Ex. (MLP Q.4.)
Martha and Nick buy a 14 inch sub for $18, and divide it using the divider-chooser method. The first four inches are ham, the next six are turkey, and the last four are beef. Martha likes ham twice as much as turkey, and likes turkey and beef equally. Nick likes beef twice as much as ham, and likes ham and turkey equally. How would they divide the sub if Martha divides, and which piece would Nick choose?
Let x be the value to Martha of one inch of turkey (or beef) sub.
Then the value of the combined length of the turkey and beef parts is
10x.
While the ham part is only 4 inches long, she likes ham twice as much, so in terms of x, the value of the ham part is 4 x 2x = 8x.
So the entire sub is worth 18x, and since it is 18 inches long, that means x = 1, meaning every inch of turkey (or beef) is worth $1, and consequently every inch of ham is worth $2.
Now fair value is clearly $9 (half of $18), so she needs to split the sub in such a way that each piece is (to her) worth $9.
The ham part is worth $8 (4 inches worth $2 each), so she needs another
$1 worth of sub. Hence she moves the knife another inch to the right
(since each inch of turkey is worth $1), and then cuts.
So s
1
is [0,5] and s
2
is [5,14].
Nick would obviously choose s
2
, as he really likes beef, and s
2
contains all the beef.

The Lone Chooser Method
The lone chooser method, developed by Arlington Fink in 1964, is in some ways the opposite of the lone divider method from Lecture 7, as here we have once chooser, and N-1 dividers among the N players.
Note:
While the method below can be generalized to any number of players, we will just consider the case where N=3. i) One person is randomly selected to be the chooser (C), with the other two being dividers (D
1
and D
2
). ii) D
1
and D
2
divide S using the divider-chooser method. iii) D
1
and D
2
then split the piece obtained in the previous step into three pieces of equal value (in their eyes). iv) The chooser then picks the best subshare from each divider (so two pieces in total). v) The dividers then keep their two remaining pieces.
Ex.
Three people, Angela, Boris, and Carlos are dividing a cake that is half vanilla-half strawberry. Their valuations of each flavor are given in the table below.
180
)
Angela $9 $27
Boris $12 $18
Carlos $24 $12
Suppose that Angela and Boris are the dividers and Carlos is the chooser. In the first division, Boris cuts the cake vertically through the center, so that the two pieces are identical, with Angela picking one piece, leaving Boris with the other.

Note:
Here comes the tricky part! Angela and Boris have to split their pieces
(which are both half vanilla-half strawberry) into three equal subpieces based on their valuations.
For Angela, we can write V
90
+ S
90
= 18 (which is half of 36). Since she likes strawberry 3 times more than vanilla, we can write V
90
+ 3V
90
= 18
Solving we get V
90
= 4.50 and hence S
90
= 13.50.
Now fair value is $6, since she is spitting a piece worth $18 into three.
Thus one piece will contain V
90
, but she needs to add another $1.50 worth of the strawberry. This represents 1/9 of 13.50, so she adds 1/9 of S
90
S
10
.
, which is
She now has S
80
remaining, worth $12 (13.50
− 1.50), and hence just splits it in half to produce the other two pieces.
For Boris, we can write V
90
+ S
90
= 15 (which is half of 30). Since he likes strawberry 1.5 times as much as vanilla, we can write V
= 9.
90
+ 1.5V
90
= 15.
Solving we get V
90
= 6, and hence S
90
Now fair value is $5, since he is splitting a piece worth $15 into three.
Thus one piece will contain 5/6 of V
90
S
90
S
40
, so V
75
, one piece will contain 5/9 of
, so S
50
, and the other piece will consist of what remains, that is V
15
and
.
Now Carlos prefers Vanilla, so he will choose the subpieces that contain the most vanilla. So he will choose the piece from Angela which is V
90
+ S
10
, and the piece from Boris which is V
75
(and be very happy!)
Finally, Angela will be left with two pieces worth $6 to her, so $12 in total.
Boris will be left with two pieces worth $5 to him, so $10 in total.
Now Carlos gets 165 degrees of vanilla and 10 degrees of strawberry. Since
V
180
= 24 (from the table), it follows that V
Similarly S
180
= 12, so S
1
1
= 0.1333…, and hence V
165
= 0.0666…, and hence S pieces worth $22.66 to him.
10
= 0.66, making his
= 22.

