easy paper

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Ch. # 6 Voting systems Different between Majority and plurality: Majority (any percent greater than 50 % is called a majority) Plurality The candidate that receives the most votes is declared the winner.

In case of a tie, a special runoff election might be held. Ex. With 50 Voters, 3 – candidates

Candidate # Votes Percent

A 13 26%

B 27 54%

C 10 20%

B : is a winner by majority

Candidate # Votes Percent

D 15 30%

E 23 46%

F 12 24%

E : is a winner by Plurality

The plurality method of voting may produce a “winner” even though a majority of voters did not vote for the “winner.” This dilemma can be avoided in several ways

To avoid this:

One common way is to eliminate the candidate with the fewest votes and then hold another election.

The voters who vote for the eliminated candidate now vote for their second choice.

If the majority is still not attained, the process is repeated until a candidate obtains a majority of votes.

The Plurality-with-Elimination Method The idea is to eliminate the candidates with the fewest first-place votes one at a time until one of them gets a majority;  Each person vote for his or her favorite candidate.

 If a candidate receives a majority of votes, that candidate is declared the winner.

 If no, the candidate with the fewest votes is eliminated and a new election is held.

 The process is repeated until a candidate receives a majority of the votes.

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Ex.

Choice A B C

# of votes 15 23 12

Using plurality method: A: 23 Most, not majority. We eliminate C, we go to next vote.

Choice A B

# of votes 21 29

B: 29, B – received a majority. B – Winner.

Rather than holding a new election after eliminating the candidate with the fewest votes, each voter rank each candidate during the first election.

Instant Runoff Method (Ranked-Choice)

 Each voter rank all the candidates; 1st. 2nd. 3rd. Choice, if a candidate receives a majority of the first choice votes, that candidate is declared the winner.

 If no candidate receives a majority, then the candidate with the fewest first choice votes is eliminated, and these votes are given to the next preferred candidate.

 If a candidate now has a majority of first choice votes, that candidate is declared the

winner.

 If no majority repeat until a candidate receive a majority.

 Because the choices are being put in an order (that is, a person must make a first choice, a second choice, and a third choice), we conclude that permutations can be used to determine the number of different rankings.

Voter preference tables: Is a tables that list different rankings of the candidates along with the number of voters who chose each specific ranking.

Ex: Consider 60 voters, 3-candidates as shown in the preference table. Majority (at least: 30+1)

# of voters 9 14 15 4 2 16

1st. Choice A A C C B B

2nd. Choice C B B A A C

3rd. Choice B C A B C A

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Round 1.

Candidate A B C

# of 1st. place 9+14=23 2+16=18 15+4=19

No Majority, we eliminate: B and these votes are given to the next preferred candidate.

# of voters 9 14 15 4 2 16

1st. Choice A A C C A C

2nd. Choice C C A A C A

Round 2.

Candidate A C

# of 1st. place 9+14+2=25 15+4+16=35

C: Is the winner. – Majority

To win an election using the instant runoff method, a candidate must capture a majority of first-choice votes.

Rather than tallying the first-choice votes only, we might want to tally the first-choice votes, the second-choice votes, the third-choice votes, and so on. This is the basis of the Borda count method.

Borda Count Method of Voting If we have an election with K candidates we will give 1 point for last place, 2 points for second to last, . . . and K points for first place. The candidate with the highest total number of points is the winner. We will call such a candidate the Borda winner.

Ex: Consider 60 voters, 3-candidates as shown in the preference table.

9 14 15 4 2 16

1st. Choice A A C C B B

2nd. Choice C B B A A C

3rd. Choice B C A B C A

K = 3 number of candidates.

Each 1st. choice vote is worth 3 – points

Each 2nd. Choice vote is worth 2 – points

Each 3rd. choice vote is worth 1 – point

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Candidate A B C

1st. Choice 9 + 14 = 23 2 + 16 = 18 15 + 4 = 19

2nd. Choice 4 + 2 = 6 14 + 15 = 29 9 + 16 = 25

3rd. Choice 15 + 16 = 31 9 + 4 = 13 14 + 2 = 16

A = (23) (3) + (6) (2) + (31) (1) = 112 points B = (18) (3) + (29) (2) + (13) (1) = 125 points C = (19) (3) + (25) (2) + (16) (1) = 123 points B: Is the winner.

C: was the winner in runoff. !! It is very important to inform voters before the election as to which system will be used.

