1. Lecture 1: Voting and Election Math
Math for Liberal Arts
Lecture 1: Voting and Election Math

2. Voting and Fairness The very essence of democracy?
Majority Rule. In an election with two candidates, the candidate with more than half the votes wins.
We call more than half the votes a majority.
But what if there are more than two choices?
What is a fair way to decide the outcome of an election?

3. MAC Election
The 37 members of the Math Anxiety Club (MAC) cast ballots for President.
There are four candidates:
Alisha (A)
Boris (B)
Carmen (C)
Dave (D)

4. Election Results Number of voters Favorite Candidate 14 11 8 4 A C D B
Who has won?

5. Capture all the voters’ preferences with a preference ballot.

6. Plurality Method
In the plurality method, the choice with the most first-place votes wins.

7. Winner: ______________
Number of voters 14 10 8 4 1
1st choice A C D B
2nd choice
3rd choice
4th choice
Candidates A B C D
1st place votes
Winner: ______________

8. Winner: ______________
Number of voters 14 10 8 4 1
1st choice A C D B
2nd choice
3rd choice
4th choice
Candidates A B C D
1st place votes
 14 4 
 11  8
Winner: ______________

9. Math Lovers Club Election
The 11 members of the Math Lovers Club choose a president from among four candidates by preference ballot.

10. Number of voters 6 2 3
1st choice A B C
2nd choice D
3rd choice
4th choice
Candidate A has 6 first place votes, a majority of the 11 voters.
A majority (more than half the votes) is automatically a plurality (the most votes).

11. Majority Criterion
If there is an alternative that is the first place choice of a majority of voters, then that alternative should be the winner of the election.
The plurality method satisfies the majority criterion.
However, there are some problems with the plurality method.

12. Problems with Plurality Method
The Bowl Games
The 100 members of the University Band must decide in which of five bowl games to march in:
Rose Bowl (R)
Hula Bowl (H)
Cotton Bowl (C)
Orange Bowl (O)
Sugar Bowl (S)

13. Number of voters 49 48 3
1st choice R H C
2nd choice S
3rd choice O
4th choice
5th choice
Most of the band members list the Rose Bowl last, yet it wins by the plurality method.

14. One-to-One Comparison
Number of voters 49 48 3
1st choice R H C
2nd choice S
3rd choice O
4th choice
5th choice
Comparison
Result
H versus R
H versus C
H versus O
H versus S

15. One-to-One Comparison
Number of voters 49 48 3
1st choice R H C
2nd choice S
3rd choice O
4th choice
5th choice
Comparison
Result
H versus R
 51 – 49 H
H versus C
H versus O
H versus S

16. One-to-One Comparison
Number of voters 49 48 3
1st choice R H C
2nd choice S
3rd choice O
4th choice
5th choice
Comparison
Result
H versus R
 51 – 49 H
H versus C
 97 – 3 H
H versus O
H versus S

17. One-to-One Comparison
Number of voters 49 48 3
1st choice R H C
2nd choice S
3rd choice O
4th choice
5th choice
Comparison
Result
H versus R
 51 – 49 H
H versus C
 97 – 3 H
H versus O
 100 – 0 H
H versus S

18. One-to-One Comparison
Number of voters 49 48 3
1st choice R H
2nd choice S C
3rd choice O
4th choice
5th choice
Comparison
Result
H versus R
 51 – 49 H
H versus C
 97 – 3 H
H versus O
 100 – 0 H
H versus S

19. Note that the Hula Bowl wins all the one-to-one comparisons with other alternatives.

20. Insincere Voting (Strategic Voting)
Actual Preferences Strategic Votes
Number of voters 49 48 3
1st choice R H C
2nd choice S
3rd choice O
4th choice
5th choice
Number of voters 49 48 3
1st choice R H
2nd choice S C
3rd choice O
4th choice
5th choice

21. Condorcet Candidate
An alternative that wins on a one-to-one comparison with every other alternative is called the Condorcet candidate.
In the Bowl Election, the Hula Bowl is the Condorcet candidate.
An election can have at most one Condorcet candidate.
An election may not have any Condorcet candidate.

22. Condorcet Criterion
If there is an alternative that wins in a one-to-one comparison between it and every other alternative, then that alternative should be the winner of the election.
The plurality method violates the Condorcet criterion.

