Violating the Condorcet Criterion

Introduction

A voting method is a mechanism for determining the winner in an election, with a series of steps explicitly listed. These methods generally have some means of scoring candidates. Each voting method is used on a preference table, which is a chart that indicates people’s relative preferences between two or more options.

Voting methods are often found as a topic in the curriculum of a recreational math or math modeling course at the high school or collegiate level. In particular, this article will use four voting methods — plurality, plurality with elimination, Borda count, and pairwise comparison — as case studies while we dissect a fairness principle.

So, what is a fairness principle? A fairness principle is an indicator of whether a voting method is fair. Admittedly, the definition of “fair” is ambiguous and often subject to philosophical scrutiny, so just bear with me. There are a number of fairness principles that can be inspected, but the universal four are called majority, condorcet, monotonicity, and independence. This particular article will focus on the condorcet (and the others are highlighted in this article) principle. The jargon for the condorcet criterion is as follows:

If one wins over each of the others when paired up, then it should win overall.

To somebody unfamiliar with voting theory, this might be a bit daunting of a description to decipher. For that reason, I replicate a portion of the text I wrote in the article I linked in the above paragraph, which interprets the above definition:

This criterion is most easily depicted with an example election involving three parties. Let’s call the parties A, B, and C. Then the condorcet criteria works as follows: run three “mini-elections:” A vs. B, A vs. C, and B vs. C. Each party participates in two of the three mini-elections. If a party manages to win both of the mini-elections they participate in, then the condorcet criteria says that party should win the overall election. In some cases there would be no party who wins two mini-elections; suppose for instance A beats B, C beats A, and B beats C. Here, the condorcet criterion is inconclusive.

Even after reading the above, you might not be completely comfortable with what the condorcet condition really means. For virtually any mathematical concept, one of the best ways to enhance understanding is to look for counterexamples. With that objective in mind, we will look at three voting methods and see how each of them can violate the condorcet condition. Each of those methods will get its own section in the article. After that, we’ll talk about a fourth voting method that always follows the condorcet condition, and explain why that’s the case.

For convenience, all examples will fall within the following scenario: a group of people are presented with a choice of three toppings for pizza. Those toppings will be pepperoni, mushrooms, and banana peppers (a personal favorite of mine). Of course, the preference tables used will be different for each section, but those will be clearly indicated every time.

Image courtesy of joshuemd via Pixabay

The Plurality Method

Let’s work towards a counterexample of where the plurality method violates the condorcet criterion. Here’s a recap of that method: each person voting in the election ranks all of the parties based on an order of preference. The party with the greatest number of first place preferences wins.

As with any voting method, to violate the condorcet principle, we need to have one party (out of three) win two miniature, head-to-head elections, but lose the overall election.

Here’s the preference table for our counterexample:

Here and henceforth pepperoni is abbreviated as P, mushroom is abbreviated as M, and of course banana peppers are abbreviated as BP.

The procedure to verify that this IS a counterexample is as follows: run three miniature head-to-head elections where we only compare two of the three pizza toppings. We will determine what won those smaller elections and see that the overall winner is in fact a different topping!

In the overall election, the point allotments are…

1st choice for pepperoni: 13+5=18

1st choice for mushrooms: 2+15=17

1st choice for banana peppers: 3+12=15

In other words, pepperoni wins by plurality. Now we run the mini-elections. The first of these will be between pepperoni and mushroom (of course, it doesn’t matter what order we record the results of the elections).

This is the same preference table as above, except the toppings involved in the miniature election are highlighted.

In the first, second, and third columns, pepperoni is above mushroom, so pepperoni gets 13+5+3=21 votes.

In the fourth, fifth, and sixth columns, mushroom is above pepperoni, so mushroom gets 12+2+15=29 votes. Hence mushroom wins this mini election.

Next we compare pepperoni to the banana pepper.

In the first, second, and fifth columns, pepperoni is above banana pepper, so pepperoni gets 13+5+2=20 votes.

In the third, fourth, and sixth columns, banana pepper is above pepperoni, so banana pepper gets 3+12+15=30 votes. As a result, banana peppers beat pepperoni one-on-one.

The last mini-election to consider is mushroom versus banana pepper.

In the first, fifth, and sixth columns, mushrooms beat banana peppers, so mushrooms get 13+2+15=30 votes.

In the remaining columns, banana peppers prevail, so they get 5 + 3 + 12 = 20 votes.

As a result, mushrooms win this matchup. This preference table is a counterexample because mushrooms win both of their mini-elections, but pepperoni is the overall winner (with plurality).

The question remains how one would think to come up with this counterexample. Even though plurality only tracks the first place votes to determine a winner, the mini-elections mean that the second place votes are very important, because some of them become first-place votes. For instance, 25 people ranked mushrooms as their second choice (add from first and fourth columns). When some of the votes immediately above those are not considered, those second place votes allow mushrooms to win the mini-elections. Then it’s just a matter of tinkering the numbers so that one of the other toppings has more first place votes.

The Plurality with Elimination Method

Now we work with a close relative of plurality, known as the plurality with elimination method. Here’s a description of that method: count the number of first place preferences for each party. If one party has over half of the first place preferences, that party is the winner. Otherwise, completely remove the lowest ranking party from the election (and the preference table) and tally the votes again.

As I’ll leave to you, the reader, to check, using plurality with elimination on the preference table from the previous section does NOT violate the condorcet criterion, so we can not recycle the previous preference table here. Instead, I introduce a new one:

To see that this is a counterexample, we’ll use the following method. We use plurality with elimination to determine the winner. Then, we test all three head-to-head matchups and see who wins those. We will then conclude that all of these results are collectively inconsistent.

