FAIR DIVISION

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FAIR DIVISION
2
If 24 candies are to be divided among 4 students,
how many should each student receive?

• Six, of course, if the 24 candies are all alike
• What if they are different sizes?
• What if some candies appeal to some students but
  not to others?
• How can this be done FAIRLY??

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What is fairness?
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Lets start with a simpler problem

• How can we divide one item between 2 players?

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The last piece of pizza
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Envy free fair division

• Every one gets what they feel is the best piece

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Envy

• a painful or resentful awareness of an advantage
  enjoyed by another, joined with a desire to
  possess the same advantage.

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What is a fair share then?

• A share that any person feels is worth 1 / nth of
  the total

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Fair Division Schemes must

• 1. Be decisive
• 2. Be internal
• 3. Assume people are ignorant of each other.
• 4. Assume people behave rationally.

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Three types of problems

• 1. Continuous
• 2. Discrete
• 3. Mixed

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ADJUSTED WINNER PROCEDURE

• PLAYERS ASSIGN 100 POINTS TO THE ITEMS THAT ARE
  TO BE DIVIDED.
• The adjusted winner procedure is a means of
  allocating items or issues to two parties in an
  equitable manner.

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PROPERTIES

• 1. EQUITABLE ALLOCATIONS.
• 2. ENVY FREE ALLOCATIONS.
• 3. PARETO-OPTIMAL
• NO OTHER ALLOCATION CAN MAKE ONE PARTY BETTER
  OFF, AND ONE WORSE OFF.

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Suppose Mike and Phil are roommates in college,
and they encounter serious conflicts during their
first week of school. Their resident advisor has
taken Discrete Math, and decides to use the
adjusted winner procedure to resolve the dispute.
The issues agreed upon, and the independently
assigned points are the following
Issue Mike's Points Phil's Points
Stereo Level 4 22
Smoking Rights 10 20
Room Party Policy 50 25
Cleanliness 6 3
Alcohol Use 15 15
Phone Time 1 8
Lights-out time 10 2
Visitor Policy 4 5
Use the adjusted winner procedure to resolve this
dispute.
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Issue Mike's Points Phil's Points
Stereo Level 4 22 ?
Smoking Rights 10 20 ?
Room Party Policy 50 ? 25
Cleanliness 6 ? 3
Alcohol Use 15 15
Phone Time 1 8 ?
Lights-out time 10 ? 2
Visitor Policy 4 5 ?
  Mike's Total Phil's Total
  66 55
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Items which received the same number of points
are given to Phil, since he has the lower point
total (this will make Phil's point total higher
than Mike's).
Issue Mike's Points Phil's Points
Stereo Level 4 22 ?
Smoking Rights 10 20 ?
Room Party Policy 50 ? 25
Cleanliness 6 ? 3
Alcohol Use 15 15 ?
Phone Time 1 8 ?
Lights-out time 10 ? 2
Visitor Policy 4 5 ?
  Mike's Total Phil's Total
  66 70
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To decide which item to transfer, we need to take
into account the relative importance of the items
to the two. We can do this by taking the ratio of
Phil's points to Mike's for the the items Phil
has, and transferring items with the lowest point
ratios first. We always put the point value from
the person we are transferring something away
from (i.e., the one with the highest point total)
in the numerator this means the ratio should
always be greater than or equal to 1.
Issue Mike's Points Phil's Points Point Ratio
Stereo Level 4 22 ? 22/4 5.5
Smoking Rights 10 20 ? 20/10 2.0
Room Party Policy 50 ? 25  
Cleanliness 6 ? 3  
Alcohol Use 15 15 ? 15/15 1
Phone Time 1 8 ? 8/1 8.0
Lights-out time 10 ? 2  
Visitor Policy 4 5 ? 5/4 1.25
  Mike's Total Phil's Total  
  66 70  
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Solution

• Since the alcohol policy has the lowest point
  ratio, it is what we will transfer. If we
  transferred this entirely from Phil to Mike, Mike
  would have more points that Phil. So we only want
  to transfer part of the alcohol policy to Mike.
  Let's let x be the amount we want to transfer to
  Mike. So Mike will have x of the alcohol policy,
  and Phil will have the remaining (1-x) of the
  policy. Since we want their points to be equal,
  we have
• 6615x 55 15(1-x)
• Which we can solve for x 0.13. The points for
  Mike and Phil are both 68 (within rounding).
• Therefore, Mike gets to decide the room party
  policy, cleanliness, lights out time and will get
  13 say in the alcohol policy. Phil gets 87 say
  in the alcohol policy, and gets to decide the
  stereo level, smoking rights, phone time, and
  visitor policy.

