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Math 103, Voting Theory, In-class materials #4

So far, we have learned about 4 voting methods (Plurality, Instant Runoff, Borda Count, and Pairwise Comparisons). We have also learned about 2 fairness criteria (Majority and Condorcet) and what it means to violate each one. Now, we will learn about the remaining 2 fairness criteria in this chapter.

Fairness criterion #3: The monotonicity criterion. An election is held using a certain voting method, and a winner is declared. For some reason, a second election (a "revote") is held, and the only changes made to any of the ballots are in favor of the original winner (pushing the original winner higher on some voters' ballots). Then the original winner should remain the winner when the revote is held.

Example: An election is held among political analysts ranking Ann, Bob, and Claire as future presidential candidates, with the following results:

st

1 choice 2nd choice 3rd choice

7 A B C

8 B C A

10 C A B

4 A C B

Total=29/2 Majority= 15 Who is the winner under instant runoff voting? C

Math 103, Voting Theory, In-class materials #4

st

1 choice 2nd choice 3rd choice

7 A B C

8 B C A

10 C A B

4 A C B

It turns out that this (above) wasn't the real election; instead it was just a straw poll. When the real election is held -- which can be thought of as a second election, or a revote -- the following is the new preference table:

st

1 choice 2nd choice 3rd choice

7 A B C

8 B C A

10 C A B

4 C A B

Precisely how does this preference table differ from that of the original election? Without determining the winner of the second election, is this a case in which the monotonicity criterion could apply?

Who is the winner of the revote under instant runoff voting? B

What conclusion follows about the instant runoff voting method and the monotonicity criterion? The IRV method violates the monotonicity sometimes. The winner moved up and still didn’t win which caused a violation.

Math 103, Voting Theory, In-class materials #4

Fairness criterion #4: The Independence of Irrelevant Alternatives ("I.I.A.") criterion. An election is held using a certain voting method, and a winner is declared. For some reason, a second election (a "revote") is held, and the only change is that one candidate who lost the original election now drops out altogether (e.g. decides to quit, or is disqualified, etc.). Then the original winner should remain the winner in the revote.

Comprehension check: What is the difference between the monotonicity criterion and the I.I.A. criterion?

Both the monotonicity and the I.I.A. criteria pertain to an original vote and a revote. For each one, there are specific circumstances in which a change in winner from the original election to the revote counts as a violation, i.e. an unfair outcome. These specific circumstances are different for the two fairness criteria:  For the monotonicity criterion, the issue is that some ballots have changed in a certain way, but  For the I.I.A. criterion, the issue is that the list of candidates has changed in a certain way.

Math 103, Voting Theory, In-class materials #4

Example: the pairwise-comparisons method is applied to the following election, a poll of 22 analysts on the contenders to host the 2020 summer Olympics. The candidates are Istanbul (I), Tokyo (T), Doha (D), Baku (B), and Madrid (M). st

1 choice 2nd choice 3rd choice 4th choice 5th choice

2 I M D T B

6 T I D M B

4 T I M B D

1 D T I M B

1 D M I T B

4 M I B D T

4 B D M T I

Find the winner under the pairwise comparisons method: I vs. T 7:15 I vs. D 16:6 I vs. B 18:4 I vs. M 13:9

T vs. D 10:12 T vs. B 14:8 T vs. M 11:11

D vs. B 14:8 B vs. M 4:18 D vs. M 12:10

I has __3__ points, T has __2.5__ points, D has _2___ points, B has __1___ points, M has __1.5___ points.

I is winner

Math 103, Voting Theory, In-class materials #4

Now suppose that Doha decides to drop out of the race. If none of the voters change their relative rankings of the candidates, what is the new preference table?

st

1 choice 2nd choice 3rd choice 4th choice

2 I M B

6 T I M B

4 T I M B

1 T I M B

1 M I T B

4 M I B T

4 B M T I

Find the winner under the pairwise comparisons method: I vs. T 7:15 I vs. B 18:4 I vs. M 13:9

T vs. B 14:8 T vs. M 11:11

B vs. M 4:18

T has 2.5 points, I has 2 points, M has 1 point, B has 0 points

A non winning

Math 103, Voting Theory, In-class materials #4

What conclusion follows about the pairwise comparisons method and the independence of irrelevant alternatives criterion?

A non winning candidate dropped out, so there is a violation since the original winner didn’t win.

Since the effect of an irrelevant alternative quitting the race (Doha, which wasn’t going to win anyway) is to change the original winner into a non-winner, this election under the pairwise comparisons method illustrates a violation of the I.I.A. criterion.

Math 103, Voting Theory, In-class materials #4

Arrow's Impossibility Theorem: Every voting system violates at least one of the four fairness criteria we have discussed.

