Brian Lins
Wednesday, May 22, 2024
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.
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.
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?
It is easy to make a voting method that passes the Condorcet criteria. Just check for a Condorcet candidate by checking every pairwise comparison.
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 |
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.
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.
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.
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.
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.
So far, we have seen some features that an ideal voting system would have:
Condorcet Criterion - A Condorcet candidate should win, if there is one.
Monotonicity Criterion - Ranking a candidate highly should never cause a candidate to lose.
No Spoilers Criterion - A third party candidate who won’t win shouldn’t change the outcome.
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
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.
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.
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?
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?
Every voting method sometimes has spoiler candidates or sometimes fails to elect a Condorcet candidate.
That doesn’t mean that all voting methods are bad or always have problems.
Most methods work pretty well most of the time.
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.
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.
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 |
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 |
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 |
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 |
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 |
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 |
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 | ✔ | ✔ | ✔ |
Which flavor wins this election using approval voting?
Fairness Criterion | |||
---|---|---|---|
Condorcet | Monotonicity | No Spoilers | |
Plurality method | ✘ | ✔ | ✘ |
Instant run-off voting | ✘ | ✘ | ✘ |
Borda count | ✘ | ✔ | ✘ |
Condorcet methods | ✔ | ✔* | ✘ |
Approval voting | ✘ | ✔ | ✔ |