23. Example with No Condorcet Candidate
The mathematics faculty at Southside University hold an election to select a chair of the department from among three candidates (A, B, and C). The 9 preference ballots are tabulated below.
Number of voters 4 3 2
1st choice A B C
2nd choice
3rd choice

24. Number of voters 4 3 2
1st choice A B C
2nd choice
3rd choice
Comparison
Result
A versus B
6 – 3 A
A versus C
4 – 5 C
B versus C
7 – 2 B
Since no candidate won all of her one-to-one comparisons, there is no Condorcet candidate.

25. Paradox of Voting
Even though individual preferences are transitive, group preferences are not always transitive.
As a group, the SU Math faculty prefer A to B and prefer B to C, but they do not prefer A to C! (Which they would if transitivity held.)
This violation of transitivity for group preferences is often called the Paradox of Voting.

26. Borda Count Method Borda point assignment for n alternatives
In the Borda Count Method each place on the ballot is assigned points. The alternative receiving the most points wins.
Last place
1 point
Next-to-last place
2 points
Third place
n-2 points
Second place
n-1 points
First place
n points
Borda point assignment for n alternatives

27. MAC Election Borda Count
Nbr of voters 14 10 8 4 1
1st choice:___ A C D B
2nd choice:___
3rd choice:___
4th choice:___

28. 2015 MLB MVP Voting Based on 14-9-8-7-6-5-4-3-2-1 scoring system
Player
Team
1st
2nd
3rd
4th
5th
6th
7th
8th
9th
10th
*Points
Bryce Harper
Nationals 30
420
Paul Goldschmidt
Diamondbacks 18 3 1 2
234
Joey Votto
Reds 6 9 4
175
Anthony Rizzo
Cubs 7
162
Andrew McCutchen
Pirates 5
139
Jake Arrieta
134
Zack Greinke
Dodgers
130
Nolan Arenado
Rockies
102
Buster Posey
Giants 84
Clayton Kershaw 49
Based on scoring system

29. Problems with Borda Count
Math Lovers Club Election
Nbr of voters 6 2 3
1st choice: 4 A B C
2nd choice: 3 D
3rd choice: 2
4th choice: 1

30. Math Lovers Club Election
Borda Counts
A = 6(4) + 0(3) + 0(2) + 5(1) = 29
B = 2(4) + 6(3) + 3(2) + 0(1) = 32
C = 3(4) + 2(3) + 6(2) + 0(1) = 30
D = 0(4) + 3(3) + 2(2) + 6(1) = 19
Why might A have a problem with this outcome?

31. Observations Disadvantages of the Borda Count Method
Borda count method violates the majority criterion.
Consequently, Borda count method violates the Condorcet criterion.
Advantages of the Borda Count Method
Borda Count Method uses all the available voter preferences, not just first choices.
Borda Count Method often produces the best compromise winner.

32. Summary
So far, we have two voting methods applied to the MAC election and two different winners!
Voting Method
Winner
Plurality
Alisha
Borda Count
Boris

33. Plurality-with-Elimination Method
Round 1
Count 1st place votes for each candidate.
If a candidate has a majority, then that candidate is the winner.
Otherwise, eliminate candidate(s) with fewest 1st place votes and simplify preference schedule.

34. Plurality-with-Elimination Method
Round 2
Count 1st place votes for each candidate.
If a candidate has a majority, then that candidate is the winner.
Otherwise, eliminate candidate(s) with fewest 1st place votes and simplify preference schedule.

35. Plurality-with-Elimination Method
Rounds 3, 4, etc.: Repeat above steps; eventually, some candidate will have a majority of 1st place votes.

36. MAC Election: Round One
A = 14 B = 4 C = 11 D = 8

37. Round 2: B dropped C: D: Nbr of voters 14 10 8 4 1 1st choice:___ A:
2nd choice:___
3rd choice:___

38. Round 2: Simplified more
Nbr of voters 14 11 12
1st choice:___ A: C: D:
2nd choice:___
3rd choice:___
A = 14 C = 11 D = 12

39. Round 3: C dropped D: Nbr of voters 14 11 12 1st choice:___ A:
2nd choice:___

40. Round 3: Simplified more
Nbr of voters 14 23
1st choice:___ A: D:
2nd choice:___
A = 14 D = 23
D Wins

41. Example: Young Mathematicians Election
The UAB Young Mathematicians elect a president of their club from among five candidates A, B, C, D, and E using the plurality-with-elimination method. There are 24 preference ballots.
Number of voters 8 6 2 3 5
1st choice A B C D E
2nd choice
3rd choice
4th choice
5th choice

42. Number of voters 8 6 2 3 5
1st choice A B C D E
2nd choice
3rd choice
4th choice
5th choice
What is the maximum number of rounds that might be needed to decide the winner of this election by the plurality-with-elimination method?

