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: Issue Mike's PointsPhil's Points Stereo Level422 Smoking Rights1020 Room Party Policy5025 Cleanliness63 Alcohol Use15 Phone Time18 Lights-out time102 Visitor Policy45 Use the adjusted winner procedure to resolve this dispute.

Issue Mike's PointsPhil's Points Stereo Level422 ♥ Smoking Rights1020 ♥ Room Party Policy50 ♥25 Cleanliness6 ♥3 Alcohol Use15 Phone Time18 ♥ Lights-out time10 ♥2 Visitor Policy45 ♥ Mike's TotalPhil's Total 6655

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 PointsPhil's Points Stereo Level422 ♥ Smoking Rights1020 ♥ Room Party Policy50 ♥25 Cleanliness6 ♥3 Alcohol Use1515 ♥ Phone Time18 ♥ Lights-out time10 ♥2 Visitor Policy45 ♥ Mike's TotalPhil's Total 6670

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 PointsPhil's PointsPoint Ratio Stereo Level422 ♥22/4 = 5.5 Smoking Rights1020 ♥20/10 = 2.0 Room Party Policy50 ♥25 Cleanliness6 ♥3 Alcohol Use1515 ♥15/15 = 1 Phone Time18 ♥8/1 = 8.0 Lights-out time10 ♥2 Visitor Policy45 ♥5/4 = 1.25 Mike's TotalPhil's Total 6670

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 = (1-x) Which we can solve for x = 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.

ItemSashaRalphFergus 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 = $ Fergus: 1/3 x $6000 = $ Kitty: $ $ $ = $ 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 $ more than their fair share! At this point, the fair division has dealt with the painting: Sasha: $ $ = $ Ralph: painting - $ $ = painting - $ Fergus: $ $ = $

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 = $ Ralph: 1/3 x $1800 = $ Kitty: $ $ $ = $ The remaining money in the kitty is split 3 ways, and each child receives an additional $ At this point, the fair division has dealt with the sculpture: Sasha: $ $77.78 = $ Ralph: $ $77.78 = $ Fergus: sculpture - $ $77.78 = sculpture - $

Final Distribution: The final result of the fair division process is the following distribution: Sasha: $ $ = $ Ralph: painting - $ $ = painting - $ Fergus: $ sculpture - $ = sculpture + $ Ralph needs to have $ 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 method Most lawyers are familiar with it

variables N = number of players S = 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: BobAnnJane 240,000210,000225,000

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

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 AndreBeaChad House150,000146,000175,000 Farm430,000425, ,000 Painting 50,00059,00057,000

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.

ChildAlanBettyConnieDougEllieTotal Min Worked

Examine Alan’s situation. Alan spent 150 out of 900 minutes on chores 150/900 = 16.7 % 150 = x

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.

PlanetABCDETotal Pop

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. stateABCDEFTotal pop164, , ,00 0 2,091, , , ,500,000

Average /250 = This is called the STANDARD DIVISOR

STANDARD QUOTA STATE POPULATION STANDARD DIVISOR Ex: A =

LOWER QUOTA ROUND DOWN SO FOR STATE A:  32

UPPER QUOTA ROUND UP FOR STATE A:  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: STATEPOPST.Q.APPORTION MENT A B C BASED ON 200 SEATS AND POP = 20000

INCREASE THE SEATS TO 201 STATEPOPST.Q.APPORTION MENT A B C

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.

ABCDETOTAL POP ST. Q LOWE R Q BASED ON 50 SEATS.

ABCDETOTAL POP ST. Q LOWE R Q

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 stateABCDEFTotal pop164, , ,0 00 2,091, , , ,50 0,000 ST. Q LOW ER Q

THIS IS NOT ENOUGH SEATS FILLED! TRY USING 49, 500 AS A STANDARD DIVISOR INSTEAD OF 50,000

stateABCDEFTotal pop164, , ,0 00 2,091, , , ,50 0,000 MOD IFIE D Q UPP ER Q

ADAM’S METHOD ALWAYS ROUNDED UP! SO GETS THE SAME # OF SEATS AS !!

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 stateABCDEFTotal pop164, , , 000 2,09 1, , , ,5 00,0 00 MOD IFIE D Q ROU ND

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 = < 3.48 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...