The Pairwise comparison Method of Voting

Each voter ranks all of the candidates; that is, each select his or her 1st. 2nd. 3rd. Choice, and so on. Each possible pairing of candidates, the candidate with the most votes receives 1 – point; if there is a tie, each candidate receives ½ point. The candidate who receives the most points is declared the winner.

• That is, if there are three candidates, X, Y, and Z, we may consider the mini-elections

or pairwise comparisons of X versus Y, X versus Z, and Y versus Z.

• We then determine the winner of each of these pairwise comparisons, and the candidate who wins the most of these is declared the winner of the overall election.

Ex.

9 14 15 4 2 16

1st. Choice A A C C B B

2nd. Choice C B B A A C

3rd. Choice B C A B C A

Because there are 3 – candidates (A, B, C) we have:

  n r

! 3! 3

! ! 2!(3 2)!

n C

r n r   

  (Pairwise)

(A – B) A = 9 + 14 + 4 = 27 B = 15 + 2 + 16 = 33 Voters preferred B over A, B – receives 1 – point.

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(A – C) A = 9 + 14 + 2 = 25 C = 15 + 4 + 16 = 35 Voters preferred C, over A, C – receives 1 – point.

(B – C) B = 14 + 2 + 16 = 32 C = 9 + 15 + 4 = 28 Voters preferred B over C, B – receives 1 – point.

So, B – receives 2 – points, C – 1 point, A – 0 point. B – Is declared the winner.

Flaws of Voting Systems The outcome of an election may depend on the voting system that is used. Political scientists and mathematician have created a list of criteria that any fair voting system should meet.

But what exactly is meant by the word fair? Political scientists and mathematicians have created a list of criteria that any “fair” voting system should meet.

Four fairness criteria are: 1. The Majority criterion

The candidate X receives a majority of votes, should be declared the winner.

2. The Head-to-Head criterion If a candidate X is favored when compare head-to-head (individually) with each other candidates, then Candidate X should be declared the winner. Some cases plurality method can violate the head-to-head criterion

3. Monotonicity criterion If candidate X wins an election and, in subsequent election, the only changes are in favor of candidate X, then candidate X should be declared the winner. The instant runoff method can in fact violate the monotonicity criterion

4. The irrelevant Alternatives criterion If candidate X wins an election and, in a recount, the only changes are that one or more losing candidates are removed from the ballot, then candidate X should be declared the winner. The pairwise comparison method can violate the irrelevant alternatives criterion.

Each of the common systems of voting can be shown to violate at least one of the four fairness criterion.

Arrow’s Impossibility Theorem It is mathematically impossible to create any system of voting (involving three or more candidates) that satisfies all four fairness criteria.

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Methods of Apportionment

Congress

 Senate: composed of 2 – senators from each state.

 House of Representatives: to determine the number of representative for a particular state by the size of population.

The process of making this decision is called apportionment. (To divide according to a plan) Consequently, several methods of apportionment have been proposed, and several different

methods have actually been used since the first apportionment in 1790.

1. Hamilton’s Method. Was proposed by Alexander Hamilton (1757–1804) and was the first plan to be approved by Congress in 1790.

2. Jefferson’s Method. Was proposed by Thomas Jefferson (1743–1826)

3. Adam’s Method. Was proposed by John Quincy Adams (1767–1848) and has never been used.

4. Webster’s Method. Was proposed by Daniel Webster (1782–1852).

5. Hill-Huntington Method. Was proposed by Joseph Hill (1860–1938) and Edward Huntington (1874–1952), and it has been used following every census from 1940 to the present.

Standard Divisor (d)

The standard divisor – d – is define as: total population

d total number of seats

 (round to two decimal place)

Standard Quota (q)

The standard quota is define as: state population

q d

 (round to two decimal place)

Quota Rule

The number assigned to each represented unit must be either integer or the standard quota

rounded down to the nearest integer or the standard quota round up to the nearest integer.

Lower quota:

The lower quota (of special state) is the standard quota rounded down to a whole number.

Upper quota:

The upper quota (of special state) is the standard quota rounded up to a whole number.

A modified divisor

Is a number that is close to the standard divisor, and it is denoted by dm

A modified divisor may be less than or greater than the standard divisor

Modified Quota

The modified quota, m

q is the ratio of a state’s population to the modified divisor;

  stat's population stat's population

modified quota round to two decimal place modified divisor

m

m

q d

  

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1. Hamilton’s Method

Hamilton’s Method

Procedure:

1. Calculate the Standard Divisor.

2. Calculate each state’s Standard Quota.

3. Initially assign each state its Lower Quota.

4. If there are surplus seats, give them, one at a time, to states in descending order of the fractional parts of their Standard Quota.

