Social Choice

Combining the preferences of many individuals into a single collective decision is far harder than it appears. This chapter studies social choice theory, characterising aggregation procedures and establishing impossibility results, most famously Arrow’s theorem, that reveal deep constraints on what any fair voting rule can achieve.

Motivating Example: Voting on exam topics¶

In the final week of term, a game theory class of 45 students receives an unexpected offer from their professor:

“You get to vote on which topic will feature most heavily on the final exam.
Each of you will submit a full ranking of the three options, and we’ll use
pairwise majority voting to decide the winner.”

The three candidate topics are:

After collecting all the votes, the professor finds that the students fall into the preference groups shown in Table 1.

Group size	1st choice	2nd choice	3rd choice
18 students	A	B	C
16 students	B	C	A
11 students	C	A	B

If the professor only looks at the first choice then Replicator dynamics will be the chosen topic of the exam. However let us examine each pairwise contest:

A vs B
- 18 students: prefer A to B and B to C
- 16 students: prefer B to C and C to A so prefer B to A
- 11 students: prefer C to A and A to B.

So A beats B, with 29 votes to 16.

B vs C
- 18 students: prefer A to B and B to C
- 16 students: prefer B to C and C to A
- 11 students: prefer C to A and A to B so prefer C to B

So B beats C, with 34 votes to 11.

C vs A
- 18 students: prefer A to B and B to C so prefer A to C
- 16 students: prefer B to C and C to A
- 11 students: prefer C to A and A to B → C beats A, with 27 votes to 18.

A majority prefers A to B, B to C, and C to A.

This cycle is called a Condorcet cycle Caritat, 1785 and it illustrates the core difficulty in aggregating preferences from a group: majority rule may not produce a consistent collective ranking. This motivates a central question in voting theory:

What properties should a “reasonable” voting rule satisfy, and can we always satisfy them all at once?

Theory¶

Definition: Preference Function¶

Let $X$ be a finite set of alternatives. A preference function for a voter is a strict linear order $\succ$ on $X$ , meaning that for all distinct alternatives $x, y, z \in X$ :

Asymmetry: if $x \succ y$ , then not $y \succ x$
Transitivity: if $x \succ y$ and $y \succ z$ , then $x \succ z$
Completeness: for all $x \ne y$ , either $x \succ y$ or $y \succ x$

We refer to a complete list of $n$ voters’ preference functions as a preference profile.

Example: Preference Function for the Exam Vote¶

For Motivating Example: Voting on exam topics the set of alternative is:

A = \{A, B, C\}

(1)

The preference profile is:

Group 1: $A \succ B \succ C$
Group 2: $B \succ C \succ A$
Group 3: $C \succ A \succ B$

A social welfare function is a rule that maps every possible preference profile over a set of alternatives $X$ to a strict linear order on $X$ .

That is, it takes individual rankings and produces a collective ranking of all the alternatives.

For Motivating Example: Voting on exam topics one social welfare function would be to rank the outcomes based on number of first choice votes, giving:

(A, B, C)

(2)

Definition: Simple Majority Rule¶

A simple majority rule is a voting rule that compares alternatives $x$ and $y$ based on how many voters prefer $x$ over $y$ :

$x$ is ranked above $y$ in the collective outcome if and only if a strict majority of voters prefer $x$ to $y$ .

This defines a binary relation over pairs, but may not yield a transitive overall ranking. When it does, the top-ranked option is called the Condorcet winner.

Theorem: Arrow’s Impossibility Theorem¶

A political centre shifting as opinion moves. — Figure 1:Where the collective ‘centre’ sits depends on how individual views are weighed against one another. Arrow’s theorem shows that no aggregation rule can reconcile a few reasonable fairness requirements all at once.

The following theorem is given without proof Arrow, 1951.

