FAIR DIVISION

FAIR DIVISION

Topics in Contemporary Mathematics

MA 103

Summer II, 2013

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Content



Fair-Division Games



Two Players: The Divider-Chooser Method



The Lone-Divider Method



The Lone-Chooser Method



The Last-Diminsher Method



The Method of Sealed Bids



The Method of Markers

2

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Basic Elements of a Fair-Division Game

The underlying elements of every fair-division game are as follows:

The goods (or “ booty ” ): This is the informal name we will give to the item or items being divided. Example cake, art, land, Chores, e.t.c.

We will use the symbol S to denote the booty.

The players: In every fair-division game there is a set of parties with the right (or in some cases the duty) to share S. They are the players in the game. Players in a fair-division game could be individuals, ethnic groups, political parties, e.t.c.

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Basic Elements of a Fair-Division Game

The value systems: The fundamental assumption we will make is that each player has an internalized value system that gives the player the ability to quantify the value of the booty or any of its parts.

This means that each player can look at the set S or any subset of S and assign to it a value – either in absolute terms ( “ to me, that ʼ s worth

$147.50

” ) or in relative terms ( “ to me, that piece is worth 30% of the total value of S ” )

5

Basic Assumptions

For the purposes of our discussion, we will make the following four assumptions:

Rationality

Each of the players is a thinking, rational entity seeking to maximize his or her share of the booty S. We will further assume that in the pursuit of this goal, a player ʼ s moves are based on reason alone (we are taking emotion, psychology, mind games, and all other nonrational elements out of the picture.)

6

Basic Assumptions

Cooperation

The players are willing participants and accept the rules of the game as binding. The rules are such that after a finite number of moves by the players, the game terminates with a division of S. (There are no outsiders such as judges or referees involved in these games – just the players and the rules.)

7

Basic Assumptions

Privacy

Players have no useful information on the other players ʼ value systems and thus of what kinds of moves they are going to make in the game. (This assumption does not always hold in real life, especially if the players are siblings or friends.)

8

Basic Assumptions

Symmetry

Players have equal rights in sharing the set S. A consequence of this assumption is that, at a minimum, each player is entitled to a proportional share of S – then there are two players, each is entitled to at least one-half of S, with three players each is entitled to at least onethird of S, and so on.

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Fair Share

Given the booty S and players P

1

, P

2

, P

3

,…, P

N

, each with his or her own value system, the ultimate goal of the game is to end up with a fair division of S.

Suppose that s denotes a share of the booty S and that P is one of the players in a fair-division game with N players. We will say that s is a fair share to player P if s is worth at least 1/Nth of the total value of S in the opinion of P. (Such a share is often called a proportional fair share, but for simplicity we will refer to it just as a fair share.)

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Fair Division Methods

We can think of a fair-division method as the set of rules that define how the game is to be played.

Thus, in a fair-division game we must consider not only the booty S and the players P

1

, P

2

, P

3

,…, P

N

(each with his or her own opinions about how S should be divided), but also a specific method by which we plan to accomplish the fair division.

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Fair Division Methods

There are many different fair-division methods known, but we will only discuss a few of the classic ones.

Depending on the nature of the set S, a fair- division game can be classified as one of three types: continuous, discrete, or mixed, and the fair-division methods used depend on which of these types we are facing.

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Types of Fair Division Games

Continuous

In a continuous fair-division game the set S is divisible in infinitely many ways, and shares can be increased or decreased by arbitrarily small amounts.

Typical examples of continuous fair-division games involve the division of land, a cake, a pizza, and so forth.

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Discrete

A fair-division game is discrete when the set S is made up of objects that are indivisible like paintings, houses, cars, boats, jewelry, and so on.

One might argue that with a sharp enough knife a piece of candy could be chopped up into smaller and smaller pieces. As a semantic convenience let ʼ s agree that candy is indivisible, and therefore dividing candy is a discrete fair-division game.

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Mixed

A mixed fair-division game is one in which some of the components are continuous and some are discrete.

Dividing an estate consisting of jewelry, a house, and a parcel of land is a mixed fair-division game.

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Fair-division methods are classified according to the nature of the problem involved. Thus, there are:

 discrete fair-division methods (used when the set S is made up of indivisible, discrete objects),

 continuous fair-division methods (used when the set S is an infinitely divisible, continuous set) and

 mixed fair-division games (used when the set S is made up of continuous and discrete parts separately).

Divider-Chooser Method

The divider-chooser method is undoubtedly the best known of all continuous fair-division methods.

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This method can be used anytime the fair-division game involves two players and a continuous set S (a cake, a pizza, a plot of land, etc.).