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Knaster Inheritance

• a means of allocating items or issues to more
  than two parties in an equitable manner. The
  downside of this procedure is that it requires
  the parties to have wads of cash handy, which is
  used to buy out other members of the group. In
  Knaster inheritance, the parties assign a dollar
  value to each item they are going to divide.

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Three children must make fair division of a
painting and sculpture left to them by their
mother. The value (in dollars) each child places
on the objects is given below. These values
should be assigned by the three children
independently of each other.
Item Sasha Ralph Fergus
Painting 4000 6300 6000
Sculpture 2300 1800 2400
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Painting

• The high bidder is awarded the painting. In this
  case, it is Ralph. Since there are three
  children, however, Ralph's fair share is only 1/3
  of what he thinks the painting is worth, which is
  6300/3 2100. Therefore, he puts into a
  temporary kitty'' (pot of money) 2/3 of what he
  bid on the painting, which would be 4200 in this
  case.
• Each of the other children withdraws from the
  kitty 1/3 of the amount they bid on the painting,
  which would be their fair share of the painting
• Sasha 1/3 x 4000 1333.33.Fergus 1/3 x
  6000 2000.00.Kitty 4200 - 1333.33 -
  2000.00 866.67.
• At this point, every child feels that they have
  received 1/3 of the value of the painting, so the
  distribution of the painting is fair. However,
  the remaining money in the kitty is split 3 ways,
  and each child receives an additional 866.67/3
  288.89, so everyone walks away feeling they got
  288.89 more than their fair share!
• At this point, the fair division has dealt with
  the painting
• Sasha 1333.33 288.89 1622.22.Ralph
  painting - 4200.00 288.89 painting -
  3911.11. Fergus 2000.00 288.89 2288.89.

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Sculpture

• The high bidder is awarded the sculpture. In this
  case, it is Fergus. Since there are three
  children, however, Fergus's fair share is only
  1/3 of what he thinks the painting is worth,
  which is 2400/3 800. Therefore, he puts into
  the kitty 2/3 of what he bid on the sculpture,
  which is 1600.
• Each of the other children withdraws from the
  kitty 1/3 of the amount they bid on the
  sculpture, which would be their fair share of the
  sculpture
• Sasha 1/3 x 2300 766.67.Ralph 1/3 x 1800
  600.00.Kitty 1600 - 766.67 - 600.00
  233.33.
• The remaining money in the kitty is split 3 ways,
  and each child receives an additional 77.78.
• At this point, the fair division has dealt with
  the sculpture
• Sasha 766.67 77.78 844.45.Ralph 600.00
  77.78 677.78. Fergus sculpture - 1600.00
  77.78 sculpture - 1522.22.

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Final Distribution

• The final result of the fair division process is
  the following distribution
• Sasha 1622.22 844.45 2466.67Ralph
  painting - 3911.11 677.78 painting -
  3233.33. Fergus 2288.89 sculpture -
  1522.22 sculpture 766.67.
• Ralph needs to have 3233.23 on hand to pay off
  Sasha and Fergus. This is a drawback of the
  Knaster inheritance procedure, since it is quite
  possible that Ralph doesn't have that much money.

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2.2 Estate Division

• Two methods
• 1. Method of sealed bids
• 2. Method of markers

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1. Method of sealed bids

• Older and much used method
• Most lawyers are familiar with it

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variables

• N number of players
• S estate to be divided up

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Step 1 Bidding

• Each player produces a sealed bid with a amount
  attached to each item they believe fair share
  (1/n th of the total).

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Step 2 Allocation

• Each item in S goes to the highest bidder.
• If it exceeds what they thought it should go for,
  they must pay the difference.
• If it falls short, the others must chip in the
  difference.

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Step 3 Dividing up the surplus

• Surplus of cash should be equally divided among
  players

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Axiom (assumption)

• Each heir is equally capable of placing a value
  on the items

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Problems (due to human nature)

• 1. Cash flow
• Must have enough or one player could get it all
• 2. Priceless items
• No one can insist on getting a favorite item.
  Everyone has to be willing to get instead of an
  item
• 3. Players know each other
• Know what each other will bid for items

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Problem 1

• Bob, Ann, and Jane wish to dissolve their
  partnership using the method of sealed bids. Bob
  bids 240,000 for the partnership, Ann bids
  210,000, and Jane bids 225,000

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Step 1 Bids

• Bob Ann Jane240,000 210,000 225,000

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Step 2Allocation

• The business goes to Bob

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Step 3Payments

• Fair shares
• Bob 80,000,owes estate 160,000
• Ann 70,000, paid by estate
• Jane 75,000, paid by estate

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Step 4Dividing the Surplus

• Estate received 160,000
• Estate paid out 145,000 (70,000 75,000)
• Surplus 15,000 (160,000 - 145,000)
• Divide it evenly among the 3.

