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 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.
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
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
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♥
Mike's Total
Phil's Total
66
70
5/4 = 1.25
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.
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
Painting
Sasha Ralph Fergus
$4000 $6300
$6000
Sculpture $2300 $1800
$2400
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 = $2466.67
Ralph: 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 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
–

2. Priceless items
–

Must have enough $ or one player could get it all
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
240,000
Ann
210,000
Jane
225,000
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?
PPWCPCPPPWWCPCPPCC

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



House
Farm
Painting
150,000
430,000
50,000
Bea
146,000
425,000
59,000
Chad
175,000
428,000
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
Min
150
Worked
Betty
Connie Doug
Ellie
Total
78
173
295
900
204
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
pop
164,60 693,60 154,00 2,091,
00
00
0
000
E
F
Total
605,00 988,00 12,500
0
0
,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.
A
940
9.4
APPORTION
MENT
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.
APPORTION
MENT
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.
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
4
9
11
16
48
LOWE 8
R Q.
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
4
9
11
16
48
LOWE 8
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
state A
pop
B
C
D
E
F
Total
164,6 693,6 154,0 2,091 605,0 988,0 12,50
000 000 00
,000 00
00
0,000
ST. Q 32.92 138.7 3.08
2
41.82 13.7
19.76 250
LOW 32
ER Q
41
19
138
3
13
246
THIS IS NOT ENOUGH SEATS
FILLED!
• TRY USING 49, 500 AS A STANDARD
DIVISOR INSTEAD OF 50,000
state A
pop
B
C
D
E
F
Total
164,6 693,6 154,0 2,091 605,0 988,0 12,50
000 000 00
,000 00
00
0,000
MOD 32.59 137.3 3.05
IFIE
5
DQ
UPP 33
138 4
ER Q
41.41 13.56 19.56 247.5
2
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
pop
B
C
D
E
164, 693, 154, 2,09 605,
6000 6000 000 1,00 000
0
MOD 32.8 138. 3.07 41.7 13.6
IFIE 5
44
4
7
DQ
ROU 33
138 3
42
14
ND
F
Total
988, 12,5
000 00,0
00
19.7 249.
2
49
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...