Chapter 1 Voting
Page 1 of 17
Notes for Chapter 1
Voting
Spring 2017 (11/9/16)
Learning Objectives: Demonstrate an understanding of
preference schedules, determine the outcome of
an election using any of (5) voting methods and
demonstrate an understanding of (4) fairness
criteria.
Chapter 1 Voting
Page 2 of 17
4 Fairness Criteria
1.) Majority Fairness Criterion - Candidate with > 1/2 1st place votes “should” win
2.) Condorcet Fairness Criterion - if a candidate is compared one-on-one with every other
candidate, then the candidate who wins the most comparisons “should” be the winner.
3.) Monotonicity Fairness Criterion - Say candidate X wins an election. If a re-election
is required, and all vote changes favor X, then the winner “should” still be X.
4.) Independence of Irrelevant Alternatives Criterion - Say candidate X wins an election.
If one of the losing candidates is removed, and the votes are recounted, X “should” still be the
winner.
Preference Ballot - lists candidates in order of voter preference:
Example: election with 37 preference ballots
Ballot
A
B
C
D
Ballot
C
B
D
A
Ballot
D
C
B
A
Ballot
B
D
C
A
14
10
8
4
Ballot
1st place A
2nd place B
3rd place C
4th place D
Ballot
C
D
B
A
1 =
37 voters
Preference Schedule
No. Voters
1st place
2nd place
3rd place
4th place
14
A
B
C
D
10
C
B
D
A
8
D
C
B
A
4
B
D
C
A
1
C
D
B
A
= 37 voters
1.) Majority is > 1/2 1st place votes
Look only at 1st place votes
> ½ (14+10+8+4+1)
A: 14
1st place votes
> ½ (37)
B: 4
1st place votes
> 18.5
C: 10 + 1 = 11 1st place votes
= 19 or more
D: 8
1st place votes
Note: no candidate has a majority (19 or more)
no winner!
2.) Plurality Method - Most 1st place votes wins
A: 14
A is WINNER
B: 4
C: 10 + 1 = 11
Ranking using Plurality Method
D: 8
A: 14 1st place
C: 11 2nd place
D: 8 3rd place
B: 4 4th place
Chapter 1 Voting
Page 3 of 17
SOLUTIONS
Practice problems page 28
2.) Given: 17 Preference Ballots
Ballot
C
A
D
B
Ballot
B
C
D
A
Ballot
A
D
B
C
Ballot
C
A
D
B
Ballot
C
A
D
B
Ballot
C
A
D
B
Ballot
A
C
D
B
Ballot
A
D
B
C
Draw: preference schedule
Ballot
B
C
D
A
Ballot
C
A
D
B
Ballot
A
D
B
C
Ballot
A
C
D
B
Ballot
B
C
D
A
Ballot
B
C
D
A
Ballot
A
D
B
C
Ballot
C
A
D
B
Ballot
B
C
D
A
2 ballots 6 ballots 5 ballots 4 ballots
A
C
B
A
C
A
C
D
D
D
D
B
B
B
A
C
-----------------------------------------------------------------------------------------------------Voters
1st
2nd
3rd
4th
5th
5
A
B
C
D
E
5
C
E
D
A
B
3
A
D
B
C
E
3
B
E
A
C
D
3
D
C
B
E
A
2
D
C
B
A
E
7a.) How many voters? 5+5+3+3+3+2= 21
7b.) What is majority? >1/2(5+5+3+3+3+2)
>1/2(21)
>10.5
= 11 or more
7c.) Which candidate had fewest last place votes?
A=3
B=5
C=0
C had fewest
D=3
There is no winner (Majority needs 11 or more)
E = 10
Chapter 1 Voting
Page 4 of 17
Voters
1st
2nd
3rd
4th
5th
5
A
B
C
D
E
5
C
E
D
A
B
3
A
D
B
C
E
3
B
E
A
C
D
3
D
C
B
E
A
2
D
C
B
A
E
17a.) Use Plurality to find winner. In case of tie, player with
fewer last place votes is winner of tie.