Problems:

 The Alabama Paradox

An increase in the total number of seats to be apportioned causes a state to lose a seat.

 The Population Paradox

An increase in a state’s population can cause it to lose a seat.

 The New States Paradox

adding a new state with its fair share of seats can affect the number of seats due other

states.

Ex:

Use Hamilton’s method to apportioned 10 – seats among the three states. (Population in thousands)

201, 000

20100 20.1 ( ) 10

d Thousans   , 94700

4.71 20100

q  

State A B C Total

Population 94.7 72.6 33.7 201

Standard quota

using d = 20.1 94.7

4.71 20.1

q   72.6

3.61 20.1

q   33.7

1.68 20.1

q  

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Lower quota 4 3 1 8

Additional seats 1 0 1 2

Number of seats 5 3 2 10

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2. Jefferson’s Method

Procedure:

1. Calculate the Standard Divisor.

2. Calculate each state’s Standard Quota.

3. Initially assign each state its Lower Quota.

4. Check to see if the sum of the Lower Quotas is equal to the correct number of seats to be apportioned.

o If the sum of the Lower Quotas is equal to the correct number of seats to be

apportioned, then apportion to each state the number of seats equal to its Lower

Quota.

o If the sum of the Lower Quotas is NOT equal to the correct number of seats to

be apportioned, then, by trial and error, find a number, dm called the Modified

Divisor to use in place of the Standard Divisor so that when the Modified

Quota, m q

, for each state (computed by dividing each State's Population by dm

instead of d ) is rounded DOWN, the sum of all the rounded (down) Modified

Quotas is the exact number of seats to be apportioned. (Note: The dm will always

be smaller than the Standard Divisor.) These rounded (down) Modified Quotas

are sometimes called Modified Lower Quotas. Apportion each state its Modified

Lower Quota.

Problem:

 Violates the Quota Rule. (However, it can only violate Upper Quota—never Lower

Quota.)

Different methods of apportionment can lead to different allocations of legislative seats.

Jefferson’s method favors larger states while Hamilton’s method favors smaller states.

Ex:

Use Jefferson’s method to apportioned 10 – seats among the three states. (Population in thousands)

State A B C Total

Population 94.7 72.6 33.7 201

Standard quota, d = 20.1 94.7 4.71

20.1 q  

72.6 3.61

20.1 q  

33.7 1.68

20.1 q  

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Modified quota (dm = 19) 4.98 3.82 1.77

Lower modified quota 4 3 1 8

Modified quota (dm = 18) 5.26 4.03 1.87

Lower modified quota 5 4 1 10

The modified divisor dm = 19 still creates two surplus seats.

The modified divisor dm = 18 works.

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3. Adams’s Method

Adams's Method

Also known as the Method of Smallest Divisors.

Procedure:

1. Calculate the Standard Divisor.

2. Calculate each state’s Standard Quota.

3. Initially assign each state its Upper Quota.

4. Check to see if the sum of the Upper Quotas is equal to the correct number of seats to be apportioned.

o If the sum of the Upper Quotas is equal to the correct number of seats to be

apportioned, then apportion to each state the number of seats equal to its Upper

Quota.

o If the sum of the Upper Quotas is NOT equal to the correct number of seats to be

apportioned, then, by trial and error, find a number, dm, called the Modified

Divisor to use in place of the Standard Divisor so that when the Modified Quota,

for each state (computed by dividing each State's Population by dm instead

of d) is rounded UP, the sum of all the rounded (up) Modified Quotas is the exact

number of seats to be apportioned. (Note: The dm will always be larger than the

Standard Divisor.) These rounded (up) Modified Quotas are sometimes called

Modified Upper Quotas. Apportion each state its Modified Upper Quota.

Problem:

 Violates the Quota Rule. (However, it can only violate Lower Quota—never Upper

Quota.)

Ex:

Use Adam’s method to apportioned 10 – seats among the three states. (Population in thousands)

Using dm = 23.8 ( dm > d)

State A B C Total

Population 94.7 72.6 33.7 201

Standard quota, d = 20.1 94.7 4.71

20.1 q  

72.6 3.61

20.1 q  

33.7 1.68

20.1 q  

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Modified quota (dm = 23) 4.12 3.16 1.47

upper modified quota 5 4 2 11

Modified quota (dm =

23.8)

3.98 3.05 1.42

upper modified quota 4 4 2 10

The modified divisor dm = 23 still creates two surplus seats. The modified divisor dm = 23.8 works.

