This document discusses voting paradoxes that can arise in elections with three or more candidates. It introduces Condorcet's concept of a Condorcet winner and loser, and describes the Condorcet paradox where pairwise elections can produce cyclic outcomes. It also discusses the Borda count method, where candidates receive points based on their ranking on each ballot. While the Borda count avoids cycles, it is not guaranteed to always select the candidate preferred by the most voters. The document analyzes different voting methods and profiles to understand when paradoxes may occur.
3. 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
4. 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
5. 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
6. 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 c3
n2 c2 c3 c1
n3 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
7. 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
8. 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
9. 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
10. 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
11. 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 w5
6
= (6
6,5
6,0) and w1
2
= (1,1
2,0), 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
12. 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
13. 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
14. 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
15. 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
16. 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
17. 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 = 1
2), 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
18. Example 2.2. Consider the vector (1
6, 0, 1
3, 0, 1
2, 0). The vector (1
6, 0, 1
3, 0, 1
2, 0) is a normalized
three-candidate profile where 1
6 of all voters are of type 1, 1
3 are of type 3, 1
2 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 = n1
n , y = n5
n , and z = n3
n
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
19. 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 Preference
Fraction of
Voters
1 A B C x = n1
n
5 B C A y = n5
n
3 C A B z = n3
n
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 (1
5, 1
10) ∈ T1 corresponds to x = 1
5, y = 1
10, z = 1 − (1
5 + 1
10) = 7
10, where the common
denominator 10 identifies one profile for (1
5, 1
10) 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
20. 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 > 1
2. 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 > 1
2 horizontal dashed
line in T1. Note that it follows from this algebra that there exists a tie between A and B (A ∼
20
21. B) whenever y = 1
2 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
22. 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 (1
2, 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 (1
2, 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
(1
3, 1
3), located in the center of the cyclic region of Triangle T1. The profile (1
3, 1
3) corresponds to
22
23. 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 (1
3, 1
3) 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 (1
4, 1
4) 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 (1
4, 1
4) 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 (1
4, 1
2) and (1
2, 1
4).
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 (2
3, 1
3) maps the outcome A B C. As expected, the profile (2
3, 1
3)
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
24. 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 1
2. 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, 1
4 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
25. 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 1
4 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+2
2 rational points in T1 with common denominator n. Therefore,
n-voters create n+2
2 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−j
n , 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
26. 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 < 1
2, y < 1
2, 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 = 1
n , then
cycles occur if and only if n is odd and x = y = n−1
2n .
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−1
2 . Hence, if n is odd, then using Equation 5, there exist
n−1
2
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
2
j=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
2
j=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
27. 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, 1
4, 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 1
4. 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 1
4 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 1
4,
so long as the total number of voters, n, is a relatively large number. For small values of n, these
1
4 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 1
10 of them. For n = 4 , this probability drops to zero,
then rebounds to 1
7 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
28. rankings with cyclic outcomes is 1
4(1 − 3
n+2 ) if n is odd and 1
4(1 − 9n−6
(n+1)(n+2)) if n is even. The
likelihood of a strict transitive ranking is 1
4(1 + 1
n+2) if n is odd and 1
4(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 = 1
2), 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 1
4 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
29. 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
30. q = (
n1 + λn3
n(1 + λ)
,
n5 + λn1
n(1 + λ)
,
n3 + λn5
n(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+λn3
n(1+λ) . This
quantity is equivalent to
n1
n(1 + λ)
+
λn3
n(1 + λ)
=
x
1 + λ
+
zλ
1 + λ
(10)
Hence, qA = x+λz
1+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
31. 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 (1
3, 1
3) ( −λ
1−2λ , 0)
A ∼ C (2-λ)x + (1 − 2λ)y = 1 − λ (1
3, 1
3) (1−λ
2−λ , 0)
B ∼ C (1+λ)x + (2 − λ)y = 1 (1
3, 1
3) ( 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
32. (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 = 1
3, 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 > 1
4(x+1) define the pairwise ranking B A, so, consequently, the profiles
below the line y = 1
4(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 = 1
3 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 (1
3, 1
3), thereby intersecting at the same point for any λ value and
32
33. 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 = 1
3 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 = 1
3 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 (1
3, 1
3) so by
focusing our attention to the plurality (λ = 0), Borda Count (λ = 1
2), 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
34. 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 = x
1
2 y = 1
3
1 y = −1
2x + 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 = −1
2x+ 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
35. 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 (1
3, 1
3). 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 → 1
2
−
, 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 → 1
2
−
, resulting in a limiting probability
of 2
9 at λ = 1
2 as opposed to 1
9, the limiting probability of the remaining three types. Therefore,
it appears that the Borda Count (λ = 1
2) 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 1
6, the
original value at λ = 0. Hence, the limiting probability of any strict election ranking (in either
35
36. the set of all profiles or the cyclic region) is 1
6 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 → 1
2
−
. Furthermore, as
λ → 1, they then change to monotonically approach the common value 1
6, 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 (1
3, 1
3), 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 (1
6, 1
6). 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 (1
6, 1
6) 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 = 1
3 at λ = 1
2. Therefore, at λ = 1
2, the profile (1
6, 1
6) now lies in the non-cyclic region denoting
the preference of type 3 voters. Hence, when using the Borda Count (λ = 1
2), the profile (1
6, 1
6)
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 (1
6, 1
6) 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
(1
6, 1
6) is A ∼ C B.
36
37. Hence, the profile (1
6, 1
6) 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, 1
2, 1.
