International Baccalaureate
Diploma Programme
Internal Assessment
The Mathematics Behind Different Voting Systems and
Arrow’s Impossibility Theorem
Subject: Mathematics Analysis and Approaches HL
Number of pages: 20
Introduction
Since general elections in my country were approaching, I was discussing the advantages and
disadvantages of different voting systems with my best friend. He told me that it has been
mathematically proven that no voting system can satisfy even just a few important criteria.
While his statement turned out to be incorrect, since it applied only to a specific type of
voting systems, it led me to discover Arrow’s Impossibility Theorem and develop an interest
in the mathematics of voting/mathematical representations of elections. I was amazed by the
fact that mathematics could be used to optimise decision-making and to improve something
that to me seemed to exist in a world completely separate from it, so I decided to dedicate my
Internal Assessment to something I am passionate about, politics and elections.
In this paper, I will use set theory to mathematically define and compare different types of
ranked voting systems by applying them to data I have gathered from simulating elections
inside my friend group, then analyze their weaknesses by using Arrow’s Impossibility
Theorem (AIT), and consider range voting as alternatives that AIT does not apply to.
What is a ranked voting system?
Voting systems are sets of rules that determine how elections are conducted and how their
outcomes are determined. In a ranked voting system, each voter submits a ranked list of all
the candidates in the election. While this intuitive definition captures the basic idea, a precise
mathematical definition is needed for this paper. To define a ranked voting system
mathematically, I will first introduce some concepts from set theory.
A binary relation over some sets 𝐴 and 𝐵 is a new set of ordered pairs (𝑎, 𝑏) consisting of
elements 𝑎 in 𝐴 and 𝑏 in 𝐵. In other words, a binary relation over two sets 𝐴 and 𝐵 is a subset
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of their Cartesian product 𝐴 × 𝐵, where the Cartesian product is the set of all ordered pairs
(𝑎, 𝑏).
For example, if we are presented with sets 𝐴 = {1, 2} and 𝐵 = {𝑋, 𝑌, 𝑍}, one possible binary
relation over 𝐴 and 𝐵 is 𝑆 = {(1, 𝑋), (1, 𝑌), (2, 𝑌)}. As we can see from this example, not all
elements of the original sets must appear in an arbitrary binary relation.
𝐴 and 𝐵 are allowed to be the same set. When 𝐴 = 𝐵, we call it a relation on set 𝐴. For
example, one binary relation on set 𝐴 = {1, 2} is 𝑆 = {(1,1), (1,2), (2,1)}.
A piece of notation often used when dealing with relations that we will use in the rest of this
paper is using 𝑎 ~ 𝑏 when we want to say that a pair (𝑎, 𝑏) belongs to the relation.
A strict total order over some set 𝐴 is a binary relation ~ which satisfies the following
criteria for all 𝑎, 𝑏, and 𝑐 in 𝐴:
1. Irreflexivity: ¬(a ~ a).1
2. Transitivity: If 𝑎 ~ 𝑏 and 𝑏 ~ 𝑐 then 𝑎 ~ 𝑐.
3. Asymmetry: If 𝑎 ~ 𝑏 then ¬(𝑏 ~ 𝑎).
4. Connectivity: If 𝑎 ≠ 𝑏 then 𝑎 ~ 𝑏 or 𝑏 ~ 𝑎.
We will now use this knowledge to mathematically define a ranked voting system.
Let 𝐶 be a set of all possible candidates for the election and let 𝑁 be the number of voters.
Also, let 𝑇(𝐶) be the set of all possible total orders of 𝐶.
For each total order 𝑇 ∈ 𝑇(𝐶), we say that if 𝑏 ~ 𝑎 for some 𝑎, 𝑏 ∈ 𝐶, then candidate 𝑎 is
ranked higher than candidate 𝑏. Because of connectivity, we know that for every pair of
candidates, one must be ranked higher than the other, and because of asymmetry we know
that no two different candidates can be ranked both above and below each other. Moreover,
1
in mathematical logic the ¬ symbol denotes negation
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transitivity ensures that if candidate 𝑎 is ranked higher than the candidate 𝑏, it is also ranked
higher than any candidate 𝑐 which is ranked below 𝑏. Finally, irreflexivity tells us that a
candidate cannot be ranked above themselves, which would make no logical sense.
Therefore, each 𝑇 corresponds to some ranked list of all candidates. Furthermore, set 𝑇(𝐴)
represents the set of all possible ranked lists of elements of 𝐴.