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4. Webster’s Method

Webster’s Method

Also known as the Webster-Willcox Method as well as the Method of Major Fractions.

Procedure:

1. Calculate the Standard Divisor.

2. Calculate each state’s Standard Quota.

3. Initially assign a state its Lower Quota if the fractional part of its Standard Quota is less than 0.5.

Initially assign a state its Upper Quota if the fractional part of its Standard Quota is

greater than or equal to 0.5. [In other words, round down or up based on the

arithmetic mean (average).]

4. Check to see if the sum of the Quotas (Lower and/or Upper from Step 3) is equal to the correct number of seats to be apportioned.

o If the sum of the Quotas (Lower and/or Upper from Step 3) is equal to the

correct number of seats to be apportioned, then apportion to each state the

number of seats equal to its Quota (Lower or Upper from Step 3).

o If the sum of the Quotas (Lower and/or Upper from Step 3) is NOT equal to the

correct number of seats to be apportioned, then, by trial and error, find a

number, dm, called the Modified Divisor to use in place of the Standard Divisor

so that when the Modified Quota, for each state (computed by dividing each

State's Population by dm instead of d) is rounded based on the arithmetic mean

(average) , the sum of all the rounded Modified Quotas is the exact number of

seats to be apportioned. Apportion each state its Modified Rounded Quota.

Problem:

 Violates the Quota Rule. (However, violations are rare and are usually associated

with contrived situations.)

Ex:

Use Webster’s method to apportioned 10 – seats among the three states. (Population in thousands)

Using dm = 21

State A B C Total

Population 94.7 72.6 33.7 201

Standard quota, d = 20.1 94.7 4.71

20.1 q  

72.6 3.61

20.1 q  

33.7 1.68

20.1 q  

10

Rounded quota 5 4 2 11

Modified quota (dm = 21) 4.51 3.46 1.60

Rounded quota 5 3 2 10

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5. Hill-Huntington Method Geometric Mean

Given two numbers x and y, the geometric mean of x and y, denoted by:

gm xy

Huntington-Hill Method

Also known as the Method of Equal Proportions.

 Current method used to apportion U.S. House

 Developed around 1911 by Joseph A. Hill, Chief Statistician of the Bureau of the

Census and Edward V. Huntington, Professor of Mechanics & Mathematics,

Harvard

 Preliminary terminology: The Geometric Mean

Procedure:

1. Calculate the Standard Divisor.

2. Calculate each state’s Standard Quota.

3. Initially assign a state its Lower Quota if the fractional part of its Standard Quota is less than the Geometric Mean of the two whole numbers that the Standard

Quota is immediately between.

Initially assign a state its Upper Quota if the fractional part of its Standard Quota

is greater than or equal to the Geometric Mean of the two whole numbers that the

Standard Quota is immediately between.

[In other words, round down or up based on the geometric mean.]

4. Check to see if the sum of the Quotas (Lower and/or Upper from Step 3) is equal to the correct number of seats to be apportioned.

o If the sum of the Quotas (Lower and/or Upper from Step 3) is equal to the

correct number of seats to be apportioned, then apportion to each state the

number of seats equal to its Quota (Lower or Upper from Step 3).

o If the sum of the Quotas (Lower and/or Upper from Step 3) is NOT equal

to the correct number of seats to be apportioned, then, by trial and error,

find a number, dm, called the Modified Divisor to use in place of the

Standard Divisor so that when the Modified Quota, for each state

(computed by dividing each State's Population by dm instead of d) is

rounded based on the geometric mean, the sum of all the rounded Modified

Quotas is the exact number of seats to be apportioned. Apportion each state

its Modified Rounded Quota.

Problem:

 Violates the Quota Rule.

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Ex.

Use Hill-Huntington method to apportioned 10 – seats among the three states. (Population in thousands)

4 5 20 4.47213 4.47 A

gm     

3 4 12 3.46 B

gm     1 2 2 1.4 C

gm    

State A B C Total

Population 94.7 72.6 33.7 201

Standard quota, d = 20.1 94.7 4.71

20.1 q  

72.6 3.61

20.1 q  

33.7 1.68

20.1 q  

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Lower quota 4 3 1 8

Upper quota 5 4 2 11

Geometric mean 4.47 3.46 1.41

Round quota 5 4 2 11

Modified quota (dm = 21.2) 4.47 3.42 1.59

Rounded quota 5 3 2 10

State A B C Total

Webster’s Apportionment 4 3 2 10

Adams’s Apportionment 5 4 2 10

Jefferson’s Apportionment 5 4 1

Hamilton’s Apportionment 5 3 2 10 Hill-Huntington’s Apportionment 5 3 2 10

Additional Seats

Once the seats of a legislature have been allocated, it might be decided that the size of the

legislature should be increased; that is, new seats might be added to an existing

apportionment. Who gets the new seats? The answer lies in the calculation of Hill-Huntington

numbers.