Plurality Method - Most 1st place votes wins
A = 5+3 = 8
B=3
C=5
D = 3+2 = 5
E=0
1st place votes
1st place votes
1st place votes
1st place votes
1st place votes
A wins
17b.) Ranking using Plurality method: 1st place
tie 2nd and 3rd
tie 2nd and 3rd
4th place
5th place
= A with 8 1st place votes
= C with 5 each
= D with 5 each
= B with 3
= E with 0
Use tie breaker to break 2nd and 3rd place tie
C has 0 last place votes, wins tie
D has 3 last place votes
C = 2nd place
D = 3rd place
Ranking: 1st place
2nd place
3rd place
4th place
5th place
=A
=C
=D
=B
=E
Chapter 1 Voting
Page 5 of 17
3.) Borda Count Method – add a number of points to each of the 1st, 2nd, 3rd, etc., place
positions
No. Voters
1st place
2nd place
3rd place
4th place
14
A
B
C
D
10
C
B
D
A
8
D
C
B
A
4
B
D
C
A
1
C
D
B
A
Procedure: - assign descending order points to the preference table
No. Voters
1st place
2nd place
3rd place
4th place
points
4
3
2
1
14
A
B
C
D
10
C
B
D
A
8
D
C
B
A
4
B
D
C
A
1
C
D
B
A
- then calculate Borda Count points for each candidate:
A: 14X4 + 10X1 + 8X1 + 4X1 + 1X1 = 79 pts.
B: 14X3 + 10X3 + 8X2 + 4X4 + 1X2 = 106 pts.
B is WINNER
C: 14X2 + 10X4 + 8X3 + 4X2 + 1X4 = 104 pts.
D: 14X1 + 10X2 + 8X4 + 4X3 + 1X3 = 81 pts.
Ranking using Borda Count Method:
B: 1st place
C: 2nd place
D: 3rd place
A: 4th place
Chapter 1 Voting
Page 6 of 17
4.) Plurality with Elimination Method - Rounds of 1st place votes until a candidate has a
majority of 1st place votes.
No. Voters
1st place
2nd place
3rd place
4th place
14
A
B
C
D
10
C
B
D
A
8
D
C
B
A
4
B
D
C
A
1
C
D
B
A
Majority is > ½ first place votes is > 14 + 10 + 8 + 4 + 1 is > 37 > 18.5 = 19 or more
2
2
Round 1 A has 14 1st place votes
B has 4 1st place votes
C has 11 1st place votes
no candidate has 19; eliminate anybody with 0 and
D has 8 1st place votes
also person with fewest 1st place votes
Eliminate B, redraw preference table
Preference table with B removed:
1st place
2nd place
3rd place
Round 2
14
A
C
D
10
C
D
A
8
D
C
A
A has 14
C has 10 + 1 =11
D has 8 +4 = 12
4
D
C
A
1
C
D
A
1st place votes
1st place votes
1st place votes
no candidate has 19; eliminate anybody
with 0 and also person with fewest
1st place votes
Eliminate C, redraw preference table
1st place
2nd place
Round 3
14
A
D
10
D
A
8
D
A
4
D
A
1
D
A
A has 14 - 1st place votes
D has 10 + 8 + 4 + 1 = 23 - 1st place votes = WINNER
Ranking using Plurality with Elimination Method:
D: 1st place (winner)
A: 2nd place
C: 3rd place (2nd eliminated)
B: 4th place (1st eliminated)
Chapter 1 Voting
Page 7 of 17
Plurality with Elimination Method - Rounds of 1st place votes until a candidate has a
majority of 1st place votes.
No. Voters
1st place
2nd place
3rd place
4th place
14
A
B
C
D
10
C
B
D
A
8
A
C
B
D
4
B
D
C
A
1
C
D
B
A
Majority is > ½ first place votes is > 14 + 10 + 8 + 4 + 1 is > 37 > 18.5
2
2
Round 1
A has 14 + 8 = 22 1st place votes
B has 4 1st place votes
C has 10 + 1 = 11 1st place votes
D has none
only one round needed
= 19 or more
A has majority ( >19 ) wins
Chapter 1 Voting
Page 8 of 17
Practice Problems page 31
23.) Borda Count Method
insert points
Choice
1ST 4 pts
2ND 3 pts
3RD 2 pts
4TH 1 pts
a.)
b.)