Let $X$ be a set of at alternatives with $|X| \geq 3$ . There exists no social welfare function that satisfies all of the following properties:

Unrestricted domain: all possible preference profiles are allowed
Pareto efficiency: if all voters prefer $x$ over $y$ , then $x$ is ranked above $y$
Independence of irrelevant alternatives (IIA): the group’s relative ranking of $x$ and $y$ depends only on how individuals rank $x$ vs $y$
Non-dictatorship: there is no single voter whose preferences always determine the outcome

In short, there is no perfect voting rule that aggregates individual preferences into a consistent group ranking while satisfying all of these fairness criteria.

In the context of Motivating Example: Voting on exam topics this theorem implies that there is no perfect way to choose. However there are two approaches that can offer a good way to obtain an outcome.

Definition: Condorcet’s Method¶

An alternative $x$ is a Condorcet winner if, for every other alternative $y$ , a strict majority of voters prefer $x$ over $y$ .

In the context of Motivating Example: Voting on exam topics we have already seen that no Condorcet winner exists.

Definition: Borda’s Method¶

Let $x$ be a finite set of alternatives. Every alternative obtains $k$ points from a voter if that voter ranks the alternative above exactly $k$ other alternatives. The total Borda score of each alternative is the sum of its points across all voters. The alternative with the highest total score is the Borda winner.

Borda’s method always produces a complete ranking, but it may fail to select a Condorcet winner when one exists.

Example: The Borda Method for the Exam Vote¶

For the Motivating Example: Voting on exam topics:

the 18 individuals in the first group give 2 points to option A and 1 point to option B.
the 16 individuals in the second group give 2 points to option B and 1 point to option C.
the 11 individuals in the third group give 2 points to option C and 1 point to option A.

Thus the Borda score of each option is:

A: $18 \cdot 2 + 11 \cdot 1=47$
B: $18 \cdot 1 + 16 \cdot 2=50$
C: $11 \cdot 2 + 16 \cdot 1=38$

The Borda winner is thus B.

Exercises¶

Exercise 4

Borda proposed the Borda Method as a response to the existence of Condorcet cycles in simply majority voting. Condorcet then presented the preference profile of Table 2.

Number of votes	1st choice	2nd choice	3rd choice
30	A	B	C
1	A	C	B
29	B	A	C
10	B	C	A
10	C	A	B
1	C	B	A

Who is the Condorcet winner (if there is one)?
Who is the Borda winner?
Why would this example be a critique of Borda’s approach?

Programming¶

The pref_voting Python library Holliday & Pacuit, 2025 provides tools for analysing preferential voting systems. In the following example, we define a preference profile based on the classroom voting scenario and use the library to explore two voting methods.

We first check for a Condorcet winner using pairwise majority comparisons. In this case, the method returns all alternatives, indicating that no Condorcet winner exists due to a cycle.

import pref_voting.profiles
import pref_voting.c1_methods

profile = pref_voting.profiles.Profile(
    [[0, 1, 2]] * 18 + [[1, 2, 0]] * 16 + [[2, 0, 1]] * 11
)
print(f"Condorcet winners: {pref_voting.c1_methods.condorcet(profile)}")

/home/runner/work/gtb/gtb/.venv/lib/python3.14/site-packages/pref_voting/profiles.py:150: SyntaxWarning: "\l" is an invalid escape sequence. Such sequences will not work in the future. Did you mean "\\l"? A raw string is also an option.
  """An anonymous profile of linear rankings of :math:`n` candidates.  It is assumed that the candidates are named :math:`0, 1, \ldots, n-1` and a ranking of the candidates is a list of candidate names.  For instance, the list ``[0, 2, 1]`` represents the ranking in which :math:`0` is ranked above :math:`2`, :math:`2` is ranked above :math:`1`, and :math:`0` is ranked above :math:`1`.
/home/runner/work/gtb/gtb/.venv/lib/python3.14/site-packages/pref_voting/profiles.py:477: SyntaxWarning: "\l" is an invalid escape sequence. Such sequences will not work in the future. Did you mean "\\l"? A raw string is also an option.
  .. warning:: Since the candidates in a Profile must be named :math:`0, 1, \ldots, n-1` (where :math:`n` is the number of candidates), you must use the candidate map returned to by the function to recover the original candidate names.
/home/runner/work/gtb/gtb/.venv/lib/python3.14/site-packages/pref_voting/profiles.py:561: SyntaxWarning: "\h" is an invalid escape sequence. Such sequences will not work in the future. Did you mean "\\h"? A raw string is also an option.
  latex_str += " & ".join([f"${rc}$" for rc in self._rcounts]) + "\\\\\hline \n"