The first step in plurality with elimination is to count the number of first place votes for each pizza topping. Those numbers are:

Pepperoni: 3+20=23 (first two columns)

Banana peppers: 10+1=11 (middle two columns)

Mushrooms: 1+15=16 (last two columns)

Notice that a total of 50 people voted in this election (as is true for all examples I have in this article), and so no pizza topping got the majority of votes (i.e. at least 25). That means we have to eliminate the least popular choice (as the name suggests), which is banana peppers in this case. The table without the banana peppers is as follows:

In some of the columns of the original table, banana peppers were first. In that case, we slid up the other preferences. For example, in the third column of the original table was banana pepper, then pepperoni, then mushroom. In the new table, that column lists pepperoni first, mushrooms second. Some of the other columns in the table behave similarly. Anyway, let’s tally again…

The number of people who prefer pepperoni to mushroom is 3+20+10=33.

The number of people who prefer mushroom to pepperoni is 1+1+15=17.

As a result, pepperoni wins this election if done by plurality with elimination. The question remains who wins the mini-elections if we pair up the pizza toppings.

This is the same preference table as before, except pepperoni is highlighted green, and mushroom is highlighted blue. From this highlighted table we conclude:

The number of people who prefer pepperoni to mushroom is 3+20+10=33.

The number of people who prefer mushroom to pepperoni is 1+1+15=17.

These numbers are actually the same as the ones we got for the table without banana peppers. The reason is because the relative preferences between pepperoni and mushrooms are the same regardless of whether banana peppers are in the table or not. Anyway, pepperoni wins this mini-election.

This is the second mini-election, between pepperoni and banana peppers. The votes are as follows:

Pepperoni is above banana pepper in the first, second, and fifth columns, with a total of 3+20+1 = 24 people preferring pepperoni to banana peppers.

At the same time, voters falling into the third, fourth, and sixth categories prefer banana peppers to pepperoni, and there are 10+1+15=26 of these. The conclusion is that banana peppers win this mini-election.

This colorful table represents the last of three mini-elections, comparing mushrooms to banana peppers.

The first, fifth, and sixth columns represent people who prefer mushrooms to banana peppers, and there are 3+1+15=19 of these people. The remaining like banana peppers more, and there are 20+10+1=31 of them. Hence, banana peppers win this mini-election.

We’ve seen that banana peppers win two out of the three head-to-head elections, but pepperoni ends up as the winner in plurality with elimination. That means we have found ourselves a counterexample. But again, let’s briefly discuss why this is a sensible counterexample from a heuristic standpoint.

Banana peppers win both of their head-to-head matchups, but the banana peppers had the least number of first-choice votes, so they are eliminated in the plurality with elimination. Ultimately it boils down to banana peppers securing 35 out of 50 second-place votes, which have an increased relevance in head-to-head elections. Since every head-to-head election omits one of the toppings, the banana pepper gets a bunch of additional votes by default. For instance, in the banana pepper versus mushroom matchup, pepperoni is ignored, so banana peppers get 20 votes from the second column of the preference table. In comparison, they only need 26 votes to win a two-topping election.

The Borda Count Method

This is the last voting method for which we’ll find a way to violate the condorcet criterion. The Borda count is described as follows: each person voting in the election ranks all of the parties based on an order of preference. The parties higher up in the preference order score more points, and the parties lower down the preference order score less. Each party gets a score, and the highest score wins.

What’s even better is that the counterexample we used for plurality with elimination in the previous section will also work for the Borda count. As a refresher, here’s the preference table:

Since this is the same preference table as before, we already know that mushrooms win two out of three mini-elections. There’s no need to redo that part of the calculation. We just need to focus on who wins with the Borda count. We’ll tally up the points for one topping at a time. First is the pepperoni.

Number of points for pepperoni: 23(3) + 11(2) + 16(1) = 107

Number of points for mushroom: 16(3) + 4(2) + 30(1) = 86

Number of points for banana peppers: 11(3) + 35(2) + 4(1) = 107

Shockingly, with the Borda count, pepperoni and banana peppers are tied! The issue is this: the condorcet fairness principle indicates that since banana peppers win both of their mini-elections, they should be the overall winner (with Borda count), but that is clearly not the case.

Concluding Remarks and Pairwise Comparison

There is one other method we’re going to quickly look at, and it’s known as the pairwise comparison method. The description of the method is as follows: run several smaller elections, each featuring a one-on-one matchup (using the original data in your preference table). Whichever party wins the most one-on-one matchups is deemed the overall winner.

This section will be shorter than the others because rather than construct a counterexample, we notice that this method actually obeys the condorcet fairness principle. That means for any preference table, the pairwise comparison method will not violate condorcet.

In fact, it might seem obvious that the pairwise comparison voting method obeys condorcet. This method relies, by its very definition, on running miniature elections and tallying which party wins the majority of them. If there is a winner determined by this method, it is precisely the party that won most of the miniature elections. If there are three parties involved in the election, then a party who wins two of these miniature elections has in fact won the majority of them. Hence the condorcet principle is followed.

What I did is provided a very conversational and informal proof that the pairwise comparison voting mechanism always obeys condorcet. Don’t go looking for counterexamples this time around, because you won’t find any.

Bibliography

Clark and Clark, “The Beautility of Math” published by Great River Learning

This textbook is used at the University of Tennessee for the entry-level mathematics course “Mathematical Reasoning,” and the content of this blog article was in part based on this text. I use the aforementioned text when I teach said course myself.

Math PhD Student University of Tennessee | Academic Sales Engineer | Writer, Educator, Researcher

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