Most of us have unwittingly used it at some time or another, and informally it is best known as the you cut–I choose method.

As this name suggests, one player, called the divider, divides S into two shares, and the second player, called the chooser, picks the share he or she wants, leaving the other share to the divider.

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Divider-Chooser Method

Under the rationality and privacy assumptions, this method guarantees that both divider and chooser will get a fair share (with two players, this means a share worth 50% or more of the total value of

S

).

Not knowing the chooser ʼ s likes and dislikes (privacy assumption), the divider can only guarantee himself a 50% share by dividing

S

into two halves of equal value (rationality assumption); the chooser is guaranteed a 50% or better share by choosing the piece he or she likes best.

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Example 1

Damian and Cleo buy jointly a raffle ticket, and as luck would have it, they win a half chocolate–half strawberry cheesecake.

Damian likes chocolate and strawberry equally well, so in his eyes the chocolate and strawberry halves are equal in value.

On the other hand, Cleo hates chocolate–– so in her eyes the value of the cake is 0% for the chocolate half and 100% for the strawberry part

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Example 1

Once again, to ensure a fair division, we will assume that neither of them knows anything about the other ʼ s likes and dislikes.

Damian volunteers to go first (the divider). He cuts the cake in a perfectly rational division of the cake based on his value system–each piece is half of the cake and to him worth one-half of the total value of the cake.

It is now Cleo ʼ s turn to choose, and her choice is obvious she will pick the piece having the larger strawberry part.

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Example 1

The final outcome of this division is that Damian gets a piece that in his own eyes is worth exactly half of the cake, but Cleo ends up with a much sweeter deal–a piece that in her own eyes is worth about two-thirds of the cake.

This is, nonetheless, a fair division of the cake–both players get pieces worth 50% or more.

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Lone-Divider Method

The first important breakthrough in the mathematics of fair division came in 1943, when Steinhaus came up with a clever way to extend some of the ideas in the divider-chooser method to the case of three players, one of whom plays the role of the divider and the other two who play the role of choosers.

Steinhaus ʼ approach was subsequently generalized to any number of players N (one divider and N – 1 choosers) by Princeton mathematician

Harold Kuhn.

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Lone-Divider Method for Three Players

Preliminaries

One of the three players will be the divider; the other two players will be choosers. Since it is better to be a chooser than a divider, the decision of who is what is made by a random draw (rolling dice, drawing cards from a deck, etc.).

We ʼ ll call the divider D and the choosers C

1

and C

2

.


Step 1 (Division)

The divider D divides the cake into three shares (s

1

, s

2

, and s

3

).

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D will get one of these shares, but at this point does not know which one.

Not knowing which share will be his is critical – it forces D to divide the cake into three shares of equal value.

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Step 2 (Bidding)

C

1

declares (usually by writing on a slip of paper) which of the three pieces are fair shares to her. Independently, C the bids.

2

does the same. These are

A chooser ʼ s bid must list every single piece that he or she considers to be a fair share (i.e., worth one-third or more of the cake)–it may be tempting to bid only for the very best piece, but this is a strategy that can easily backfire.

To preserve the privacy requirement, it is important that the bids be made independently, without the choosers being privy to each other ʼ s bids.

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Step 3 (Distribution)

Who gets which piece? The answer, of course, depends on which pieces are listed in the bids. For convenience, we will separate the pieces into two types:



C-pieces (these are pieces chosen by either one or both of the choosers) and



U-pieces (these are unwanted pieces that did not appear in either of the bids).

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Step 3 (Distribution) continued

Expressed in terms of value, a U-piece is a piece that both choosers value at less than 33 1/3% of the cake, and a C-piece is a piece that at least one of the choosers (maybe both) value at 33 1/3% or more.

Depending on the number of C-pieces, there are two separate cases to consider.

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Case 1

When there are two or more C-pieces, there is always a way to give each chooser a different piece from among the pieces listed in her bid. (The details will be covered later.)

Once each chooser gets her piece, the divider gets the last remaining piece. At this point every player has received a fair share, and a fair division has been accomplished.

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Case 1

(Sometimes we might end up in a situation in which C

1

likes C

2

ʼ s piece better than her own and vice versa. In that case it is perfectly reasonable to add a final, informal step and let them swap pieces–this would make each of them happier than they already were.)


Case 2

When there is only one

C

-piece, we have a bit of a problem because it means that both choosers are bidding for the very same piece.

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The solution requires a little more creativity. First, we take care of the divider D–to whom all pieces are equal in value– by giving him one of the pieces that neither chooser wants.


