Voting: Issues, Problems, and Systems, Continued. Voting II 1/27

Transcription

1 Voting: Issues, Problems, and Systems, Continued Voting II 1/27

2 Last Time Last time we discussed some elections and some issues with plurality voting. We started to discuss another voting system, the Borda count. We will review and continue discussing that method. We ll also look at two other systems, Sequential Pairwise Voting and the Hare System. But first, let s look at a video about Homer voting in the 2012 Presidential Election Voting Voting II 2/27

3 The Borda Count With the Borda count voters rank order the candidates. If there are n candidates, each first place vote is worth n 1 points, each second place vote is worth n 2 points, and so on, down to 0 points for a last place vote. The person who received the most points wins the election. Voting II 3/27

4 The Avengers Vote Let s review the vote we did on Monday. The candidates were: Hulk (A), Thor (B), Captain America (C), Iron Man (D), and Black Widow (E). Voting II 4/27

5 Using Borda with the Marvel Comics Ballot Rank the five Marvel Comic characters. With your clicker, enter your top-ranked character. The characters are Iron Man (A), Captain America (B), The Hulk (C), Thor (D), and Black Widow (E). Now enter your second-ranked character. Enter your third-ranked character. Enter your fourth-ranked character. Now enter your five-ranked character. In a bit we ll see who won the election with the Borda count. Voting II 5/27

6 Finishing Monday s Example Let s review the example we considered on Monday: There are three candidates, which we will list as A, B, and C. Suppose that 60% lists A first, B second, and C third, and the remaining 40% lists C first, B second, and A third. We ll assume there are 10 ballots. Number of Voters Rank 6 4 First A C Second B B Third C A Voting II 6/27

7 Number of Voters Rank 6 4 First A C Second B B Third C A A receives 2 points for each of her 6 first place votes, and 0 for each of the 4 third place votes. Her total is then 12 points. B receives 1 point each for all of his 10 second place votes. He then has a total of 10 points. Voting II 7/27

8 Number of Voters Rank 6 4 First A C Second B B Third C A C receives 2 points for each of her 4 first place votes and 0 for each of her 6 third place votes. Her total is 8 points. A then wins the election under the Borda count. This is the same outcome as in plurality voting, since A received the most first place votes. Voting II 8/27

9 This example shows a flaw in the Borda count. Suppose that the 4 voters who rated C > B > A instead vote B > C > A. This gives the following chart: Number of Voters Rank 6 4 First A B Second B C Third C A Note that nobody switched their preference between A and B. Voting II 9/27

10 Clicker Question Q Who gets elected with this set of votes? Number of Voters Rank 6 4 First A B Second B C Third C A A Now B has = 14 points, while A has = 12 and C has 1 4 = 4, so B is elected. Voting II 10/27

11 This example shows that the Borda count does not satisfy the Independence of Irrelevant Alternatives: It is impossible for a non-winning candidate B to change to winner unless at least one voter reverses the order in which they listed B and the winner. Voting II 11/27

12 Sequential Pairwise Voting In this voting system, voters rank candidates as in the Borda count. The winner is determined by comparing pairs of candidates. The loser between a comparison is eliminated and the winner is then compared to the next candidate. The last person remaining is then the winner of the election. This system requires ordering the candidates to decide the order of comparison. We ll illustrate how this system works with an example. Voting II 12/27

13 Suppose there are 4 candidates and 3 votes: Number of Voters Rank First A C B Second B A D Third D B C Fourth C D A Suppose we order the candidates alphabetically. Voting II 13/27

14 Number of Voters Rank First A C B Second B A D Third D B C Fourth C D A We first see that A is preferred over B in 2 of 3 ballots, so B is eliminated. We then compare A with the next candidate, which is C. Q Who is eliminated when we compare A and C? A Since C is preferred over A in 2 of 3 ballots, A is eliminated. Voting II 14/27

15 Number of Voters Rank First A C B Second B A D Third D B C Fourth C D A Finally, D is preferred over C in 2 of 3 ballots, so C is eliminated. Thus, D wins the election. The problem is that all voters prefer B to D, even though D was elected. Voting II 15/27