Hill-Huntington Number

The Hill-Huntington Number for a state, Denoted by HHN ,  

 

2 '

1

state s population HHN

n n 



Where: n = state’s current number of seats.

Ex:

The populations of A, B, and C and the apportionment of the ten seats via the Hill-Huntington

Method are given in the Figure

State A B C Total

Population 94.7 72.6 33.7 201

# of seats 5 3 2 10

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Suppose the high council decides to add an additional seat; (the council should now consist of

eleven seats). Use Hill-Huntington numbers to determine which realm should receive the new seat.

Beginning with A, we calculate the Hill-Huntington number for each realm:

2 2

94.7 94.7 298.936333... 298.93633

5(5 1) 30 A

HHN     

Similarly with B, we calculate the Hill-Huntington number for each realm:

2 2

72.6 72.6 439.23

3(3 1) 12 B

HHN    

Similarly with C, we calculate the Hill-Huntington number for each realm:

2 2

33.7 33.7 189.28

2(2 1) 6 C

HHN    

Because B, has the highest HHN (439.23 is greater than 298.23 or 189.28)

B, should receive the new seat.

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Doing an Apportionment Problem

1. Standard Divisor: (round to two decimal place) total population

d total number of seats



2. Standard Quota: (round to two decimal place) state population

q d



3. Round Quotas: Use the appropriate Rounding Rule to get the Rounded Quotas from the Standard Quotas:

 For HM, the Rounding Rule is to always round DOWN to Lower Quotas.

 For JM, the Rounding Rule is to always round DOWN to Lower Quotas.

 For AM, the Rounding Rule is to always round UP to Upper Quotas.

 For WM, the Rounding Rule is to round UP if the computed Quota is greater than or equal to the

arithmetic mean, and round DOWN if not.

 For HHM, the rounding rule is to round UP if the computed Quota is greater than or equal to the

geometric mean, and round DOWN if not.

4. Apportionment (for each state)  For HM: If Lower Quotas do not apportion enough seats, add the proper number of seats by giving

an additional seat to those states whose Divisors have the largest fractional parts.

 All other methods: If Rounded Quotas do not apportion enough seats, modify the divisor and use this

Modified Divisor to compute Modified Quotas and then go back to STEP 3 using these Modified

Quotas instead of the Standard Quotas.

Some Problems with Apportionments Methods • Violating Quota • Paradoxes

The Alabama Paradox An increase in the total number of seats to be apportioned causes a state to lose a seat.

The Alabama Paradox first surfaced after the 1870 census. With 270 members in the House of Representatives, Rhode Island got 2 representatives but when the House size was increased to 280, Rhode Island lost a seat. After the 1880 census. W. Seaton (chief clerk of U. S. Census Office) computed apportionments for all House sizes between 275 and 350 members. He then wrote a letter to Congress pointing out that if the House of Representatives had 299 seats, Alabama would get 8 seats but if the House of Representatives had 300 seats, Alabama would only get 7 seats.

The Population Paradox An increase in a state’s population can cause it to lose a seat.

The Population Paradox was discovered around 1900, when it was shown that a state could lose seats in the House of Representatives as a result of an increase in its population. (Virginia was growing much faster than Maine--about 60% faster--but Virginia lost a seat in the House while Maine gained a seat.)The New States Paradox: Adding a new state with its fair share of seats affects the number of seats apportioned to other states.

The New States Paradox Adding a new state with its fair share of seats can affect the number of seats due other states.

The New States Paradox was discovered in 1907 when Oklahoma became a state. Before Oklahoma became a state, the House of Representatives had 386 seats. Comparing Oklahoma's population to other states, it was clear that Oklahoma should have 5 seats so the House size was increased by five to 391 seats. The intent was to leave the number of seats unchanged for the other states. However, when the apportionment was mathematically recalculated, Maine gained a seat (4 instead of 3) and New York lost a seat (from 38 to 37).