Voting: Issues, Problems, and Systems, Continued 9 March 2012 - - PowerPoint PPT Presentation

Voting: Issues, Problems, and Systems, Continued

9 March 2012

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Last Time

We’ve discussed several voting systems and conditions which may or may not be satisfied by a system. We’ll review them. But first, let’s look at some clips from reporting of the 2000 presidential election.

1

Florida called for Gore

2

Gore declared the winner in Pennsylvania

3

The vote in Florida is now too close to call

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Plurality Voting

The method we use in the U.S. to decide most elections is called Plurality

• Voting. In this method, a voter chooses one candidate, and the candidate

with the most votes wins. It is nearly the most simple method of deciding elections.

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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. The person who received the most points wins the election.

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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.

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The Hare System

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 is then elected. Once a candidate is elected, 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 have C > A. That is, C would be listed first and A second.

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The Four Conditions

Condorcet winner criterion: The Condorcet winner (preferred over all

• thers head to head), if there is one, always wins the election.

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The Four Conditions

Condorcet winner criterion: The Condorcet winner (preferred over all

• thers head to head), if there is one, always wins the election.

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.

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The Four Conditions

Condorcet winner criterion: The Condorcet winner (preferred over all

• thers head to head), if there is one, always wins the election.

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. Pareto: If everybody prefers one candidate to another, then the latter is not elected.

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The Four Conditions

Condorcet winner criterion: The Condorcet winner (preferred over all

• thers head to head), if there is one, always wins the election.

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. Pareto: If everybody prefers one candidate to another, then the latter is not elected. Monotonicity: If no voter were to switch his/her preference between the winner and another, then the outcome of the election would be the same.

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The following table summarizes how each system behaves with respect to each of the conditions. A yes means the system satisfies the condition. CWC IIA Pareto Mono Plurality no no yes yes Borda no no yes yes Sequential yes no no yes Hare no no yes no CWC = Condorcet Winner Condition IIA = Independence of Irrelevant Alternatives Mono = Monotonicity Condition

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Back to the 1998 Minnesota Gubernatorial Election

How would the 1998 Minnesota gubernatorial election come out if

• ther voting systems were used?

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Back to the 1998 Minnesota Gubernatorial Election

How would the 1998 Minnesota gubernatorial election come out if

• ther voting systems were used?

To answer this we’d need to have had candidates rank ordered. Let’s suppose the voters would rank the three candidates, Coleman, Humphrey, and Ventura, as follows. This table is not based on exit

• polls. I made up the table, trying to make a reasonable guess as to

what would have happened.

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Back to the 1998 Minnesota Gubernatorial Election

How would the 1998 Minnesota gubernatorial election come out if

• ther voting systems were used?

To answer this we’d need to have had candidates rank ordered. Let’s suppose the voters would rank the three candidates, Coleman, Humphrey, and Ventura, as follows. This table is not based on exit

• polls. I made up the table, trying to make a reasonable guess as to

what would have happened. Recall that Ventura received 37% of the first place votes, while Coleman received 35% and Humphrey 28%. 20% 25% 8% 10% 20% 17% H C H C V V C H V V C H V V C H H C

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Using the Borda Count

With the Borda count, since there are three candidates, we assign 2 points for first place votes, 1 point for second place votes, and 0 points for third place votes. To make it easier, we pretend there are 100 votes, so the percentage refers to the number of votes.

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Using the Borda Count

With the Borda count, since there are three candidates, we assign 2 points for first place votes, 1 point for second place votes, and 0 points for third place votes. To make it easier, we pretend there are 100 votes, so the percentage refers to the number of votes. 20% 25% 8% 10% 20% 17% H C H C V V C H V V C H V V C H H C

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Using the Borda Count

With the Borda count, since there are three candidates, we assign 2 points for first place votes, 1 point for second place votes, and 0 points for third place votes. To make it easier, we pretend there are 100 votes, so the percentage refers to the number of votes. 20% 25% 8% 10% 20% 17% H C H C V V C H V V C H V V C H H C first place votes 2nd 3rd total points C 35 40 25 110 H 28 42 30 98 V 37 18 45 92

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We computed Coleman’s totals by: 35 · 2 + 40 · 1 + 25 · 0 = 70 + 40 = 110. The other totals were computed similarly. Thus, with the Borda count, Coleman wins the election.

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Sequential Pairwise Voting

20% 25% 8% 10% 20% 17% H C H C V V C H V V C H V V C H H C

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Sequential Pairwise Voting

20% 25% 8% 10% 20% 17% H C H C V V C H V V C H V V C H H C If we use sequential pairwise voting, and order them C, H, V, then Coleman is preferred to Humphrey 55% to 45%, so Humphrey is

• eliminated. Since Coleman is also preferred 55% to 45% over

Ventura, so Coleman is elected in this system.