We define a ranked voting system as a function
𝐹: 𝑇(𝐶)𝑁 → 𝑇(𝐶)
which maps an 𝑁-tuple2 of total orders of 𝐶 to just one. We will denote this N-tuple by
(𝑉1 , 𝑉2 , ⋯ , 𝑉𝑁 ), where 𝑉𝑖 corresponds to the preference list of the voter 𝑖, and we will call it a
preference profile. In other words, a ranked voting system is any function that takes 𝑁 ranked
lists of candidates as input and outputs a single ranked candidate list, which determines the
result of the election.
Now that we have a mathematical definition of this type of voting systems, let’s take a look at
some of the most popular ranked voting systems.
Plurality Voting (First Past the Post)
One of the simplest voting systems is a plurality voting. In this system, the candidates are
ranked based on the number of people who ranked them above all other candidates.
Before we formally define plurality voting, we will first introduce the concepts of the
maximal element and cardinality.
The maximal element of a total order of some set 𝐴 is an element 𝑥 such that 𝑦 ~ 𝑥 for every
𝑦 ≠ 𝑥 in 𝐴. Another way of thinking about it is substituting the relation ~ with “less than”. In
2
An 𝑁-tuple is an ordered list of 𝑁 elements.
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that case, the maximal element would be the greatest element in 𝐴. This is a particularly
intuitive way, since the most common usage of the term “maximal element” is precisely for
denoting the greatest element in a set.
The cardinality of a set 𝐴, denoted by |𝐴|, is the number of elements in 𝐴. For example, if
𝐴 = {1, 3, 7} then |𝐴| = 3.
We will now use these concepts to define a plurality voting system.
Let 𝑣𝑖 be the maximal element of 𝑉𝑖 for each 𝑖 ∈ {1, 2, ⋯ , 𝑁}. Notice that the maximal
element here is the candidate ranked the highest by voter 𝑖.
Now, let’s look at the following function which takes some candidate 𝑐 ∈ 𝐶 as input:
𝑓(𝑐) = |{ 𝑖 ∈ {1, 2, ⋯ , 𝑁} ∣ 𝑣𝑖 = 𝑐 }|
The function 𝑓 outputs the cardinality of a set of indices of voters who have candidate 𝑐 as
the maximal element in their preference list. In other words, 𝑓 gives us the number of voters
who ranked 𝑐 above all other candidates.
We define plurality voting as a ranked voting system that produces a total order of
candidates that satisfies the following for every two candidates 𝑐𝑖 and 𝑐𝑗 :
𝑐𝑖 ~ 𝑐𝑗 if and only if 𝑓(𝑐𝑖 ) < 𝑓(𝑐𝑗 ).
Note that this definition does not account for cases where two candidates have the same
number of people who ranked them first. In fact, voting systems, as defined in this paper, do
not allow candidates to share a place in the final list. One possible solution for this is to, in
case of a tie, randomly select an ordering of tied candidates. We will use this method to solve
this problem for all ranking systems defined in this paper, using RANDOM.ORG as our
random number generator of choice.
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This means that the condition above is not completely true, since two candidates can belong
to a relation even if their values of 𝑓 are equal. Therefore, we will replace it with the
following pair of conditions:
𝑐𝑖 ~ 𝑐𝑗 if 𝑓(𝑐𝑖 ) < 𝑓(𝑐𝑗 ),
(𝑐𝑖 ~ 𝑐𝑗 or 𝑐𝑗 ~ 𝑐𝑖 ) if 𝑓(𝑐𝑖 ) = 𝑓(𝑐𝑗 ).
Note that this makes plurality voting a collection of multiple ranked voting systems, not a
unique system.
Instant Runoff
Another popular voting system is instant runoff. This system is more sophisticated than
simple plurality voting, and it consists of several rounds. In each round, the candidates are
ranked based on the number of people who ranked them above all other candidates, just like
in plurality voting. Then, if there are two or more candidates left, the one that was ranked first
by the least number of voters is eliminated and they get removed from all voting lists so that
their votes get redistributed to remaining candidates. The eliminated candidate is ranked
above than all the candidates eliminated before them and below all remaining candidates in
the final list. The last remaining candidate is the winner of election.
Based on this description, we can define instant runoff as a ranked voting system that
outputs a total order of the set of candidates 𝐶 according to the following algorithm:
At each round of the voting process, let 𝑆 be the set of candidates still in contention
for the election and 𝑅 be the output of instant runoff. Initially, 𝑆 = 𝐶 and 𝑅 = ∅.