Condorcet winners: [0, 1, 2]

Next, we apply the Borda count, which assigns points based on rankings and selects the alternative with the highest total score.

import pref_voting.scoring_methods

print(f"Borda winners: {pref_voting.scoring_methods.borda(profile)}")

/home/runner/work/gtb/gtb/.venv/lib/python3.14/site-packages/pref_voting/scoring_methods.py:57: SyntaxWarning: "\s" is an invalid escape sequence. Such sequences will not work in the future. Did you mean "\\s"? A raw string is also an option.
  """The **Borda score** of a candidate is calculated as follows: If there are :math:`m` candidates, then the Borda score of candidate :math:`c` is :math:`\sum_{r=1}^{m} (m - r) * Rank(c,r)` where :math:`Rank(c,r)` is the number of voters that rank candidate :math:`c` in position :math:`r`. The Borda winners are the candidates with the largest Borda score in the ``profile`` restricted to ``curr_cands``.

Borda winners: [1]

Notable Research¶

The foundational work of Condorcet and Borda Caritat, 1785Borda, 1781 established key ideas in social choice theory, culminating in Arrow’s impossibility theorem Arrow, 1951. Arrow’s result shows that no voting rule can simultaneously satisfy a set of seemingly reasonable fairness criteria.

These challenges deepen with the Gibbard–Satterthwaite theorem, which demonstrates that any non-dictatorial voting rule with at least three alternatives is vulnerable to strategic manipulation Gibbard, 1973Satterthwaite, 1975. This perhaps helps explain why some open source software projects, such as Python, historically adopted a Benevolent Dictator For Life (BDFL) model of governance O’Mahony, 2007, with Guido van Rossum serving in that role until 2018.

Encouragingly, recent work has shown that although manipulation is theoretically possible, it can be computationally difficult in practice Conitzer & Sandholm, 2002Faliszewski et al., 2010.

Other modern approaches focus on multiwinner voting rules that aim for fair and proportional representation Elkind et al., 2017Bredereck et al., 2018. These rules are increasingly relevant in applications such as committee selection, participatory budgeting, and recommendation systems.

These developments show that while Arrow’s theorem rules out a perfect solution, there are many meaningful and principled ways to navigate the trade-offs, especially when one considers computation, uncertainty, or broader notions of fairness.

A helpful overview of these developments is provided in Pacuit, 2019.

Conclusion¶

This chapter explored the central challenge of social choice: how to aggregate individual preferences into a fair and consistent group decision. Starting from a motivating classroom example, we saw how simple majority rule can lead to intransitive group preferences, known as Condorcet cycles, even when every individual’s ranking is rational.

We then introduced the formal framework of preference functions and social welfare functions, leading to Arrow’s impossibility theorem. This result shows that no voting rule can satisfy a basic set of fairness properties when there are at least three alternatives.

Despite this impossibility, we studied two influential responses: the Condorcet method, which seeks an alternative that beats all others in pairwise comparisons, and Borda’s method, which aggregates preferences using a scoring rule. While each method has strengths and limitations, they both offer principled approaches to navigating the trade-offs inherent in group decision making.