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Sequential Pairwise Voting

20% 25% 8% 10% 20% 17% H C H C V V C H V V C H V V C H H C If we use sequential pairwise voting, and order them C, H, V, then Coleman is preferred to Humphrey 55% to 45%, so Humphrey is

• eliminated. Since Coleman is also preferred 55% to 45% over

Ventura, so Coleman is elected in this system. It turns out that, in this election, the order in which we list the candidates does not affect the outcome

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Hare System

If we use the Hare system, then Humphrey is eliminated since he received the fewest first place votes, 28%, compared to 35% and 37% for the other two candidates. We then reinterpret the ballot by eliminating Humphrey. This gives

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Hare System

If we use the Hare system, then Humphrey is eliminated since he received the fewest first place votes, 28%, compared to 35% and 37% for the other two candidates. We then reinterpret the ballot by eliminating Humphrey. This gives 20% 25% 8% 10% 20% 17% C C V C V V V V C V C C

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After reinterpreting the ballots, Coleman then has 55% first place votes to Ventura’s 45%, so Coleman would win under the Hare system.

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After reinterpreting the ballots, Coleman then has 55% first place votes to Ventura’s 45%, so Coleman would win under the Hare system. So, in all of the systems, other than plurality voting, Coleman would win the election.

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The 1992 Presidential Election Revisited

Let’s make an estimate of how the voting would have been conducted if voters ranked the three main candidates, and determine who would be elected in each of the four systems we have studied.

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The 1992 Presidential Election Revisited

Let’s make an estimate of how the voting would have been conducted if voters ranked the three main candidates, and determine who would be elected in each of the four systems we have studied. I’ve made an attempt to predict how voters would rank the candidates, in order to come up with the following table.

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The 1992 Presidential Election Revisited

Let’s make an estimate of how the voting would have been conducted if voters ranked the three main candidates, and determine who would be elected in each of the four systems we have studied. I’ve made an attempt to predict how voters would rank the candidates, in order to come up with the following table. Rank 30% 13% 12% 25% 6% 13% First Clinton Clinton Bush Bush Perot Perot Second Bush Perot Clinton Perot Clinton Bush Third Perot Bush Perot Clinton Bush Clinton

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Borda count

We assign 2 points for each first place vote and 1 point for each second place vote. To make the arithmetic easier, we interpret the percentages as votes.

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Borda count

We assign 2 points for each first place vote and 1 point for each second place vote. To make the arithmetic easier, we interpret the percentages as votes. Clinton: 43 first place votes and 18 second place votes. Total points, 43 · 2 + 18 · 1 = 104

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Borda count

We assign 2 points for each first place vote and 1 point for each second place vote. To make the arithmetic easier, we interpret the percentages as votes. Clinton: 43 first place votes and 18 second place votes. Total points, 43 · 2 + 18 · 1 = 104 Bush: 37 first place votes and 43 second place votes. Total points, 37 · 2 + 43 · 1 = 117

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Borda count

We assign 2 points for each first place vote and 1 point for each second place vote. To make the arithmetic easier, we interpret the percentages as votes. Clinton: 43 first place votes and 18 second place votes. Total points, 43 · 2 + 18 · 1 = 104 Bush: 37 first place votes and 43 second place votes. Total points, 37 · 2 + 43 · 1 = 117 Perot: 19 first place votes and 38 second place votes. Total points, 19 · 2 + 38 · 1 = 76 Since Bush has the most points, he would be elected.

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Pairwise Sequential Voting

Rank 30% 13% 12% 25% 6% 13% First Clinton Clinton Bush Bush Perot Perot Second Bush Perot Clinton Perot Clinton Bush Third Perot Bush Perot Clinton Bush Clinton

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Pairwise Sequential Voting

Rank 30% 13% 12% 25% 6% 13% First Clinton Clinton Bush Bush Perot Perot Second Bush Perot Clinton Perot Clinton Bush Third Perot Bush Perot Clinton Bush Clinton Let us first order the candidates Bush, Clinton, Perot. We compare Clinton and Bush head to head. Bush is preferred to Clinton by 51%

• f the voters. Therefore, Clinton is eliminated. Bush is then

compared to Perot head to head, and Bush is preferred. Perot is then eliminated.

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Pairwise Sequential Voting

Rank 30% 13% 12% 25% 6% 13% First Clinton Clinton Bush Bush Perot Perot Second Bush Perot Clinton Perot Clinton Bush Third Perot Bush Perot Clinton Bush Clinton Let us first order the candidates Bush, Clinton, Perot. We compare Clinton and Bush head to head. Bush is preferred to Clinton by 51%

• f the voters. Therefore, Clinton is eliminated. Bush is then

compared to Perot head to head, and Bush is preferred. Perot is then eliminated. Bush is the one remaining, so he is elected. Other orderings of the candidates turn out to produce the same result.

