Fairness Criteria

Brian Lins

Wednesday, May 22, 2024

Problems with Borda Count

Jean Charles Borda proposed the Borda count method in 1770. It has the advantage that spoiler candidates are much less likely to be a problem. But Nicholas de Condorcet was very critical of the Borda count method.

Jean Charles Borda (left) and Nicholas de Condorcet (right)

Condorcet’s Criterion

According to Condorcet, if a candidate would beat all the others in head-to-head match-ups, then they should win the election.

A candidate that would beat every other candidate in a one-on-one run-off is called a Condorcet candidate. Not every election has a Condorcet candidate, but when there is it seems reasonable that they should win.

Borda Count Fails the Condorcet Criterion

In fact, Borda count doesn’t even always elect the majority candidate! Suppose 25 kids vote on the best ice cream flavor.

Number of Voters
13 6 5 1
Chocolate 1st 3rd 3rd 2nd
Vanilla 2nd 2nd 1st 1st
Strawberry 3rd 1st 2nd 3rd

Exercise. Is there a Condorcet candidate? Why?

Exercise. What flavor does Borda count pick?

Condorcet Methods

It is easy to make a voting method that passes the Condorcet criteria. Just check for a Condorcet candidate by checking every pairwise comparison.

Checking for a Condorcet Candidate

With 5 candidates, there are 10 possible pairwise match-ups. That could take a long time to check. Save time by only checking match-ups until every candidate has lost one (or you find a Condorcet candidate).

Number of Voters
33 30 25 20 18
Italian     2nd         5th         1st         2nd         3rd    
Mexican     1st         4th         5th         4th         1st    
Thai     3rd         1st         4th         5th         2nd    
Chinese     4th         2nd         3rd         1st         4th    
Indian     5th         3rd         2nd         3rd         5th    

Mexican vs. Italian

Number of Voters
33 30 25 20 18
Italian     2nd         5th         1st         2nd         3rd    
Mexican     1st         4th         5th         4th         1st    
Thai     3rd         1st         4th         5th         2nd    
Chinese     4th         2nd         3rd         1st         4th    
Indian     5th         3rd         2nd         3rd         5th    

Mexican wins.

Mexican vs. Thai

Number of Voters
33 30 25 20 18
Italian     2nd         5th         1st         2nd         3rd    
Mexican     1st         4th         5th         4th         1st    
Thai     3rd         1st         4th         5th         2nd    
Chinese     4th         2nd         3rd         1st         4th    
Indian     5th         3rd         2nd         3rd         5th    

Mexican wins.

Mexican vs. Chinese

Number of Voters
33 30 25 20 18
Italian     2nd         5th         1st         2nd         3rd    
Mexican     1st         4th         5th         4th         1st    
Thai     3rd         1st         4th         5th         2nd    
Chinese     4th         2nd         3rd         1st         4th    
Indian     5th         3rd         2nd         3rd         5th    

Chinese wins.

Chinese vs. Indian

Number of Voters
33 30 25 20 18
Italian     2nd         5th         1st         2nd         3rd    
Mexican     1st         4th         5th         4th         1st    
Thai     3rd         1st         4th         5th         2nd    
Chinese     4th         2nd         3rd         1st         4th    
Indian     5th         3rd         2nd         3rd         5th    

Chinese wins.

Chinese vs. Italian

So far, only Chinese hasn’t lost, but that changes if you compare it with Italian.

Number of Voters
33 30 25 20 18
Italian     2nd         5th         1st         2nd         3rd    
Mexican     1st         4th         5th         4th         1st    
Thai     3rd         1st         4th         5th         2nd    
Chinese     4th         2nd         3rd         1st         4th    
Indian     5th         3rd         2nd         3rd         5th    

So there is no Condorcet candidate, since every restaurant loses at least one match-up.

What Makes a Voting Method Fair?

So far, we have seen some features that an ideal voting system would have:

Arrow’s Impossibility Theorem

Unfortunately, even though we can find methods that satisfy some of the fairness criteria, it is impossible to satisfy them all. This is a consequence of a famous theorem that was proved by the economist Kenneth Arrow in 1951

What is Impossible?

Here is one version of Arrow’s theorem (there are other more complicated versions that say more).

Arrow’s Impossibility Theorem. No voting method can satisfy both the Condorcet criterion and the no spoilers criterion when there are 3 or more candidates.


Important:

Every voting method either (i) sometimes has spoiler candidates or (ii) sometimes fails to elect a Condorcet candidate. But usually those bad things don’t happen.

Why is it Impossible?

Consider an election with three front runner candidates (assume any other candidates are ranked lower by all of the voters and so don’t matter) and just 9 voters.

Number of Voters
4 3 2
Candidate A 1st 3rd 2nd
Candidate B 2nd 1st 3rd
Candidate C 3rd 2nd 1st

Exercise. Show that there is no Condorcet candidate. Why not?

A Cycle of Preferences

In the last example A beats B and B beats C, but C beats A.

Exercise. Why is it impossible for a Condorcet method to pick a winner of this election without one of the losers being a spoiler candidate?

Important to Understand

The Story So Far

Fairness Criterion
Condorcet Monotonicity No Spoilers
Plurality method
Instant run-off voting
Borda count
Condorcet methods* /*

*There are actually several Condorcet methods, not all satisfy the Monotonicity criterion.

Approval Voting

It would be nice to have a method that doesn’t have spoiler candidates. Several voting methods have that property including one very simple method.

Approval Voting Example

The preference table below does not have enough information to figure out the approval voting winner. Why not?

Number of Voters
13 6 5 1
Chocolate 1st 3rd 3rd 2nd
Vanilla 2nd 2nd 1st 1st
Strawberry 3rd 1st 2nd 3rd

Approval Voting Example

Suppose that the first 13 voters split into two groups. Ten of them are okay with chocolate and vanilla, but 3 only like chocolate. Then that column could split into:

Number of Voters
13 6 5 1
Chocolate 1st 3rd 3rd 2nd
Vanilla 2nd 2nd 1st 1st
Strawberry 3rd 1st 2nd 3rd

Approval Voting Example

Suppose that the first 13 voters split into two groups. Ten of them are okay with chocolate and vanilla, but 3 only like chocolate. Then that column could split into:

Number of Voters
10 3 6 5 1
Chocolate 3rd 3rd 2nd
Vanilla 2nd 1st 1st
Strawberry 1st 2nd 3rd

Approval Voting Example

The next group of voters might split evenly into two groups, one that approves their top two flavors, and one that only wants their top choice.

Number of Voters
10 3 6 5 1
Chocolate 3rd 3rd 2nd
Vanilla 2nd 1st 1st
Strawberry 1st 2nd 3rd

Approval Voting Example

The next group of voters might split evenly into two groups, one that approves their top two flavors, and one that only wants their top choice.

Number of Voters
10 3 3 3 5 1
Chocolate 3rd 2nd
Vanilla 1st 1st
Strawberry 2nd 3rd

Approval Voting Example

The next 5 voters all like every flavor about equally; and the one guy at the end only approves of vanilla.

Number of Voters
10 3 3 3 5 1
Chocolate 3rd 2nd
Vanilla 1st 1st
Strawberry 2nd 3rd

Approval Voting Example

The next 5 voters all like every flavor about equally; and the one guy at the end only approves of vanilla.

Number of Voters
10 3 3 3 5 1
Chocolate
Vanilla
Strawberry

Exercise.

Which flavor wins this election using approval voting?

The Story with Approval Voting

Fairness Criterion
Condorcet Monotonicity No Spoilers
Plurality method
Instant run-off voting
Borda count
Condorcet methods *
Approval voting