Please show all of your work for full credit.
MATH 105 Practice Final Exam – Solutions
A.(50 points total) Voting Systems
Consider the preference schedule given below for the questions in this section.
Number of voters 14
1st
A
B
2nd
3rd
C
D
4th
10
C
B
D
A
8
D
C
B
A
4
B
D
C
A
1
C
D
B
A
1.(10points) Find the full ranking of candidates using plurality.
Just use the first-place votes. A has 14, C has 10 + 1 = 11, so comes second, then
D and then B.
2.(20points) Find the full ranking of candidates using the Borda count method.
Award 4 points for 1st place, 3 for 2nd and so on. Then count the points for each
candidate. This gives
A
B
C
D
14 · 4 + 10 · 1 + 8 · 1 + 4 · 1 + 1 · 1 = 79
14 · 3 + 10 · 3 + 8 · 2 + 4 · 4 + 1 · 2 = 106
14 · 2 + 10 · 4 + 8 · 3 + 4 · 2 + 1 · 4 = 104
14 · 1 + 10 · 2 + 8 · 4 + 4 · 3 + 1 · 3 = 81
so the ranking is B, C, D, A. Note how different the ranking is compared to
plurality!
3.(15points) Find the full ranking of candidates using plurality with elimination.
First B is eliminated, giving 4 votes to D (the 2nd preference of those four voters).
Now D has 8 + 4 = 12 votes, more than C, so C is eliminated next. The C-votes
all go to D (because B has already been eliminated). So D has now 12 + 11 = 23
votes and wins! Ranking is D, A, C, B (reverse order of elimination).
4.(5 points) Identify whether or not there is a Condorcet candidate. If there is, which candidate is it?
Head-to-head comparisons show C wins against A 23 : 14, against B 19 : 18, and
against D 25 : 12. So C is a Condorcet candidate.
B.(50 points total) Power
Use the following weighted voting system for the questions in this section.
[10 : 5, 4, 2, 2]
5. (20 points) Find the power distribution using Banzhaf power.
The winningn coalitions are just
{P1 , P2 , P3 }, {P1 , P2 , P4 }, {P1 , P2 , P3 , P4 }.
Critical players are underlined. We count for each player how often that player is
critical, getting the counts
B1 = 3,
B2 = 3,
B3 = 1,
B4 = 1.
The sum of these numbers is 8. So the Banzhaf power distribution is
3
β1 = ,
8
3
β2 = ,
8
1
β3 = ,
8
1
β4 = .
8
6. (20 points) Find the power distribution using Shapley-Shubik power.
We could list all 4! = 24 sequential coalitions and underline the pivotal player in
each. Eg
(P2 , P3 , P1 , P4 )
has weights (4, 2, 5, 2) and so P1 is pivotal. Or we can spend some brainpower!
There are 3! = 6 sequential coalitions where P1 comes last, and P1 is then pivotal.
There are 6 where P2 comes last, and P2 is then pivotal.
When P3 and P4 come last, they are never pivotal (there are 6 + 6 = 12 sequential
coalitions like that). Suppose P3 comes last. With these weights, the player in
3rd position will automatically be pivotal. This gives 2 sequential coalitions each
to P1 , P2 , P4 . Repeat this thought exercise with the roles of P3 and P4 reversed,
and you get counts
SS1 = 6 + 2 + 2 = 10,
SS2 = 6 + 2 + 2 = 10,
SS3 = 2,
SS4 = 2
(check that they add up to 4! = 24). So the Shapley-Shubik power distribution is
σ1 =
10
,
24
σ2 =
10
,
24
σ3 =
2
,
24
σ4 =
2
.
24
7. (10 points) Is there any player who has veto power? Why or why not?
P1 and P2 have veto power, since without each one of them a coalition can only
have at most 9 votes.
C.(50 points total) Sharing
8. (20 points) A sandwich has two flavors, meatball and vegetarian. Adam and Bill wish to split
the sandwich using the divider chooser method with Adam as the divider (making
a single cut). Adam dislikes vegetarian and Bill has no preference. What shares
do Adam and Bill receive?
Call the meatball part M and the vegetarian part V. A fair cut for Adam would
be M/2 versus M/2 + V (Adam cut split the V part any which way, but he is
supposed to only make one cut, so he’ll have to give all of the V part to one slice.
Bill just goes by size, so he will choose Adam’s second slice, and Adam gets just
M/2.
9. (20 points) Suppose a third person wishes to share the sandwich. What method(s) could be
used to split the sandwich between the three players. Briefly describe how the
distribution of the sandwich would play out using one of your methods.