Case 2

After D gets his piece, the two pieces left (the Cpiece and the remaining U-piece) are recombined into one piece that we call the B-piece.

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We can revert to the divider-chooser method to finish the fair division: one player cuts the B-piece into two pieces; the other player chooses the piece she likes better.

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Case 2

This process results in a fair division of the cake because it guarantees fair shares for all players.

We know that D ends up with a fair share by the very fact that D did the original division, but what about C

1

and C

2

?

The key observation is that in the eyes of both C

1

and C

2

the B-piece is worth more than two-thirds of the value of the original cake (think of the

B-piece as 100% of the original cake minus a U-piece worth less than

33 1/3%), so when we divide it fairly into two shares, each party is guaranteed more than one-third of the original cake.

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Example 2: Lone Divider with 3 Players (Case 1, Version 1)

Dale, Cindy, and Cher are dividing a cake using Steinhaus ʼ s lone-divider method.

They draw cards from a well-shuffled deck of cards, and Dale draws the low card (bad luck!) and has to be the divider.


Dale divides the cake into three pieces s

1

, s

2

, and s

3

.

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Table shows the values of the three pieces in the eyes of each of the players.


We can assume that Cindy ʼ s bid list is {s

1

{s

1

, s

3

}.

, s

3

} and Cher ʼ s bid list is also

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The C-pieces are s

1

and s

3

. There are two possible distributions. One distribution would be: Cindy gets s

1

, Cher gets s

3

, and Dale gets s

2

. s

An even better distribution (the optimal distribution) would be: Cindy gets

3

, Cher gets s

1

, and Dale gets s

2

.

In the case of the first distribution, both Cindy and Cher would benefit by swapping pieces, and there is no rational reason why they would not do so.

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Thus, using the rationality assumption, we can conclude that in either case the final result will be the same:

Cindy gets s

3

, Cher gets s

1

, and Dale gets s

2

.

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We ʼ ll use the same setup as in Example 2–- Dale is the divider,

Cindy and Cher are the choosers.



1

, s

2

, and s

3

.

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Here Cindy ʼ s bid list is {s

2

} only, and Cher ʼ s bid list is {s

1

} only.

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This is the simplest of all situations, as there is only one possible distribution of the pieces: Cindy gets s gets s

3

.

2

, Cher gets s

1

, and Dale

Example 4: Lone Divider with 3 Players (Case 2)



1

, s

2

, and s

3

.

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Here Cindy ʼ s and Cher ʼ s bid list consists of just {s

3

}.

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The only C-piece is s

3

. Cindy and Cher talk it over, and without giving away any other information agree that of the two U-pieces, s

1

is the least desirable, so they all agree that Dale gets s

1

.

Dale doesn ʼ t care which of the three pieces he gets, so he has no rational objection.

The remaining pieces (s method.

2

and s

3

) are then recombined to form the Bpiece, to be divided between Cindy and Cher using the divider-chooser


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One of them divides the B-piece into two shares, the other one chooses the share she likes better.

Regardless of how this plays out, both of them will get a very healthy share of the cake:

Cindy will end up with a piece worth at least 40% of the original cake

(the B-piece is worth 80% of the original cake to Cindy), and Cher will end up with a piece worth at least 45% of the original cake

(the B-piece is worth 90% of the original cake to Cher).

The Lone-Divider Method for More Than Three

Players

42

The first two steps (i.e division and bidding) of this method are a straightforward generalization of Steinhaus’s lone-divider method for three players, but the distribution step requires some fairly sophisticated mathematical ideas and is rather difficult to describe in full generality, so we will only give an outline here and will illustrate the details with a couple of examples for N = 4 players.


Players

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Preliminaries

One of the players is chosen to be the divider D, and the remaining N – 1 players are all going to be choosers.

As always, it’s better to be a chooser than a divider, so the decision should be made by a random draw.

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Players


The divider D divides the set S into N shares s

1

, s

2

, s

3

, ..., s

N

.

D is guaranteed of getting one of these shares, but doesn’t know which one.

45


Players


Each of the N – 1 choosers independently submits a bid list consisting of every share that he or she considers to be a fair share

(i.e., worth 1/Nth or more of the booty S).

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Players


The bid lists are opened. Much as we did with three players, we will have to consider two separate cases, depending on how these bid lists turn out.

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Players

Case 1.

If there is a way to assign a different share to each of the N – 1 choosers, then that should be done. (Needless to say, the share assigned to a chooser should be from his or her bid list.)

The divider, to whom all shares are presumed to be of equal value, gets the last unassigned share.

At the end, players may choose to swap pieces if they want.

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Players

Case 2.