Caution: This theorem does NOT assert that democracy is inherently unfair! It asserts that if we take certain insights about fairness and make them mathematically precise as fairness criteria, then no voting method can never violate all of them (in the sense of never producing a violation of any of them). In other words, in any election there is always a risk that something will go wrong – in a narrow, mathematically precise sense of “wrong” – but there is no reason why something must go wrong in any one particular election.

Remark: Arrow’s theorem, first published in the 1950s, is a mathematical result, not an empirical one. It is a general statement, proved mathematically, not inferred from a large number of actual elections; it therefore has no counterexamples, and will have no counterexamples, no matter how hard we look. Moreover, Arrow’s theorem is an example of ongoing discovery in mathematics. New discoveries are made every year, often very significant ones.

Math 103, Voting Theory, In-class materials #4

Review of never violating and sometimes violating fairness criteria: In each of the following scenarios, indicate if a fairness criterion has been violated, and if so, say which criterion. Read each scenario carefully!

1. Ann has more first choice votes than any other candidate, but Bob wins the election. Doesn't violate because it doesn't mean Ann has the majority. There is not enough information so it doesn’t violate anything. 2. Ann has more first choice votes than all other candidates combined, but Bob wins the election. This violates the majority criterion and Condorcet because the majority candidate is always a Condorcet candidate Ann should have won but Bob won.

3. An election is held, and Bob is the winner. For some reason Ann is disqualified, and a revote is held. In the revote, Claire is the winner. Violation of of IAA criterion. A non winning candidate dropped and someone else won that wasn't the original winner.

4. An election is held, and Bob is the winner. For some reason Bob is disqualified, and a revote is held. In the revote, Claire is the winner. No violation because Bob is out and he was the winner so the election had to of changed.

5. Claire defeats every opponent in head to head competition, but Dave wins the election. Violates Condorcet because she won every head to head and still lost.

Math 103, Voting Theory, In-class materials #4

6.An election is held and Dave is the winner. For some reason a second election (a revote) is held, in which some voters change their ballots from what had been 1st choice Claire 2nd choice Ann 3rd choice Dave 4th choice Bob in the first election, to 1st choice Ann 2nd choice Dave 3rd choice Bob 4th choice Claire in the second election. When the votes are counted in the second election, Ann turns out to be the winner. No violation because no one dropped out, and a revote didn't affect the original winner

7. Claire defeats Ann and Dave, and ties Bob in head to head competition; Claire never loses any head to head competition. Dave wins the election. This does not violate the Condorcet because there is a tie and there is no Condorcet candidate.

Math 103, Voting Theory, In-class materials #4

We have now studied elections with 3, 4, or 5 candidates. To begin to transition to the next topic in the course, we consider how many different preference ballots are possible, i.e. how many different ways there are to rank the candidates on a single preference ballot, if there are N candidates.

Claim: If there are 3 candidates, then there are (3)(2)(1) = 3! = "3 factorial" = 6 possible preference ballots.

If there are 4 candidates, then there are (4)(3)(2)(1) = 4! = 24 possible preference ballots.

If there are 5 candidates, then there are (5)(4)(3)(2)(1) = 5! = 120 possible preference ballots.

If there are N candidates, then there are (N)(N – 1)(N – 2)...(3)(2)(1) = N! = "N factorial" possible preference ballots.

Math 103, Voting Theory, In-class materials #4

Why does this make sense? Idea: Write out all 6 preference ballots with candidates A, B, and C: A B C

A C B

B A C

B C A

C A B

C B A

If D joins the election as a candidate, there are  6 preference ballots resulting from taking the 6 you just wrote out and putting D as first choice: D A B C

D A C B

D B A C

D B C A

D C A B

D C B A

 another 6 preference ballots resulting from taking the 6 you just wrote out and putting D as second choice, A D B C

A D C B

B D A C

B D C A

C D A B

C D B A

 another 6 preference ballots resulting from taking the 6 you just wrote out and putting D as third choice, A B D C

A C D B

B A D C

B C D A

C A D B

C B D A

Math 103, Voting Theory, In-class materials #4

 and another 6 preference ballots resulting from taking the 6 you just wrote out and putting D as fourth choice. A B C D

A C B D

B A C D

B C A D

C A B D

C B A D

That makes a total of 4 times 6 preference ballots, i.e. 4(3!) = (4)(3)(2)(1) = 4! = 24 different possible preference ballots.

The same line of reasoning shows that once you have written out all 4! preference ballots for 4 candidates, putting a fifth candidate into the election multiplies the number of possible preference ballots by 5.

The same line of reasoning shows that once you have written out all (N–1)! preference ballots for N – 1 candidates, putting one more candidate into the election multiplies the number of possible preference ballots by N, hence N! possible preference ballots.

How many different preference ballots are possible if there are 10 candidates? 10! What if there are 20 candidates? 20!...