C Roettger, Spring 16
Math 105 – Exam 2A – solutions
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 $6, the blueberry part $4, and the raspberry part $2.
Diego thinks the apple part is worth $6, the blueberry part $3, and the raspberry part $9.
Chico will be the Lone Chooser, and he thinks both apple and blueberry are
worth $5, but hates raspberry 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 apple, one slice blueberry + raspberry.
Both worth $ 6 in her opinion.
b) Diego chooses the latter (worth $3 + $9 = $12 in his opinion, more than
the apple part).
c) So Dora gets the entire apple 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 raspberry in equal ninths, and combines all the blueberry with 1/9 raspberry, then makes two slices with 4/9 raspberry (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 raspberry 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 raspberry
(gives him 5/3 + 5 = $6.67). Dora gets two thirds of the apple, so $4. Diego
gets 8/9 raspberry, $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.
Windows Floor Dishes Total Fair share
Alan
50
20
65
135
45
45
25
50
120
40
Bethany
Carrie
55
40
55
150
50
b) Alan vacuums the floor, Bethany cleans windows and washes dishes,
Carrie 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 $40, works for $95, so she gets $55.
Carrie thinks a share is worth $50, and she has to pay that $50.
d) Total income for the ’chores account’ is $25 + $50 = $75 from Alan and
Carrie. Total outgoing is $55 for the indefatigable Bethany. 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
Pear Inc.
A B C D |E F |G H
I J |K
L
Sumsang
A B C D |E F G |H I J |K
L
AD&D
A B C |D E F |G H
I J K |L
a) Which company gets the rights to which letters in the first allocation of assets/letters?
Macrohard Corp. gets LM.
Pear Inc.
gets EF.
Sumsang
gets HIJ.
AD&D
gets ABC.
M
M
M
M
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 Sumsang’s, so
Sumsang gets HIJ, and similarly Macrohard gets LM.
b) DGK 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, 2, 5, 4, 7, 1)?
b) Why is player P2 pivotal in the sequential coalition (3, 4, 5, 6, 1, 2, 7)?
c) How many sequential coalitions also have player P2 in 4th place, the players
P3 , P4 , P5 (with weights 5,4,3) before P2 and players P1 , P6 , P7 (with weights
7,2,1) after P2 ?
These are all the possibilities that P2 , P3 , P4 , P5 vote Yes, P2 is pivotal, and
the rest votes No. No need to write down all the possibilities, but show your
work.
Solution a) P3 (has weight 5).
b) Because 3 + 4 + 5 = 12 < 15 but P2 (adds weight 6) puts them over the
quota, 3 + 4 + 5 + 6 = 18 ≥ 15.
c) There are 3! = 6 ways to shuffle the three players P3 , P4 , P5 before P2 .
Similarly, there are 6 ways to shuffle the three players after P2 . All of these
ways can be combined, giving a total of 6 × 6 = 36 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 = 13 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
2
β1 = ,
4
2
β2 = ,
4
β3 =
0
= 0.
4
d) For [13 : 10, 9, 3], you get a new winning coalition {10, 3}. Also, 9 is
no longer critical in {10, 9, 3}. The Banzhaf counts become 3/1/1 and your
power increases to 1/5.
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.