Example 2.19. Consider the point ( 3
12, 7
12). 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 = 1
3 at λ = 1
2, the profile ( 3
12, 7
12) experiences two
different wλ election rankings rather than one. That is, note that for for λ = 0, the profile ( 3
12, 7
12)
is on the right side of the boundary line denoting the pairwise tie A ∼ C, yet for λ = 1
2, the profile
( 3
12, 7
12) is now on the left side of the boundary line denoting the pairwise tie A ∼ C. Therefore,
the profile ( 3
12, 7
12) 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 = 1
2. As λ → 1 , we see that clockwise rotation of
the boundary lines does not change the election ranking for the profile ( 3
12, 7
12); the election ranking
remains B C A at λ = 1.
Hence, the profile ( 3
12, 7
12) 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
38. 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 = 1
2 setting, i.e., the Borda Count, we see that the pairwise
38
39. A B C ranking denoted by region 1 is accompanied by a w1
2
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 w1
2
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 1
6 for λ = 0, 1. For the Borda Count the limiting
probability for either setting is 2
9 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 → 1
2.
These results show that even with only three types of voter preferences, conflict can arise
39
40. 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
41. 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 Preference
Fraction of
Voters
2 A C B x = n2
n
5 B C A y = n5
n
4 C B A z = n4
n
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
42. 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 = n2
n , y = n5
n , and z = n4
n 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 ( 6
15, 4
15) in T2, implying that z = 5
15.
To represent the {A,B} pairwise vote, notice that A beats B if and only if x > 1
2: in Fig. 4b,
these generalized profiles are to the right of the x = 1
2 vertical dashed line in T2. Similarly, B beats
C if and only if y > 1
2: these generalized profiles are above the horizontal dashed line in T2. The
final A C relationship also has x > 1
2. 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
43. 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 < 1
2; 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 1
2 of the profiles in T2 to have a type 4
outcome while about 1
4 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
44. 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 (1
2, 1
2) ( λ
1+λ, 0)
A ∼ C (2-λ)x + (1 − λ)y = 1 (1, −1) ( 1
2−λ, 0)
B ∼ C (1- 2λ)x + 2(1 − λ)y = 1 − λ (0, 1
2) ( 1−λ
1−2λ, 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
45. candidate pairs of Section 2.3 were all satisfied by the same point, (1
3, 1
3), 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 (1
2, 1
2), the rotation point for the A ∼ C
boundary line is (1, −1), and the rotation point for the B ∼ C boundary line is (0, 1
2). These three
rotation points are indicated by the solid dots in Figure 5 for the plurality (λ = 0), the Borda
Count (λ = 1
2), 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, ( λ
1+λ , 0), approaches the point (1
2, 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 ( 1
2−λ , 0) goes to infinity (is undefined) for λ = 1
2, the C ∼ B boundary line has no
x-intercept point at λ = 1
2. 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
46. 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 (1
3, 1
3) so the point (1
3, 1
3)
is the tie point for this election rule. However, once λ reaches the value 1
2, the three boundary
lines rotate so that now they intersect at the point (0, 1
2). 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 = 1
2), 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 (1
3, 1
3), it moves to the T2
boundary at (1
2, 1
2) when λ = 1
2, and it vanishes at infinity (there is no three-way tie point!) when
46
47. 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 w1
2
, 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
48. 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 = 1
2 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
49. (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 = 1
4 while the dashed lines
indicate the λ = 1
4 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 w1
4
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 w1
4
ranking; that is, we can select a point that is
strictly inside one of the w1
4
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 < 1
2. In contrast, we learn
49
50. from Figure 6a that because the y-value approaches infinity as λ → 1 and x → 2
3, 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 < 1
2, 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 w1
4
profile regions. If p has a type 4 election
outcome for w1
4
, 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 w1
4
,
then based on the λ number of points assigned to the middle-ranked candidate, where 0 ≤ λ ≤ 1
4,
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
51. 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 1
12 − 1
9ln(2) ≈ .0063,
the limiting probability of this peculiar behavior (where any candidate can “win”) is twice this
value because the area of T2 is 1
2. Hence, with limiting probability 1
6 − 2
9ln(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 1
4), the limiting probability for this
behavior is 1
3 − 4
9ln(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, 1
12 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, or
1
12
1
4
[2]. Again, using a similar simple
area computation we find that if C is the Condorcet (pairwise) winner, then with probability 1
3,
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 = 1
2) 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 1
2 ≤ 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 = 1
2) winner. But
51
52. 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 1
3. 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 1
6 − 2
9ln2, 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 1
3 − 4
9ln2. The election
phenomenom where each candidate is bottom-ranked with some wλ procedure has limiting probability
1
6 − [1
6 − 2
9ln2] = 2
9ln2.
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 1
6 [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 6
3 = 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
53. 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
54. References
[1] Arrow, K., A. Sen, and K. Suzumura, “Handbook of Social Choice and Welfare”, Chapter
27. Web. 1 Nov. 2014., in Elsevier
[2] Donald G. Saari, “Complexity and the Geometry of Voting”, Mathematical Modeling of
Voting 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 Basic
Geometry of Voting Springer, 1995. 2-42. Print.
[4] Donald G. Saari, Fabrice Valognes, “Geometry, Voting, and Paradoxes”, Mathematics
Magazine 71.4 (1998), pp. 243-59.
[5] Gerhlein, William V., and Dominique Lepelley, “Studies in Choice and Welfare”, in
Voting Paradoxes and Group Coherence Vol. 7. Springer, 2011. 385. Print.
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