Method	Summary	Strength	Limitation
Simple Majority	Pairwise majority voting	Intuitive and widely used	Can produce cycles
Condorcet	Chooses the option that beats all others in pairwise votes	Respects majority preferences	May not always exist
Borda	Points based on rank positions	Always yields a complete ranking	Fails to select Condorcet winner

Solution to Exercise 1 #

The preference profile over $X = \{A, B, C\}$ is:

3 voters: $A \succ B \succ C$
2 voters: $B \succ C \succ A$
2 voters: $C \succ A \succ B$

There are 7 voters in total.

a. Pairwise majority contests:

A vs B:

3 voters prefer $A$ to $B$ (group 1).
2 voters prefer $B$ to $A$ (group 2).
2 voters prefer $A$ to $B$ (group 3: $C \succ A \succ B$ , so $A \succ B$ ).

Total: $A$ preferred by $3+2=5$ voters, $B$ preferred by 2 voters.

$A$ beats $B$ with 5 votes to 2.

B vs C:

3 voters prefer $B$ to $C$ (group 1: $A \succ B \succ C$ ).
2 voters prefer $B$ to $C$ (group 2: $B \succ C \succ A$ ).
2 voters prefer $C$ to $B$ (group 3).

Total: $B$ preferred by $3+2=5$ voters, $C$ preferred by 2 voters.

$B$ beats $C$ with 5 votes to 2.

C vs A:

3 voters prefer $A$ to $C$ (group 1).
2 voters prefer $C$ to $A$ (group 2).
2 voters prefer $C$ to $A$ (group 3).

Group 1 ( $A\succ B\succ C$ ) prefers $A$ to $C$ ; groups 2 ( $B\succ C\succ A$ ) and 3 ( $C\succ A\succ B$ ) both prefer $C$ to $A$ . Total preferring $A$ : 3; total preferring $C$ : $2+2=4$ .

$C$ beats $A$ with 4 votes to 3.

Summary of pairwise results:

$A \succ_{\text{maj}} B$ (5 to 2)
$B \succ_{\text{maj}} C$ (5 to 2)
$C \succ_{\text{maj}} A$ (4 to 3)

b. Is there a Condorcet winner?

A Condorcet winner must beat all other alternatives in pairwise majority comparisons.

$A$ beats $B$ but loses to $C$ . Not a Condorcet winner.
$B$ beats $C$ but loses to $A$ . Not a Condorcet winner.
$C$ beats $A$ but loses to $B$ . Not a Condorcet winner.

There is no Condorcet winner.

c. Is the majority preference relation transitive?

The majority relation gives $A \succ_{\text{maj}} B$ , $B \succ_{\text{maj}} C$ , but $C \succ_{\text{maj}} A$ (not $A \succ_{\text{maj}} C$ ).

Transitivity would require $A \succ_{\text{maj}} C$ , but we have the opposite. Therefore the majority preference relation is not transitive. This is a Condorcet cycle: $A$ beats $B$ , $B$ beats $C$ , $C$ beats $A$ .

import pref_voting.profiles
import pref_voting.c1_methods

# Encode: A=0, B=1, C=2
profile = pref_voting.profiles.Profile(
    [[0, 1, 2]] * 3 + [[1, 2, 0]] * 2 + [[2, 0, 1]] * 2
)
condorcet_winner = pref_voting.c1_methods.condorcet(profile)
print("Condorcet winner(s):", condorcet_winner)

Condorcet winner(s): [0, 1, 2]

# Pairwise margin matrix
import numpy as np

# voters[i] = list of preferences in order (most to least preferred)
votes = [[0, 1, 2]] * 3 + [[1, 2, 0]] * 2 + [[2, 0, 1]] * 2
n_alt = 3
margin = np.zeros((n_alt, n_alt), dtype=int)
for v in votes:
    for i in range(n_alt):
        for j in range(i + 1, n_alt):
            x, y = v[i], v[j]
            margin[x][y] += 1
            margin[y][x] -= 1

labels = ["A", "B", "C"]
print("Pairwise margins (row beats column by this many votes):")
for i in range(n_alt):
    for j in range(n_alt):
        if i != j:
            print(f"  {labels[i]} vs {labels[j]}: {margin[i][j]:+d}")

Pairwise margins (row beats column by this many votes):
  A vs B: +3
  A vs C: -1
  B vs A: -3
  B vs C: +3
  C vs A: +1
  C vs B: -3

Solution to Exercise 2 #

The preference profile is:

4 voters: $A \succ B \succ C$
3 voters: $B \succ C \succ A$
2 voters: $C \succ A \succ B$

There are 9 voters in total.

a. Borda scores:

With three alternatives, each voter assigns 2 points to their first choice, 1 point to their second choice, and 0 points to their last choice.