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Bottom line

• Bob gets business, pays out a total of 155,000 
• Anne gets 75,000
• Jane gets 80,000

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Method 2 Method of Markers

• This method gets around the problem of having to
  come up with money .
• Start the process by stringing out all the items
  in S in a long line.

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Steps

• Step 1. Bidding. Each player secretly divides the
  line of items into N segments, each of which she
  considers a fair share. This is easily done by
  positioning markers.
• Step 2. Allocation. Find the leftmost marker in
  the line, give the player whose marker it is
  everything to the left of it, and remove the rest
  of her markers. Then find the first marker in the
  second group of markers, give the player whose
  marker it is every thing between it and her first
  marker, and remove the rest of her markers.
  Continue the process until everybody has her fair
  share.
• Step 3. Dividing the Surplus. Again there will
  usually be some leftovers, and these can be
  distributed by chance. If there are many
  leftovers, we can even use the method of markers
  again.

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• Leftovers If there are more leftovers than
  players, use the Method of Markers again.
  Otherwise, use a lottery.
• Necessary conditions
• There must be many more items than Players.
• Every Player must be able to divide the array of
  items into segments of equal value.

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Problem 2

• Alice, Beth, and Carol want to divide what is
  left of a can of mixed nuts. There are 6 cashews,
  9 pecan halves, and 3 walnut halves to be
  divided. The women's value systems are as
  follows.
• Alice does not care at all about pecans of
  cashews but loves walnuts.
• Beth Likes walnuts twice as much as pecans and
  really does not care for cashews.
• Carol likes all the nuts but likes cashews twice
  as much as the others.

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• If the nuts are lined up as shown below, where
  would each woman, playing rationally, place her
  markers?
• P P W C P C P P P W W C P C P P C C

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• P P W C P C P P P W W C P C P P C
  C
• B B
• C C
• A A

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• Alice's marker is leftmost, so she gets two
  pecans, a walnut, and a cashew. She is satisfied
  with this take since it contains one-third of the
  total value (to her) of S. We give her her nuts,
  remove her markers, and send her home happy (or
  at least satisfied).
• The first marker in the second group of markers
• belongs to Beth, so she gets everything back to
  her first marker. Her take is a walnut, a cashew,
  and two pecans, which she considers a fair share.
  Notice that there is a pecan left over.
• Finally, Carol gets every thing to the right of
  her last marker--three cashews and two pecans,
  which is a fair share in her value system. Here
  we have lots of leftovers--a walnut, a cashew,
  and a pecan. That makes the total of the
  leftovers a walnut, a cashew, and two pecans
  after everybody has received what she considers a
  fair share! These leftovers can be distributed
  randomly as a bonus.

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Problem 3

• Three heirs (Andre, Bea, and Chad) wish to divide
  up an estate consisting of a house, a small farm,
  and a painting, using the method of sealed bids.

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Step 1The Bids

• Andre Bea Chad
• House 150,000 146,000 175,000
• Farm 430,000 425,000 428,000
• Painting 50,000 59,000 57,000

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Step 2The Allocation

• Chad gets the house
• Andre gets the farm
• Bea gets the painting

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Step 3The Payments

• Fair share Andre 210,000
• Bea 210,000
• Chad 220,000
• Chad gets 45,000 from the estate
• Andre pays the estate 220,000
• Bea gets 151,000 from the estate

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Step 4Dividing the leftovers

• Estate has 220,000, estate pays 196,000 
• Leftover 24,000 (each player gets 8,000)

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Problem 4

•       Amanda, Brian, and Charlene are heirs to an
  estate that includes a house, a boat, a car, and
  75,000 in cash. Each submits a bid for the
  house, boat, and car. Bids are summarized in the
  following table
•                          House             Boat
                 Car
• Amanda                100,000            3000
                5000
• Brian                    92,000            5000
                8000
• Charlene                89,000            4000
                9000

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2.3 Apportionment

• Special type of discrete fair division
• Indivisible objects divided among a set of
  players
• Objects are the same, but now the players are
  entitled to different size shares
• Legislative body
• Seats are objects and states are players
• Based on population

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Problem 1

• Mom has 50 identical pieces of candy to divide up
  among her 5 kids. The candy will be apportioned
  based on time spent on chores.

