Math 115
Quiz 1
Feb 4, 2013 Name: Key
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9 People were asked to rank their preferences among 4 candidates. All questions refer to this preference schedule:
1. Show that b would win every comparison in the method of head-to-head comparisons.
In head to head comparison,
2
voters prefer a to b, but 7 prefer b to a;
3 voters prefer c to b, but 6
prefer b to c;
4 voters prefer d to b, but 5
prefer b to d;
(Since this is all that was asked for, there is no need to do the other comparisons)
2. Rank the candidates using the recursive Plurality method
D has the most 1st place votes.
Without d, b would have 4 1st place votes and that would win, so b gets 2nd place.
Without d or b, it would be 5 votes for c to 4 for a, so c gets 3rd place and a gets 4th. So the ranking is
d-b-c-a
3. Rank the candidates using the extended Borda count method.
With 9 voters and 4 candidates, there are a total of 9*(4)(5)/2 = 90 points.
A gets 2*4+5*2 + 2*1 = 20
B gets 9*3 = 27
C gets 3*4+4*2+2*1 = 22
D gets 4*4+5*1 = 21.
A quick check shows this adds to 90, which helps confirm that no error was made.
So the ranking is b-c-d-a
4. a. For this preference schedule, does the Borda count violate the Majority criterion? Explain
No candidate received a majority of the 1st place votes, so no matter what the outcome it couldn’t violate the majority criterion. So NO it didn’t.
b. For this preference schedule, does the plurality method violate the Condorcet criterion? Explain
As seen in #1, there is a Condorcet winner – B won all of its head-to-head comparisons. The plurality method did not select B as the winner – it had d – so yes, this did violate the Condorcet criterion.