A: $4 \times 2 + 3 \times 0 + 2 \times 1 = 8 + 0 + 2 = 10$
B: $4 \times 1 + 3 \times 2 + 2 \times 0 = 4 + 6 + 0 = 10$
C: $4 \times 0 + 3 \times 1 + 2 \times 2 = 0 + 3 + 4 = 7$

Borda scores: $A = 10$ , $B = 10$ , $C = 7$ .

The Borda method produces a tie between $A$ and $B$ . (Without a tie-breaking rule, neither is uniquely selected.)

b. Condorcet winner:

A vs B:

4 voters prefer $A$ (group 1).
3 voters prefer $B$ (group 2).
2 voters prefer $A$ (group 3: $C \succ A \succ B$ , so $A \succ B$ ).

$A$ preferred by $4+2=6$ , $B$ preferred by 3. $A$ beats $B$ 6 to 3.

A vs C:

4 voters prefer $A$ (group 1).
3 voters prefer $C$ (group 2).
2 voters prefer $C$ (group 3).

$A$ preferred by 4, $C$ preferred by $3+2=5$ . $C$ beats $A$ 5 to 4.

B vs C:

4 voters prefer $B$ (group 1).
3 voters prefer $B$ (group 2).
2 voters prefer $C$ (group 3).

$B$ preferred by $4+3=7$ , $C$ preferred by 2. $B$ beats $C$ 7 to 2.

Pairwise results: $A$ beats $B$ ; $B$ beats $C$ ; $C$ beats $A$ . This is again a Condorcet cycle. There is no Condorcet winner.

c. Do Borda and Condorcet select the same winner?

There is no Condorcet winner here (due to the cycle). The Borda method yields a tie between $A$ and $B$ . Since no Condorcet winner exists, we cannot directly compare the two methods in terms of who they “select”, but we can note that the Borda method at least produces a complete ranking and is decisive (up to ties), while the Condorcet method fails to identify a winner in this case.

import pref_voting.profiles
import pref_voting.c1_methods
import pref_voting.scoring_methods

profile = pref_voting.profiles.Profile(
    [[0, 1, 2]] * 4 + [[1, 2, 0]] * 3 + [[2, 0, 1]] * 2
)
print("Condorcet winner:", pref_voting.c1_methods.condorcet(profile))
print("Borda winner:", pref_voting.scoring_methods.borda(profile))

Condorcet winner: [0, 1, 2]
Borda winner: [0, 1]

# Manual Borda score computation
groups = [(4, [0, 1, 2]), (3, [1, 2, 0]), (2, [2, 0, 1])]
n_alt = 3
borda_scores = [0, 0, 0]
for count, ranking in groups:
    for rank, alt in enumerate(ranking):
        borda_scores[alt] += count * (n_alt - 1 - rank)
labels = ["A", "B", "C"]
for label, score in zip(labels, borda_scores):
    print(f"Borda score of {label}: {score}")

Borda score of A: 10
Borda score of B: 10
Borda score of C: 7

Solution to Exercise 3 #

The preference profile is:

5 voters: $A \succ B \succ C$
4 voters: $B \succ C \succ A$
3 voters: $C \succ A \succ B$

There are 12 voters in total.

a. Original Borda scores:

With three alternatives each voter assigns 2 points to their top choice, 1 to second, 0 to last.