There is a standoff- in other words, there are two choosers both bidding for just one share, or three choosers bidding for just two shares, or K choosers bidding for less than K shares.

This is a much more complicated case, and what follows is a rough sketch of what to do. To resolve a standoff, we first set aside the shares involved in the standoff from the remaining shares.

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Players

Case 2.

Likewise, the players involved in the standoff are temporarily separated from the rest.

Each of the remaining players (including the divider) can be assigned a fair share from among the remaining shares and sent packing.

All the shares left are recombined into a new booty S to be divided among the players involved in the standoff, and the process starts all over again.

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We have one divider, Demi, and three choosers, Chan, Chloe, and Chris.

Step 1 (Division) Demi divides the cake into four shares; s

1 and s

4

.

, s

2

, s

3

,


Chan’s bid list is {s

1

, s

3

};

Chloe’s bid list is {s

3

} only;

Chris’s bid list is {s

1

, s

4

}.

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The bid lists are opened. It is clear that for starters Chloe must get s

3 there is no other option.

–

This forces the rest of the distribution: s

1

must then go to Chan, and s goes to Chris. Finally, we give the last remaining piece, s

2

, to Demi.

4

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This distribution results in a fair division of the cake, although it is not entirely “envy-free”

Chan wishes he had Chloe’s piece (35% is better than 30%) but Chloe is not about to trade pieces with him, so he is stuck with s

1

.

(From a strictly rational point of view, Chan has no reason to gripe–he did not get the best piece, but got a piece worth 30% of the total, better than the 25% he is entitled to.)

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Step 2 (Bidding)

Chan’s bid list is {s

4

};

Chloe’s bid list is {s

2

, s

3

} only;

Chris’s bid list is {s

4

}.

54



The bid lists are opened, and the players can see that there is a standoff brewing on the horizon–Chan and Chris are both bidding for s

4

. s

The first step is to set aside and assign Chloe and Demi a fair share from s

1

2

, s

2

, and s

4

. Chloe could be given either s

,of course, but it’s not for her to decide.)

2

or s

3

. (She would rather have

55


Step 3 (Distribution) s

A coin toss is used to determine which one. Let’s say Chloe ends up with

3

.

Demi could be now given either s up with s

1

.

1

or s

2

. Another coin toss, and Demi ends

The final move is to recombine s

2

and s

4

into a single piece to be divided between Chan and Chris using the divider-chooser method.

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Since (s

2

+ s

4

) is worth 60% to Chan and 58% to Chris, regardless of how this final division plays out they are both guaranteed a final share worth more than 25% of the cake.

Mission accomplished! We have produced a fair division of the cake.

The Lone-Chooser Method

This is also an extension of the divider-chooser method and was proposed by a mathematician A.M.Fink in 1964.

57

In this method one player plays the role of chooser, all the other players start out playing the role of dividers.

For this reason, the method is known as the lone-chooser method.

The Lone-Chooser Method for 3 Players

58

Preliminaries

We have one chooser and two dividers.

Let’s call the chooser C and the dividers D

1 who is what by a random draw.

and D

2

. As usual, we decide

59



D

1

and D

2

divide S between themselves into two fair shares. To do this, they use the divider-chooser method. Let ʼ s say that D ends up with s

2

.

1

ends up s

1

and D

2


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Step 2 (Subdivision)

Each divider divides his share into three subshares. Thus, D

1 into three subshares, which we will call s

1a divides s

2

, s

1b

, and s

into three subshares, which we will call s

1c

2a

, s

2b

divides s

. Likewise, D

, and s

2c

.

2

1

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Step 3 (Selection)

The chooser C now selects one of D

1

ʼ s three subshares and one of D

2

ʼ s three subshares (whichever she likes best).

These two subshares make up C ʼ s final share. D

1 remaining two subshares from s

1

,

then keeps the and D

2

keeps the remaining two subshares from s

2.


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Why is this a fair division of S? D

1

ends up with two-thirds of s least one- third–a fair share. The same argument applies to D

2

1

.

. To D is worth at least one-half of the total value of S, so two-thirds of s

1

1

, s

is at

1

What about the chooser’s share? We don ʼ plus a one-third or better share of s

2

(s

1

+ s

2

) and thus a fair share of the cake.

t know what s

1

and s

2

are each worth to C, but it really doesn’t matter– a one-third or better share of s

1

equals a one-third or better share of

63

Example 7: Lone Chooser with 3 Players

David, Dinah, and Cher are dividing an orange-pineapple cake using the lone-chooser method. The cake is valued by each of them at $27, so each of them expects to end up with a share worth at least $9.

