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Study Guides > Mathematics for the Liberal Arts Corequisite

The Fairness Criteria

Learning Outcomes

  • Apply the fairness criteria to evaluate the fairness of contests whose outcome is determined using the voting methods of the previous two sections.
 

Fairness Criteria

The fairness criteria are statements that seem like they should be true in a fair election.
  • Condorcet Criterion: If a choice is favored over every other candidate in one on one comparisons, that choice should be the winner.
  • Majority Criterion: If a choice has a majority of first-place votes, that choice should be the winner.
  • Monotonicity Criterion: If voters change their votes to increase the preference for a choice, it should not harm that choice’s chances of winning.
  • Independence of Irrelevant Alternatives (IIA) Criterion: If a non-winning choice is removed from the ballot, it should not change the winner of the election.
In this section, we will consider examples that exploit some of the ways in which our voting methods fail to meet various fairness criteria.  We have already hinted at some of these issues in previous examples.

Important Note

Even though a criterion may be violated by a voting method in one particular contest scenario, it does not mean that voting method always violates the criterion.  It simply demonstrates that the voting method has the potential to violate the criterion in certain election scenarios.
 

What's Wrong with the Plurality Method?

The following video includes a a review of the plurality method including an example that illustrates this method's ability to violate the Condorcet fairness criterion. https://youtu.be/r-VmxJQFMq8   The next example gives another illustration of an election in which the plurality winner is not the Condorcet Winner.

EXAMPLE

Consider a city council election in a district that is historically 60% Democratic voters and 40% Republican voters. Even though city council is technically a nonpartisan office, people generally know the affiliations of the candidates. In this election there are three candidates: Don and Key, both Democrats, and Elle, a Republican. A preference schedule for the votes looks as follows:
342 214 298
1st choice Elle Don Key
2nd choice Don Key Don
3rd choice Key Elle Elle
Notice that a total of [latex]342+214+298=854[/latex] voters participated in this election. Computing percentage of first place votes:
  • Don: 214/854 = 25.1%
  • Key: 298/854 = 34.9%
  • Elle: 342/854 = 40.0%
So in this election, there is no majority winner.  It appears that the Democratic voters split their vote over the two Democratic candidates, allowing the Republican candidate Elle to win under the plurality method with only 40% of the vote.   Analyzing this election further, we calculate the one on one comparisons:  
  • Elle vs Don: 342 prefer Elle; 512 prefer Don: Don is preferred
  • Elle vs Key: 342 prefer Elle; 512 prefer Key: Key is preferred
  • Don vs Key: 556 prefer Don; 298 prefer Key: Don is preferred
  So even though Don had the smallest number of first-place votes in the election, he is the Condorcet Winner, being preferred in every one-to-one comparison with the other candidates.  Thus, the plurality method, in this scenario violates the Condorcet Criterion.   This example is worked out in the following video. https://youtu.be/x6DpoeaRVsw?list=PL1F887D3B8BF7C297
   

What’s Wrong with the Borda Count Method?

Let's consider the example from the last section in which a group of mathematicians are meeting for a conference.  Perhaps you already noticed a potential flaw of the Borda Count from that example. In particular, Seattle had a majority of first-choice votes, yet lost the election!  This example demonstrates that the Borda Count method has the potential to violate the Majority Criterion.  We will also note in this example the Borda Count method's potential to violate the Condorcet Criterion.

Recall: Example

In this example, our group of mathematicians are getting together for a conference. The members are coming from four cities: Seattle, Tacoma, Puyallup, and Olympia.A map of the Puget Sound. Seattle, Tacoma, Puyallup, and Olympia are marked. Their approximate locations on a map are shown to the right. The votes for where to hold the conference were:
51 25 10 14
1st choice Seattle Tacoma Puyallup Olympia
2nd choice Tacoma Puyallup Tacoma Tacoma
3rd choice Olympia Olympia Olympia Puyallup
4th choice Puyallup Seattle Seattle Seattle
Verify that the Borda Count method violates the Majority Criterion in this scenario.

Answer: Recall that using the Borda Count method, the points awarded to each city were:

  • Seattle: [latex]253[/latex] points
  • Tacoma: [latex]325[/latex] points
  • Puyallup: [latex]194[/latex] points
  • Olympia: [latex]228[/latex] points
Thus, under the Borda Count method, Tacoma was the winner of this vote.  However, since Seattle has the majority of first-choice votes, the Borda Count method violates the Majority Criterion in this scenario. Also, notice that Seattle won 51 out of 100 votes (or 51%) thus winning not only under the plurality method, but also under majority rule.  This automatically means that the Condorcet Criterion will also be violated, as Seattle would have been preferred by 51% of voters in any head-to-head comparison.

 
  While we have demonstrated the Borda Count method's ability to violate at least two of the fairness criteria, there are some important benefits to this method.  In particular, Borda can sometimes choose a more broadly acceptable option over the one with majority support and for this reason is sometimes described as a consensus-based voting system. In the previous example, Tacoma is probably the best compromise location. This is a different approach than plurality and instant run-off voting that focus on first-choice votes; Borda Count considers every voter’s entire ranking to determine the outcome. Because of this consensus behavior, Borda Count, or some variation of it, is commonly used in awarding sports awards. Variations are used to determine the Most Valuable Player in baseball, to rank teams in NCAA sports, and to award the Heisman trophy.

