FAIR DIVISION

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??

What is fairness?

Let’s start with a simpler problem:

How can we divide one item between 2 players?

The last piece of pizza…

Envy free fair division

• Every one gets what they feel is the best piece

Envy:

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

What is a fair share then?

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

Fair Division Schemes must:

• 1. Be decisive

• 2. Be internal

• 3. Assume people are ignorant of each other.

• 4. Assume people behave rationally.

Three types of problems:

• 1. Continuous

• 2. Discrete

• 3. Mixed

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.

PROPERTIES

• 1. EQUITABLE ALLOCATIONS.

• 2. ENVY FREE ALLOCATIONS.

• 3. PARETO-OPTIMAL:

NO OTHER ALLOCATION CAN MAKE ONE PARTY BETTER OFF, AND ONE WORSE OFF.

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:

IssueMike'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.

IssueMike'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

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).

IssueMike'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

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.

IssueMike'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  

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:

• 66+15x = 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.

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.

Item Sasha Ralph Fergus

Painting $4000 $6300 $6000

Sculpture $2300 $1800 $2400

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.

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.

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.

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.

2.2 Estate Division

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

1. Method of sealed bids

Older and much used methodMost lawyers are familiar with it

variables

N = number of playersS = estate to be divided up

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).

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.

Step 3: Dividing up the surplus

Surplus of cash should be equally divided among players

Axiom (assumption)

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

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

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

Step 1: Bids:

Bob Ann Jane240,000 210,000225,000

Step 2:Allocation:

The business goes to Bob

Step 3:Payments:

Fair shares: Bob = $80,000,owes estate $160,000 Ann = $70,000, paid by estate Jane = $75,000, paid by estate

Step 4:Dividing 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.

Bottom line:

Bob gets business, pays out a total of $155,000 

Anne gets $75,000 Jane gets $80,000

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.

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.

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.

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.

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

P P W C P C P P P W W C P C P P C C

B B

C C

A A

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.

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.

Step 1:The Bids

Andre BeaChad

House 150,000 146,000 175,000 Farm 430,000 425,000 428,000 Painting 50,000 59,000 57,000

Step 2:The Allocation

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

Step 3:The 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

Step 4:Dividing the leftovers

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

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

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

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.

Child Alan Betty Connie Doug Ellie Total

Min

Worked

150 78 173 204 295 900

Examine Alan’s situation.

Alan spent 150 out of 900 minutes on chores

150/900 = 16.7 %

150 = x

900 50

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.

Planet A B C D E Total

Pop. 150 78 173 204 295 900

This problem led to the Great Compromise of the Constitution

Proposed solutions:– Hamilton– Jefferson– Quincy Adams– Webster

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 Jefferson’s method was adopted instead. Jefferson’s was used for 50 years until it was replaced by Webster’s method, which in turn was replaced by Huntington-Hill’s in 1941.

Hamilton’s Method

1. Calculate each state’s 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.

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

Average

12500000/250 = 50 000

This is called the STANDARD DIVISOR

STANDARD QUOTA

STATE POPULATION

STANDARD DIVISOR

Ex: A 1646000 = 32.92

50000

LOWER QUOTA

ROUND DOWN SO FOR STATE A: 32.92 32

UPPER QUOTA

ROUND UP FOR STATE A: 32. 92 33

PROBLEM 1

CALCULATE THE UPPER AND LOWER QUOTA FOR EACH PLANET.

WHAT HAPPENS?

PROBLEM 2

USE HAMILTON’S METHOD TO FIND THE APPORTIONMENT.

QUOTA RULE

YOU ARE EITHER LUCKY OR UNLUCKY!

STATE’S APPORTIONMENT SHOULD BE ITS UPPER OR LOWER QUOTA ONLY.

PROBLEMS WITH HAMILTON’S 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

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

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.

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

BASED ON 50 SEATS.

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

PROBLEM

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

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

JEFFERSON’S METHOD

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

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

THIS IS NOT ENOUGH SEATS FILLED!

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

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

ADAM’S METHOD

• ALWAYS ROUNDED UP!

• SO 12.217 GETS THE SAME # OF SEATS AS 12.968!!

WEBSTER’S METHOD

• STEPS:

• 1. FIND A MODIFIED DIVISOR d SUCH THAT THE TOTAL IS THE EXACT NUMBER OF SEATS.

• 2. ROUND THE CONVENTIONAL WAY.

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

HUNTINGTON-HILL’S METHOD

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

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

√ a ·b = √ 3 ·4 = 3. 464 < 3.48

SO STATE GETS 4 SEATS!

APPORTIONMENT IMPOSSIBILITY THEOREM

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

2.5 THE CONTINUOUS CASE

• The Divider-Chooser Method

• The Lone Divider Method

• The Lone Chooser Method

• The Last Diminisher Method

Divider - Chooser Method

• One divides, the other chooses.

• Why is/isn’t this fair?

• Which is better, to be the divider or the

chooser?

What if the quantity can’t 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.

What if there are more than two players involved?

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

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

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 X’s three shares and one of Y’s.

Last Diminisher Method

• X first takes what he considers to be his fair share. He passes this on to Y.

• If Y thinks X’s 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.

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.

Final Problem:

Farmer’s Field Problem

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 Geometer’s Sketchpad to solve this one...