■ David likes pineapple and orange the same.

■ Dinah likes orange but hates pineapple.

■ Cher likes pineapple twice as much as she likes orange.


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After a random selection, Cher gets to be the chooser.

Step 1 (Division)

David and Dinah start by dividing the cake between themselves using the divider-chooser method. After a coin flip, David cuts the cake into two pieces.


65

Step 1 (Division) continued

Since Dinah doesn’t like pineapple, she will take the share with the most orange.


David divides his share into three subshares that in his opinion are of equal value (all the same size).

Note: O is used to denote degrees.


66

Step 2 (Subdivision) continued

Dinah also divides her share into three smaller subshares that in her opinion are of equal value.

(Remember that Dinah hates pineapple. Thus, she has made her cuts in such a way as to have one-third of the orange in each of the subshares.)


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It’s now Cher’s turn to choose one sub-share from David’s three and one subshare from Dinah’s three. She will choose one of the two pineapple wedges from

David’s subshares and the big orange-pineapple wedge from

Dinah’s subshares.


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The final fair division of the cake is shown. David gets a final share worth

$9, Dinah gets a final share worth $12, and Cher gets a final share worth

$14. David is satisfied, Dinah is happy, and Cher is ecstatic.

The Lone-Chooser Method for N Players

69

In the general case of N players, the lone-chooser method involves one chooser C and N – 1 dividers D

1

, D

2

,…, D

N–1

.

As always, it is preferable to be a chooser than a divider, so the chooser is determined by a random draw.

The method is based on an inductive strategy. If you can do it for three players, then you can do it for four players; if you can do it for four, then you can do it for five; and we can assume that we can use the lonechooser method with N – 1 players.


70


D

1

, D

2

,…, D

N–1

divide fairly the set S among themselves, as if C didn’t exist. This is a fair division among N – 1 players, so each one gets a share he or she considers worth at least of 1/(N –1)th of S.


Each divider subdivides his or her share into N sub-shares.

71



The chooser C finally gets to play. C selects one sub-share from each divider – one subshare from D

1

, one from D

2

, and so on.

At the end, C ends up with N – 1 subshares, which make up C’s final share, and each divider gets to keep the remaining N – 1 subshares in his or her subdivision.

When properly played, the lone-chooser method guarantees that everyone, dividers and chooser alike, ends up with a fair share.

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The Last-Diminisher Method

The last-diminisher method was proposed by Polish mathematicians

Stefan Banach and Bronislaw Knaster in the 1940s.

The basic idea behind this method is that throughout the game, the set S is divided into two pieces: a piece currently “owned” by one of the players ( we will call that piece the C-piece and the player claiming it the “claimant ” ) and the rest of S, ”owned” jointly by all the other players. We will call this latter piece the R-piece and the remaining players the “non-claimants .”


The tricky part of this method is that the entire arrangement is temporary –

73 as the game progresses, each non-claimant has a chance to become the current claimant (and bump the old claimant back into the non-claimant group) by making changes to the C-piece and consequently to the Rpiece.

Thus, the claimant, the non-claimants, the C-piece, and the R-piece all keep changing throughout the game.

74


Preliminaries

Before the game starts the players are randomly assigned an order of play (this can be done by rolling a pair of dice).

We will assume that P

1

plays first, P

2

second,…, P will play in this order throughout the game.

N

last, and the players

The game is played in rounds, and at the end of each round there is one fewer player and a smaller S to be divided.

75


Round 1

P

1

kicks the game off by “cutting” for herself a 1/Nth share of S (i.e., a share whose value equals 1/Nth of the value of S).

This will be the current C-piece, and P

1

is its claimant.

P

1

does not know whether or not she will end up with this share, so she must be careful that her claim is neither too small (in case she does) nor too large (in case someone else does).


Round 1

P

2

comes next and has a choice: pass (remain a nonclaimant) or diminish the C-piece into a share that is a 1/Nth share of S.

76

Obviously, P

2

can be a diminisher only if he thinks that the value of the current C-piece is more than 1/Nth the value of S.

If P

2

diminishes, several changes take place: P

2 claimant; P

1

becomes the new

is bumped back into the nonclaimant group.

77

The Last-Diminisher Method

Round 1 the old R -piece to form a new, larger best to visualize it).

R -piece (there is a lot going on here and it ʼ s

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The Last-Diminisher Method

Round 1

P

3

comes next and has exactly the same opportunity as P

2

: pass or diminish the current C-piece.

If P

3

passes, then there are no changes and we move on to the next player. If P

3

diminishes (only because in her value system the current Cpiece is worth more than 1/Nth of S), she does so by trimming the Cpiece to a 1/Nth share of S.