Ex.
Consider the same people, with the same preferences and valuations.
This time, in the first division, Angela splits the cake into one piece that is 120 degrees of strawberry, and another that is 60 degrees of strawberry and 180 degrees of vanilla.
Now Boris must choose between the two.
The strawberry piece (S
120
) is worth 2/3 of S which is $12.
The mixed piece (S
60
+ V
180
, and hence 2/3 of 18,
180
) is worth $6 (since it is half of S
120
, which we just calculated) plus $12 (see table), so $18 in total.
So he chooses the mixed piece (leaving Angela with S
120
).
Note:
Here comes the tricky part (though here it’s not so tricky!)
Angela and Boris have to split their pieces into three equal subpieces based on their valuations.
Angela just splits S
120
into three pieces of S
40
.
Boris values his piece at $18, so S
60
+ V
180
= 18. Fair value is hence $6
(18 divided by 3). However since we know from the table that V
180 which means that S
60
.
= 6, and so the division is simple: V
90
, V
90
= 12,
, and
S
60
Now Carlos gets to choose. Angela’s pieces are identical, so he gets S
40 from her. He prefers vanilla, and so he picks V
90
from Boris.
The valuations for each player are then calculated as in the previous example.

The Last Diminisher Method
Designed by Stefan Banach and Bronislaw Knaster in 1948, the last diminisher method has several rounds before eventually there are only two players left – at which point the divider-chooser method is used. i) The N players are randomly assigned an order of play: P
1
, …, P
N ii) P
1
cuts a share that has a value (to P temporarily, P
1
owns the piece cut.
1
) of 1/N of the total. At least iii) P
2 by P
1
then either passes (and P that in the eyes of P
2
1
retains ownership) or accepts a smaller piece
is worth 1/N of the total, thus diminishing the piece cut
(and the ownership passes, at least temporarily, to P iv) This continues until either P
N
passes or diminishes the piece (by a tiny fraction, essentially zero) and keeps what remains.
2
). v) At the end of each round someone will own a piece, and the process continues until there are two players remaining, at which point the dividerchooser method is used.
Ex. A cake valued at $30 is divided among five player using the last diminisher method. P
1
makes the first cut, at which point the valuations of that piece in the eyes of the others is given by the table below.
P
2
P
3
P
4
P
5
So in round 1, the following will occur:
P
2
will diminish ($7 represents more than fair value, which is 30/5 = $6)
P
3
P
4
P
5
will pass
will diminish ($6.50 represents more than fair value, which is 30/5 = $6)
will pass
So at the end of the round P
4
Round 2 then starts with P
1
will get a piece, worth $6 to P
4
.
again, with each player valuing the total cake remaining at least $24. (This is always the Total – Fair Value)

Ex.
A piece of land with a value of $300,000 is divided among four players using the last diminisher method. P
1
makes the first cut, at which point the valuations of that piece in the eyes of the others is given by the table below.
P
2
P
3
P
4
Valuation $65,000 $85,000 $80,000
So in round 1, the following will occur:
P
P
3
2
will pass
will diminish ($85,000 represents more than fair value, which is 300,000/4
= $75,000)
P
4
will diminish ($80,000 represents more than fair value, which is 300,000/4
= $75,000)
So at the end of the round P
4
will get a piece, worth $80,000 to P
4
(since P will essentially reduce by zero given that P
4
Round 2 then starts with P
1 remaining at least $225,000 (Total – Fair Value).
4
is the last player in the round). again, with each player valuing the total cake
The Method of Sealed Bids
Designed by Steinhaus and Knaster in 1948, the method of sealed bids is particularly effective for the settling of inheritance and divorce issues. i) Each player privately submits a bid for each item in a sealed envelope. The envelopes are opened and a table constructed, including the fair shares. ii) The items are allocated according to who bid the most for them.
Note: No money changes hands at this point. iii) The players calculate the value of the goods allocated to them minus their fair share. If the amount is positive they pay the difference in cash. If it is negative they receive the difference in cash. iv) Any cash payments received go into a pool, which is then divided evenly among all the players.

Ex.
Four players, Art, Betty, Carla, and Dave are bidding on the items listed below as part of an inheritance settlement.
Picasso $280,000 $240,000 $234,000 $190,000
Total $540,000 $520,000 $492,000 $440,000
Looking at the bids, the items are allocated in the following way:
Log Cabin – Betty
Rolls Royce – Dave
Picasso – Art
According to step iii)
Art owes $145,000 ($280,000 – $135,000)
Betty owes $120,000 ($250,000 – $130,000)
Carla receives $123,000 (since she won none of the items)
Dave receives $58,000 ($52,000 – $110,000)
Once the money is paid, there is a surplus of $84,000.
Hence each player receives $21,000.
Ex. Al and Betty are getting a divorce. The only thing of value they own is the house. They agree to use the method of sealed bids to reach a settlement, and bid the amounts shown below.
Player Al Betty
Fair Value $170,000 $182,000
Betty gets the house since she bid the most.
According to step iii)
Al receives $170,000 ($0 – $170,000)
Betty owes $182,000 ($364,000 – $182,000)
There is a $12,000 surplus, which they split equally. So in effect Betty pays
Al $176,000 for his share of the house, and he gets $176,000 in cash.