C Roettger, Spring 16
Problem 1 Dora, Diego and Chico the squirrel are sharing a delicious pie.
It is equal parts apple, blueberry, and raspberry. Dora thinks the apple part is worth $2, the blueberry part $4, and the raspberry part $6.
Diego thinks the apple part is worth $9, the blueberry part $3, and the raspberry part $6.
Chico will be the Lone Chooser, and he thinks both raspberry and blueberry are worth $5, but hates apple pie.
a) Describe a cut that Dora would make to fairly divide the pie between her and Diego.
b) Which of Dora’s slices would Diego choose?
c) How do they go on to finish the fair division problem with the Lone
Chooser method?
d) Which value does each one get, in their own opinion?
Solution a) Dora makes one slice raspberry, one slice blueberry + apple.
Both worth $ 6 in her opinion.
b) Diego chooses the latter (worth $3 + $9 = $12 in his opinion, more than the raspberry part).
c) So Dora gets the entire raspberry part of the pie. Dora and Diego subdivide their slices into three shares that they think are fair. Dora just cuts equal thirds. Diego cuts the apple in equal ninths, and combines all the blueberry with 1/9 apple, then makes two slices with 4/9 apple (to figure this out, first compute the total, as was done in part b). Divide by three, so each of Diego’s slices should have $4, meaning the apple makes two entire slices and a little bit more, namely $4 + $4 + $1).
d) Chico chooses one of Dora’s slices and the blueberry with 1/9 apple (gives him 5/3 + 5 = $6.67). Dora gets two thirds of the raspberry, so $4. Diego gets 8/9 apple, $8.
Problem 2 Alan, Bethany and Carrie have to clean their apartment. They have to clean the windows, vacuum the floor, and do the dishes. They want to apply what they have learned and submit sealed bids for each chore, as follows.

a) What is the price of a fair share of the work, according to each of the housemates?
b) How should they allocate the chores?
c) How much money does each of them have to pay, or get paid, in the first settlement?
d) How much surplus is left over afterwards, to be divided equally?
Solution a) Below are the totals for each person, and dividing by three we get the price of a fair share in each person’s opinion.
Alan
Bethany
Windows Floor Dishes Total Fair share
50
55
20
40
65 135
55 150
45
50
Carrie 45 25 50 120 40 b) Alan vacuums the floor, Carrie cleans windows and washes dishes,
Bethany does nothing. This is because each chore goes to the lowest bidder .
c) Alan thinks a share is worth $45, he works for $20, so has to pay $25.
Bethany thinks a share is worth $50, and she has to pay that $50.
Carrie thinks a share is worth $40, works for $95, so she gets $55.
d) Total income for the ’chores account’ is $25 + $50 = $75 from Alan and
Bethany. Total outgoing is $55 for the indefatigable Carrie. Leaves a surplus of $20.
Problem 3 Four major technology companies are settling an intellectual property lawsuit about owning the letters of the alphabet. Your job is only to oversee the division of the first half, A–M. The companies use the method of markers to divide the letters A–M into fair shares. Each company has placed markers as follows.
Macrohard Corp.
A B C D | E F G H | I J | K L M
Pear Inc.
Sumsang
A B C D | E F | G H I J | K L M
A B C D | E F G | H I J K | L M
AD&D A B C | D E F | G H I J K | L M a) Which company gets the rights to which letters in the first allocation of assets/letters ?
Macrohard Corp.
gets IJ.
Pear Inc.
gets EF.
Sumsang
AD&D gets LM.
gets ABC.

b) Which letters are left over as surplus afterwards? (to be given to charity).
Solution a) First, ABC goes to AD&D because AD&D’s first marker is earlier (more to the left) of everyone else’s. AD&D’s markers are then removed. Earliest marker for the 2nd share is Pear Inc.’s, so Pear gets EF.
Pear’s markers are removed. Earliest marker for 3rd share is Macrohard’s, so Macrohard gets IJ, and similarly Sumsang gets LM.
b) DGHK are left over.
Problem 4 Consider the weighted voting system described by
[15 : 7 , 6 , 5 , 4 , 3 , 2 , 1] .
In the problem below, all sequential coalitions are written just by listing the weights, not the players.
a) Which player is pivotal in the sequential coalition (6 , 3 , 5 , 2 , 4 , 7 , 1)?
b) Why is player P
2 pivotal in the sequential coalition (7 c) How many sequential coalitions also have player P
2
, 5 , 6 , 1 , 3 , 4 , 2)?
in 3rd place, the players
P
1
, P
3
(with weights 7,5) before P
2
4,3,2,1) after P
2
?
and players P
4
, P
5
, P
6
, P
7
(with weights
These are all the possibilities that P
1
, P
2
, P
3 vote Yes, P
2 is pivotal, and the rest votes No. No need to write down all the possibilities, but show your work.
Solution a) P
6
(has weight 2).
b) Because 7 + 5 = 12 < 15 but P
2
7 + 5 + 6 = 18 ≥ 15.
(adds weight 6) puts them over the quota, c) There are 2! = 2 ways to shuffle the two players P
1
, P
3 before P
2
. Similarly, there are 4! = 24 ways to shuffle the four players after P
2
. All of these ways can be combined, giving a total of 2 × 24 = 48 sequential coalitions satisfying the given conditions.
Problem 5 You are Howe, the junior partner in the law firm Dewey, Cheatum, & Howe. Decisions are made according to the weighted voting system
[15 : 10 , 9 , 3].
a) List all winning coalitions in this weighted voting system.
b) Underline the critical players in each winning coalition.
c) Compute the Banzhaf power index for each player in this weighted voting system.

d) Bonus question : Dewey himself has called you into his private office.
Dewey and Cheatum offer to increase your voting power by lowering the quota while leaving all the weights the same. They offer to lower it to q = 14 for $1000 or to q = 12 for $2000. Which offer is better, and why?
Solution a) ’Coalitions’ means the order does not matter, it’s the Banzhaf scenario! so we only have two winning coalitions,
{ 10 , 9 , 3 } , { 10 , 9 } .
b) Underline every 10 and every 9 in the previous part - without either one of the big players Dewey and Cheatum, the quote is not reached. You (Howe) are, sadly, never critical.
c) First, the critical count is 2/2/0 for Dewey, Cheatum, and you. The total of these numbers is 2 + 2 + 0 = 4. So the Banzhaf power indices are
β
1
=
2
4
, β
2
=
2
4
, β
3
=
0
4
= 0 .
d) For [12 : 10 , 9 , 3], you get new winning coalitions { 10 , 3 } and { 9 , 3 } . Also,
9 is no longer critical in { 10 , 9 , 3 } . The Banzhaf counts become 2 / 2 / 2 and your power increases to 1/3.
For the quota q = 14, the coalitions and critical players stay the same, you still would have Banzhaf power 0 (you are a ’dummy’), and that is certainly not worth paying $1000 for.