Find the candidate 𝑐 ∈ 𝑆 with the lowest value of 𝑓(𝑐), where 𝑓 is the function that
we defined in the previous section. If there are more than once such 𝑐, choose one at
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random. Eliminate 𝑐 from 𝑆 and redistribute its votes to the remaining candidates
according to each voter's next highest ranked candidate still in contention. Essentially,
this means that we replace 𝑆 with 𝑆 − {𝑐} and remove all elements containing 𝑐 from
each 𝑉𝑖 , so the next round is done with these reduced preference lists. For each 𝑘 ∈ 𝑆
except for 𝑘 = 𝑐, add pair (𝑐, 𝑘) to 𝑅. This ensures that 𝑐 is ranked below all
remaining candidates in the final output.
Repeat this process until only one candidate remains in 𝑆.
Note that in the end, for every two candidates 𝑥 and 𝑦, either (𝑥, 𝑦) or (𝑦, 𝑥) is in 𝑅, where
the first element of this pair is the candidate that was eliminated before first. This ensures that
𝑅 is a total order of 𝐶.
Borda Count
The last ranked voting system we will be examining is the Borda count. This system is much
different from the previous two, and it assigns a score to each candidate. More precisely, for
each preference list 𝑉𝑖 , Borda count assigns 0 points to the lowest ranked candidate, 1 point
to the second lowest ranked one, and so on, assigning |𝐶| − 1 points to the highest ranked
voter. It then ranks all the candidates by their total number of points, called the Borda score.
In order to formally define this voting system, we need to mathematically express the ranking
of some candidate in the preference list of given voter. We do so by using the following
function:
𝑟𝑎𝑛𝑘(𝑐, 𝑖) = |{ 𝑑 ∈ 𝐶 ∣ (𝑐, 𝑑) ∈ 𝑉𝑖 }| + 1.
As we can see, the 𝑟𝑎𝑛𝑘(𝑐, 𝑖) gives us the number of candidates ranked above 𝑐 in the
preference list of voter 𝑖. Moreover, this number is increased by 1, so that the candidate
ranked the highest has rank 1, the second highest ranked candidate has rank 2, and so on.
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We can define Borda score as the following function which takes some candidate 𝑐 ∈ 𝐶 as
input:
𝐵(𝑐) = ∑𝑁
𝑖=1(|𝐶| − 𝑟𝑎𝑛𝑘(𝑐, 𝑖)).
Finally, we define Borda count as any ranked voting system that produces a total order of
candidates that satisfies the following for every two candidates 𝑐𝑖 and 𝑐𝑗 :
𝑐𝑖 ~ 𝑐𝑗 if 𝐵(𝑐𝑖 ) < 𝐵(𝑐𝑗 ),
(𝑐𝑖 ~ 𝑐𝑗 or 𝑐𝑗 ~ 𝑐𝑖 ) if 𝐵(𝑐𝑖 ) = 𝐵(𝑐𝑗 ).
Election Simulation
In order to compare plurality voting, instant runoff, and Borda count, I ran a few mock
elections with five of my friends. Each one of us was asked to submit a ranked list of all six
of us, based on how likely we considered each of us to be to successfully accomplish certain
task. The task was different for each election to make the data more diverse. Moreover, in
order to avoid getting biased data, the order of every person in their own final ranking list
was generated randomly.
I will analyze results of the one of these elections here, which was chosen to highlight certain
situations that can happen when using ranked voting systems.
𝑽𝟏
𝑽𝟐
𝑽𝟑
𝑽𝟒
𝑽𝟓
𝑽𝟔
1
3
2
3
3
6
2
2
5
5
1
5
4
5
3
1
3
5
1
5
1
4
4
1
2
6
2
4
5
2
4
6
2
3
6
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6
6
6
4
4
1
3
Table 1. All the data obtained from MOCK elections
Table 1 shows the ranking lists of each voter. The rows in this table correspond to different
rankings of candidates in all lists, while the columns correspond to different lists. The
candidates are enumerated from 1 to 6, where candidate 𝑖 and voter 𝑉𝑖 are in fact the same
person.
We will first generate results of this election using plurality voting, by computing the values
of the function 𝑓 we defined above.
Candidate C
3
2
6
4
1
5
f (c)
3
2
1
0
0
0
Table 2. Election results computed by plurality voting function
Table 2 shows the resulting list produced by plurality voting, as well as values of 𝑓 for each
candidate. As we can see from the table, voters 1, 4, and 5 all have the value of 0, so their
relative order was randomly assigned.
We will now use instant runoff to generate the result of our election. Notice that in instant
runoff, eliminating candidates with zero votes redistributes no votes to remaining candidates,
so no matter in which order we eliminate voters 1, 4, and 5, we will end up in the state
depicted in Table 3. In this case, we ended up eliminating candidates 4, 5, and 1 in that order.