The rest of the routine is always the same: The trimmed piece is added to the R-piece, and the previous claimant (P nonclaimant group.

1

or P

2

) is bumped back into the

79


Round 1

The round continues this way, each player in turn having an opportunity to pass or diminish.

The last player P

N

can also pass or diminish, but if he chooses to diminish, he has a certain advantage over the previous players–knowing that there is no player behind him who could further diminish his claim.

In this situation if P

N

chooses to be a diminisher, the logical move would be to trim the tiniest possible sliver from the current C-piece–a sliver so small that for all practical purposes it has zero value.

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Round 1

(Remember that a player’s goal is to maximize the size of his or her share.) We will describe this move as “trimming by 0%,” although in practice there has to be something trimmed, however small it may be.

At the end of Round 1, the current claimant, or last diminisher , gets to keep the C-piece (it’s her fair share) and is out of the game. The remaining players (the nonclaimants) move on to the next round, to divide the R-piece among themselves.

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Round 1

At this point everyone is happy–the last diminisher got his or her claimed piece, and the non-claimants are happy to move to the next round, where they will have a chance to divide the R-piece among themselves.

82


Round 2

The R-piece becomes the new S, and a new version of the game is played with the new S and the N – 1 remaining players.

This means that the new standard for a fair share is a value of

1/(N–1)th or more of the new S.

At the end of this round, the last diminisher gets to keep the current Cpiece and is out of the game.

83


Rounds 3, 4, and so on

Repeat the process, each time with one fewer player and a smaller S, until there are just two players left.

At this point, divide the remaining piece between the final two players using the divider-chooser method.

84

Continuous versus Discrete

We will now discuss discrete fair-division methods – methods for dividing a booty S consisting of indivisible objects.

Discrete fair division is harder to achieve than continuous fair division because there is a lot less flexibility in the division process, and discrete fair divisions that are truly fair are only possible under a limited set of conditions.

Thus, it is important to keep in mind that while discrete methods have limitations, they still are the best methods we have available.

85


The method of sealed bids was originally proposed by Hugo

Steinhaus and Bronislaw Knaster around 1948.

The best way to illustrate how this method works is by means of an example.

86

Example: Settling Grandma’s Estate

In her last will and testament, Grandma plays a little joke on her four grandchildren (Art, Betty, Carla, and Dave) by leaving just three items–

• a cabin in the mountains,

• a vintage 1955 Rolls Royce, and

• a Picasso painting,

with the stipulation that the items must remain with the grandchildren

(not sold to outsiders) and must be divided fairly in equal shares among them. How can we possibly resolve this conundrum?

The method of sealed bids will give an ingenious and elegant solution.

87

Example: Settling Grandma ʼ s Estate


Each of the players makes a bid (in dollars) for each of the items in the estate, giving his or her honest assessment of the actual value of each item.

To satisfy the privacy assumption, it is important that the bids are done independently, and no player should be privy to another player’s bids before making his or her own.

The easiest way to accomplish this is for each player to submit his or her bid in a sealed envelope. When all the bids are in, they are opened .

88



Table shows each player’s bid on each item in the estate.

89


Step 2 (Allocation)

Each item will go to the highest bidder for that item. (If there is a tie, the tie can be broken with a coin flip.)

In this example the cabin goes to Betty, the vintage Rolls Royce goes to

Dave, and the Picasso painting goes to Art.

Notice that Carla gets nothing. Not to worry–it all works out at the end!

(In this method it is possible for one player to get none of the items and another player to get many or all of the items. Much like in a silent auction, it’s a matter of who bids the highest.)

90


Step 3 (First Settlement)

It’s now time to settle things up. Depending on what items (if any) a player gets in Step 2, he or she will owe money to or be owed money by the estate.

To determine how much a player owes or is owed, we first calculate each player’s

fair-dollar share

of the estate.

A player’s

fair-dollar share

is found by adding and dividing the total by the number of players.

that player’s bids

91


Step 3 (First Settlement)

The last row of Table shows the fair-dollar share of each player

92

Example 3.9 Settling Grandma ʼ s Estate


For example, Art’s bids on the three items add up to $540,000.

Since there are four equal heirs, Art realizes he is only entitled to onefourth of that–his fair-dollar share is therefore $135,000.

93



The fair-dollar shares are the baseline for the settlements; if the total value of the items that the player gets in Step 2 is more than his or her fair-dollar share, then the player pays the estate the difference.

If the total value of the items that the player gets is less than his or her fair-dollar share, then the player gets the difference in cash.

Here are the details of how the settlement works out for each of our four players.

