Lecture 3: Head-To-Head Comparisons, Condorcet Winner.

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Lecture 3: Head-To-Head Comparisons, Condorcet Winner. If we know the preference rankings of the voters, we can compare the votes of any two candidates to see which one would win in a plurality election in the absence of the other candidates. Such a comparison is called a Head-To-Head comparison of the two candidates. Example There are 3 candidates for the position of President for the Notre Dame Squash Club. The preferences of the five members of the selection committee are shown in the following table: Presidential preference rankings #Voters 1 2 1 1 Colley 1 2 2 3 Henry 2 1 3 2 Taylor 3 3 1 1 (a) Which candidate would win in a head-to-head comparison between Colley and Henry? (b) Which candidate would win in a head-to-head comparison between Colley and Taylor? (c) Which candidate would win in a head-to-head comparison between Henry and Taylor? In the above example, Henry beat all of the other candidates in a head-to-head comparison. Such a candidate is an example of a Condorcet winner, named after the Marquis de Condorcet (1743-1794). Condorcet Winner A candidate who is the winner of a head-to-head comparison with every other candidate is called a Condorcet winner. If a candidate beats or ties with every other candidate in a head-to-head comparison, that candidate is called a weak Condorcet winner. For any election, there may or may not be a Condorcet winner or a weak Condorcet winner. Example In the example above, Henry is a Condorcet winner. Example Check that there is no Condorcet winner in the election for which the preferences of the voters are shown below. #Voters 1 1 1 Alpha 1 3 2 Beta 2 1 3 Gamma 3 2 1 1

In the following example, we see that a Condorcet winner may not even make it to the runoff in a plurality election with a runoff. Example In a poll to determine player of the year for the North Bend Tennis Club, the ranking preferences of the 100 voters are shown below. #Voters 9 28 11 23 23 6 Federer 1 1 2 3 2 3 Williams 2 3 1 1 3 2 Nadal 3 2 3 2 1 1 In the following example, we see that a Condorcet winner may not be the winner in a Borda count. Example (Purely Fictitious) The following teams participated in 11 events in the Celtic Games 2009: Ireland, Wales, Scotland, Brittany and Cornwall. The rankings of the teams in the events and the number of events in which each given ranking occurred are shown in the table below. It remains for the tournament organizers to determine which team should be declared the winner of the tournament (Who should be Number 1?) #Events 2 1 2 1 1 4 Brittany 1 3 5 5 5 5 Cornwall 2 1 1 3 3 4 Ireland 3 2 2 1 2 3 Scotland 4 4 3 2 1 1 Wales 5 5 4 4 4 2 (a) Who has the top Borda count? (d) Which option, if any, is the Condorcet winner? 2

Notes 1. We have seen above that the Condorcet winner may not win under the Borda Count method and may not even make the runoff if the plurality with runoff method is used. Many argue that if a Condorcet winner exists, then that candidate should be declared the winner. One could regard the roles of all other candidates as spoilers; One can draw a loose analogy with horse racing or long-distance track and bicycle events where a rabbit or pacer with no chance of winning is entered in order to increase the chances of winning for the lead contender on the team. (There are some interesting examples where the rabbit has gone on to win the race). 2. For three candidates, with voters voting randomly, the probability that there is a Condorcet winner is always above 90%. As the number of candidates (and the number of voters) increases, the probability of a Condorcet winner decreases; for six candidates it is always above 68%. In reality, there are often strong and weak candidates and voting is not random, this increases the liklihood of a Condorcet winner. 3. We can shorten our calculations to find a Condorcet winner. We can begin with two strong candidates and compare them. If one wins, then we can eliminate the loser of that comparison from future calculations. We compare the winner to a third candidate, eliminate the loser from our calculations and compare the winner to a fourth candidate, etc, etc, etc... This reduces the number of calculations significantly. Single Peaked Preferences Consider the following example where voters vote for candidates Alan, Bart, Colin, Dan, and Ed: Voters > α β γ Colin 3 1 2 Bart 2 2 1 Alan 1 3 5 Dan 4 4 3 Ed 5 5 4 We can graph the preferences of each voter as follows: We say the graph is single-peaked if it has a single peak. 3