Candidate 𝒄
𝒇(𝒄)
3
3
2
2
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1
6
Table 3. Election results calculated using the instant runoff function
As we can see from the table, at this round candidate 6 has the lowest value of 𝑓, so they
were eliminated and their vote got distributed to candidate 2, resulting in situation depicted in
Table 4.
Candidate 𝒄
𝒇(𝒄)
3
3
2
3
Table 4
At this round, candidates 3 and 2 are tied for the first place, so the tie was broken by
randomly eliminating candidate 3. The final list produced by instant runoff is presented in
Table 5.
Rank
1
2
3
4
5
6
candidate
2
3
6
1
5
4
Table 5
Finally, we will generate results by using Borda count.
Candidate C
5
3
2
1
4
6
B (C)
22
19
16
15
9
9
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Table 6. Election results obtained via Borda count
Table 6 shows the resulting ranked list, as well as the Borda score for each candidate.
Candidates 4 and 6 were tied, so their relative order was randomly determined.
If we compare results produced by these three methods, we see that the resulting lists are
substantially different, despite being results of the same election.
For example, the winners in each list are different, being candidates 3, 2, and 5 for plurality
voting, instant runoff, and Borda count, respectively. It is important to note that candidate 3
had even odds of winning in instant runoff as well. Still, this election had only 6 voters and 6
candidates, and we can easily imagine an election of much larger scale where a big number of
candidates is tied for first place, giving each one of them a very small chance of winning.
Moreover, while the resulting lists for plurality voting and instant runoff are fairly similar and
they could have ended up being the same by chance, the list for Borda count is substantially
different. For example, it favours candidate 5 even though 𝑓(5) = 0, because they were
either second or third choice of every voter. It also ranks candidate 6 below every other
candidate due to them being in the bottom half of almost every voter’s list, even though they
have a higher value of 𝑓 than half of candidates.
This raises the question of which one, if any, of these voting systems should we trust. The
answer does not appear to be obvious, and in practice, different ranked voting systems are
used in different circumstances. Most elections in United States use plurality voting, the
President of India and winners of Oscars are chosen through instant runoff voting, and some
members of the National Assembly of Slovenia, as well as winners of many sport awards, are
determined using Borda count.
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However, it would be good if we had some way to pinpoint the flaws of each one of these
voting systems. Fortunately, there is a theorem in social choice theory which help us do that.
Arrow’s Impossibility Theorem
A particularly important result for mathematics of voting is Arrow’s Impossibility Theorem
(AIT), discovered by Kenneth J. Arrow in 1950. The theorem says that no ranked voting
system can satisfy the following three criteria:
1. Pareto efficiency: If every voter prefers candidate 𝐴 over candidate 𝐵, then the
results will also prefer 𝐴 over 𝐵 as.
2. Independence of irrelevant alternatives: If every voter’s preference between 𝐴 and
𝐵 stay the same, then the order of 𝐴 and 𝐵 in the resulting list will stay the same even
if voters’ preferences about any other pairs (which might include 𝐴 or 𝐵) change.
3. Non-dictatorship: There is no voter such that their preference between any two
candidates determine the order of these candidates in the resulting list regardless of
other voters.
All three of these traits are considered desirable for a voting system, so we can use this
theorem to point out flaws in previously defined systems. However, we will first present a
more formal statement of the Arrow’s theorem.
Theorem (Arrow’s Impossibility Theorem): There is no ranked voting system 𝐹 with more
than two candidates that satisfies the following three conditions, called Pareto efficiency
(PE), independence of irrelevant alternatives (IIA), and non-dictatorship (ND), respectively:
1. If candidate 𝐴 is ranked above candidate 𝐵 in each of 𝑉1 , 𝑉2 , ⋯ , 𝑉𝑁 , then 𝐴
will be ranked above 𝐵 in 𝐹(𝑉1 , 𝑉2 , ⋯ , 𝑉𝑁 ) as well.
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2. For any two preference profiles (𝑉1 , 𝑉2 , ⋯ , 𝑉𝑁 ) and (𝑉′1 , 𝑉′2 , ⋯ , 𝑉′𝑁 ) such
that for each voter 𝑖, candidate 𝐴 is ranked above 𝐵 by either both or none of
𝑉𝑖 and 𝑉′𝑖 , it is true that candidate 𝐴 is ranked above 𝐵 by either both or none
of 𝐹(𝑉1 , 𝑉2 , ⋯ , 𝑉𝑁 ) and 𝐹(𝑉′1 , 𝑉′2 , ⋯ , 𝑉′𝑁 ).