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Child Alan Betty Connie Doug Ellie Total
Min Worked 150 78 173 204 295 900
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Examine Alans situation.

• Alan spent 150 out of 900 minutes on chores
• 150/900 16.7
• 150 x
• 900 50

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Problem 2.

• Five planets have signed a peace treaty Alanos,
  Betta, Conii, Dugos, Ellisium. They decide to
  join forces and form an Intergalactic
  Confederation. They will be ruled by a congress
  of 50 delegates, and each of the planets will get
  delegates based on their population.

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Planet A B C D E Total
Pop. 150 78 173 204 295 900
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This problem led to the Great Compromise of the
Constitution

• Proposed solutions
• Hamilton
• Jefferson
• Quincy Adams
• Webster

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The first 2 methods were proposed after the 1790
census to determine seats for congress.

• Alexander Hamilton proposed his method, but
  Washington vetoed it (first veto ever!!). So
  Jeffersons method was adopted instead.
  Jeffersons was used for 50 years until it was
  replaced by Websters method, which in turn was
  replaced by Huntington-Hills in 1941.

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Hamiltons Method

• 1. Calculate each states standard quota.
• 2. Give each state its lower quota.
• 3. Give the surplus seats one at a time to the
  states with the largest decimals until they are
  gone.

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Problem 3

• The Congress of Williamsonium has 240 seats to be
  shared by 6 states.

state A B C D E F Total
pop 164,6000 693,6000 154,000 2,091,000 605,000 988,000 12,500,000
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Average

• 12500000/250 50 000

This is called the STANDARD DIVISOR
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STANDARD QUOTA

• STATE POPULATION
• STANDARD DIVISOR
• Ex A 1646000 32.92
• 50000

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LOWER QUOTA

• ROUND DOWN
• SO FOR STATE A 32.92 ? 32

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UPPER QUOTA

• ROUND UP
• FOR STATE A 32. 92 ? 33

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PROBLEM 1

• CALCULATE THE UPPER AND LOWER QUOTA FOR EACH
  PLANET.
• WHAT HAPPENS?

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PROBLEM 2

• USE HAMILTONS METHOD TO FIND THE APPORTIONMENT.

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QUOTA RULE

• YOU ARE EITHER LUCKY OR UNLUCKY!
• STATES APPORTIONMENT SHOULD BE ITS UPPER OR
  LOWER QUOTA ONLY.

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PROBLEMS WITH HAMILTONS METHOD

• 1. ALABAMA PARADOX

STATE POP ST.Q. APPORTIONMENT
A 940 9.4 10
B 9030 90.3 90
C 10030 100.3 100
BASED ON 200 SEATS AND POP 20000
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INCREASE THE SEATS TO 201
STATE POP ST.Q. APPORTIONMENT
A 940 9.45 9
B 9030 90.75 91
C 10030 100.80 101
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PROBLEM 2 POPULATION PARADOX

• IN 1900, STATE X COULD LOSE SEATS TO STATE Y EVEN
  IF THE POPULATION OF X HAD GROWN AT A FASTER RATE
  THAN Y.

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BASED ON 50 SEATS.
A B C D E TOTAL
POP 150 78 173 204 295 900
ST. Q. 8.3 4.3 9.61 11.3 16.38 50
LOWER Q. 8 4 9 11 16 48
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A B C D E TOTAL
POP 150 78 181 204 296 909
ST. Q. 8.25 4.29 9.96 11.22 16.28 50
LOWER Q. 8 4 9 11 16 48
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PROBLEM

• E LOSES ONE SEAT TO B, WHOSE POPULATION REMAINED
  THE SAME!!

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NEW STATES PARADOX

• A STATE ALREADY IN THE UNION CAN LOSE A SEAT WHEN
  A NEW STATE IS ADMITTED, EVEN IF THE HOUS SIZE
  INCREASED BY THE NUMBER OF NEW SEATS THE STATE
  RECEIVES

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JEFFERSONS METHOD

• USES AN ADJUSTED STANDARD DIVISOR AND THE UPPER
  QUOTA, BUT FOLLOWS SAME STEPS AS HAMILTONS METHOD

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PROBLEM 1 REVISIT THE PLANETS
state A B C D E F Total
pop 164,6000 693,6000 154,000 2,091,000 605,000 988,000 12,500,000
ST. Q 32.92 138.72 3.08 41.82 13.7 19.76 250
LOWER Q 32 138 3 41 13 19 246
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THIS IS NOT ENOUGH SEATS FILLED!