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Hare System

We first eliminate the candidate with the fewest first place votes. This is Perot, so he is eliminated. We next reinterpret each vote to rank only Clinton and Bush. This yields the following table.

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Hare System

We first eliminate the candidate with the fewest first place votes. This is Perot, so he is eliminated. We next reinterpret each vote to rank only Clinton and Bush. This yields the following table. Original vote Rank 30% 13% 12% 25% 6% 13% First Clinton Clinton Bush Bush Perot Perot Second Bush Perot Clinton Perot Clinton Bush Third Perot Bush Perot Clinton Bush Clinton

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Hare System

We first eliminate the candidate with the fewest first place votes. This is Perot, so he is eliminated. We next reinterpret each vote to rank only Clinton and Bush. This yields the following table. Original vote Rank 30% 13% 12% 25% 6% 13% First Clinton Clinton Bush Bush Perot Perot Second Bush Perot Clinton Perot Clinton Bush Third Perot Bush Perot Clinton Bush Clinton Reinterpreted vote Rank 30% 13% 12% 25% 6% 13% First Clinton Clinton Bush Bush Clinton Bush Second Bush Bush Clinton Clinton Clinton Bush

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In the reinterpreted vote, Clinton receives 49% of the first place votes and Bush 51%. Therefore, Clinton is eliminated.

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In the reinterpreted vote, Clinton receives 49% of the first place votes and Bush 51%. Therefore, Clinton is eliminated. Thus, Bush is elected with the Hare system.

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In the reinterpreted vote, Clinton receives 49% of the first place votes and Bush 51%. Therefore, Clinton is eliminated. Thus, Bush is elected with the Hare system. Thinking about the various voting systems we have considered, and the fact that all of them violate a reasonable condition, leads one to ask the following question:

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In the reinterpreted vote, Clinton receives 49% of the first place votes and Bush 51%. Therefore, Clinton is eliminated. Thus, Bush is elected with the Hare system. Thinking about the various voting systems we have considered, and the fact that all of them violate a reasonable condition, leads one to ask the following question: Is there a voting system which satisfies all four conditions we have introduced?

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The 2000 Presidential Election

Since Gore won the popular vote, he would have won with plurality voting.

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The 2000 Presidential Election

Since Gore won the popular vote, he would have won with plurality voting. Gore would also have won with sequential pairwise voting, and with the Hare system.

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The 2000 Presidential Election

Since Gore won the popular vote, he would have won with plurality voting. Gore would also have won with sequential pairwise voting, and with the Hare system. It is less clear to me who would have won with the Borda count.

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Arrow’s Theorem

Kenneth Arrow, who won the Nobel prize in Economics in 1972, proved in 1951 that all voting systems have flaws. More precisely, he proved that there does not and cannot exist a voting system which satisfies both the Condorcet Winner condition and the Independence

• f Irrelevant Alternatives.

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Arrow’s Theorem

Kenneth Arrow, who won the Nobel prize in Economics in 1972, proved in 1951 that all voting systems have flaws. More precisely, he proved that there does not and cannot exist a voting system which satisfies both the Condorcet Winner condition and the Independence

• f Irrelevant Alternatives.

Arrow’s theorem says, therefore, that no matter what voting system we use, there will be flaws with it.

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Arrow’s Theorem

Kenneth Arrow, who won the Nobel prize in Economics in 1972, proved in 1951 that all voting systems have flaws. More precisely, he proved that there does not and cannot exist a voting system which satisfies both the Condorcet Winner condition and the Independence

• f Irrelevant Alternatives.

Arrow’s theorem says, therefore, that no matter what voting system we use, there will be flaws with it. To determine what voting system to use in a given situation, we should then consider what advantages and disadvantages systems have, and what are our priorities. We can then try to pick a reasonable system that will reflect the priorities.

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Arrow’s Theorem

Kenneth Arrow, who won the Nobel prize in Economics in 1972, proved in 1951 that all voting systems have flaws. More precisely, he proved that there does not and cannot exist a voting system which satisfies both the Condorcet Winner condition and the Independence

• f Irrelevant Alternatives.

Arrow’s theorem says, therefore, that no matter what voting system we use, there will be flaws with it. To determine what voting system to use in a given situation, we should then consider what advantages and disadvantages systems have, and what are our priorities. We can then try to pick a reasonable system that will reflect the priorities. In practice, this is a complicated political problem.

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Homework #6, due next Friday

Assignment 6 is on the course website. It is due by next Friday.

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Homework #6, due next Friday

Assignment 6 is on the course website. It is due by next Friday. Next week we will discuss Apportionment. More specifically, we will discuss how seats in the U.S. House of Representatives have been and are assigned, or apportioned, among the states.

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Quiz Question

True or false: Every voting system has some flaw. A True B False

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