6
C
D
A
B
5
A
D
C
B
4
B
D
C
A
2
B
A
C
D
2
C
B
A
D
2
C
B
D
A
2
C
D
B
A
C: 6X4 + 5X2 + 4X2 + 2X2 + 2X4 + 2X4 + 2X4
D: 6X3 + 5X3 + 4X3 + 2X1 + 2X1 + 2X2 + 2X3
A: 6X2 + 5X4 + 4X1 + 2X3 + 2X2 + 2X1 + 2X1
B: 6X1 + 5X1 + 4X4 + 2X4 + 2X3 + 2X3 + 2X2
= 70
= 59
= 50
= 51
C is
winner
Ranking using Borda Count Method:
C: 1st place (winner)
D: 2nd place
B: 3rd place
A: 4th place
34.) Plurality with Elimination method
Choice
1ST
2ND
3RD
4TH
6
A
C
D
B
6
B
C
A
D
5
B
C
D
A
4
D
A
C
B
3
A
C
D
B
3
B
A
C
D
a.) Majority is >1/2 of (6+6+5+4+3+3) is > 1/2 of 27 is > 13.5 = 14 or more
Round 1: A gets 6+3=9
B gets 6+5+3 = 14
C gets 0
D gets 4
1st place votes
B wins in 1st round,
no more rounds needed
Round 2: not needed
b.)
Ranking using Plurality with Elimination method:
B: 1st place (winner)
A: 2nd place
D: 3rd place
C: 4th place
Chapter 1 Voting
Page 9 of 17
37.) Plurality with Elimination method
1st place
2nd place
3rd place
4th place
5th place
8
B
E
A
C
D
7
C
E
D
A
B
5
A
B
C
D
E
4
D
C
B
E
A
3
A
D
E
C
B
2
D
B
C
A
E
Majority is >1/2 of (8+7+5+4+3+2) is > 1/2 of 29 is > 14.5 = 15 or more
Round 1: B gets 8
1st place votes
E gets 0
A gets 5+3=8
C gets 7
D gets 4+2=6
8
1 place B
2nd place A
3rd place C
4th place
5th place
st
7
C
A
B
5
A
B
C
4
C
B
A
3
A
C
B
no one gets 15
eliminate E and D
redraw
2
B
C
A
Round 2: B gets 8+2=10
A gets 5+3=8
C gets 7+4=11
no one gets 15
eliminate A
redraw
8 7 5 4 3 2
1 place B C B C C B
2nd place C B C B B C
3rd place
4th place
5th place
st
Round 3: B gets 8+ 5+2=15
C gets 7+ 4+3=14
b.)
B
wins
Ranking using Plurality with Elimination method:
B: 1st place (winner)
C: 2nd place
A: 3rd place (3rd eliminated)
D: 4th place (2nd eliminated)
E: 5th place (1st eliminated)
Chapter 1 Voting
Page 10 of 17
Pairwise Comparison Method is the Condorcet Criteria method….satisfies the Majority,
Condorcet, and Monotonicity fairness criteria.
Hint: no. comparisons to make: (N-1)(N) ; where N = no. candidates
2
A
B
C
D
AvsB
BvsC
CvsD
AvsC
(N-1)(N) = (4-1)(4) = 6 comparisons
2
2
BvsD
AvsD
In each one/one or pairwise comparison, give winner 1 point, loser 0 points. If tie, give
each 1/2 point
No. Voters
1st place
2nd place
3rd place
4th place
14
A
B
C
D
10
C
B
D
A
8
D
C
B
A
4
B
D
C
A
1
C
D
B
A
Comparison 1.) A vs B
14 10
8
4
1
14 23
B wins; give B 1 point
No. Voters
1st place
2nd place
3rd place
4th place
14
A
B
C
D
10
C
B
D
A
8
D
C
B
A
4
B
D
C
A
1
C
D
B
A
Comparison 2.) C vs D
14
8
10
4
1
25 12
C wins; give C 1 point
Chapter 1 Voting
Page 11 of 17
No. Voters
1st place
2nd place
3rd place
4th place
14
A
B
C
D
10
C
B
D
A
8
D
C
B
A
4
B
D
C
A
1
C
D
B
A
The other 4 comparisons:
Comp 3.) B vs C Comp 4.) B vs D
14 10
14 8
4
8
10 1
1
4
18 19
28 9
give C 1 point
Comp 5.) A vs C
14 10
8
4
1
14 23
give B 1 point
C vs A no, this is a repeat
C vs B no, this is a repeat
C vs D no, this is a repeat
D vs A no, this is a repeat
D vs B no, this is a repeat
D vs C no, this is a repeat
give C 1 point
All other comparisons
are repeats, don’t use
Summarize: B gets +1+1 = +2
C gets +1+1+1= +3 = winner
D gets +1
A gets none
Ranking by Pairwise comparison: C 1st place
B 2nd place
D 3rd place
A 4th place
Comp 6.) A vs D
14 10
8
4
1
14 23
give D 1 point
Chapter 1 Voting
Page 12 of 17
Another example: illustrates tied comparisons…give each ½ point.