A: $5 \times 2 + 4 \times 0 + 3 \times 1 = 10 + 0 + 3 = 13$
B: $5 \times 1 + 4 \times 2 + 3 \times 0 = 5 + 8 + 0 = 13$
C: $5 \times 0 + 4 \times 1 + 3 \times 2 = 0 + 4 + 6 = 10$

Borda scores: $A = 13$ , $B = 13$ , $C = 10$ .

The Borda winner is a tie between $A$ and $B$ .

b. Manipulation: one voter changes from $B \succ C \succ A$ to $B \succ A \succ C$ :

The new profile is:

5 voters: $A \succ B \succ C$
3 voters: $B \succ C \succ A$ (one fewer)
1 voter: $B \succ A \succ C$ (the manipulating voter)
3 voters: $C \succ A \succ B$

New Borda scores:

A: $5 \times 2 + 3 \times 0 + 1 \times 1 + 3 \times 1 = 10 + 0 + 1 + 3 = 14$
B: $5 \times 1 + 3 \times 2 + 1 \times 2 + 3 \times 0 = 5 + 6 + 2 + 0 = 13$
C: $5 \times 0 + 3 \times 1 + 1 \times 0 + 3 \times 2 = 0 + 3 + 0 + 6 = 9$

New Borda scores: $A = 14$ , $B = 13$ , $C = 9$ .

c. Does the outcome change?

Originally $A$ and $B$ were tied (both at 13). After the manipulation, $A$ wins outright with 14 points, while $B$ drops to 13.

The manipulating voter prefers $B$ to $A$ , yet their manipulation caused $A$ to win (or more precisely, broke the tie in $A$ ’s favour, away from $B$ ). This shows that the manipulation was not profitable for this voter; it was inadvertent harm.

However, the fact that changing a ballot in a way that does not affect the voter’s stated preference between $B$ and $C$ (they still rank $B$ first) can nonetheless change the winner illustrates the Borda method’s vulnerability to strategic voting. A voter who, in a different scenario, wanted to harm $A$ could promote $C$ (an irrelevant alternative) to second place, boosting $C$ ’s score and potentially tipping the outcome. This violates the Independence of Irrelevant Alternatives property and is the mechanism through which manipulation operates.

More broadly, the Borda method is susceptible to strategic manipulation because the relative ranking of any two alternatives can be influenced by how a voter ranks a third, “irrelevant” alternative. This is precisely what the Gibbard-Satterthwaite theorem predicts: any non-dictatorial voting rule over three or more alternatives is manipulable.

import pref_voting.profiles
import pref_voting.scoring_methods

# Original profile: A=0, B=1, C=2
profile_original = pref_voting.profiles.Profile(
    [[0, 1, 2]] * 5 + [[1, 2, 0]] * 4 + [[2, 0, 1]] * 3
)
print("Original Borda winner:", pref_voting.scoring_methods.borda(profile_original))

Original Borda winner: [0, 1]

# Manipulated profile: one voter switches from B>C>A to B>A>C
profile_manipulated = pref_voting.profiles.Profile(
    [[0, 1, 2]] * 5 + [[1, 2, 0]] * 3 + [[1, 0, 2]] * 1 + [[2, 0, 1]] * 3
)
print("Manipulated Borda winner:", pref_voting.scoring_methods.borda(profile_manipulated))

Manipulated Borda winner: [0]

# Manual score verification
def borda_scores(groups, n_alt=3):
    scores = [0] * n_alt
    for count, ranking in groups:
        for rank, alt in enumerate(ranking):
            scores[alt] += count * (n_alt - 1 - rank)
    return scores

original = borda_scores([(5,[0,1,2]),(4,[1,2,0]),(3,[2,0,1])])
manipulated = borda_scores([(5,[0,1,2]),(3,[1,2,0]),(1,[1,0,2]),(3,[2,0,1])])
labels = ["A","B","C"]
print("Original scores:", {l:s for l,s in zip(labels, original)})
print("Manipulated scores:", {l:s for l,s in zip(labels, manipulated)})

Original scores: {'A': 13, 'B': 13, 'C': 10}
Manipulated scores: {'A': 14, 'B': 13, 'C': 9}

Solution to Exercise 4 #

The preference profile is:

Number of votes	1st	2nd	3rd
30	A	B	C
1	A	C	B
29	B	A	C
10	B	C	A
10	C	A	B
1	C	B	A

Total voters: $30+1+29+10+10+1 = 81$ .