• TRY USING 49, 500 AS A STANDARD DIVISOR INSTEAD
  OF 50,000

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state A B C D E F Total
pop 164,6000 693,6000 154,000 2,091,000 605,000 988,000 12,500,000
MODIFIED Q 32.59 137.35 3.05 41.41 13.56 19.56 247.52
UPPER Q 33 138 4 42 14 20 250
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ADAMS METHOD

• ALWAYS ROUNDED UP!
• SO 12.217 GETS THE SAME OF SEATS AS 12.968!!

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WEBSTERS METHOD

• STEPS
• 1. FIND A MODIFIED DIVISOR d SUCH THAT THE TOTAL
  IS THE EXACT NUMBER OF SEATS.
• 2. ROUND THE CONVENTIONAL WAY.

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TRY USING 50,100
state A B C D E F Total
pop 164,6000 693,6000 154,000 2,091,000 605,000 988,000 12,500,000
MODIFIED Q 32.85 138.44 3.07 41.74 13.67 19.72 249.49
ROUND 33 138 3 42 14 20 250
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HUNTINGTON-HILLS METHOD

• TWO CLASSMATES AT HARVARD WHO THOUGHT THE OTHER
  METHODS WERE UNFAIR!

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HILL METHOD

• LIKE WEBSTERS, BUT USES A CUT OFF POINT.
• EX STATE WITH MOD. Q 3.48
• WEBSTER ROUND DOWN TO 3.5
• (3 4) / 2 3.5
• HILL CUT OFF POINT WOULD BE
• v a b v 3 4 3. 464 lt 3.48
• SO STATE GETS 4 SEATS!

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APPORTIONMENT IMPOSSIBILITY THEOREM

• ANY APPORTIONMENT THEOREM WILL ALLOW THE ALABAMA,
  THE POPULATION PARADOXES, OR VIOLATE THE QUOTA
  RULE!

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2.5 THE CONTINUOUS CASE

• The Divider-Chooser Method
• The Lone Divider Method
• The Lone Chooser Method
• The Last Diminisher Method

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Divider - Chooser Method

• One divides, the other chooses.
• Why is/isnt this fair?
• Which is better, to be the divider or the chooser?

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What if the quantity cant be divided?

• Example a brother and a sister inherit their
  parents home. Both want it.
• Solution 1 Sell it and split the profit.
• Solution 2 Each decides how much it is worth to
  them by, in effect, bidding on it. The highest
  bidder wins the house and then owes the other 50
  of his bid.

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What if there are more than two players involved?

• For example, three students are to divide a
  quantity of Coke fairly.

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Lone Divisor Method

• One player (X) divides the set into three groups
  he considers to be fair.
• The other two players (Y,Z) then declare which of
  the three groups they consider to be fair
  sections.
• The distribution is done as follows
• If Y and Z agree all are fair, each player takes
  a group.
• If Y and Z think two of the three are fair, X
  gets the third Y and Z each get one of those
  left they claim is fair.
• If Y and Z think only one of the three is fair, X
  is given one of the other two. The two remaining
  pieces are recombined Y and Z divide using
  Divider-Chooser

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Lone Chooser Method

• Suppose X,Y, and Z are three players.
• X and Y divide the set into two fair groups,
  using the Divider-Chooser method.
• Both X and Y divide their groups into 3 equal
  shares
• Z, the lone chooser, selects one of Xs three
  shares and one of Ys.

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Last Diminisher Method

• X first takes what he considers to be his fair
  share. He passes this on to Y.
• If Y thinks Xs share is OK, he passes it on to
  Z, etc.
• If Y thinks X has taken too much, he diminishes
  the share until it is fair, then passes it on.
• The last person to diminish the share takes it
• This is continued until there are two players
  left and they use the Divider-Chooser method.

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Moving Knife Method

• Consider dividing a loaf.
• An impartial party slowly begins moving a knife
  from left to right above the loaf.
• At any point, any player can call CUT when he
  thinks the knife will cut a fair share.
• Whoever calls CUT gets the piece.
• The process is continued until there are two
  players left.

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Final Problem

• Farmers Field Problem

93
How can a farmer divide his field into smaller
sections of equal area if he has...

• A triangular field and two children
• A triangular field and three children
• A quadrilateral field and two children
• A quadrilateral field and three children
• A field in any polygonal shape and any number of
  children
• Use the Geometers Sketchpad to solve this one...