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

44. A = 10 B = 6 D = 3 E = 5 Number of voters 8 6 2 3 5 1st choice A B D E
2nd choice
3rd choice
4th choice
A = 10 B = 6 D = 3 E = 5

45. A = 10 B = 6 E = 8 Number of voters 8 6 2 3 5 1st choice A B E
2nd choice
3rd choice
A = 10 B = 6 E = 8

46. A = 10 E = 14 E WINS Number of voters 8 6 2 3 5 1st choice A E
2nd choice
A = 10 E = 14
E WINS

47. And the Best Actor Oscar goes to…
To determine the magic number for nomination, take the total number of ballots received for a particular category and divide it by the total possible nominees plus one.
Example: 600 potential ballots for the Best Actor category, divide that by 6 (5 possible nominees plus 1), thus making the magic number for the category 100 ballots to become an official nominee.

48. The starts based on a voter’s first choice selection until someone reaches the magic number.
The ballots that named him as a first choice are then all set aside, and there are now four spots left for the Best Actor category.
The actor with the fewest first-place votes is automatically knocked out, and those ballots are redistributed based on the voters' second place choices
The counting continues, and actors or different categories rack up redistributed votes until all five spots are filled.

49. Problems with Plurality-with-Elimination
Results of the Young Mathematicians Student of the Year Award Election
Number of voters 7 8 10 4
1st choice A B C
2nd choice
3rd choice
A = 11 B = 8 C = 10

50. Problems with Plurality-with-Elimination
Number of voters 7 8 10 4
1st choice A C
2nd choice
A = 11 C = 18

51. Because of election irregularities, the original election is declared void. Meanwhile, candidate C convinces the 4 voters represented by the last column of the preference schedule that she is better than candidate A. They switch their preference order to C, A, B. The reelection results are as follows.

52. Problems with Plurality-with-Elimination
Results of the Young Mathematicians Student of the Year Award Election
Number of voters 7 8 10 4
1st choice A B C
2nd choice
3rd choice
A = 7 B = 8 C = 14

53. Problems with Plurality-with-Elimination
Number of voters 7 8 10 4
1st choice B C
2nd choice
B = 15 C = 14
C added first place votes, but lost the election

54. Monotonicity Criterion
If an alternative X is the winner of an election, and, in a reelection, all the voters who change their preferences do so in a way that is favorable only to X, then X should still be the winner of the election.
The plurality-with-elimination method violates the monotonicity criterion.
The plurality-with-elimination method also violates the Condorcet criterion.

55. Fairness Criterion Voting Method Satisfied Violated Plurality
Majority Monotonicity
Condorcet
Borda Count
Monotonicity
Majority Condorcet
Plurality-with-
Elimination
Majority
Monotonicity Condorcet

56. Plurality-with- Elimination
Voting Method
Winner
Plurality
Alisha
Borda Count
Boris
Plurality-with- Elimination
Dave
In the MAC election, we have used three voting methods and have three different winners!

57. Method of Pairwise Comparisons
Match each candidate on a one-to-one basis with every other candidate.
Points Suppose X is compared with Y.
If X wins the comparison, then X gets 1 point and Y gets 0.
If Y wins the comparison, then Y gets 1 point and X gets 0.
In case of a tie, both X and Y get ½ point.
Winner. After all pairwise comparisons have been made, the candidate with the most points is the winner.

58. Pairwise Comparison
This is similar to a round-robin tournament, where each player plays every other player to decide the winner.
The method of pairwise comparisons is one of several so-called Condorcet methods. It is also called Copeland’s Method.

59. MAC Election Number of voters 14 10 8 4 1 1st choice A C D B
2nd choice
3rd choice
4th choice

60. MAC Election
There are four candidates in the MAC election (A, B, C, D)
There are 6 pairwise comparisons that must be made:
A vs B B vs C C vs D
A vs C B vs D
A vs D

61. Nbr of voters 14 10 8 4 1 1st A C D B 2nd 3rd 4th Comparison Result
Points
A vs B
A vs C
A vs D
B vs C
B vs D
C vs D

62. Nbr of voters 14 10 8 4 1 1st A C D B 2nd 3rd 4th
Comparison
Result
Points
A vs B
 14 – 23
 B = 1
A vs C
 C = 1
A vs D
 D = 1
B vs C
 18 – 19
B vs D
 28 – 9
C vs D
 25 – 12
A = 0 B = 2 C = 3 D = 1
C Wins

