Chapter 11

A simple majority is a number equaling more than 50 percent of the total.

0.50(158,394,605) = 79,197,302.5

Yes, Joe Biden secured a majority of the votes because he had more than 79,197,302.5 votes.

Yes, Joe Biden won the majority.

When no candidate has a majority of the votes, the one with the most votes is said to have a plurality. In the Maryland race, the candidate with the most votes (Tony Campbell) only secured 29.22 percent of the votes. To win by a majority of votes is winning by more than 50 percent. Tony Campbell only won by a plurality.

Tony Cambell won a plurality of votes in the Republican primary.

There was no winner because there was a tie. Kelly must be removed because Kelly had the least number of votes and there will be a third election.

Kelly must be removed in the runoff.

Step 1: Determine the number of votes necessary to have a majority. There are six members, so 50 percent of 6 is 0.5(6) = 3. A majority is at least one more than 50 percent, which is 4 or more votes.

Step 2: In the first round, A and B have 2 votes. C and D have 1 vote. No one has the majority. C and D tie for the least number of votes, so eliminate them both.

Step 3: Reorder the votes with the remaining options.

Option B now has 4 votes, which wins the majority.

Option B.

The numbers in the top row tell you the number of ballots.

4 + 6 + 4 + 7 = 21

$4+6+4+7=21$

Look at the Yellow row. There are no 1s in the Yellow row. A 1 would indicate that someone chose yellow as their favorite color, so the answer is “none.”

None

There are six colors. The rank of 1 indicates a favorite color. The rank of 6 indicates a least favorite color. Look at the Blue row and the Green row. There is only one column where there is a 1 in the Blue row (indicating a favorite color) and a 6 in the Green row (indicating a least favorite color). The number at the top of that column is 4. That means that there are four ballots with those choices.

4

Using ranked-choice voting, Yoda is determined to be the winner.

Step 1: Determine the number of votes needed to achieve a majority.

Add the number of ballots.

7 + 6 + 10 + 8 + 4 + 5 + 6 + 7 + 2 + 3 = 58

0.50(58) = 29

To win a majority requires 30 or more votes.

Step 2: Count the number of first-place votes for each candidate. If a candidate has a majority, then you are done. Otherwise, eliminate the candidate with the fewest votes and repeat this step.

Han Solo: 10 + 2 = 12

Princess Leia: 6 + 7 = 13

Luke Skywalker: 8 + 3 = 11

Chewbacca: 5

Yoda: 7 + 6 = 13

R2-D2 = 4

No one won. Eliminate R2-D2.

Second Round:

Han Solo: 10 + 4 + 2 = 16

Princess Leia: 6 + 7 = 13

Luke Skywalker: 8 + 3 = 11

Chewbacca: 5

Yoda: 7 + 6 = 13

Still, no one has the majority. Eliminate Chewbacca.

Third Round:

Han Solo: 10 + 4 + 2 = 16

Princess Leia: 6 + 7 = 13

Luke Skywalker: 8 + 3 = 11

Yoda: 7 + 5 + 6 = 18

Still, no one has the majority. Eliminate Luke Skywalker.

Fourth Round:

Han Solo: 10 + 4 + 2 = 16

Princess Leia: 6 + 7 + 3 = 16

Yoda: 7 + 8 + 5 + 6 = 26

Still, no one has the majority. Eliminate both Han Solo and Princess Leia. Only Yoda remains and would have the majority.

Yoda wins.

Step 1: For each cell in the table, multiply the number of votes times (6 minus the rank assigned). The 6 comes from the number of colors in the table.

For the color blue, find the points for each column:

First column: 4 × (6 – 1) = 4 × 5 = 20

Second column: 6 × (6 – 2) = 6 × 4 = 24

Third column: 4 × (6 – 1) = 4 × 5 = 20

Fourth column: 7 × (6 – 4) = 7 × 2 = 14

Step 2: Add the points for each column.

20 + 24 + 20 + 14 = 78

Blue received 78 points.

Blue received 78 points.

Ranked Choice Method:

Step 1: Determine the number of votes needed to achieve a majority.

Add the number of ballots.

12 + 19 + 27 + 29 + 31 +21 = 139

0.50(139) = 69.5

To win a majority requires 70 or more votes.

Step 2: Count the number of first-place votes for each candidate. If a candidate has a majority, then you are done. Otherwise, eliminate the candidate with the fewest votes and repeat this step.

A: 12 + 19 = 31

B: 27 + 31 = 58

C: 29 + 21 = 50

No one has the majority. Eliminate Candidate A.

Second Round:

B: 12 + 27 + 31 = 70

C: 19 + 29 + 21 = 69

Candidate B wins the majority vote using the ranked-choice method.

Candidate B would be considered the winner using the ranked-choice voting method.

To save subtracting all the time, a first-place vote earns 2 points. A second-place vote wins 1 point. A last place vote wins no points.

Candidate A: 12(2) + 19(2) + 27(1) + 29(1) + 31(0) + 21(0) = 118

Candidate B: 12(1) + 19(0) + 27(2) + 29(0) + 31(2) + 21(1) = 149

Candidate C: 12(0) + 19(1) + 27(0) + 29(2) + 31(1) + 21(2) = 150

Candidate C has the most points, so Candidate C wins.

Candidate C would be considered the winner using the Borda count method.

Different candidates won using the two methods. Candidate B won by a difference of just 1 vote in the ranked-choice method. Candidate C won by just 1 point in the Borda count method. When the difference between the candidates is so small, the choice of voting method changes who wins.

