Loading in 2 Seconds...
Loading in 2 Seconds...
For All Practical Purposes. Chapter 9: Social Choice: The Impossible Dream Lesson Plan. Voting and Social Choice Majority Rule and Condorcet’s Method Other Voting Systems for Three or More Candidates Plurality Voting Borda Count Sequential Pairwise Voting Hare System
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
Mathematical Literacy in Today’s World, 7th ed.
© 2006, W.H. Freeman and Company
Example: Selecting a winner of an election using a good procedure that will result in an outcome that “reflects the will of the people”
Throughout the chapter, we assume the number of voters is odd (to help simplify and focus on the theory). Furthermore, in the real world, the number of voters is often so large that ties seldom occur.
May’s Theorem – Among all two-candidate voting systems that never result in a tie, majority rule is the only one that treats all voters and both candidates equally and is monotone.
Monotone means that if a new election were held and a single voter were to change his or her ballot from voting for the losing candidate to voting for the winning candidate (and everyone else voted the same), the outcome would be the same.
Condorcet’s Voting Paradox – With three or more candidates, there are elections in which Condorcet’s method yields no winners.
A beats B, 2 out of 3; and
B beats C, 2 out of 3; and
C beats A, 2 out of 3 — No winner!
Rank Number of Voters (3)
First A B C
Second B C A
Third C A B
Example: Find the plurality vote of the 3 candidates and 13 voters.
The candidate with the most first-place votes wins. Count each candidate’s first-place votes only. (A has the most.)
A = 5, B = 4, C = 4 A wins.
It is assumed if Al Gore was pitted against any one of the other three candidates, (Bush, Buchanan, Nader), Gore would have won.
Manipulabilityoccurs when voters
misrepresents their preference
rather than “throw away” their vote.
First-place vote is worth n − 1 points, second-place vote is worth n − 2 points, and so on down to…Last place vote is worth n − n = 0, zero points.
Example: Total the Borda score of each candidate.
A = 2 + 2 + 2 + 0 + 0 = 6 B = 1 + 1 + 1 + 2 + 2 = 7 C = 0 + 0 + 0 + 1 + 1 = 2
B has the most, B wins.
Another way: Count the occurrences of letters below the candidate—for example, there are 7 “boxes” below B
Example showing that Borda count fails to satisfy IIA:
B went from loser to winner and did not switch with A!
Suppose the last two voters change their ballots (reverse the order of just the losers). This should not change the winner.
Original Borda Score: A=6, B=5, C=4
New Borda Score: A= 6, B=7, C=2
Example: Who would be the winner using the agenda A, B, C, D for the following preference list ballots of three voters?
Using the agenda A, B, C, D, start with A vs. B and record (with tally marks) who is preferred for each ballot list (column).
A vs. B
A vs. C
C vs. D
Candidate D wins for this agenda.
A wins; B is deleted.
C wins; A is deleted.
D wins; C is deleted.
Different agenda orders can change the outcomes. For example, agenda D, C, B, A results in A as the winner.
For the Hare system, delete the candidate with the least first-place votes:
A = 5, B = 4, and C = 4
Since B and C are tied for the least first place votes, they are both deleted and A wins.
In the previous example, A won. For the last voter, move A up higher on the list (A and B switch places). Round 1: B is deleted. Round 2: C moves up to replace B on the third column. However, C wins—this is a glaring defect!