1st place
2nd place
3rd place
4th place
5th place
15
A
B
C
D
E
3
B
A
C
D
E
6
D
E
C
A
B
12
E
B
A
D
C
Must make (5-1)(5) = (4)(5) = 10 valid
2
2
comparisons
1.) A vs B:
15 3
6 12
21 15
A gets 1
2.) A vs C:
15 6
3
12
30 6
A gets 1
3.) A vs D:
15 6
3
12
30 6
A gets 1
4.) A vs E:
15 6
3 12
18 18
A gets ½
E gets ½
5.) B vs C:
15 6
3 12
12
30
6
B gets 1
6.) B vs D:
15 6
3
12
30 6
B gets 1
7.) B vs E:
15 6
3 12
18 18
B gets ½
E gets ½
8.) C vs D:
15 6
3 12
18 18
C gets ½
D gets ½
9.) C vs E:
15 6
3 12
18 18
C gets ½
E gets ½
10.) D vs E:
15 12
3
6
24 12
D gets 1
All other comparisons between A, B, C, D, E will be duplicates
Summary: A gets 1+1 +1+1/2 = 3-1/2
B gets 1+1+1/2
= 2-1/2
C gets 1/2 + 1/2
=1
D gets 1 +1/2
= 1-1/2
E gets 1/2+1/2+1/2 = 1-1/2
Ranking: A WINS!
B 2nd
E 3rd
tied for
D 4th
3rd and 4th
C last
need a tie
breaker rule
to break this
Chapter 1 Voting
Page 13 of 17
Examples of Fairness Criteria violations
Borda violates Majority Criterion:
st
1 place
2nd place
3rd place
4th place
4
3
2
1
6
A
B
C
D
2
B
C
D
A
3
C
D
B
A
Majority > ½ (6+2+3) = 6 or more
A: 6 Winner
B: 2
C: 3
violation
Borda: A: 6X4 + 2X1 + 3X1 = 29
B: 6X3 + 2X4 + 3X2 = 32 Winner
C: 6X2 + 2X2 + 3X4 = 28
D: 6X1 + 2X2 + 3X3 = 19
Plurality violates Condorcet Criterion(= PWC):
1st place
2nd place
3rd place
4th place
5th place
49
R
H
F
O
S
48
H
S
O
F
R
3
F
H
S
O
R
Plurality
R: 49 Winner
H: 48
F: 3
violation
PWC: Must make (5-1)(5) = (4)(5) = 10 comparisons
2
2
R vs H = + 49 - 48 - 3 = -2
R vs F = + 49 - 48 - 3 = -2
R vs O = + 49 - 48 - 3 = -2
R vs S = + 49 - 48 - 3 = -2
H vs F = + 49 + 48 - 3 = +94
H vs O = + 49 + 48 + 3 = +100
H vs S = + 49 + 48 + 3 = +100
F vs O = + 49 - 48 + 3 = +4
F vs S = + 49 - 48 + 3 = +4
O vs S = + 49 - 48 - 3 = -2
H gets 1 point
F gets 1 point
O gets 1 point
S gets 1 point
H gets 1 point
H gets 1 point
H gets 1 point
F gets 1 point
F gets 1 point
S gets 1 point
H: 4 points Winner
F: 3 points
O: 1 point
S: 2 points
R: 0 points
Chapter 1 Voting
Page 14 of 17
Plurality with Elimination violates Monotonicity Criterion:
7
A
B
C
8 10
B C
C A
A B
2
A
C
B
Majority > ½ (27) = 14 or more
1st place
2nd place
3rd place
7 8 10
A C C
C A A
2
A
C
Round 2
1st place
2nd place
Round 1
A has 9
1st place votes
C has 10
1st place votes
B has 8
1st place votes
no candidate has 14; eliminate B
A has 9
C has 18…. C wins!
Wait a minute, election is not valid because poll inspectors were late arriving (or some other
procedural reason), must….DO OVER. While waiting for voting to commence again, the 2
voters in the last column decide to switch their votes to favor the winner (C) because they hoped
to be a “friend” of the winner C.