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.

Step 1: Create a grid with each actor on the horizontal and vertical headers. Indicate a win with a cross and a tie with a slash.

For instance, let’s find SP: Add the number of ballots in which Scorsese ranks higher than Phillips:

2,100 + 1,900 + 1,100 = 5,100

To find SM, add the number of ballots in which Scorsese ranks higher than Mendes:

2,400 + 2,100 + 1,100 = 5,600

To find ST, add the number of ballots in which Scorsese ranks higher than Tarantino.

2,400 + 2,100 + 1,100 = 5,600

To find SB, add the number of ballots in which Scorsese ranks higher than Bong.

2,100

To find PS, add the number of ballots in which Phillips ranks higher than Scorsese.

2,400 + 1,200 = 3,600

To find PM, add the number of ballots in which Phillips ranks higher than Mendes.

2,400 + 2,100 + 1,200 + 1,100 = 6,800

To find PT, add the number of ballots in which Phillips ranks higher than Tarantino.

2,400 + 1,200 = 3,600

To find PB, add the number of ballots in which Phillips ranks higher than Bong.

2,400

To find MS, add the number of ballots in which Mendes ranks higher than Scorsese.

1,900 + 1,200 = 3,100

To find MP, add the number of ballots in which Mendes ranks higher than Phillips.

1,900

To find MT, add the number of ballots in which Mendes ranks higher than Tarantino.

It does not happen, so record a zero.

To find MB, add the number of ballots in which MB ranks higher than Bong.

Again, it does not happen, so record a zero.

Continue to fill in the rest of the table.

It should look like this.

Step 2: Add the total number of ballots: 2,400 + 2,100 + 1,900 + 1,200 + 1,100 = 8,700

Half of that is 4,350.

Anyone who did win over 4,350 votes is a loser. Put an X on the losers. Put a slash on the ties, who have exactly 4,350 (there are not any).

Each win (X) is one point. Each tie (slash) is a half point. Each loss is zero points.

Step 3: The winner is the one with the most points. Bong Joon-ho has 4 points.

Bong Joon-ho won according to the pairwise comparison method.

The winner is Bong Joon-ho with 4 points using the pairwise comparison method.

Compare PQ 1 to QP 5. Because 1 is less than 5, PQ 1 is the loser. Cross out PQ 1.

Compare PR 4 to RP 2. Because 2 is less than 4, RP 2 is the loser. Cross out RP 2.

Compare PS 6 to SP 0. Because 0 is less than 6, SP 0 is the loser. Cross it out.

Continue comparing in this manner until you have completed the table. Because there are no 3s in the table, there will be no ties.

A win earns 1 point, a tie earns a half point, and a loss earns nothing.

P has 4 wins (1 point each) for 4 points.

Q has 3 wins (1 point each) for 3 points.

R has 2 wins (1 point each) for 2 points.

S has 2 wins (1 point each) for 2 points.

T has 2 wins (1 point each) for 2 points.

Z has 2 wins (1 point each) for 2 points.

Candidate P wins with 4 points. P is not a Condorcet winner because P lost to Q.

Candidate P wins. Candidate P is not a Condorcet candidate because P lost to Candidate Q.

A Condorcet candidate wins every pairwise matchup. Candidate C had to win 25 pairwise matchups with the other letters to be the Condorcet candidate. Thus, Candidate C has 25 points.

Candidate C won 25 points.

The greatest number of points another candidate can earn is 24 points. Another candidate could win 24 matchups with everyone but C.

The greatest number of points another candidate could win is 24 points.

To meet the qualification to be a Condorcet candidate, you must win every pairing. That excludes the possibility of a tie or a loss.

No, Candidate C cannot lose or tie.

Count the number of Yes votes earned by each option.

Rainbow China: 4

Dough Boys Pizza: 6

Taco City: 3

Caribbean Flavor: 3

The option with the most Yes votes wins.

Dough Boys Pizza wins with the most points.

Dough Boys Pizza is the winner for dinner.

Count the number of Yes votes earned by each option. Anything with a score of 1, 2, or 3 counts as a Yes.

Rainbow China: 5

Dough Boys Pizza: 5

Taco City: 3

Caribbean Flavor: 2

The option with the most Yes votes wins.

Rainbow China and Dough Boys Pizza have a tie this time.

There is a tie between Rainbow China and Dough Boys Pizza.

100 percent of Pat Buchanan supporters would approve of George W. Bush.

100 percent of Ralph Nader supporters would approve Al Gore.

72 percent of Libertarians would approve of George W. Bush.

28 percent of Libertarians would approve Al Gore (as was roughly the known percentage at the time according to the Cato Institute).

50 percent of the supporters of other candidates would approve of George Bush, whereas 50 percent would approve of Al Gore.

100 percent of Al Gore supporters would approve of Ralph Nader.

50 percent of George Bush supporters would approve of Pat Buchanan.

50 percent of 2,912,790 is 1,456,395.

50 percent of George Bush supporters would approve of Harry Brown.

George W. Bush: 2,912,790 + 17,484 + 11,819 + 3340 = 2,945,433

Al Gore: 2,912,253 + 97,488 + 4,596 + 3,340 = 3,017,677

Ralph Nader: 97,488 + 2,912,253 = 3,009,741

Pat Buchanan: 17,484 + 1,456,395 = 1,473,879

Harry Brown: 11,819 + 4,596 + 1,456,395 = 1,472,810

Other Candidates: 6,680

Al Gore would win.

