Fairness Criteria

 

Majority Criterion:  If a candidate receives a majority of the first place votes, that candidate should win the election.

The plurality, plurality-with-elimination, and pairwise comparisons methods all satisfy this criterion (think about why).

However, look at the following example with 3 candidates and 5 voters:

 

Preference Schedule:

 

Number of voters

3

2

points

1st choice

A

B

2

2nd choice

B

C

1

3rd choice

C

A

0

 

A receives  3 * 2 + 2 * 0 =  6 points.

B receives  3 * 1 + 2 * 2 =  7 points.

C receives  3 * 0 + 2 * 1 =  2 points

 

So B wins by the Borda count method, although A receives a majority of the first place votes.

 

Thus, the Borda count method doesn’t satisfy the majority criterion.

 

(There are many preference schedules for which a candidate receiving a majority also wins by the Borda count method, however there is at least one example where it fails).


Monotonicity Criterion:  If a candidate wins an election, and then we change some of the ballots, but only so as to increase the ratings on those ballots of the winning candidate, then that candidate should still win. (DUH!)

The plurality, Borda count, and pairwise comparisons methods all satisfy this criterion (think about why).

However, look at the following example with 3 candidates and 5 voters:

 

Preference Schedule:

 

Number of voters

6

5

3

7

1st choice

A

B

B

C

2nd choice

C

A

C

B

3rd choice

B

C

A

A

 

A has the fewest first place votes, so is eliminated.  The preference schedule for the remaining candidates is:

 

Number of voters

8

13

1st choice

B

C

2nd choice

C

B

 

So C wins by the plurality-with-elimination (easily, in fact). 

 

 

 

 

Now we’ll help C out by changing some of the ballots in C’s favor.  Switch C and B in the third column:

 

Number of voters

6

5

3

7

1st choice

A

B

C

C

2nd choice

C

A

B

B

3rd choice

B

C

A

A

 

This time B has the fewest first place votes, so is eliminated.  The preference schedule for the remaining candidates is:

 

Number of voters

11

10

1st choice

A

C

2nd choice

C

A

 

So A wins by the plurality-with-elimination. 

Thus plurality-with-elimination fails to satisfy the monotonicity criterion. Another example is given in the text on page 13.

So far we’ve seen drawbacks to the Borda count and plurality-with-elimination methods.  Do the plurality and pairwise comparison methods have any drawbacks?

 

 

 

 

Condorcet Criterion:  If a candidate wins all pairwise comparisons, it should win the election.  (The “cet” in Condorcet is pronounced “say”).

The pairwise comparisons method satisfies this criterion (think about why).

However, look at the following example with 3 candidates and 7 voters:

 

Preference Schedule:

 

Number of voters

3

2

4

1st choice

A

B

C

2nd choice

B

A

A

3rd choice

C

C

B

 

C wins by the plurality method.  However, A beats both B and C in pairwise comparisons.  Thus, the plurality method does not satisfy the Condorcet criterion.  Actually the Borda count and plurality-with-elimination methods fail to satisfy this criterion also (the book gives examples).

Only the pairwise comparisons method satisfies all three of the fairness criteria we’ve looked at; it is looking good right now!

 

 

 

 

 

 

Criterion of Independence of Irrelevant Alternatives:  If a candidate wins, and then one of the losing candidates is eliminated, then the original winner still wins.

This criterion can be violated by all of the voting methods discussed in the text. An example for pairwise comparisons is given in the text (it’s kind of complicated, so I’ll let you read it at your leisure).  Here is a simpler example showing that the Borda method fails to satisfy this criterion:

 

Preference Schedule:

 

Number of voters

2

2

3

points

1st choice

A

B

C

3

2nd choice

D

A

B

2

3rd choice

C

D

A

1

4th choice

B

C

D

0

 

Check that A receives 13 points, B 12, C 11 and D 6.  So A is the Borda method winner.  D is an “irrelevant alternative”; suppose we eliminate D:

 

Number of voters

2

2

3

points

1st choice

A

B

C

2

2nd choice

C

A

B

1

3rd choice

B

C

A

0

 

Now A receives 6 points, B 7 and C 8. So C now wins.

 

We see that none of the election methods we have studied satisfies all four fairness criteria.  Is there some more ingenious election method, which does?

 

 

Arrow’s Impossibility Theorem:  It is impossible to devise an election method, which satisfies all four fairness criteria.

 

Kenneth Arrow was an economist with a background in mathematics. He proved this theorem in 1952, as part of his Ph.D. Thesis.  For this and related work, he received the Nobel Prize in Economics in 1972.