DO OVER:
1 place
2nd place
3rd place
7
A
B
C
1st place
2nd place
7 8
B B
C C
st
8 10
B C
C A
A B
10
A
B
2
C
A
B
A has 7 1st place votes
C has 12 1st place votes
B has 8 1st place votes
2
A
B
A has 12 1st place votes
B has 15 1st place votes…B wins!
no
candidate
has 14;
eliminate A
What happened? C won. Then, in a revote with all vote changes favoring the
winner, C should win according to the Monotonicity Fairness Criteria, but it
didn’t
Chapter 1 Voting
Page 15 of 17
Pairwise Comparison violates Independence of Irrelevant Alternatives Criterion:
No.voters
1st place
2nd
3rd
4th
5th
2
A
D
C
B
E
6
B
A
C
D
E
4
B
A
D
E
C
1
C
B
A
D
E
1
C
D
A
B
E
4
D
A
E
C
B
4
E
C
D
B
A
A vs. B etc.
A vs. C etc.
A won 3 comparisons
B won 2 ½
“
C won 2
“
D won 1 ½
“
E won 1
“
skipped details
A WINS
Wait a minute, election not valid because one of the losers (C) not eligible to participate.
Remove C and recalculate…
No.voters
1st place
2nd
3rd
4th
2
A
D
B
E
6
B
A
D
E
4
B
A
D
E
1
B
A
D
E
1
D
A
B
E
4
D
A
E
B
A vs. B etc.
A vs. D etc.
skipped details
4
E
D
B
A
A won 2 comparisons
B won 2 ½
“
D won 1 ½
“
E won 0
“
B wins
What happened? A won. When a losing candidate is removed and Pairwise
comparison method recalculated, B wins. PWC violated Condorcet criteria.
Arrow’s Impossibility Theorem - there is no voting method that will satisfy all the fairness
criteria all the time.
Chapter 1 Voting
Page 16 of 17
Practice Problems: page 32
44a.) Pairwise comparison method
Choice
1ST
2ND
3RD
4TH
6
A
C
D
B
6
B
C
A
D
5
B
C
D
A
4
D
A
C
B
3
A
C
D
B
AvsB: 13 votes to 14 votes
AvsC: 16 votes to 11 votes
AvsD: 18 votes to 9 votes
BvsC: 14 votes to 13 votes
BvsD: 14 votes to 13 votes
CvsD: 23 votes to 4 votes
3
B
A
C
D
B wins
A wins
A wins
B wins
B wins
C wins
(N-1)(N)/2 = (3)(4)/2 = 12/2 = 6 head to head
comparisons
B gets 1 point
A gets 1 point
A gets 1 point
B gets 1 point
B gets 1 point
C gets 1 point
A=2
B = 3 wins
C=1
D=0
b.) Ranking by Pairwise comparison: B 1st place
A 2nd place
C 3rd place
D 4th place
51.) Borda Count vs Condorcet Fairness Criteria.
st
1 place
2nd place
3rd place
4th place
4
3
2
1
6
A
B
C
D
2
B
C
D
A
3
C
D
B
A
Condorcet Criteria is
Pairwise Comparison method
Borda Count method
A: 6X4+2X1+3X1= 29
B: 6X3+2X4+3X2= 32
C: 6X2+2X3+3X4= 30
D: 6X1+2X2+3X3= 19
B
wins
AvsB: 6 votes to 5 votes A gets 1 point
AvsC: 6 votes to 5 votes A gets 1 point
AvsD: 6 votes to 5 votes A gets 1 point
BvsC: 8 votes to 3 votes B gets 1 point
BvsD: 8 votes to 3 votes B gets 1 point
CvsD: 11 votes to 0 votes C gets 1 point
A
wins
Borda Count method violated Condorcet Fairness Criteria (which is the Pairwise Comparison
method)
(selected B)
(selected A)
Chapter 1 Voting
Page 17 of 17
53.) Plurality vs Independent Irrelevant Alternatives Criteria
1st
2nd
3rd
4th
5th
49
R
H
F
O
S
48
H
S
O
F
R
3
F
H
S
O
R
Plurality = most 1st place votes wins
R = 49
H = 48
F = 3
R wins
O= 0
S= 0
What if F (loser) is eliminated, IIAC says R should still win.
1st
2nd
3rd
4th
5th
49
R
H
O
S
48
H
S
O
R
3
H
S
O
R
Plurality = most 1st place votes wins
R = 49
H = 48 + 3 = 51
F gone
H wins….Plurality violated IIAC
O= 0
S= 0