Contemporary Mathematics

# Chapter 11

11.1
1.
Yes, Joe Biden won the majority.
11.2
1.
Tony Cambell won a plurality of votes in the Republican primary.
11.3
1.
Hearn and Lim tied. Yes, there must be a third election.
2.
Kelly must be removed in the runoff.
11.4
1.
Option B.
11.5
1.
$4 + 6 + 4 + 7 = 21$
2.
None
3.
4
11.6
1.
Using ranked-choice voting, Yoda is determined to be the winner.
11.7
1.
11.8
1.
Candidate B would be considered the winner using the ranked-choice voting method.
2.
Candidate C would be considered the winner using the Borda count method.
3.
Different candidates won. It appears that the vote counts were so close that a small shift in either direction could change the results of either method.
11.9
1.

2.
Bong Joon-ho won according to the pairwise comparison method.
3.
Yes, the winner is a Condorcet candidate.
11.10
1.
Candidate P wins. Candidate P is not a Condorcet candidate because P lost to Candidate Q.
11.11
1.
Candidate C won 25 points.
2.
The greatest number of points another candidate could win is 24 points.
3.
No, Candidate C cannot lose or tie.
11.12
1.
Dough Boys Pizza is the winner for dinner.
11.13
1.
There is a tie between Rainbow China and Dough Boys Pizza.
11.14
1.
Al Gore is the winner.
11.15
1.
Animal Kingdom
2.
Animal Kingdom
3.
Magic Kingdom
11.16
1.
Using the plurality voting method, there would be a tie between Option B and Option C.
2.
Using the ranked-choice voting method, Option B would be the winner.
3.
No, there was not a majority candidate in Round 1.
11.17
1.
Using the pairwise comparison voting method, Option B wins.
2.
Using the Borda count method, Option A and Option B tie.
3.
Yes. The majority criterion fails for Borda count.
11.18
1.
Yes, because Option A is a Condorcet candidate.
11.19
1.
Option A. Yes, the Condorcet criterion is satisfied.
2.
Option A. Yes, the Condorcet criterion is satisfied.
3.
Option B. No, the Condorcet criterion is not satisfied.
11.20
1.
Standard Poodle.
2.
Standard Poodle.
3.
This election does not violate the monotonicity criterion.
4.
Increasing the ranking for a winner of a Borda count election on a ballot will increase that candidate’s Borda score while decreasing another candidate’s Borda score, but leaving the remaining candidates’ Borda score unchanged. So, a Borda count election will never violate the monotonicity criterion.
11.21
1.
2.
Golden retriever wins the election.
3.
This election violates the monotonicity criterion.
4.
It is possible that the monotonicity criterion would be met in other ranked-choice election scenarios, but overall, the ranked-choice voting method is said to fail the monotonicity criterion even if it failed in only one scenario.
11.22
1.
Chihuahua.
2.
Yorkshire Terrier.
3.
Yes, this election violates IIA.
11.23
1.
Jim wins.
2.
Pam wins.
3.
Yes, this is a violation of IIA.
11.24
1.
B $\frac{2}{3}$; 0.67, C $\frac{2}{3}$; 0.67, D $\frac{2}{3}$; 0.67, and E $\frac{2}{3}$; 0.67
2.
B $\frac{3}{2}$; 1.5, C $\frac{3}{2}$; 1.5, D $\frac{3}{2}$; 1.5, and E $\frac{3}{2}$; 1.5
3.
Yes, the constant ratio is $\frac{3}{2}$ pencils per desk.
11.25
1.
42 pencils would be allocated.
2.
42 pencils would be allocated.
3.
34 desks
11.26
1.
IL 700,000; OH 700,000; MI 700,000; GA 800,000; NC 700,000
2.
IL 0.0000014; OH 0.0000014; MI 0.0000014; GA 0.0000013; NC 0.0000014
3.
IL 0.000001; OH 0.000001; MI 0.000001; GA 0.000001; NC 0.000001
4.
The ratio of State Population to Representative Seats seems to be either 700,000 or 800,000 to 1. There does appear to be a constant ratio of about 0.000001 to 1 of Representative Seats to State Population when rounding to six decimal places. This is the same as the top five states.
11.27
1.
60,449.4
11.28
1.
The states are the Hernandez family and the Higgins family. The seats are the pieces of candy. The house size is 313. The state populations are three in the Hernandez family and four in the Higgins family. The total population is 7.
2.
The standard divisor is the ratio of the number of children to the number of pieces of candy. 0.0224 children per piece of candy.
11.29
1.
0.98
11.30
1.
Family Family’s Standard Quota
Hernandez 175.6667 candies
Higgins 234.2222 candies
Ho 117.1111 candies
Total 527

The sum is very close to 527.

11.31
1.
6, 5, 6, 7, 5, 6
2.
35
3.
No.
11.32
1.
It is not possible to give a fractional part of a gift card. Also, traditional rounding to the nearest integer results in 4 gift cards for each student, which leaves one extra gift card.
11.33
1.
30, 18, 20
2.
68
3.
2
11.34
1.
The final Hamilton apportionment is Fictionville as follows: 1, Pretendstead 3, Illusionham 5, and Mythbury 26.
11.35
1.
Method V violates the quota rule because State F receives 4 seats instead of 5 or 6.
11.36
1.
53.33
2.
480
3.
96
4.
Answers may vary. With a modified divisor of 0.100, the modified quota is 120.
5.
The modified quota from part 3 was the smallest, because the divisor was the largest of the three. Dividing the same number by a larger value gives a smaller result.
11.37
1.

