditional) probability 16 [2]. In contrast, the Borda Count allows

only four strict rankings and the antiplurality method allows only two; the limiting probability of all

strict outcomes in the Borda Count and the antiplurality vote are not equal. [4]

4 Conclusion

The results from Section 2, the Condorcet example setting, and Section 3, the beverage example

setting, prove that surprising inconsistencies arise even if the profiles are restricted to only three

specified voter types. But, they are not alone; all profile classes with only three rankings allow

conflict among the pairwise and positional election rankings. What simplifies the analysis is that

while there are(63

)= 20 different situations to examine, the actual number is sharply reduced by

using symmetries, e.g., changing the names of the candidates assigned to each Figure 1 vertex does

52

not change the theoretical conclusions. Therefore, although we have discussed only 2 of these 20

possible cases, by exploiting the symmetry admitted by voting, we can extend the analysis of these

two cases to cover 14 of the 20 cases instead. Since, in view of Figure 1, we have covered the

setting where three types are one ranking region away from one another and we have covered the

setting where two of the ranking regions are adjacent to one another, the last remaining setting

(which extends to cover the remaining 6 cases) involves the situation where all three of our ranking

regions are next to one another, e.g., voters come from types 1, 2, and 3. However, there are no

real surprises that follow from this analysis so, for our purposes, we have completed the analysis

behind three-candidate voting paradoxes restricted to only three specified voter types.

In conclusion, the beverage example exhibits voters’ preferences where milk, the “winning”

alternative, loses when compared with each of the other choices. If this were the only division

of voters’ preferences causing such troublesome outcome, this example could be dismissed as a

curious anomaly. The discomforting fact is that such election behavior is not exceptional. A

quintessential example of this paradoxical behavior is the Condorcet paradox. Given that the

preference ordering of every voter is transitive, the Condorcet Paradox shows us that the preference

ordering of the majority of the voters may nevertheless be intransitive. Hence, one might wonder

whether these problems are isolated anomalies that can be safely ignored, or issues that must be

seriously considered. In this paper, we answer these questions by using the geometry of voting,

specifically the techniques used by Donald G. Saari and Fabrice Valognes in their article titled

“Geometry, Voting, and Paradoxes”.

53

References

[1] Arrow, K., A. Sen, and K. Suzumura, “Handbook of Social Choice and Welfare”, Chapter27. Web. 1 Nov. 2014., in Elsevier

[2] Donald G. Saari, “Complexity and the Geometry of Voting”, Mathematical Modeling ofVoting Systems and Elections: Theory and Applications, in Elsevier 48.9-10 (2008), pp. 1335-336.

[3] Donald G. Saari, ”Geometry for Positional and Pairwise Voting.” Berlin, Germany, in BasicGeometry of Voting Springer, 1995. 2-42. Print.

[4] Donald G. Saari, Fabrice Valognes, “Geometry, Voting, and Paradoxes”, MathematicsMagazine 71.4 (1998), pp. 243-59.

[5] Gerhlein, William V., and Dominique Lepelley, “Studies in Choice and Welfare”, inVoting Paradoxes and Group Coherence Vol. 7. Springer, 2011. 385. Print.

54