3. There exists no single voter 𝑖 such that for all possible preference profiles
(𝑉1 , 𝑉2 , ⋯ , 𝑉𝑁 ), the equality 𝐹(𝑉1 , 𝑉2 , ⋯ , 𝑉𝑁 ) = 𝑉𝑖 holds. In other words, there
exists no single voter 𝑖 such that 𝑋 > 𝑌 in 𝑉𝑖 implies 𝑋 > 𝑌 in
𝐹(𝑉1, 𝑉2 , ⋯ , 𝑉𝑁 ) for any two candidates 𝑋 and 𝑌.
There are many proofs of this theorem, most notable ones being the 1950’s original proof by
Arrow himself and the John Geanakoplos’ proof from 2005. In this section, we will present a
variation of a more recent proof, presented by Ning Niel Yu in 2012. It was chosen because
of its compactness and simplicity.
In the proof, we will use 𝑋 > 𝑌 to say that candidate 𝑋 is ranked above 𝑌. Note that if 𝑋 > 𝑌
and 𝑌 > 𝑍, then 𝑋 is ranked above 𝑍 as well, so we can merge these two inequalities into
𝑋 > 𝑌 > 𝑍. The motivation for introducing this notation is that, when comparing relative
ranking of different candidates for different preference lists, it can become hard for the reader
to follow. Hopefully it will address this problem.
Proof: Let’s suppose some ranked ranking system 𝐹 satisfied both PE and IIA. We will show
that this system must violate non-dictatorship.
Consider an arbitrary preference profile 𝑃 = (𝑉1 , 𝑉2 , ⋯ , 𝑉𝑁 ) in which candidate 𝐴 > 𝐵 in
each preference list 𝑉𝑖 . PE guarantees that 𝐹(𝑃) will rank 𝐴 above 𝐵.
We will perform a sequence of 𝑁 steps: At step 𝑖, we will swap the places of 𝐴 and 𝐵 in 𝑉𝑖 ,
leaving other preference lists unchanged. This will produce a new preference profile 𝑃𝑖 =
(𝑉′1 , 𝑉′2 , ⋯ , 𝑉 ′ 𝑖 , 𝑉𝑖+1 , ⋯ , 𝑉𝑁 ). After the sequence is over, we will obtain a new preference
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profile 𝑃𝑁 = (𝑉′1 , 𝑉′2 , ⋯ , 𝑉′𝑁 ) where every 𝑉′𝑖 is identical to 𝑉𝑖 , except for the places of 𝐴
and 𝐵 being swapped. This means that in the new list, 𝐵 > 𝐴 in every preference list, so PE
guarantees us that 𝐹( 𝑃𝑁 ) will rank 𝐵 above 𝐴.
Clearly, at some step 𝑖, we obtained a preference profile 𝑃𝑖 = (𝑉′1 , 𝑉′2 , ⋯ , 𝑉 ′ 𝑖 , 𝑉𝑖+1 , ⋯ , 𝑉𝑁 )
such that 𝐹(𝑃𝑖 ) ranks 𝐵 above 𝐴. Let 𝑖 be the first such step. We will refer to voter 𝑖 as the
(𝑨, 𝑩)-pivotal voter and we will define the number 𝑛𝐴𝐵 = 𝑖. IIA tell us that this number does
not depend on the choice of 𝑃.
𝑽𝟏
…
𝑽𝒏𝑨𝑩 −𝟏
𝑽𝒏𝑨𝑩
𝑽𝒏𝑨𝑩 +𝟏
…
𝑽𝑵
B
…
B
A
A
…
A
C
…
C
B
B
…
B
A
…
A
C
C
…
C
Table 7.
Let’s now choose an arbitrary candidate 𝐶, different from 𝐴 and 𝐵. We will now examine an
arbitrary preference profile 𝑃′ = (𝑉1 , 𝑉2 , ⋯ , 𝑉𝑁 ) depicted in Table 7. Here 𝐵 > 𝐶 > 𝐴 in the
first 𝑛𝐴𝐵 − 1 preference lists, while 𝐴 > 𝐵 > 𝐶 in the rest. As we can see, 𝐵 > 𝐶 in every
preference list, so 𝐵 > 𝐶 in 𝐹(𝑃′) as well. On the other hand, notice that each voter’s
preference between 𝐴 and 𝐵 is the same in 𝑃′ and 𝑃𝑛𝐴𝐵 −1 from before, so 𝐴 > 𝐵 in 𝐹(𝑃′) by
the definition of 𝑛𝐴𝐵 . Therefore, 𝐴 > 𝐵 > 𝐶 in 𝐹(𝑃′).