94


Art

As we have seen, Art’s fair dollar share is $135,000.

At the same time, Art is getting a Picasso painting worth (to him)

$280,000, so Art must

$135,000).

pay

the estate the difference of $145,000 ($280,000 –

The combination of getting the $280,000 Picasso painting but paying

$145,000 for it in cash results in Art getting his fair share of the estate.

95


Betty

Betty’s fair-dollar share is $130,000. Since she is getting the cabin, which she values at $250,000, she must pay the estate the difference of

$120,000. By virtue of getting the $250,000 house for $120,000, Betty ends up with her fair share of the estate.

Carla

Carla’s fair-dollar share is $123,000. Since she is getting no items from the estate, she receives her full $123,000 in cash. Clearly, she is getting her fair share of the estate.

96


Dave

Dave’s fair-dollar share is $110,000. Dave is getting the vintage Rolls, which he values at $52,000, so he has an additional $58,000 coming to him in cash. The Rolls plus the cash constitute Dave’s fair share of the estate.

97


At this point each of the four heirs has received a fair share, and we might consider our job done, but this is not the end of the story–there is more to come (good news mostly!).

If we add Art and Betty’s payments to the estate and subtract the payments made by the estate to Carla and Dave, we discover that there is a surplus of $84,000! ($145,000 and $120,000 came in from Art and

Betty; $123,000 and $58,000 went out to Carla and Dave.)

98


Step 4 (Division of the Surplus)

The surplus is common money that belongs to the estate, and thus to be divided equally among the players. In our example each player’s share of the $84,000 surplus is $21,000.

99


Step 5 (Final Settlement)

The final settlement is obtained by adding the surplus money to the first settlement obtained in Step 3.

Art:

Gets the Picasso painting and pays the estate $124,000–the original

$135,000 he had to pay in Step 3 minus the $21,000 he gets from his share of the surplus.

(Everything done up to this point could be done on paper, but now, finally, real money needs to change hands!)

100


Step 5 (Final Settlement)

Betty :

Gets the cabin and has to pay the estate only $99,000 ($120,000 – $21,000).

Carla :

Gets $144,000 in cash ($123,000 + $21,000).

Dave :

Gets the vintage Rolls Royce plus $79,000 ($58,000 + $21,000).

101


The method of sealed bids works so well because it tweaks a basic idea in economics.

In most ordinary transactions there is a buyer and a seller, and the buyer knows the other party is the seller and vice versa.

In the method of sealed bids, each player is simultaneously a buyer and a seller, without actually knowing which one until all the bids are opened. This keeps the players honest and, in the long run, works out to everyone’s advantage.

102


At the same time, the method of sealed bids will not work unless,

1). Each player must have enough money to play the game.

If a player is going to make honest bids on the items, he or she must be prepared to buy some or all of them, which means that he or she may have to pay the estate certain sums of money.

If the player does not have this money available, he or she is at a

definite disadvantage in playing the game.

2). Each player must accept money (if it is a sufficiently large amount) as a substitute for any item. This means that no player can consider any of the items priceless.

103


The method of markers is a discrete fair-division method proposed in

1975 by a mathematician William F. Lucas.

The method has the great virtue that it does not require the players to put up any of their own money. On the other hand, unlike the method of sealed bids, this method cannot be used effectively unless

(1)

(2) there are many more items to be divided than there are players in the game and

the items are reasonably close in value.

104


In this method we start with the items lined up in a random but fixed sequence called an array.

Each of the players then gets to make an independent bid on the items in the array.

A player’s bid consists of dividing the array into segments of consecutive items (as many segments as there are players) so that each of the segments represents a fair share of the entire set of items.

105


For convenience, we might think of the array as a string.

Each player then cuts” the string into N segments, each of which he or she considers an acceptable share.

Notice that to cut a string into N sections, we need N – 1 cuts.

In practice, one way to make the “cuts” is to lay markers in the places where the cuts are made. Thus, each player can make his or her bids by placing markers so that they divide the array into N segments.

106


To ensure privacy, no player should see the markers of another player before laying down his or her own.

The final step is to give to each player one of the segments in his or her bid.

The easiest way to explain how this can be done is with an example.

107

Example: Dividing the Halloween Leftovers

Alice, Bianca, Carla, and Dana want to divide the Halloween leftovers shown in the Figure among themselves. There are 20 pieces, but having each randomly choose 5 pieces is not likely to work well– the pieces are too varied for that.

108


Arrange the 20 pieces randomly in an array.

109



Each child writes down independently on a piece of paper exactly where she wants to place her three markers. (Three markers divide the array into four sections.)

