Your Turn
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.
11.5
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. Blue received 78 points.
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. Labrador retriever wins the election.
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.
11.28
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.
11.30
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.