Al Gore is the winner.

Plurality considers only the top choice of each voter. The percentages for first-place votes show that the Animal Kingdom wins.

Animal Kingdom: 35%

Magic Kingdom: 25%

Epcot: 30%

Hollywood Studios: no first-place votes

Animal Kingdom

A park needs more than 50 percent of the first-place votes to win in the ranked-choice method.

Animal Kingdom: 35%

Magic Kingdom: 25%

Epcot: 30%

Hollywood Studios: no first-place votes

No one wins the first round. Eliminate Hollywood Studios and redo the ranks.

Second Round:

Animal Kingdom: 35%

Magic Kingdom: 25%

Epcot: 30%

No one wins. Eliminate Magic Kingdom and redo the ranks.

Third Round:

Animal Kingdom: 35% + 25% = 60%

Epcot: 30%

Animal Kingdom wins.

Animal Kingdom

First place gets (4 – 1) = 3 points, second place gets 2 points, third place gets 1 point, and fourth place gets 0 points.

Animal Kingdom: 35(3) + 25(1) + 30(0) = 130

Magic Kingdom: 35(2) + 25(3) + 30(1) = 175

Epcot Center: 35(1) + 25(0) + 30(3) = 125

Hollywood Studios: 35(0) + 25(3) + 30(3) = 165

Magic Kingdom wins.

Magic Kingdom

Option A: 1 vote

Option B: 2 votes

Option C: 2 votes

Option D: 1 vote

There is a tie between Options B and C.

Using the plurality voting method, there would be a tie between Option B and Option C.

There are five voters.

50 percent of 5 is 2.5.

The majority is 3 votes or more.

First Round:

Option A: 1 vote

Option B: 2 votes

Option C: 2 votes

Option D: 1 vote

No one wins. Eliminate A and D.

Second Round:

Option B: 3 votes

Option C: 2 votes

Option B wins.

Using the ranked-choice voting method, Option B would be the winner.

The majority criterion states that the winner should be in favor with the majority of the voters. That is not the case here because there was not a majority winner in the first round.

No, there was not a majority candidate in Round 1.

AB 2 vs BA 3. BA wins, so cross out AB.

AC 3 vs CA 2. AC wins, so cross out CA.

AD 4 vs DA 1. AD wins, so cross out DA

BC 3 vs CB 2. BC wins, so cross out CB.

BD 3 vs DB 2. BD wins, so cross out DB.

CD 3 vs DC 2. CD wins, so cross out DC.

B wins.

Using the pairwise comparison voting method, Option B wins.

First place earns (4 – 1) or 3 points.

Second place earns (4 – 2) or 2 points.

Third place earns (4 – 3) or 1 point.

Fourth place earns (4 – 4) or 0 points.

Option A: 2 + 2 + 1 + 2 + 2 = 9

Option B: 3 + 0 + 3 + 0 + 3 = 9

Option C: 1 + 3 + 0 + 3 + 1 = 8

Option D: 0 + 1 + 2 + 1 + 0 = 4

Option A and B tie.

Using the Borda count method, Option A and Option B tie.

There are five votes.

50 percent of five is 2.5.

A majority would need 3 first-place votes.

In a simple ballot, Option B would win the majority criterion because Option B had 3 first-place votes. The majority criterion would fail in the Borda count method because Options A and B tied.

Yes. The majority criterion fails for Borda count.

An X represents crossing out a cell.

Option A is a Condorcet candidate because Option A won every matchup.

Yes, because Option A is a Condorcet candidate.

Option A: 3 votes

Option B: 2 votes

Option C: 0 votes

Option A wins by having the most votes. The Condorcet criterion is satisfied.

The pairwise comparison had already shown that Option A was a Condorcet candidate.

Option A. Yes, the Condorcet criterion is satisfied.

First Round:

Total votes: 5

50 percent of 5 is 2.5.

To win the majority, an option needs 3 votes or more.

Option A: 3 votes

Option B: 2 votes

Option C: 0 votes

Option A has enough votes to satisfy winning by a majority. The Condorcet criterion is satisfied.

Option A. Yes, the Condorcet criterion is satisfied.

In Borda count, the first place out of three earns (3 – 1) or 2 points.

Second place earns (3 – 2) or 1 point.

Third place earns (3 – 3) or 0 points.

Option A: 3(2) + 2(0) = 6

Option B: 3(1) + 2(2) = 7

Option C: 3(0) + 2(1) = 2

Option B wins. No, the Condorcet criterion is not satisfied. The Condorcet candidate did not win.

Option B. No, the Condorcet criterion is not satisfied.

The total number of votes: 130.

50 percent of 130 is 65.

To win the majority, you need at least one more than 65, which is 66.

First Round:

Miniature Poodle: 53

Yorkshire Terrier: 24 + 11 = 35

Chihuahua: 42

No one has 66, so eliminate the Yorkshire Terrier.

Second Round:

Miniature Poodle: 53 + 11 = 64

Chihuahua: 42 + 24 = 66

The Chihuahua wins.

Chihuahua.

The total number of votes: 130.

50 percent of 130 is 65.

To win the majority, you need at least one more than 65, which is 66.

First Round:

Yorkshire Terrier: 53 + 24 + 11 = 88

Chihuahua: 42

The Yorkshire Terrier wins.

Yorkshire Terrier.