64. Best won-lost-tied percentage in games played within the division.
TO BREAK A TIE WITHIN A DIVISION
If two or more clubs in the same division finish with identical won-lost-tied percentages, the following steps will be taken until a champion is determined.
Head-to-head (best won-lost-tied percentage in games between the clubs).
Best won-lost-tied percentage in games played within the division.
Best won-lost-tied percentage in common games.
Best won-lost-tied percentage in games played within the conference.
Strength of victory.
Strength of schedule.
Best combined ranking among conference teams in points scored and points allowed.
Best combined ranking among all teams in points scored and points allowed.
Best net points in common games.
Best net points in all games.
Best net touchdowns in all games.
Coin toss

65. Summary of MAC Election Results
Voting Method
Winner
Plurality
Alisha
Borda Count
Boris
Plurality-with- Elimination
Dave
Pairwise Comparisons
Carmen
In the MAC election, we have used four voting methods and have four different results!

66. Problems with Pairwise Comparisons
Number of voters 1 4 3
1st choice A C E
2nd choice B D
3rd choice
4th choice
5th choice

67. A Wins Comparison Result Points A vs B 5 – 4 A = 1 A vs C A vs D
 5 – 4
 A = 1
A vs C
A vs D
A vs E
 1 – 8
 E = 1
B vs C
 7 – 2
 B = 1
A = 3
B = 2
C = 2
D = 2
E = 1
B vs D
 4 – 5
 B = 1
B vs E
 5 – 4
B = 1 
C vs D
C = 1 
C vs E
D vs E
 D = 1
A Wins

68. Because of an election irregularity, the votes have to be recounted.
Meanwhile, B, C, and D drop out of the election. The new preference schedule is much simpler.
Nbr of voters 1 8
1st A E
2nd
E Wins

69. Independence of Irrelevant Alternatives Criterion
If an alternative X is the winner of an election, and one or more of the other alternatives are removed and the ballots recounted, then X should still be the winner of the election.
The method of pairwise comparisons violates the independence of irrelevant alternatives criterion.
The method of pairwise comparisons satisfies the majority, Condorcet, and monotonicity criteria.

70. Conflict with Borda Count
An even more serious problem with pairwise comparisons is revealed by the preceding example. Let us apply the Borda count method to it.
Numberof voters  1  4  3
1st choice: A C E
2nd choice: 4 B D
3rd choice: 3
4th choice: 2
5th choice: 1
A = 1(5)+4(4)+4(1) = 25
B = 1(4)+7(3)+1(2) = 27
C = 4(5)+1(3)+4(1) = 27
D = 4(4)+1(3)+4(2)= 27
E = 4(5)+4(2)+1(1)= 29

71. Observations
A comes in last by Borda count, exactly the opposite order as with pairwise comparisons
Pairwise comparisons: A, B-C-D tied, E.
The Borda count winner E is preferred to the pairwise comparisons winner A by 8 of the 9 voters.
This overwhelming margin of 8 to 1 is picked up by Borda count, but is missed by the method of pairwise comparisons.
We call A a disliked winner since 8 of the 9 voters prefer E

72. More Problems with Pairwise Comparisons Example: Deadlock
Number of voters 3 2 4
1st choice A B C
2nd choice
3rd choice
Comparison
Result
Points
A vs B
 7 – 2
A = 1
A vs C
 3 – 6
C = 1
B vs C
 5 – 4
B = 1

73. Observations Advantages of the method of pairwise comparisons
Pairwise comparisons satisfies
The Condorcet criterion
The majority criterion
The monotonicity criterion
Pairwise comparisons uses all the information provided by the voters.

74. Observations Disadvantages of the method of pairwise comparisons
Violates the independence of irrelevant alternatives criterion.
Often is indecisive, producing a deadlock.
Can conflict spectacularly with Borda count, producing a disliked winner.
The number of comparisons mounts fast as the number of candidates increases.

75. Fairness Criterion Voting Method Satisfied Violated Plurality
Majority Monotonicity
Condorcet
Independence Irrelevant Alternatives
Borda Count
Monotonicity
Majority
Plurality-with-
Elimination
Pairwise
Comparisons

76. Assumptions Voting is between two alternatives: “yes” or “no”
The motion passes if the sum of the “yes” votes exceeds the quota.
The motion fails if the sum of the “yes” votes does not exceed the quota.
A motion cannot be deadlocked. It either passes or fails.
Everyone must vote (no abstentions).
There are 3 or more players.