1. Who is the Condorcet winner?

We check each pairwise majority contest.

A vs B:

Voters preferring $A$ over $B$ : groups ranking $A$ above $B$ :

30 voters ( $A \succ B \succ C$ )
1 voter ( $A \succ C \succ B$ )
10 voters ( $C \succ A \succ B$ )

Total preferring $A$ : $30+1+10=41$ .

Voters preferring $B$ over $A$ :

29 voters ( $B \succ A \succ C$ )
10 voters ( $B \succ C \succ A$ )
1 voter ( $C \succ B \succ A$ )

Total preferring $B$ : $29+10+1=40$ .

$A$ beats $B$ with 41 to 40.

A vs C:

Voters preferring $A$ over $C$ :

30 voters ( $A \succ B \succ C$ )
1 voter ( $A \succ C \succ B$ )
29 voters ( $B \succ A \succ C$ )

Total: $30+1+29=60$ .

Voters preferring $C$ over $A$ :

10 voters ( $B \succ C \succ A$ )
10 voters ( $C \succ A \succ B$ )
1 voter ( $C \succ B \succ A$ )

Total: $10+10+1=21$ .

$A$ beats $C$ with 60 to 21.

B vs C:

Voters preferring $B$ over $C$ :

30 voters ( $A \succ B \succ C$ )
29 voters ( $B \succ A \succ C$ )
10 voters ( $B \succ C \succ A$ )

Voters preferring $C$ over $B$ :

1 voter ( $A \succ C \succ B$ )
10 voters ( $C \succ A \succ B$ )
1 voter ( $C \succ B \succ A$ )

Total preferring $B$ : $30+29+10=69$ . Total preferring $C$ : $1+10+1=12$ .

$B$ beats $C$ with 69 to 12.

$A$ beats both $B$ and $C$ , so $A$ is the Condorcet winner.

2. Who is the Borda winner?

With three alternatives, each voter assigns 2 points to their 1st choice, 1 point to their 2nd choice, 0 to their 3rd choice.

Group (count)	A points	B points	C points
30 ( $A\succ B\succ C$ )	$30\times2=60$	$30\times1=30$	$30\times0=0$
1 ( $A\succ C\succ B$ )	$1\times2=2$	$1\times0=0$	$1\times1=1$
29 ( $B\succ A\succ C$ )	$29\times1=29$	$29\times2=58$	$29\times0=0$
10 ( $B\succ C\succ A$ )	$10\times0=0$	$10\times2=20$	$10\times1=10$
10 ( $C\succ A\succ B$ )	$10\times1=10$	$10\times0=0$	$10\times2=20$
1 ( $C\succ B\succ A$ )	$1\times0=0$	$1\times1=1$	$1\times2=2$

Total Borda scores:

A: $60+2+29+0+10+0 = 101$
B: $30+0+58+20+0+1 = 109$
C: $0+1+0+10+20+2 = 33$

The Borda winner is $B$ with 109 points.

3. Why is this a critique of Borda’s method?

$A$ is the Condorcet winner: it beats every other alternative in pairwise majority votes (41 to 40 over $B$ ; 60 to 21 over $C$ ). A majority of voters prefers $A$ to $B$ (41 out of 81).

Yet the Borda method selects $B$ as the winner.

This is Condorcet’s critique of the Borda method: the Borda method can fail to elect the Condorcet winner even when one exists. The method gives weight to the intensity of preferences (how far apart alternatives are ranked) rather than just pairwise majority relationships. Here, $B$ benefits from being ranked second by many voters (the 30 who rank $A$ first and the 10 who rank $C$ first), accumulating enough second-place votes to overtake $A$ ’s lead in direct comparisons.