Yes, this violates IIA, or Independence of Irrelevant Alternatives Criterion. This means that the introduction or removal of a third candidate should not change or reverse the rankings of the original two candidates relative to one another. In particular, if a losing candidate is removed from a race or a new candidate is added, the winner should not change.

Yes, this election violates IIA.

First place earns (4 – 1) or 3 points.

Second place earns (4 – 2) or 2 points.

Third place earns (4 – 3) or 1 point.

Fourth place earns (4 – 4) or 0 points.

Jim Halpert: 9(3) + 11(2) + 7(0) + 6(2) + 3(0) = 61

Pam Beesley-Halpert: 9(0) + 11(3) + 7(2) + 6(0) + 3(1) = 50

Dwight Schrute: 9(2) + 11(1) + 7(1) + 6(3) + 3(2) = 60

Michael Scott: 9(1) + 11(0) + 7(3) + 6(1) + 3(3) = 45

Jim Halpert wins.

Jim wins.

Divide the first quantity by the second quantity.

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

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

$\frac{\text{Pencils}}{\text{Desks}}=\frac{3}{2}=\frac{p}{28}$

$\begin{array}{c}2p=84\\ p=42\end{array}$

You need to allocate 42 pencils.

42 pencils would be allocated.

$\frac{\text{Desks}}{\text{Pencils}}=\frac{2}{3}=\frac{28}{p}$

$\begin{array}{c}2p=84\\ p=42\end{array}$

You need to allocate 42 pencils.

42 pencils would be allocated.

$\frac{\text{Desks}}{\text{Pencils}}=\frac{2}{3}=\frac{d}{51}$

$\begin{array}{c}3d=102\\ d=34\end{array}$

You need to allocate 34 desks.

34 desks

Make a ratio of $\frac{\text{State Population}}{\text{Representative Seats}}$ for each state to the nearest hundred thousand.

IL 700,000; OH 700,000; MI 700,000; GA 800,000; NC 700,000

IL 0.0000014; OH 0.0000014; MI 0.0000014; GA 0.0000013; NC 0.0000014

IL 0.000001; OH 0.000001; MI 0.000001; GA 0.000001; NC 0.000001

The standard divisor is the total population divided by the house size. Divide the population by the 65 representative seats.

$\frac{3,929,214}{65}\approx <!--\; \approx \; -->60,449.4$

60,449.4

The states are the recipients of the apportioned resource (the Hernandez family and the Higgins family).

The seats are the units of the resource being apportioned (the pieces of candy).

The house size is the total number of seats to be apportioned (the 313 pieces of candy).

The state population is the measurement of the state’s size (the size of the two families, which is three in the Hernandez family and four in the Higgins family).

The total population is the sum of the state populations $\text{(3 + 4 = 7)}$.

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.

The standard divisor is the total population divided by the house size.

In this scenario, the total population is the number of children in the two families (7).

The house size is the number of pieces of candy (313).

$\frac{7}{313}\approx <!--\; \approx \; -->0.0224$ children per piece of candy.

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.

The standard divisor is the total population divided by the house size.

Standard Divisor: $\frac{3,929,214}{65}\approx <!--\; \approx \; -->60,449.4$

Standard Quota: $\frac{\text{State Population}}{\text{Standard Divisor}}\text{ seats = }\frac{59,000}{60,449.4}\text{ seats = 0}\text{.98 seats}$

0.98

The standard divisor is the total population divided by the house size.

In this scenario, the total population is the number of children in the two families $\text{(3 + 4 + 2 = 9)}$.

The house size is the number of pieces of candy (527).

$\text{Standard Divisor = }\frac{\text{Total Population}}{\text{House Size}}=\frac{\text{9 children}}{527\text{ candies}}\approx <!--\; \approx \; -->0.0170778$ children per piece of candy.

Add the three standard quotas: $\text{175}\text{.67 + 234}\text{.22 + 117}\text{.11 = 527}$

This sum is 527. Even if you round to whole numbers, which any parent would when counting candy, the sum is 527.

$\text{176 + 234 + 117 = 527}$

The sum is very close to 527.

The total population of students: $\text{30 + 25 + 28 + 32 + 24 + 27 = 166}$ students. $\text{Standard Divisor = }\frac{\text{Total Population}}{\text{House Size}}=\frac{\text{166 students}}{\text{34 laptops}}\approx <!--\; \approx \; -->4.88\text{ students per laptop}$

6, 5, 6, 7, 5, 6

The total population of students: $\text{30 + 25 + 28 + 32 + 24 + 27 = 166}$ students.$\text{Standard Divisor = }\frac{\text{Total Population}}{\text{House Size}}=\frac{\text{166 students}}{\text{34 laptops}}\approx <!--\; \approx \; -->4.88\text{ students per laptop}$

Sum of last column: $\text{6 + 5 + 6 + 7 + 5 + 6 = 35}$

35

The total population of students: $\text{30 + 25 + 28 + 32 + 24 + 27 = 166}$ students.$\text{Standard Divisor = }\frac{\text{Total Population}}{\text{House Size}}=\frac{\text{166 students}}{\text{34 laptops}}\approx <!--\; \approx \; -->4.88\text{ students per laptop}$

Sum of last column: $\text{6 + 5 + 6 + 7 + 5 + 6 = 35}$

No. The rounded values added up to 35. There are only 34 available laptops. This is a problem.

No.

You cannot give out fractional gift cards. Round the gift card amounts using traditional rounding and add the results.