Now, let’s examine a preference profile 𝑃′′ obtained by swapping the places of 𝐴 and 𝐵 in
preference list 𝑛𝐴𝐵 , in which we now have 𝐵 > 𝐴 > 𝐶. Since the order of 𝐴 and 𝐶 did not
change in any preference list, we know that 𝐴 > 𝐶 in 𝐹(𝑃′′) by PE. However, each voter’s
preference between 𝐴 and 𝐵 is the same in 𝑃′′ and 𝑃𝑛𝐴𝐵 , so now 𝐵 > 𝐴 in 𝐹(𝑃′′). Therefore,
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𝐵 > 𝐴 > 𝐶 in 𝐹(𝑃′′). An important observation is that this conclusion was completely
determined by the voters’ preferences between 𝐴 and other two candidates.
𝑽𝟏
…
𝑽𝒏𝑨𝑩 −𝟏
?
…
?
B
…
B
?
…
?
A
…
A
𝑽𝒏𝑨𝑩
𝑽𝒏𝑨𝑩 +𝟏
…
𝑽𝑵
A
…
A
?
…
?
A
B
…
B
C
?
…
?
B
Table 8.
Now consider the collection of preference profiles depicted in Table 8, in which the question
marks denote the possible positions of 𝐶 relative to 𝐴 and 𝐵. We can see that no matter how
we choose the positions of 𝐶 in each list, all voters’ preferences between 𝐴 and 𝐶, as well as
𝐴 and 𝐵, stay the same as in 𝑃′′. By IIA, the same arguments for 𝐵 > 𝐴 and 𝐴 > 𝐶 hold for
these preference profiles, so for any profile 𝑃 ∗ belonging to this collection, we have 𝐵 > 𝐴 >
𝐶 in 𝐹(𝑃∗ ). Note that this collection contains all possible configurations of voters’
preferences between 𝐵 and 𝐶 where 𝐵 > 𝐶 for voter 𝑛𝐴𝐵 . Using IIA, we can now conclude
that whenever 𝐵 > 𝐶 for voter 𝑛𝐴𝐵 , our voting system 𝐹 will also rank 𝐵 above 𝐶, regardless
of other voters’ preferences. We will call voter 𝑛𝐴𝐵 a dictator over 𝑩𝑪.
Note that this result was completely independent from our choice of 𝐴, 𝐵, and 𝐶, so a more
general claim holds: For any three candidates 𝑋, 𝑌, and 𝑍, voter 𝑛𝑋𝑌 is a dictator over 𝑌𝑍.
We will now do the same thing from the start of the proof where we considered an arbitrary
preference profile, but we will swap 𝐴 and 𝐵 for 𝐶 and 𝐵, so now in our profile we have 𝐶 >
𝐵 in each preference list. We will perform the same sequence of 𝑁 steps where we swap the
positions of 𝐵 and 𝐶 in 𝑉𝑖 at step 𝑖, eventually obtaining a preference profile for which 𝐵 > 𝐶
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in each preference list. We will define (𝑪, 𝑩)-pivotal voter and 𝑛𝐶𝐵 analogously to (𝑨, 𝑩)pivotal voter and 𝑛𝐴𝐵 . Notice that, since the voter 𝑛𝐴𝐵 is a dictator over 𝐵𝐶, step 𝑛𝐴𝐵 will
produce a preference profile 𝑃𝑛𝐴𝐵 such that 𝐵 > 𝐶 in 𝐹(𝑃𝑛𝐴𝐵 ). This means that:
𝑛𝐶𝐵 ≤ 𝑛𝐴𝐵 .
However, we can now do the same thing, but swap the order of 𝐶 and 𝐵, which means that
we start with a preference profile such that 𝐵 > 𝐶 in each preference list, perform the 𝑁 steps
described above, and end up with a profile such that 𝐶 > 𝐵 in each list. Notice that in this
case, before step 𝑛𝐴𝐵 , we have 𝐵 > 𝐶 for voter 𝑛𝐴𝐵 , so 𝐹 will also rank 𝐵 above 𝐶 before this
step. This implies that:
𝑛𝐴𝐵 ≤ 𝑛𝐵𝐶 .
