Math 115

 

1.  The Math 115 executive committee consists of a president (P), vice president (V), treasurer (T) and secretary (S).  The committee rules state that P gets 4 votes, V gets 3, T gets 2, and S gets 1.  A simple majority is necessary to pass the motion.

a.  How many votes are there?

            10

 

b.  What is the quota?

            6

 

c.  How many different coalitions are possible?

            24-1=15

 

 

d.  List all possible coalitions, and find the weight of each. 

{P}=4  

{V}=3                         

{T}=2                                     

{S}=1

{P, V}=7

{P, T}=6

{P,S}=5

{V,T}=5

{V,S}=4

{T,S}=3

{P, V, T}=9

{P, V, S}=8

{P, T, S}=7

{V, T, S}=6

{P, V, T, S}=10

 

e.  For each winning coalition, underline each member that is critical.

 

f.  Count how many times each person is a critical member of a coalition.

            P: 5

            V: 3

            T: 3

            S:  1

g.  What is the Banzhaf power distribution?

P: 5/12 ≈ 41.7%

V, T = 3/12 = 25%

S = 1/12 ≈ 8.3%

 

h.  Part of your answer to part (g) should be a little surprising (in light of the power ratings).  Explain in practical terms why it works out this way.

 

V and T have the same power, even though they have different numbers of votes.  It works out this way because with this quota, there is no situation in which V’s 3rd vote can ever be of any use, and so V is critical the exact same number of times as T
                                                                                                                                                         p.2

2.  Suppose the 115 executive committee was considering changing the quota to 7. 

 

a. What would the new Banzhaf power distribution be?

{P}=4                                       Now, we have P critical 5 times, V 3 times, T once and S once, so:

{V}=3                                      P = 50%, V=30%, T and S 10% each 

{T}=2                                      (Also, note that P now has veto power)

{S}=1

{P, V}=7

{P, T}=6

{P,S}=5

{V,T}=5

{V,S}=4

{T,S}=3

{P, V, T}=9

{P, V, S}=8

{P, T, S}=7

{V, T, S}=6

{P, V, T, S}=10

 

 

b.  If the each committee member wants to have as much power as possible (which is realistic, don’t you think?), who would vote to change the quota to 7, and who would vote to keep the quota at 6?

 

P, V and S all gain power at the expense of T, so they would all vote to make the switch.

 

 

 

Some rather startling changes in power can happen if we make the quota even higher.  You may want to investigate a quota of 8 on your own, but for now we'll look at 9 and 10:

3.  What is the Banzhaf power distribution if the quota is 9?  Explain, in clear everyday English, why it is this way.

 

Without listing all of the coalitions, we note that with quota 9, a winning coalition must be either unanimous, {P, V, T, S}, OR contain all but one vote, which must be  {P, V, T}.

In each case, P, V and T are critical, so the three of them are now equally powerful, with 1/3≈33.3% of the power each (and veto power!), while S is now a dummy.

 

 

4.  What about if the quota is 10?  Again, explain as clearly as possible, why it is this way.  (Hint:  You should be able to answer this without any tough computations!)

 

This time, the only winning coalition is the one with ALL voters, so everyone must be critical, and everyone must have veto power.  Since everyone is critical the same number of times (once each), the power distribution is 25% apiece.

 

 

 

What can’t the quota be even higher than 10?

 

Nothing could ever pass, since quota would never be reached!