$\text{4 + 4 + 4 = 12}$

There are 13 cards. This leaves out one gift card.

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.

The total population $\text{ = 15 + 9 + 10 = 34}$ acres

$\text{Standard Divisor = }\frac{\text{Total Population}}{\text{House Size}}=\frac{\text{34 acres}}{\text{70 lights}}\approx <!--\; \approx \; -->0.4857\text{ acres per light}$

Hamilton’s method is to always round down, which results in the lower quota.

30, 18, 20

68

Add the lower quotas: $\text{30 + 18 + 20 = 68}$.

2

Add the lower quotas: $\text{30 + 18 + 20 = 68}$.

There are 70 lights available.

This is a difference of 2.

Use Hamilton’s method.

Step 1: Find the standard divisor.

$\text{Standard Divisor = }\frac{\text{Total Population}}{\text{House Size}}=\frac{\text{1,593,000}}{\text{35 seats}}\approx <!--\; \approx \; -->45,514.28571\text{ people per seat}$

Step 2: Find each state’s standard quota.

Step 3: Give each state the state’s lower quota (with each state receiving at least 1 seat). The lower quota always rounds down.

Step 4: Give each remaining seat one at a time to the states with the largest fractional parts of their standard quotas until no seats remain.

Add up the lower quotas: $\text{1 + 2 + 4 + 26 = 33}$.

There are two more seats to allot. The largest fraction parts are 0.64 for Illusionham and 0.57 for Pretendstead. Give them the extra seats.

Fictionville: 1

Pretendstead: 3

Illusionham: 5

Mythbury: 26

Step 5: Check the solution by confirming that the sum of the modified quotas equals the house size.

$\text{1 + 3 + 5 + 26 = 35}$

The final apportionment:

Fictionville: 1

Pretendstead: 3

Illusionham: 5

Mythbury: 26

The final Hamilton apportionment is Fictionville as follows: 1, Pretendstead 3, Illusionham 5, and Mythbury 26.

If a scenario exists in which a particular apportionment allocates a value greater than the upper quota or less than the lower quota, then that apportionment violates the quota rule and the apportionment method that was used violates the quota rule.

State F’s lower quota is 5 and upper quota is 6.

State F violates the quota rule because it receives 4 seats.

4 is lower than the lower quota of 5.

The other states are compliant with the quota rule.

Method V violates the quota rule because State F receives 4 seats instead of 5 or 6.

$\text{Standard Quota = }\frac{\text{State Population}}{\text{Standard Divisor}}\text{ = }\frac{\text{12}}{\text{0}\text{.225}}\text{ = 53}\text{.33 seats}$

53.33

Decrease 0.225 by 0.2:

0.225 – 0.2 = 0.025, the new standard divisor

$\text{Standard Quota = }\frac{\text{State Population}}{\text{Standard Divisor}}\text{ = }\frac{\text{12}}{\text{0}\text{.025}}\text{ = 480 seats}$

480

Increase 0.025 by 0.10.

0.025 + 0.10 = 0.125, the new standard divisor

$\text{Standard Quota = }\frac{\text{State Population}}{\text{Standard Divisor}}\text{ = }\frac{\text{12}}{\text{0}\text{.125}}\text{ = 96 seats}$

96

Step 1: Find the standard divisor.

$\text{Standard Divisor = }\frac{\text{Total Population}}{\text{House Size}}=\frac{\text{1,593,000}}{\text{35 seats}}\approx <!--\; \approx \; -->45,514.28571\text{ people per seat}$

Step 2: Find each state’s standard quota.

Step 3: Give each state the state’s lower quota (with each state receiving at least 1 seat). The lower quota always rounds down. Find the sum, too.

Add up the lower quotas: $\text{1 + 2 + 4 + 26 = 33}$.

Step 4: If the sum of the lower quotas equals the number of seats, you are done. If the sum of the lower quotas is less than the number of seats, reduce the standard divisor. If the sum is greater, increase the standard divisor. Return to Step 2 using the modified divisor.

Step 5: Repeat the process of finding the quotas. Find each state’s modified quota, lower quota, and the sum of the lower quotas based on the modified divisor.

Second Round:

Find each state’s standard quota using 43,000 people per seat.

Give each state the state’s lower quota (with each state receiving at least 1 seat). The lower quota always rounds down. Find the sum, too.

Because the sum (34) is still lower than the number of seats (35), reduce the standard divisor again.

Third Round:

Find each state’s standard quota using 42,500 people per seat.

Give each state the state’s lower quota (with each state receiving at least 1 seat). The lower quota always rounds down. Find the sum, too.

Now the sum of the lower quotas is equal to the number of seats.

Step 6: The new sum of the lower quotas equals the house size. The apportionment is complete.

$\text{1 + 2 + 4 + 28 = 35}$

Fictionville: 1

Pretendstead: 2

Illusionham: 4

Mythbury: 28

The modified divisor you chose may be different, but the final apportionment seat numbers will be the same for everyone who uses Jefferson’s method.

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

Use Adams’s method.

Step 1: Find the standard divisor.

$\text{Standard Divisor = }\frac{\text{Total Population}}{\text{House Size}}=\frac{\text{1,593,000}}{\text{35 seats}}\approx <!--\; \approx \; -->45,514.28571\text{ people per seat}$

Step 2: Find each state’s standard quota.

Step 3: Give each state the state’s upper quota (with each state receiving at least 1 seat). The upper quota always rounds up. Find the sum, too.