16 Clicker Question Q Suppose we order the candidates B, C, D, A. Who wins the election? Number of Voters Rank First A C B Second B A D Third D B C Fourth C D A A We compare B to C, and C is eliminated. We then compare B to D, and D is eliminated. We then compare B to A, and B is eliminated. Candidate A then wins the election. Voting II 16/27

17 If we use the symbol > to indicate one candidate is preferred by another, in the ballot we have A > B, C > A, A > D B > C, B > D, D > C Written another way, we have A > B > D > A, which doesn t seem to make sense. It would seem to indicate that we should have both A > D and D > A. Voting II 17/27

18 This voting system then violates the Pareto condition: If everybody prefers one candidate to another, then the latter is not elected. The example violates this because everybody prefers B to D, but D was elected (with the original vote). As we ve seen, in this example the order of the candidates matters. There are more sophisticated variants of this voting system, but all have flaws. Voting II 18/27

19 The Hare System Thomas Hare was a political scientist in England in the 19th century. The voting system that is named after him is used to elect members of the Australian House of Representatives and the President of Ireland. The method does not require additional rounds of voting, but acts like a series of runoff elections. The Hare system is convenient for electing committees, when more than one candidate is to be elected, but can be used to elect a single person. Here is some information on the system from sof.uchicago.edu: Voting II 19/27

20 The University of Chicago Information about Hare Purpose. The Hare System is intended to secure the representation of every shade of the electorate s opinion in direct proportion to its numerical strength. What it seeks to rectify. Under the usual form of voting for a list of people for a committee or representative body where several are to be chosen, a bare majority of the votes or even a plurality is sufficient to elect. The outstanding example of this system is the method used in this country for presidential electors. Equally glaring is the inequality where but one person is chosen to office in a representative assembly. Voting II 20/27

21 The following example of an election in Indiana indicates this: Party Votes Representatives Elected Democratic 291, Republican 166,698 0 Progressives 127,041 0 Others 55,807 0 In this instance, while 349,546 voters, a majority, went without representation, a minority elected all the representatives. This occurs with considerable frequency in American legislative elections. Voting II 21/27

22 In the Hare system, candidates are ranked. The candidate (or candidates) with the fewest number of first place votes is eliminated. This process continues until one candidate remains, and the remaining candidate is then elected. When a candidate is removed, we remove that candidate from each ballot. For example, if B is eliminated, then a ballot which lists C > B > A would then be reinterpreted to just have C > A. That is, C would be listed first and A second. Voting II 22/27

23 For example, suppose that an election between 3 candidates has the following votes: Number of Votes Rank First A C B A Second B B C B Third C A A C Then B is eliminated, having fewer first place votes (3) than A (6) and C (4). The ballots are reinterpreted to eliminate B. Voting II 23/27

24 Original: Number of Votes Rank First A C B A Second B B C B Third C A A C Reinterpreted: Number of Votes Rank First A C C A Second C A A C When B is eliminated, the ballots are reinterpreted to see how A is ranked compared to C. In this case, C is preferred by 7 people while A is by 6. Then A is eliminated, and C is elected. Voting II 24/27

25 This system also has flaws. In the previous example, the rightmost column has 1 vote, listing A > B > C. If that voter changed their mind to B > A > C, then the preference between A and C is not changed. However, we would have the following vote tally. Original: New Vote: Number of Votes Rank First A C B A Second B B C B Third C A A C Number of Votes Rank First A C B B Second B B C A Third C A A C Voting II 25/27

26 New Vote: Number of Votes Rank First A C B B Second B B C A Third C A A C Now, A has 5 first place votes while B and C have 4 each. They are both eliminated, and A is then elected. The Hare system thus violates the Monotonicity Criterion: If no voter were to switch his/her preference between the winner and another, then the outcome of the election would be the same. Voting II 26/27

27 Next Time The four systems we have considered, Plurality voting, the Borda count, Sequential Pairwise voting, and the Hare system, each violates at least one of the following conditions: the Condorcet winner condition, independence of irrelevant alternatives, the Pareto condition, and monotonicity. We will review these voting systems and these conditions next time. We will also see how the 1998 Minnesota gubernatorial and 1992 Presidential elections could come out using the various voting systems. Voting II 27/27