Combining the two inequalities together, we have:
𝑛𝐶𝐵 ≤ 𝑛𝐴𝐵 ≤ 𝑛𝐵𝐶
(1)
Once again, this result was completely independent from our initial choice of the
triplet (𝐴, 𝐵, 𝐶), so we have a more general result: For any three candidates 𝑋, 𝑌, and 𝑍, the
following inequality holds:
𝑛𝑍𝑌 ≤ 𝑛𝑋𝑌 ≤ 𝑛𝑌𝑍 .
This means that by considering a triplet (𝐴, 𝐶, 𝐵), we can do the same thing we did above to
obtain the following inequality:
𝑛𝐵𝐶 ≤ 𝑛𝐴𝐶 ≤ 𝑛𝐶𝐵
(2)
From inequalities (1) and (2), we obtain the following:
𝑛𝐵𝐶 = 𝑛𝐶𝐵 = 𝑛𝐴𝐵
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(3)
Moreover, we can easily extend this to all other pivotal voters. For example, we can first
substitute 𝐶 for any other voter 𝑍 different from 𝐴 and 𝐵, and obtain 𝑛𝐵𝑍 = 𝑛𝑍𝐵 = 𝑛𝐴𝐵 for
all 𝑍. We can then swap 𝐴 and 𝐵 to obtain 𝑛𝐴𝑍 = 𝑛𝑍𝐴 = 𝑛𝐵𝐴 = 𝑛𝐴𝐵 for all voters 𝑍, and
then substitute 𝐴 for any voter 𝑋 different from 𝐵 to obtain the following:
𝑛𝑋𝑍 = 𝑛𝑍𝑋 = 𝑛𝐵𝑋 = 𝑛𝑋𝐵 = 𝑛𝐴𝐵
(4)
Finally, equality (4) implies 𝑛𝑋𝑍 = 𝑛𝐴𝐵 for any two voters 𝑋 and 𝑍, meaning that the same
voter is the pivotal voters for all pairs of candidates.
However, we know that for any pair of candidates, there exists some pivotal voter who is
dictator over that pair. Since we have just concluded that all pivotal voters are in fact the
same voter, we now conclude that there exists a voter who is a dictator over all possible pairs
of candidates, which violates ND. ∎
What AIT has to say about our ranked voting systems
Since all three voting systems that we have examined so far are ranked voting systems,
Arrow’s Impossibility Theorem applies to all of them. While the theorem may not tell us
which of these voting systems we should use, it tells us where to look for flaws.
Starting with plurality voting, it is obvious that there is no dictator, since in a situation where
every voters except for one ranks candidate 𝐴 above everyone else, this candidate will be
ranked first by plurality voting no matter what the last voter’s preferences are.
However, the version of plurality voting violates PE. We can imagine a scenario where both
𝑓(𝐴) = 0 and 𝑓(𝐵) = 0, but every voter prefers 𝐴 to 𝐵. Still, since their order is arbitrarily
determined, it is possible for candidate 𝐵 to end up above 𝐴 in the final list.
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More importantly, plurality voting violates IIA. Imagine a scenario where candidate 𝐴 is
ranked first by a majority of voters, while candidate 𝐵 is ranked first by the rest. In this
scenario, plurality voting will rank 𝐴 above all other candidates, including 𝐵. However,
consider an alternative scenario where every voter who prefers 𝐴 ranks candidate 𝐶 first
instead and the rest of the lists stay the same, meaning that every voter’s preference between
𝐴 and 𝐵 stay the same. However, now we have 𝑓(𝐴) = 0 < 𝑓(𝐵), so plurality voting will
rank 𝐵 above 𝐴 despite the fact that their relative order in every voting list is the same as in
the first scenario, which violates IIA.
Next, we move on to instant runoff. For the same reasons as plurality voting, the version of
instant runoff we defined in this paper satisfies ND and violates PE.
It turns out that instant voting violates IIA as well. In fact, the example used for plurality
voting works here as well. Since in the first scenario every voter ranks either 𝐴 or 𝐵 first,
other candidates will be eliminated without changing values of 𝑓(𝐴) and 𝑓(𝐵), before
eliminating 𝐵. In the end, 𝐴 will still end up winning the election. However, since in the
second scenario every voter ranks either 𝐶 or 𝐵 first, voter 𝐴 will get eliminated with the rest
of candidates whose value of 𝑓 is zero before 𝐵 gets eliminated, which means that 𝐵 will be
ranked above 𝐴 by instant runoff, even though the relative order of 𝐴 and 𝐵 is the same in
both scenarios, which violates IIA.
Finally, we have Borda count. It satisfies ND for the same reason plurality and instant runoff
do, but it also satisfies PE. Since a candidate gets more points for being ranked higher in any
specific preference list, if every voter ranks 𝐴 above 𝐵 then we know that 𝐴 will get more
points from each preference list than 𝐵, making its final Borda score higher.