Add up the upper quotas: $\text{2 + 3 + 5 + 27 = 37}$

Step 4: If the sum of the upper quotas equals the number of seats, you are done. If the sum of the upper quotas is less than the number of seats, reduce the standard divisor. If the sum is greater, increase the standard divisor. Return to Step 2 using the modified divisor.

Step 5: Repeat the process of finding the quotas. Find each state’s modified quota, upper quota, and the sum of the upper quotas based on the modified divisor.

Second Round:

Find each state’s standard quota using 48,000 people per seat.

Give each state the state’s upper quota (with each state receiving at least 1 seat). The upper quota always rounds up. Find the sum, too.

Now the sum of the upper quotas is equal to the number of seats.

Step 6: The new sum of the upper quotas equals the house size. The apportionment is complete.

$\text{2 + 3 + 5 + 25 = 35}$

Fictionville: 2

Pretendstead: 3

Illusionham: 5

Mythbury: 25

The modified divisor you chose may be different, but the final apportionment seat numbers will be the same for everyone who uses Adams’s method.

Fictionville 2, Pretendstead 3, Illusionham 5, and Mythbury 25.

Use Webster’s method.

Step 1: Find the standard divisor.

$\text{Standard Divisor = }\frac{\text{Total Population}}{\text{House Size}}=\frac{\text{1,593,000}}{\text{35 seats}}\approx <!--\; \approx \; -->45,514.28571\text{ people per seat}$

Step 2: Find each state’s standard quota.

Step 3: Round each state’s quota to the nearest whole number (with each state receiving at least 1 seat). Use traditional rounding. Find the sum, too.

Add up the rounded quotas: $\text{2 + 3 + 5 + 26 = 36}$.

Step 4: If the sum of the rounded quotas equals the number of seats, you are done. If the sum of the rounded quotas is less than the number of seats, reduce the standard divisor. If the sum is greater, increase the standard divisor. Return to Step 2 using the modified divisor.

Step 5: Repeat the process of finding the quotas. Find each state’s modified quota, rounded quota, and the sum of the rounded quotas based on the modified divisor.

Second Round:

Find each state’s standard quota using 46,820 people per seat.

Give each state the state’s rounded quota (with each state receiving at least 1 seat). Use traditional rounding. Find the sum, too.

Now the sum of the rounded quotas is equal to the number of seats.

Step 6: The new sum of the rounded quotas equals the house size. The apportionment is complete.

$\text{2 + 2 + 5 + 26 = 35}$

Fictionville: 2

Pretendstead: 2

Illusionham: 5

Mythbury: 26

The modified divisor you chose may be different, but the final apportionment seat numbers will be the same for everyone who uses Webster’s method.

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

The largest state is Mythbury. Their citizens would likely favor Jefferson’s method of apportionment most because they received the most seats by that method. They would likely favor Adams’s and Webster’s methods the least because they received the least number of seats by those methods.

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.

States A and B had no change.

State C lost 3.

State A loses 0, seats State B loses 0, and State C loses 3.

State A loses 0 seats, State B loses 1, and State C loses 4.

State A lost nothing.

State B lost 1.

State C lost 4.

State C, the largest state, loses the most representatives each time the divisor is increased.

The switch from 11,500 to 12,000:

States A and B had no change.

State C lost 3.

The switch from 12,000 to 12,500:

State A lost nothing.

State B lost 1.

State C lost 4.

Overall, state C had the greatest losses each time the divisor was increased.

$\text{Standard Divisor = }\frac{\text{Total Population}}{\text{House Size}}=\frac{\text{76,212,168}}{\text{356 seats}}\approx <!--\; \approx \; -->214,079.1236\text{ people per seat}$

$\text{Standard Quota = }\frac{\text{State Population}}{\text{Standard Divisor}}\text{ = }\frac{\text{539,700 people}}{214,079.1236\text{ people per seat}}\text{ = 2}\text{.521 seats}$

The standard divisor is 214,079.1236 citizens per seat. The standard quota for Colorado is 2.5210 seats.

$\text{Standard Divisor = }\frac{\text{Total Population}}{\text{House Size}}=\frac{\text{76,212,168}}{\text{357 seats}}\approx <!--\; \approx \; -->213,479.4622\text{ people per seat}$

$\text{Standard Quota = }\frac{\text{State Population}}{\text{Standard Divisor}}\text{ = }\frac{\text{539,700 people}}{213,479.4622\text{ people per seat}}\text{ = 2}\text{.528 seats}$

The standard divisor is 213,479.4622 citizens per seat. The standard quota for Colorado is 2.5281 seats.

The house size increased from 356 to 357 seats. The standard quota increased from 2.521 seats to 2.528 seats.

The standard quota increased.

Use Hamilton’s method for 323 seats.

Step 1: Find the standard divisor.

Total population: $\text{56,700 + 38,500 + 4,200 + 600 = 100,000}$

$\text{Standard Divisor = }\frac{\text{Total Population}}{\text{House Size}}=\frac{\text{100,000}}{\text{323 seats}}\approx <!--\; \approx \; -->309.5975\text{ votes per seat}$

Step 2: Find each state’s standard quota.

Step 3: Give each state the state’s lower quota (with each state receiving at least 1 seat). The lower quota always rounds down.

Step 4: Give each remaining seat one at a time to the states with the largest fractional parts of their standard quotas until no seats remain.

Add up the lower quotas: $\text{183 + 124 + 13 + 1 = 321}$.