Each state would receive the following seats: Fictionville 1, Pretendstead 2, Illusionham 4, and Mythbury 28.

11.38
1.
Fictionville 2, Pretendstead 3, Illusionham 5, and Mythbury 25.
11.39
1.

The apportionment is Fictionville 2, Pretendstead 2, Illusionham 5, and Mythbury 26.

11.40
1.
The largest state is Mythbury. The citizens would likely favor the Jefferson method of apportionment most since they received the most seats by that method. They would likely favor the Adams and Webster methods of apportionment least because they received the least number of seats by those method.
2.
As a group the other three states received nine seats by either the Hamilton method, seven seats by the Jefferson Method, and ten seats by either the Adams method or the Webster method. They would likely favor the Adams method and Webster method most and favor the Jefferson methods least.
11.41
1.
State A loses 0, seats State B loses 0, and State C loses 3.
2.
State A loses 0 seats, State B loses 1, and State C loses 4.
3.
State C, the largest state, loses the most representatives each time the divisor is increased.
11.42
1.
The standard divisor is 214,079.1236 citizens per seat. The standard quota for Colorado is 2.5210 seats.
2.
The standard divisor is 213,479.4622 citizens per seat. The standard quota for Colorado is 2.5281 seats.
3.
The standard quota increased.
4.
It must have been the case that either the fractional part 0.5281 ranked lower amongst the other fractional parts of the state quotas than the fractional part 0.5210 did, or there were fewer remaining seats to be distributed, or both.
11.43
1.
The final apportionment is: A 183, B 124, C 14, and D 2, which sums to 323.
2.
The final apportionment is: A 184, B 125, C 13, and D 2, which sums to 324.
3.
Yes, this demonstrates the Alabama paradox because State C receives 14 seats if the house size is 323, but only 13 seats if the house size is 324.
11.44
1.
Hospital C lost a respirator while hospital A gained a seat, but hospital C has a higher growth rate than hospital A.
11.45
1.
The Hamilton reapportionment is: 19 for Mudston, 13 for WallaWalla, and 6 for Dilberta. This is an example of the population paradox because WallaWalla lost a seat to Dilberta, even though WallaWalla’s population grew by 2.24 percent while Dilberta’s only grew by 1.56 percent.
11.46
1.
90
2.
95
3.
The original state of Beaversdam lost a seat to the original state of Beruna when the new state of Chippingford was added.
11.47
1.
The reapportionment gives 38 seats to Neverwood, 46 seats to Mermaids Lagoon, and 9 seats to Marooners Rock. This is an example of the new-states paradox because the original state Mermaids Lagoon lost a seat to the original state Neverwood when the new state was added to the union.
11.48
1.

${\text{Original Standard Divisor}} \approx \frac{{76,000,000}}{{391}} \approx 194,400$

${\text{New Population}} \approx 76,000,000 + 300,000 = 76,300,000$

${\text{New House Size}} \approx \frac{{{\text{76,300,000}}}}{{194,400}} \approx 392$

There are $392 - 391 = 1$ new seats to be apportioned to New Mexico.

11.49
1.
Only Hamilton’s method violates the population paradox.

1.
Answers may vary. Example: Ranked-choice, Borda count, and pairwise comparison.
2.
True.
3.
True
4.
Borda count
5.
Pairwise comparison
6.
Hare
7.
In two-round voting, only the top two candidates from Round 1 move on to Round 2, and there are only two rounds. In ranked-choice voting, all candidates except those in last place move on to the next round, and there can be many rounds of voting.
8.
(Independence of) Irrelevant Alternatives Criterion
9.
Pairwise comparison
10.
Borda count method
11.
Pairwise comparison
12.
All of them: Plurality, ranked-choice, pairwise comparison, and Borda count
13.
True, because a majority candidate is always the Condorcet candidate.
14.
False, because the ranked-choice method violates the Condorcet criterion, but it doesn’t violate the majority criterion.
15.
No. Arrow’s Impossibility Theorem does not apply to approval voting because it is not a ranked voting system
16.
False. Arrow’s Impossibility Theorem says that no ranked voting system is perfect and that voter profiles may arise that will lead to a violation of one or more fairness criteria, but it does not guarantee that those voter profiles will occur or are even likely to occur.
17.
True. Candidates are rated as approved or not approved, and voters can give multiple candidates the same rating.
18.
Student 1 is correct.
19.
Both are correct.
20.
Neither are correct.
21.
Neither are correct.
22.
Student 2 is correct.
23.
Student 2 is correct.
24.
Both are correct.
25.

The Hamilton method.

26.

The Hamilton method.

27.

The Jefferson method.

28.

29.

30.

The Hamilton and Jefferson methods.

31.

Webster’s method.

32.

Larger

33.
34.
35.
Quota rule violation
36.
None of these
37.
38.
39.
40.
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