What’s Wrong with Run-off Methods?

Next, we consider election scenarios in which run-off methods violate, in the first example, the Condorcet Criterion and in the second example, the Monotonicity Criterion.

Examples

Example 1: Returning to the first example on this page, we consider again our City Council Election.  Use a run-off method to find the winner of this election.  (Recall that IRV and single run-off are the same when we have only three choices.)  Then, verify that a run-off method used on this election scenario would violate the Condorcet Criterion.
342 214 298
1st choice Elle Don Key
2nd choice Don Key Don
3rd choice Key Elle Elle

Answer: In this election, Don has the smallest number of first place votes, so Don is eliminated in the first round. The 214 people who voted for Don have their votes transferred to their second choice, Key.

342 512
1st choice Elle Key
2nd choice Key Elle
  So Key is the winner under our run-off method.  We immediately notice that in this election scenario, run-off method violates the Condorcet Criterion, since we determined earlier that Don was the Condorcet winner.  On the other hand, the temptation has been removed for Don’s supporters to vote for Key; they now know their vote will be transferred to Key, not simply discarded.

  The following video works out this example in which we find that the IRV method violates the Condorcet Criterion in an election for a city council seat. https://youtu.be/BCRaYCU28Ro?list=PL1F887D3B8BF7C297   Example 2: Suppose a student club at Colorado Mesa University has three candidates for Club President: Adams, Brown, and Carter.  A preference ballot is used and the preference schedule is shown below.  First, determine a winner using a run-off method.  Note that in this election scenario, the Monotonicity Criterion will be violated.  
37 22 12 29
1st choice Adams Brown Brown Carter
2nd choice Brown Carter Adams Adams
3rd choice Carter Adams Carter Brown
 

Answer: In this election, Carter would be eliminated in the first round since he has the lowest number of first-choice votes.  Since his 29 votes would go to Adams, then Adams would be the winner with 66 votes to 34 for Brown.   Now suppose that the results were announced, but election officials accidentally destroyed the ballots before they could be certified, and the votes had to be recast. Wanting to “jump on the bandwagon,” 10 of the voters who had originally voted in the order Brown, Adams, Carter change their vote to favor the presumed winner, changing those votes to Adams, Brown, Carter.

47 22 2 29
1st choice Adams Brown Brown Carter
2nd choice Brown Carter Adams Adams
3rd choice Carter Adams Carter Brown
In this re-vote (still using a run-off), Brown will be eliminated in the first round, having the fewest first-place votes. After transferring votes, we find that Carter will win this election with 51 votes to Adams’ 49 votes! Even though the only vote changes made favored Adams, the change ended up costing Adams the election. This scenario demontrates that run-off methods can violate the Monotonicity Criterion.

  Example 2 is explained in the following video. https://youtu.be/NH78zNXHKUs?list=PL1F887D3B8BF7C297
 

What’s Wrong with the Pairwise Comparisons Method?

The idea behind the Independence of Irrelevant Altenartive Criterion is this: suppose a voting method yields a winner and one of the losing choices is no longer available (disqualified, say) then, applying the same preferences and the same voting method should still produce the same winner.

Example

Consider the preference schedule shown below.  Show that the Independence of Irrelevant Alternatives Criterion is violated by the pairwise comparisons method used on this preference schedule.
11 7 6 9 3
1st choice B D C A D
2nd choice A B A C C
3rd choice C C D D B
4th choice D A B B A

Answer: To apply the pairwise comparisons method, we will write out all of the head-to-head results.

  • A vs B: 15 to 21 (B wins 1 point)
  • A vs C: 20 to 16 (A wins 1 point)
  • A vs D: 26 to 10 (A wins 1 point)
  • B vs C: 18 to 18 (Tie: each earns 1/2 point)
  • B vs D: 11 to 25 (D wins 1 point)
  • C vs D: 26 to 10 (C wins 1 point)
We have that A wins 2 points, B and C each earn 1.5 points, and D wins 1 point.  Since A has won the most head-to-head results (two), A is the winner using the pairwise comparisons method.   Now, let's remove D (a non-winning choice) from the election and repeat the pairwise comparisons.
11 7 6 9 3
1st choice B C A
2nd choice A B A C C
3rd choice C C B
4th choice A B B A
  • A vs B: 15 to 21 (B wins 1 point)
  • A vs C: 20 to 16 (A wins 1 point)
  • B vs C: 18 to 18 (Tie: each earns 1/2 point)
Now, A wins just 1 point, B wins 1.5, and C wins 0.5, thus, making B the winner.

 

Important Note (Worth repeating!)

Even though a criterion may be violated by a voting method in one particular contest scenario, it does not mean that voting method always violates the criterion.  It simply demonstrates that the voting method has the potential to violate the criterion in certain election scenarios.

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