Remaining seats to allot: 2

The largest fractional part is 0.938 for Party D and 0.566 for C. Give the seats to Party C and D.

Step 5: Check the solution by confirming that the sum of the modified quotas equals the house size.

$\text{184 + 125 + 14 = 323}$

The final apportionment is: A 183, B 124, C 14, and D 2, which sums to 323.

Use Hamilton’s method for 323 seats.

Step 1: Find the standard divisor.

Total population: $\text{56,700 + 38,500 + 4,200 + 600 = 100,000}$

$\text{Standard Divisor = }\frac{\text{Total Population}}{\text{House Size}}=\frac{\text{100,000}}{\text{324 seats}}\approx <!--\; \approx \; -->308.6420\text{ votes per seat}$

Step 2: Find each state’s standard quota.

Step 3: Give each state the state’s lower quota (with each state receiving at least 1 seat). The lower quota always rounds down.

Step 4: Give each remaining seat one at a time to the states with the largest fractional parts of their standard quotas until no seats remain.

Add up the lower quotas: $\text{183 + 124 + 13 + 1 = 321}$.

Remaining seats to allot: 3

The largest fractional part is 0.944 for D, 0.740 for B, 0.708 for A. Give A, B, and D a seat.

Step 5: Check the solution by confirming that the sum of the modified quotas equals the house size.

$\text{184 + 125 + 13 + 2 = 324}$

The final apportionment is: A 184, B 125, C 13, and D 2, which sums to 324.

The Alabama paradox occurs when an increase in the house size reduces a state’s quota. Party C lost a seat despite an increase in the house size.

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.

Hospital C lost a respirator while hospital A gained a seat, but hospital C has a higher growth rate than hospital A.

The population paradox occurs when a state with an increasing population loses a seat while a state with a decreasing population either retains or gains seats. More generally, the population paradox occurs when a state with a higher population growth rate loses seats while a state with a lower population growth rate retains or gains seats.

Hospital C lost a respirator while hospital A gained a seat, but hospital C has a higher growth rate than hospital A.

Use Hamilton’s method.

Step 1: Find the standard divisor.

Total population: $\text{866,000 + 626,000 + 256,000 = 1,748,000}$

$\text{Standard Divisor = }\frac{\text{Total Population}}{\text{House Size}}=\frac{\text{1,748,000}}{\text{38 seats}}=46,000\text{ people per seat}$

Step 2: Find each state’s standard quota.

Step 3: Give each state the state’s lower quota (with each state receiving at least 1 seat). The lower quota always rounds down.

Step 4: Give each remaining seat one at a time to the states with the largest fractional parts of their standard quotas until no seats remain.

Add up the lower quotas: $\text{18 + 13 + 5 = 36}$.

There are two more seats to allot. The largest fractional parts are 0.825 for Mudstown and 0.608 for WallaWalla. Give them the extra seats.

Step 5: Check the solution by confirming that the sum of the modified quotas equals the house size.

$\text{19 + 14 + 5 = 38}$

Now, redo for the changed population:

Step 1: Find the standard divisor.

Total population: $\text{921,000 + 640,000 + 260,000 = 1,821,000}$

$\text{Standard Divisor = }\frac{\text{Total Population}}{\text{House Size}}=\frac{1,821,000}{\text{38 seats}}\approx <!--\; \approx \; -->47921.05263\text{ people per seat}$

Step 2: Give each remaining seat one at a time to the states with the largest fractional parts of their standard quotas until no seats remain.

Add up the lower quotas: $\text{19 + 13 + 5 = 37}$.

There is one more seat to allot. The largest fractional part is 0.426 for Dilberta. Give Dilberta the extra seat.

Step 3: Check the solution by confirming that the sum of the modified quotas equals the house size.

$\text{19 + 13 + 6 = 3}$

With the original population numbers, WallaWalla had 14 seats. Even though WallaWalla had a higher growth rate than Dilberta, Dilberta gained a seat and WallaWalla lost a seat.

This violates the population paradox.

The population paradox occurs when a state with an increasing population loses a seat while a state with a decreasing population either retains or gains seats.

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.

The house size is the number of seats.

$\text{39 + 51 = 90}$ seats

The original house size is 90 seats.

90

The house size is the number of seats.

$\text{40 + 50 + 5 = 95}$

The new house size is 95 seats.

95

The existing state of Beaversdam lost a seat to the existing state of Beruna when the new state of Chippingford was added.

This is an example of the new-state paradox, which occurs when the addition of a new state is accompanied by an increase in seats to maintain the standard ratio of population to seats, but one of the existing states loses a seat in the resulting reapportionment.

The original state of Beaversdam lost a seat to the original state of Beruna when the new state of Chippingford was added.

Use Hamilton’s method for the inclusion of the new state.

Step 1: Find the standard divisor.

Total population: $\text{760,000 + 943,000 + 190,000 = 1,893,000}$

$\text{Standard Divisor = }\frac{\text{Total Population}}{\text{House Size}}=\frac{1,893,000}{\text{93 seats}}\approx <!--\; \approx \; -->20,354.839\text{ people per seat}$

Step 2: Give each remaining seat one at a time to the states with the largest fractional parts of their standard quotas until no seats remain.

Add up the lower quotas: $\text{37 + 46 + 9 = 92}$.

There is one more seat to allot. The largest fractional part is 0.338 for Neverwood. Give Neverwood the extra seat.

Step 3: Check the solution by confirming that the sum of the modified quotas equals the house size.