We see that in the graph above, the preferences of voters α and β give single-peaked graphs, whereas the preferences of voter γ do not. Note that we can choose an ordering of the candidates above to get single-peaked graphs for all voter preferences. In any election, if the graphs of all voters preference rankings have a single peak with respect to some fixed ordering of the candidates, we say the preference rankings are single-peaked with respect to that order. Black s Theorem If there exists an ordering of the candidates for which preference rankings are singlepeaked, then the election has a weak Condorcet or a Condorcet winner. A weak Condorcet winner can occur only if the number of voters is even and two or more candidates tie head-to-head against each other and defeat the rest of the candidates. (Note: if the number of voters is odd, there must be a Condorcet winner). Note It can also be shown that If we have n voters, where n is odd and we label the peaks from left to right as P i, 1 i n on a diagram where the preference rankings are single-peaked with distinct peaks, the Condorcet winner is the candidate corresponding to the median peak, P n+1. 2 Example Find the Condorcet winner of the above election/contest using the diagram where the preference rankings are single-peaked. Example Can we find a diagram where the preference rankings are single-peaked for the following example #Voters 1 1 1 Alpha 1 3 2 Beta 2 1 3 Gamma 3 2 1 4

Example (a) Use the following graph to determine the Condorcet winner of the contest. Rank 1 2 3 4 5 Alan Bart Col. Dan Ed Cadidate (b) Construct a table showing voter preferences from the graph and determine the contestant with the highest Borda count. Condorcet Completion Process In a contest where the Condorcet winner, if there is one, emerges victorious, a back-up procedure is necessary for a situation where no Condorcet winner emerges. A number of methods are available. Black 1963 If no Condorcet winner exists, use Borda s method to decide the election. Nanson Round 1. : Use Borda count to eliminate all candidates with below average Borda count. Round 2. : Modify the original ranking of the voters to create new rankings for the remaining candidates. Calculate new Borda Counts and eliminate those with below average Borda count. Round 3: Continue in this way until there is one candidate left. Note if there is a Condorcet winner, that candidate will also be the winner using Nanson s method. Copeland s method This is a simple method often used to decide a Round Robin Tournament, where every candidate plays every other candidate exactly once. With Copeland s method, candidates are ordered by the number of pairwise victories minus the number of pairwise defeats. This method is easily understood and easy to calculate, however it often leads to ties and puts more emphasis on the number of victories and defeats rather than their magnitude. Method of Pairwise Comparisons As with Copeland s method, candidates are compared in pairs. The one who wins gets one point for a win from such a comparison, 1/2 point for a tie and 0 points for a loss. The points from each pairwise comparison are added and the winner is the candidate with the most points. 5

Example Use Copeland s method, Nanson s method and the method of pairwise comparisons to decide the winner of the Nation s Cup in the Celtic games 2010, the result of which are shown below (there were only ten events in the 2010 games), #Events 3 3 2 1 1 Brittany 5 1 3 5 3 Cornwall 3 4 1 4 1 Ireland 1 2 4 2 5 Scotland 2 3 5 3 4 Wales 4 5 2 1 2 Example In a round robin soccer tournament, there are 10 teams, two of which are much stronger than the other teams. Discuss methods of deciding the winner, so all teams have some chance of winning. Example The Icelandia State University hockey team is on a road trip. An important decision needs to be made: Where to go for dinner? In the past, this has led to some heated arguments, so this time they decide to have an election. The choices boil down to three restaurants: Hunan (H), Pizza Palace (P) ands Danny s (D). The decision is to be made using the method of pairwise comparisons. The table below shows the results of the voting by the 11 players on the squad. #Voters 4 2 5 H 1 3 2 P 2 1 3 D 3 2 1 (a) Show that using the method of pairwise comparisons, we get a three way tie. In general there is no set way to break a tie, but it is important to establish the rules on how ties will be broken before the election begins. (b) If ties should be broken by counting first place votes, which restaurant wins? (c) If ties should be broken with a Borda count, which restaurant wins? 6