The bids are opened, and the results are shown on the next slide.

The A -labels indicate the position of Alice’s markers ( her second marker, and A

3

her third and last marker).

A

1

denotes her first marker, A

2

110



111



Alice’s bid means that she is willing to accept one of the following as a fair share of the candy:

(1) pieces 1 through 5 (first segment),

(2)

(3)

pieces 6 through 11 (second segment), pieces 12 through 16 (third segment), or

(4)

(5) pieces 17 through 20 (last segment).

Bianca’s bid is shown by the break up the array into four segments that are fair shares; Carla’s bid (

C

B

-markers and indicates how she would

-markers) and Dana’s bid (

D

-markers).

112


Step 2 (Allocations)

This is the tricky part, where we are going to give to each child one of the segments in her bid. Scan the array from left to right until the first

marker

comes up. Here the first

first marker

is Bianca’s

B

1

.

first

113



This means that Bianca will be the first player to get her fair share consisting of the first segment in her bid (pieces 1 through 4).

114



Bianca is done now, and her markers can be removed since they are no longer needed. Continue scanning from left to right looking for the first

second marker

.

Here the first is Carla’s C

2

second marker

, so Carla will be the second player taken care of.

115



Carla gets the second segment in her bid (pieces 7 through 9). Carla’s remaining markers can now be removed.

116



Continue scanning from left to right looking for the first

Here there is a tie between Alice’s

A

3

and Dana’s

D

3

.

third marker

.

117



As usual, a coin toss is used to break the tie and Alice will be the third player to go–she will get the third segment in her bid (pieces 12 through

16).

118



Dana is the last player and gets the last segment in her bid (pieces 17 through 20,).

119



At this point each player has gotten a fair share of the 20 pieces of candy. The amazing part is that there is leftover candy!

120


Step 3 (Dividing the Surplus)

The easiest way to divide the surplus is to randomly draw lots and let the players take turns choosing one piece at a time until there are no more pieces left .

Here the leftover pieces are 5, 6, 10, and 11 The players now draw lots;

Carla gets to choose first and takes piece 11. Dana chooses next and takes piece 5.

Bianca and Alice receive pieces 6 and 10, respectively.

121

The Method of Markers Generalized

The ideas behind the last Example can be easily generalized to any number of players. We now give the general description of the method of markers with N players and M discrete items.

Preliminaries

The items are arranged randomly into an array. For convenience, label the items 1 through M, going from left to right.

122



Each player independently divides the array into by placing N – 1 markers along the array.

N segments (segments 1, 2, the fair shares of the array in the opinion of that player.

. . . , N

These segments are assumed to represent

)

123



Scan the array from left to right until the first player owning that marker (let’s call him

P first marker

is located. The

1

) goes first and gets the first segment in his bid. (In case of a tie, break the tie randomly.)

P

1

ʼ s markers are removed, and we continue scanning from left to right, looking for the first

second marker

.

124



The player owning that marker (let’s call her

P

2

) goes second and gets the second segment in her bid. Continue this process, assigning to each player in turn one of the segments in her bid. The last player gets the last segment in her bid.

Step 3 (Dividing the Surplus)

The players get to go in some random order and pick one item at a time until all the surplus items are given out.

125

The Method of Markers: Limitation

Despite its simple elegance, the method of markers can be used only under some fairly restrictive conditions: it assumes that every player is able to divide the array of items into segments in such a way that each of the segments has approximately equal value.

This is usually possible when the items are of small and homogeneous value, but almost impossible to accomplish when there is a combination of expensive and inexpensive items; an example is dividing fairly 19 candy bars plus an iPod using the method of markers!.

FAIR DIVISION

FAIR DIVISION

Content

Basic Assumptions

Types of Fair Division Games

Types of Fair Division Games

S

S

Lone-Divider Method for Three Players

C

The Lone-Divider Method for More Than Three

Players

The Lone-Divider Method for More Than Three

Players

The Last-Diminisher Method

fair-dollar share

fair-dollar share

pay

C

B

D

marker

first marker

B

first

second marker

second marker

A

D

third marker

P first marker

P

second marker

P

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FAIR DIVISION

FAIR DIVISION

Content

Basic Assumptions

Types of Fair Division Games

Types of Fair Division Games

S

S

Lone-Divider Method for Three Players

C

The Lone-Divider Method for More Than Three

Players

The Lone-Divider Method for More Than Three

Players

The Last-Diminisher Method

fair-dollar share

fair-dollar share

pay

C

B

D

marker

first marker

B

first

second marker

second marker

A

D

third marker

P first marker

P

second marker

P

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