$\text{38 + 46 + 9 = 93}$, the same as the new house size.

This violates the new state paradox. The existing state of Neverwood lost a seat to the existing state of Mermaids Lagoon when the state of Marooners Rock was added.

This is an example of the new-state paradox, which occurs when the addition of a new state is accompanied by an increase in seats to maintain the standard ratio of population to seats, but one of the existing states loses a seat in the resulting reapportionment.

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.

$\text{Original Standard Divisor}\approx <!--\; \approx \; -->\frac{76,000,000}{391}\approx <!--\; \approx \; -->194,400$

$\text{New Population}\approx <!--\; \approx \; -->76,000,000+300,000=76,300,000$

$\text{New House Size}\approx <!--\; \approx \; -->\frac{\text{76,300,000}}{194,400}\approx <!--\; \approx \; -->392$

There are $392-<!--\; -\; -->391=1$ new seats to be apportioned to New Mexico.

$\text{Standard Divisor = }\frac{\text{Total Population}}{\text{House Size}}=\frac{\text{76,000,000}}{\text{391 seats}}\approx <!--\; \approx \; -->194,400\text{ people per seat}$

New total population $\text{ = 76,000,000 + 300,000 = 76,300,000}$ people.

To estimate the new house size, rearrange the equation for the standard divisor.

$\text{New House Size = }\frac{\text{New Total Population}}{\text{Standard Divisor}}=\frac{76,300,000}{194,400\text{ people per seat}}\approx <!--\; \approx \; -->392\text{ seats}$

New House Size – Old House Size = 392 – 391 = 1 new seat to be apportioned to New Mexico.

The original standard divisor: $194,400\text{ people per seat}$

The new population: 76,300,000 people

The new house size: 392 seats

The number of seats New Mexico should receive: 1 seat

Student 1:

Student 2:

$A=B÷<!--\; ÷\; -->\frac{A}{B}$

This is the same as

Student 1 is correct.

Student 1 is correct.

Student 1:

$A=B÷<!--\; ÷\; -->\frac{B}{A}$

This is the same as

Student 2:

$B=A÷<!--\; ÷\; -->\frac{A}{B}$

This is the same as

Both students are correct.

Both are correct.

Student 1:

$\frac{A}{B}=\frac{110}{55}=2$

Student 1 is wrong.

Student 2:

$\frac{B}{A}=\frac{55}{110}=\frac{1}{2}$

$\frac{1}{0.5}=2$

Student 2 is wrong.

Both students are incorrect.

Neither are correct.

Given: $\frac{A}{B}=0.75$

Student 1:

$\frac{A}{B}=\frac{4}{3}\approx <!--\; \approx \; -->1.3333$

Student 1 is wrong.

Student 2:

B to A should be $\frac{1}{0.75}\approx <!--\; \approx \; -->1.33333$.

Student 2 says B to A is $\frac{3}{4}=0.75$.

Student 2 is wrong.

Both students are wrong.

Neither are correct.

Student 1 is incorrect. $\text{Standard Divisor = }\frac{\text{Total Population}}{\text{House Size}}$

Student 2 is correct. $\text{Standard Divisor = }\frac{\text{Total Population}}{\text{House Size}}$. The total population is the sum of the state populations. The house size is the total number of seats available.

Student 2 is correct.

Student 1 is incorrect. The House Size should be in the denominator. $\text{Standard Divisor = }\frac{\text{Total Population}}{\text{House Size}}$

Student 2 is correct. $\text{Standard Quota = }\frac{\text{State Population}}{\text{Standard Divisor}}$

Student 2 is correct.

Because the Standard Quota = $\frac{\text{State Population}}{\text{Standard Divisor}}\text{ seats }$ and the $\text{Standard Divisor = }\frac{\text{Total Population}}{\text{House Size}}$,

both students are correct. Student 1 substituted the Standard Divisor expression into the Standard Quota formula.

Both are correct.

The Adams method.

Adams’s method favors smaller states. One of the Your Turn exercises showed this tendency. Here is a reminder:

The first change from 11,500 to 12,000:

States A and B had no change.

State C lost 3.

The second change from 12,000 to 12,500:

State A lost nothing.

State B lost 1.

State C lost 4.

Overall, state C had the greatest losses each time the divisor was increased.

The Alabama paradox occurs when an increase in the house size reduces a state’s quota. This is a violation of the Alabama paradox because your neighborhood fire station suffered a reduction in fire trucks when the house size (number of fire trucks) increased.

The Alabama paradox

The population paradox occurs when a state with an increasing population loses a seat while a state with a decreasing population either retains or gains seats. More generally, the population paradox occurs when a state with a higher population growth rate loses seats while a state with a lower population growth rate retains or gains seats.

This is a violation of the population paradox because the Chapel Run school numbers went up, but they lost resources.

Population paradox

State C’s standard quota is 5.71. The lower quota is 5 and the upper quota is 6. If a scenario exists in which a particular apportionment allocates a value greater than the upper quota or less than the lower quota, then that apportionment violates the quota rule and the apportionment method that was used violates the quota rule. State C received 4 seats, which is lower than the lower quota.

Quota rule violation

This is an example of the new-state paradox, which occurs when the addition of a new state is accompanied by an increase in seats to maintain the standard ratio of population to seats, but one of the existing states loses a seat in the resulting reapportionment.

An existing community of Cocoa lost a seat to another existing community when a new community was added.

New-states paradox