In Minnesota in the 1998 governatorial race, Reform Party candidate Jesse "The Body" Ventura (former professional wrestler and radio shock-jock) claimed a stunning victory over Minnesota Attorney General Skip Humphrey (Democrat) and St. Paul Mayor Norm Coleman (Republican). Ventura won under plurality by receiving 37% of the votes. Post-election surveys indicate (that when viewed as percentages) the voters perferences were similar to the following table.
Number of Voters
Rank 35 28 20 17 1 N S J J 2 S N N S 3 J J S N
Problem Set 5 :What would be the outcome of this election under majority rules? Under plurality? Under Borda count? Which candidate is ranked first by the largest number of voters? Which candidate is ranked last by the largest number of voters? In a head-to-head race between Skip and Norm, who would win? What about between Skip and Jesse? What about Norm and Jesse? Does anything about your answers strike you as being odd?End of Problem Set 5
The properties of winning or losing to other candidates in head-to-head races are usually attributed to Marie Jean Antoine Nicolas de Caritat, the Marquis de Condorcet, (or just Condorcet for short) who was a french mathematician and contemporary of Borda. Consider the following terms
- A candidate in an election who would defeat every other candidate in a head-to-head race is said to be a Condorcet winner.
- A candidate in an election who would lose to every other candidate in a head-to-head race is said to be a Condorcet loser.
- A voting system that will always elect a Condorcet winner, when it exist, is said to satisfy the Condorcet winner criterion (CWC).
- A voting system that will never elect a Condorcet loser, when it exist, is said to satisfy the Condorcet loser criterion (CLC).
We've already seen in Problem Set 5 that plurality does not satisfy CLC or CWC. Instead of using plurality why don't we just always find the Condorcet winner and always elect them? Well, consider the following example (known as Condorcet's paradox in a race between candidates A,B,C with three voters.
Number of Voters
Rank 1 1 1 1 A B C 2 B C A 3 C A B
Notice that in a head-to-head race between A and B, that A wins. In a head-to-head race between B and C, B wins. Also, in a head-to-head race between C and A, C wins. This is why we just qualify the definitions of CLC and CWC, as shown by this example the Condorcet winner and Condorcet loser, need not exist.
We've seen now that plurality violates the CWC and by changing the number of voter's in the previous example to 3,2,2 you can see that the Borda count violates CWC (verify on your own). So let's try to find a new voting system that will satisfies the CWC.
One such voting system is Sequential Pairwise Voting where the sociatal preference order is found as follows.
- Fix an ordering (also called an agendaof the candidates (choosen however you please, ex A,D,B,C,F,E)
- Have the first two compete in a head-to-head (majority rules) race, the winner of this race will then face the 3rd candidate on the list in a head-to-head race, the winner of that race will face the next candidate... continue until the the last candidate in the ordering is in a head-to-head race with the winner of the previous head-to-head and the winner of that race is declared the winner of the general election.
- The societal preference order then starts with the winner (say C) with everyone else tied, i.e. C>A=B=D=E=F.
This brings us to...
Problem Set 6 :Given any agenda, could a Condorcet winner ever lose in the sequential pairwise voting system. So, have we found a voting system that satisfies CWC? Does sequential pairwise voting satisfy CLC?
Recall the race for the Graduate Student Representative from the previous section.
Number of Voters
Rank 12 7 5 3 1 P S J L 2 S J L J 3 J L P S 4 L P S P
>From previous activities we learnt that using the plurality method produced the sociatal preference order P>S>J>L, and Borda count produced S>J>P>L. Now, who would win this election under sequential pairwise voting with agenda P,S,J,L?
Find an agenda under which Peter would win. Do the same for Sergio and for Liz.
Suppose that all the voters switched J and L on there preference orders. Use the original and new preference schedule, with agenda S,L,P,J, to show that sequential pairwise voting is not neutral.End of Problem Set 6
References and Further Reading:
[1]Jonathan K. Hodge and Richard E. Klima.The Mathematics of Voting and Elections: A Hands-On Approach. American Mathematical Society, Providence, R.I., 2005.