Chapter 1 Practice Problems and Possible Solutions

Part I Multiple Choice

Use the preferece schedule below to answer Questions 1-4.

Number of voters	8	6	2	3	5
1st choice	A	B	C	D	E
2nd choice	B	D	A	E	A
3rd choice	C	E	E	A	D
4th choice	D	C	B	C	B
5th choice	E	A	D	B	C

1. Using the Borda Count method the winner of this election is:
   1. A
   2. B
   3. E
   4. D
   5. None of the above

   Solution

   The Borda count is 83, 79, 72, 69, and 57 for A, B, D, E, and C in that order. Thus the winner is A and the answer choice 1

2. Using the Plurality method the winner of the election is:
   1. A
   2. B
   3. C
   4. E
   5. None of the above

   Solution

   Using pularlity A gets 8 first-place votes as opposed to 6, 5, 3, 2 for B, E, D, and C respectively. This mean A also wins this method. The choice is then 1

3. Usingthe Pairwise Comparisons method the winner of the election is:
   1. A
   2. B
   3. a tie between B and A
   4. D
   5. None of the above

      Solution

      Here we have 10 comparisions to check and A wins against all but that against E; B wins all head-to-heads except that against A, C looses to all; and D looses to A and B, but beats C and E. The final tally then is A: 3 wins, B: 3 wins, C: 0 wins, D: 2 wins and E: 2 wins. This gives a tie between A and B for a winner. Therefore, the choice is 3

4. Using the Plurality-with-elimination method the winner of the election is:
   1. A
   2. E
   3. B
   4. D
   5. None of the above

   Solution

   Round 1, C gets eliminated. Round 2, D gets eliminated. Round 3, B gets eliminated. Round 4, A gets eliminated and E wins the election. Thus the choice is 2.

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5. What is the total number of pairwise comparisons possible in an election among 20 candidates:
   1. 20
   2. 210
   3. 150
   4. 190
   5. None of the above

   Solution

   The answer is found by taking the calculation 20(19)/2 = 190. Thus the correct choice is 4.

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   Questions 6-11 refer to the following preference schedule. The Mathematics For All Club is having an election for a president. The candidates are Amber, Bill and Jeniere. Each of the members is asked to submit a preference ballot. Here is the result:

   Number of Voters	5	4	4	2
   First Choice	Amber	Bill	Jeniere	Bill
   Second Choice	Jeniere	Jeniere	Amber	Amber
   Third Choice	Bill	Amber	Bill	Jeniere

6. How many members of the Mathematics for All Club submitted their ballots?
   1. 10
   2. 4
   3. 15
   4. 20
   5. None of the above

   Solution

   Add the numbers in the first row to get 15. The correct answer is 3

7. Who is the winner under the plurality method?
   1. Jeniere
   2. Amber
   3. Bill
   4. No winner
   5. None of the above

   Solution

   Bill runs away with the win in the plurality method.

8. Who is the winner under the plurality-with-elimination method?
   1. Amber
   2. Bill
   3. a tie between Bill and Jeniere
   4. Jeniere
   5. None of the above

   Solution

   Jeniere gets eliminated in the first round. Leaving Bill and Amber, and Amber wins this.

9. Using the Borda Count method the winner of this election is:
   1. Bill
   2. Amber
   3. a tie between Jeniere and Amber
   4. Jeniere
   5. None of the above

   Solution

   The Borda Count is
   • Amber : 5(3) + 6(2) + 4(1) = 31
   • Bill: 6(3) + 0(2) + 9(1) = 27
   • Jeniere: 4(3) + 9(2) + 2(1) = 32

   This method is won by Jeniere.

10. Do we have a Condorcet winner under the pairwise comparison method?
    1. Yes
    2. No
    3. It is impossible to know
    4. None of the above

    Solution

    Yes there is a Condorcet winner and it is Jeniere. She wins both head-to-head matches with Bill and Amber. Thus the answer is Yes.

11. Which candidate comes in second place under the extended plurality-with-elimination method of ranking?
    1. Amber
    2. Bill
    3. a tie between Bill and Amber
    4. Jenier
    5. None of the above

    Solution

    Bill came in second under this method. This is because Jeniere got eliminated in the 1st round and Amber won the election.

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    Part II Show all your work

    Given the following preference schedule:

    Number of voters	5	3	8	7	3
    1st choice	C	D	E	B	A
    2nd choice	B	A	B	A	D
    3rd choice	D	C	A	C	C
    4th choice	E	B	C	D	E
    5th choice	A	E	D	E	B

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12. Find the winner of the election using the Borda count method which assigns 5, 4, 3, 2, 1 point(s) for a first, second, third, fourth, and fifth choice in that order. If there is a tie indicate so.

    Solution

    For A :3 (5) + 10 (4) + 8 (3) + 5 (1) = 15 + 40 + 24 + 5 = 84 
    
    For B 7 (5) + 13 (4) + 3 (2) + 3 (1) = 35 + 52 + 6 + 3 = 96
    
    For C : 5 (5) + 13 (3) + 8 (2) = 25 + 39 + 16 = 80 
    
    For D: 3 (5) + 3(4) + 5 (3) + 7 (2) + 8 (1) = 15 + 12 + 15 + 14 + 8 = 64
    
    For  E	8 (5) + 8 (2) + 10 (1) = 40 + 16 + 10 = 66
    

    Thus by Borda count method the ranking is B, A, C, E, D. Therefore B is the winner.

13. Find the winner of the election using the pairwise comparisons method, if there is a tie indicate so.

    There are 10 comparisons we need to check

    A vs B	A vs C		A vs D		A vs E		B vs C
    B wins	A wins		A wins		A wins		B wins
    
    B vs D	B vs E		C vs D		C vs E		D vs E
    B wins	B wins		C wins		C wins		D wins
    

    If we assign 1 point for winning, 1/2 point for a tie, and 0 for loosing, we see that A has 3 points; B has four points, C has 2 points, D has 1 point and E has 0 points. Thus, the winner is B.

14. Find the winner of the election using the plurality-with-elimination method, if there is a tie indicate so.

    Round 1

    Since D and A has the lowest first place votes, 3 each, one of them has to be eliminated in the first round. Since D lost to A in a pairwise competition, D will be the one to be eliminated. (Had we opted for the elimination of A in the first round instead of D, the final outcome will not change; check it for yourselves.)

    Number of voters	5	3	8	7	3
    1st choice	C	A	E	B	A
    2nd choice	B	C	B	A	C
    3rd choice	E	B	A	C	E
    4th choice	A	E	C	E	B

    Round 2

    C has the lowest first place votes and is the one to be eliminated.

    Number of voters	5	3	8	7	3
    1st choice	B	A	E	B	A
    2nd choice	E	B	B	A	E
    3rd choice	A	E	A	E	B

    Round 3

    A has the lowest first place votes and is thus the one to get eliminated in this round.

    Number of voters	5	3	8	7	3
    1st choice	B	B	E	B	E
    2nd choice	E	E	B	E	B

    Since B has the majority of the first place votes 15, B wins this election as well.

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    Part III Discussion Problems:

15. Address the problems/inconsistencies, positive aspects and a typical situation when the Borda Count method may be used.

    Solution

    • Problems
      • violates majority criterion
      • violates Condorcet criterion

    • Positive aspects
      • uses all the information of a voter

    • Typical use
      • Ranking of teams
      • Heisman trophy winner
      • ranking students using GPA