However, Borda count also violates IIA. Consider the following scenario: There are 3
candidates and 5 voters. In first 3 lists 𝐶 > 𝐴 > 𝐵, while 𝐵 > 𝐶 > 𝐴 in the remaining 2 lists.
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In this scenario, we have 𝐵(𝐴) = 3 and 𝐵(𝐵) = 4, so Borda count will rank 𝐵 above 𝐴.
Now imagine a scenario where in the first 3 lists 𝐴 > 𝐶 > 𝐵, while the remaining 2 lists stay
the same. While the relative order of 𝐴 and 𝐵 stayed the same for all lists, now we have
𝐵(𝐴) = 6 and 𝐵(𝐵) = 4, so Borda count will rank 𝐵 above 𝐴, which violates IIA.
It turned out that all three of our voting systems violate IIA and that two of them violate PE
as well3. In fact, because of AIT we know that all ranked voting systems we come up with
will share the same fate. Therefore, it makes sense to look at for an alternative to ranked
voting systems.
Range Voting
Range voting is a voting system in which voters assign scores to candidates rather than
ranking them. This makes it a non-ranked voting system that satisfies all three Arrow’s
criteria. We will define range voting and discuss its pros and cons compared to ranked voting
systems.
Let 𝑆 be a non-empty set of real numbers, which represents the set of possible scores. Note
that this set does not have to be finite, since 𝑆 = ℕ and 𝑆 = ℝ are valid choices.
We will define a function
𝑠: {1, 2, ⋯ , 𝑁} × 𝐶 → 𝑆,
where 𝑠(𝑖, 𝑐) represents a score voter 𝑖 assigns to candidate 𝑐. We will call 𝑠 a scoring
function.
Finally, let 𝑉𝑖 (𝐶) = (𝑠(𝑖, 1), 𝑠(𝑖, 2), ⋯ , 𝑠(𝑖, |𝐶|)) be a scoring list for voter 𝑖.
3
Although there are versions of both plurality and instant runoff that do satisfy PE, so Borda count does not
necessarily have the advantage.
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We
define
range
voting
as
a
function
𝑅: (𝑉1 , 𝑉2 , ⋯ , 𝑉𝑁 ) → 𝑇(𝐶),
which maps an 𝑁-tuple of scoring lists to a total order of 𝐶, where 𝐶 is the set of all
candidates and 𝑁 is the number of voters.
Range voting must also satisfy the following two criteria:
𝑐𝑖 ~ 𝑐𝑗 if 𝑇𝑆(𝑐𝑖 ) < 𝑇𝑆(𝑐𝑗 ),
(𝑐𝑖 ~ 𝑐𝑗 or 𝑐𝑗 ~ 𝑐𝑖 ) if 𝑇𝑆(𝑐𝑖 ) = 𝑇𝑆(𝑐𝑗 ),
where we define the total score 𝑇𝑆 to be
𝑁
𝑇𝑆(𝑐) = ∑ 𝑠(𝑖, 𝑐).
𝑖=1
Apart from satisfying PE, IIA, and ND, range voting has other advantages. For example, it
allows voters to communicate their preferences more accurately than ranked voting systems,
since voters can express not just if they prefer some candidate over others, but also by how
much.
However, range voting has problems as well. While ties are expected to happen less often
than for ranked voting, there is still no satisfactory way to resolve them. Range voting is also
easier to exploit, since voters can assign much more extreme scores to candidates they prefer
the most/least than they would if they were honest. While some sorts of strategic voting are
possible in ranked voting as well, they are much harder to perform successfully than in range
voting.
Therefore, while range voting may be better than ranked voting, it still suffers from many of
the same problems.
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Conclusion
In conclusion, Arrow's Impossibility Theorem tells us that no ranked voting system is perfect,
as all of them violate at least one of the three criteria Pareto efficiency, independence of
irrelevant alternatives, and non-dictatorship. We examined three ranked voting systems
(plurality voting, instant runoff, and Borda count), and found that all three violate IIA, while
the versions of two of them that we used also violate PE. As an alternative, we looked at
range voting, which is a non-ranked voting system that satisfies all three Arrow's criteria.
Range voting allows voters to assign scores to candidates rather than ranking them, which
allows for more accurate expression of preferences. However, range voting also suffers from
problems, such as the inability to resolve ties and the potential for exploitation. Overall, there
appears to be no perfect voting system, and any system will have its own set of advantages
and flaws. It is up to society to decide which voting system best serve them in different
situations.
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