This example showed Condorcet that the Borda method violates the Condorcet criterion, a property that many would argue is a minimal requirement of a reasonable voting rule.

import pref_voting.profiles
import pref_voting.c1_methods
import pref_voting.scoring_methods

# A=0, B=1, C=2
profile = pref_voting.profiles.Profile(
    [[0, 1, 2]] * 30
    + [[0, 2, 1]] * 1
    + [[1, 0, 2]] * 29
    + [[1, 2, 0]] * 10
    + [[2, 0, 1]] * 10
    + [[2, 1, 0]] * 1
)

print("Condorcet winner:", pref_voting.c1_methods.condorcet(profile))
print("Borda winner:", pref_voting.scoring_methods.borda(profile))

Condorcet winner: [0]
Borda winner: [1]

# Manual Borda score verification
groups = [
    (30, [0, 1, 2]),
    (1,  [0, 2, 1]),
    (29, [1, 0, 2]),
    (10, [1, 2, 0]),
    (10, [2, 0, 1]),
    (1,  [2, 1, 0]),
]
scores = [0, 0, 0]
for count, ranking in groups:
    for rank, alt in enumerate(ranking):
        scores[alt] += count * (2 - rank)
print("Borda scores. A:", scores[0], "B:", scores[1], "C:", scores[2])

Borda scores. A: 101 B: 109 C: 33

References¶

Marie Jean Antoine Nicolas de Caritat, M. de C. (1785). Essai sur l’application de l’analyse à la probabilité des décisions rendues à la pluralité des voix. Imprimerie Royale.
Arrow, K. J. (1951). Social Choice and Individual Values. Wiley.
Holliday, W. H., & Pacuit, E. (2025). pref_voting: The Preferential Voting Tools package for Python. Journal of Open Source Software, 10(105), 7020. 10.21105/joss.07020
de Borda, J.-C. (1781). Mémoire sur les élections au scrutin. Histoire de l’Académie Royale Des Sciences.
Gibbard, A. (1973). Manipulation of voting schemes: a general result. Econometrica: Journal of the Econometric Society, 587–601.
Satterthwaite, M. A. (1975). Strategy-proofness and Arrow’s conditions: Existence and correspondence theorems for voting procedures and social welfare functions. Journal of Economic Theory, 10(2), 187–217.
O’Mahony, S. (2007). The governance of open source initiatives: what does it mean to be community managed? Journal of Management & Governance, 11, 139–150.
Conitzer, V., & Sandholm, T. (2002). Complexity of manipulating elections with few candidates. AAAI/IAAI, 314–319.
Faliszewski, P., Hemaspaandra, E., & Hemaspaandra, L. A. (2010). Using complexity to protect elections. Communications of the ACM, 53(11), 74–82.
Elkind, E., Faliszewski, P., Skowron, P., & Slinko, A. (2017). Properties of multiwinner voting rules. Social Choice and Welfare, 48, 599–632.
Bredereck, R., Faliszewski, P., Igarashi, A., Lackner, M., & Skowron, P. (2018). Multiwinner elections with diversity constraints. Proceedings of the AAAI Conference on Artificial Intelligence, 32(1).
Pacuit, E. (2019). Voting methods. Stanford Encyclopedia of Philosophy.

Motivating Example: Voting on exam topics¶

Theory¶

Definition: Preference Function¶

Example: Preference Function for the Exam Vote¶

Definition: A Social Welfare Function¶

Example: Social Welfare Function for the Exam Vote¶

Definition: Simple Majority Rule¶

Theorem: Arrow’s Impossibility Theorem¶

Definition: Condorcet’s Method¶

Definition: Borda’s Method¶

Example: The Borda Method for the Exam Vote¶

Exercises¶

Programming¶

Notable Research¶

Conclusion¶

Solutions¶