65
Math 115 – Homework Project #2
Name:
KEY
Show as much work as possible for each of the problems below.

1. The members of the Jedi Academy are electing a new group leader. They are to choose from
among 4 candidates – A (Anakin), B (Jacen), C (Jaina), and D (Tenel Ka). The preference
schedule is shown below.
# Voters
1st Choice
2nd Choice
3rd Choice
4th Choice
4
A
C
B
D
1
B
A
D
C
9
C
D
A
B
8
A
B
D
C
5
D
C
B
A
(a) Who will be the leader if Master Skywalker uses the plurality-with-elimination method
to count the votes? (10)
C = Jaina
(b) Who will be the leader if Master Skywalker uses the method of pairwise comparisons to
count the votes? (10)
D = Tenel Ka
2. In the weighted voting system [95: 65, 35, 30, 25], determine which player(s) are dictators, are
dummies, have veto power. (10)
No dictator, P1 has veto power, P4 is a dummy
3. Consider the weighted voting system [q: 5, 3, 2, 2, 1, 1].
(a) What is largest possible valid value of q?
(5)
14
(b) What is the smallest possible valid value of q?
(5)
8
(c) If no player is to have veto power, what is the largest possible value of q?
(5)
9
4. Find the Banzhaf power distribution of the system [9: 5, 5, 4, 2, 1]. Is it fair? Explain.
(10)
P1 = P2 = P3 = 33.3%, P4 = P5 = 0%; not fair because P4 and P5 have no power
5. Find the Shapley-Shubik power distribution of the system [7: 4, 3, 2, 1]. Is it fair? Explain.
P1 = 58.3%, P2 = 25%, P3 = P4 = 8.3%; fair because all have some power
(10)
SOLUTIONS
1. (a) Note: majority = 14
round 1: A = 12, B = 1, C = 9, D = 5; eliminate B
round 2: A = 13, C = 9, D = 5; eliminate D
round 3: A = 13, C = 14; C wins
(b) A beats B (21 to 6)
A loses to C (13 to 14)
A loses to D (13 to 14)
B loses to C (9 to 18)
B loses to D (13 to 14)
C loses to D (13 to 14)
A = 1pt, B = 0pts, C = 2pts, D = 3pts, and so D, Tenel Ka, wins
2. (a) There is no dictator since no weight is greater than or equal to the quota
(b) P1 has veto power because the quota cannot be achieved without his 65 votes.
(c) P4 is a dummy since his 25 votes never are needed to reach the quota
3. (a) largest possible quota is the total number of votes = 5+3+2+2+1+1=14
(b) smallest possible quota is a majority of the votes = 8 (7 would allow ties)
(c) If no one is to have veto power, then specifically P1, the player with the most votes, cannot
have veto power. Thus the number of votes the other players have must achieve the quota.
So q <= 3+2+2+1+1 = 9.
4. The winning coalitions are shown below with critical player(s) underlined.
{P1, P2}
{P1, P2, P3}
{P1, P2, P3, P4}
{P1, P3}
{P1, P2, P4}
{P1, P2, P3, P5}
{P2, P3}
{P1, P2, P5}
{P1, P2, P4, P5}
{P1, P3, P4}
{P1, P3, P4, P5}
{P1, P3, P5}
{P2, P3, P4, P5}
{P2, P3, P4}
{P1, P2, P3, P4, P5}
P1 
8
100  33. 3 %
24
 
P4  0%
P2 
{P2, P3, P5}
8
100  33. 3 %
24
 
P5  0%
P3 
Since some players have no power, this is not fair.

8
100  33. 3 %
24
 
5. Each sequential coalition is listed below, and the pivotal player is underlined.
<P1, P2, P3, P4>
<P2, P1, P3, P4>
<P3, P1, P2, P4>
<P4, P1, P2, P3>
<P1, P2, P4, P3>
<P2, P1, P4, P3>
<P3, P1, P4, P2>
<P4, P1, P3, P2>
<P1, P3, P2, P4>
<P2, P3, P1, P4>
<P3, P2, P1, P4>
<P4, P2, P1, P3>
<P1, P3, P4, P2>
<P2, P3, P4, P1>
<P3, P2, P4, P1>
<P4, P2, P3, P1>
<P1, P4, P2, P3>
<P2, P4, P1, P3>
<P3, P4, P1, P2>
<P4, P3, P1, P2>
<P1, P4, P3, P2>
<P2, P4, P3, P1>
<P3, P4, P2, P1>
<P4, P3, P2, P1>
P1 
14
100  58. 3 %
24
P3 
2
100  8.3 %
24
P2 
6
100  25%
24
P4 
2
100  8.3 %
24
 
 
 
 
Since everyone has power, this system is fair.