We’ll use Lone Divider (the alternative is Lone Chooser – note that the problem
asks you to name both!). Let’s say Adam is the lone chooser. Adam makes three
fair slices according to his value system:
M
M
M
M +V =
+
+
+V .
3
3
3
Then Bill and the third person ’Theo’ submit their bids for acceptable slices. Bill
will find only the third slice acceptable. If Theo finds the second slice acceptable,
he’ll get it and Adam will get the first. If Theo does not, meaning he only finds the
third slice acceptable, there is a standoff. In that case, give Adam the first slice
and recombine the second and third slice. Then use the basic Divider-Chooser
method on the recombined slices to find fair shares for Bill and Theo.
10. (10 points) At an estate sale there are three main items. A cabinet, a table, and a bookshelf.
Three people are putting bids on the items as shown in the table below. Describe
the distribution of assets using the method of sealed bids.
Player cabinet table bookshelf
A
100
200
50
B
100
100
100
C
200
100
50
First, A gets the table, B gets the bookshelf, C gets the cabinet because they are
the highest bidder for these items. The total value of items for A is 350, for B is
300, for C is 350. So a fair share for A is worth 116.67, for B 100 and for C 116.67
(in dollars, rounded to nearest cent). Therefore A pays
200 − 116.67 = 83.33
and B pays nothing (also gets no money at this stage). C also pays 83.33. The
surplus is then 166.66 which is divided equally in the last stage, so each player
gets 55.55 (again, rounded to the nearest cent).
D.(30 points total) Apportionment
Consider a health service that provides nurses across four different districts (A,B,C,
and D). There are 50 nurses to go around to these districts and the populations are
given below
District Population
A
40
30
B
C
60
D
20
11.(10 points) Find the apportionment of nurses using Hamilton’s method.
Total population is
40 + 30 + 60 + 20 = 150.
Then the standard divisor is SD = 150/50 = 3. The standard quotas and lower
quotas are
A
B
C
D
District
13.33 10.00 20.00 6.67
lower quota 13
10
20
6
So far, this allocates 13 + 10 + 20 + 6 = 49 nurses. Since district D has the largest
remainder 0.67, it gets the one extra nurse.
12.(20 points) Find the apportionment of nurses using Adams’ method.(Hint: Change your divisor by .1)
Again, the standard quotas and the upper quotas:
A
B
C
D
District
standard quota 13.33 10.00 20.00 6.67
14
10
20
7
upper quota
This allocates 51 nurses, so 1 too many! we need to make the divisor bigger. Take
the hint and try the divisor 3.1 instead of 3. The modified quotas become
District
A
B
C
D
modified quota
12.9 9.68 19.35 6.45
upper modified quota 13
10
20
7
and the upper modified quotas add up to 50 as desired! This ends up as exactly
the same as the result of Hamilton’s method - coincidence.
E.(20 points total) Game Theory
−2 5
13.(20 points) For the payoff matrix
give the optimal strategies and the value of the
0 −1
game.
1. We look for the biggest number in each column. This is 0 in column 1, 5 in
column 2. Circle them.
2. We look for the smallest number in each row. This is -2 in column 1, -1 in
column 2. Underline them.
3. No number is both underlined and circled – no saddle point! So mixed strategies are best for both players.
4. Suppose Player A chooses strategy 1 with probability p and Player B chooses
strategy 1 with probability q. Then they choose the other strategy with probability 1 − p and 1 − q, respectively. The expected value of the game is
Epq = −2pq + 5p(1 − q) − 1(1 − p)(1 − q).
Player A does not know what B will do. If q = 1, the expected payoff (as always,
from B to A) is
Ep1 = −2p
meaning actually a payment from A to B (a loss for A). In case q = 0, the expected
payoff is
Ep0 = 5p − (1 − p) = 6p − 1.
The best payoff would be if B chooses q = 0, then A would choose p = 1 IF HE
KNEW! But since he cannot know, this might also lead to a −2 since B might
just choose q = 1. The safe play is to choose the value of p where the two lines
meet (graph Ep1 and Ep0 !). This happens when
−2p = 6p − 1
1 = 8p
so for p = 1/8. The expected value of the game is then −2/8, or a payment of
1/4 units from A to B.
5. For the best strategy for B, look at cases p = 1:
E1q = −2q + 5(1 − q) = 5 − 7q.
and p = 0:
E0q = −1(1 − q) = −1 + q.
These two lines meet where 5 − 7q = −1 + q, so 6 = 8q or q = 6/8 = 3/4. You
get the same answer of −1/4 for the value of the game.
(15 point bonus) What is the most useful thing that you learned in this course? How will you apply
it in your daily life?
Election Rules matter more than I thought. Fair sharing is possible. Apportionment rules can still rig the outcome even if they look fair.