Voting and Elections

VIII. Solutions\Hints to Problem Sets


Solutions/Hints to Problem Set 1 :
AnonymousNeutralMonotone
DictatorshipNoYesYes
Imposed RuleYesNoYes
Minority RuleYesYesNo


End of Solutions/Hints to Problem Set 1


Solutions/Hints to Problem Set 2 :There are n+1 questions at first, number those questions 1 to n+1. Now answer them in order and the first question with answer yes (say the ith one) the quote will be i-1. If the answer is no, to all of them then the quote is greater than the number of voters. Also, the property we have is anonymity.

For the next set of questions, use anonymity for the first and third, use monotonicity for the second and fourth, and for the fifth use nuetrality.


End of Solutions/Hints to Problem Set 2


Solutions/Hints to Problem Set 3 :Majority rules gives a four way tie. Plurality gives P>S>J>L. With n candidates there are n! different possible individual preference orders ( n! means n*n-1*n-2*...*3*2*1)

As for the critic, just take n candidates and n+1 voters. Now assume only 2 voters rank A first, then assume every other voter puts a different candidate in rank one and every other one ranks A last. Now let n get arbitrarily large, and you're done.

End of Solutions/Hints to Problem Set 3


Solutions/Hints to Problem Set 4 : For the GSR election using Borda count we get S>J>P>L.

For showing that Borda count violates the majority criterion. You can a preference schedule with 13 voters prefering A>B>C, 11 voters prefering B>A>C, and 1 voter prefering B>A>C.

Plurality and Borda count both satisfy anonymity, neutrality, and monotocity.

Plurality is not a quota system, and since we are dealing with elections with more that two candidates it does not contradict the lemma.

End of Solutions/Hints to Problem Set 4


Solutions/Hints to Problem Set 5 :Majority rule results in a tie. Plurality gives J. Borda count gives N. Largest number of first place votes is N. Largest number of last place votes is J. Head-to-head with S and N gives N>S. Head-to-head between S and J gives S>J. Head-to-head between N and J gives N>J. This should seem odd, since J would lose in a head-to-head with either candidate.

End of Solutions/Hints to Problem Set 5


Solutions/Hints to Problem Set 6 : SPV satisfies both CWC and CLC. (So the answer to these to questions is no)

With SPV with agenda P,S,J,L... J wins. Agenda L,J,S,P gives P. Agenda S,J,P,L gives Liz. Agenda P,L,J,S gives S.

With agenda S,L,P,J, J wins. If every voter switches there preference for J and L, J still wins but if SVP were neutral then L would have won.

End of Solutions/Hints to Problem Set 6


Solutions/Hints to Problem Set 7 :Black's system satisfies anonymity, neutrality, monotonicity, the majority criterion, and CWC. Under Black's System for this election we would get D>P>W. Since Wayne is ranked last his exclusion shouldn't change the outcome of the election, however with his exclusion we get P>D.

End of Solutions/Hints to Problem Set 7


Solutions/Hints to Problem Set 8 :Under Instant Runoff we get P>K>R. If those judges swap there preferences then we get R>P>K and K experiences a drop in ranking thus Instant Runoff violates IIA. Instant Runoff satisfies anonymity, neutrality, MC, and it violates monoticity and CWC.

End of Solutions/Hints to Problem Set 8


Solutions/Hints to Problem Set 9 :Under approval voting we get P>James>John. For the voters that approved of Peter and James a,b,c are all possible, while d and e are not. Approval violates universality by restricting the kind of preference order are permissible.

Since in individual preference orders for approval voting there is only one ">" we can conclude that this did not effect the number of approval votes that A and B received. Hence, approval voting satisfies IIA.

End of Solutions/Hints to Problem Set 9


Solutions/Hints to Problem Set 10 :With the two ballots G>S, Greg would receive 1 more point that Sharon. On the ballot with S>C>G, Sharon receives 2 more points that Greg. On the last ballot Greg receives 3 more points that Sharon. Hence, we have G>S.

Borda count does satisfy IBI, Just think of the scoring system used in the Borda count and relate this to the definition of intensity. Also, the Borda count satisfies universality, unanimity, and IBI, and is not a dictatorship. However, it is not a contradiction to Arrow's theorem since IBI is weaker that IIA (meaning that there are more voting systems that satisfy IBI than IIA).

End of Solutions/Hints to Problem Set 10


Solutions/Hints to Problem Set 11 :Lets call voter 2 Sharon to make this easier to write (so we have Doug, Sharon, Elisabeth). with [101:101,97,2] Doug could be by himself or with Sharon or with Elisabeth or with both of them. Doug is distinguished because he is in every winning coalition and absent from every loosing coalition.

With [103:101,97,2] winning combinations are the same except now Doug needs a comdrad. He is in every winning coalition.

With [105:101,97,2] the only two winning combinations are (Doug, Sharon, and Elisabeth) and (Doug and Sharon). In this weighted voting system Elisabeth is not necessary.

For a) we can have 2,2 or 2,2,1. For b) we can have 3,2 or 3,1 or 3,2,1. For c) we can have 3,2 or 3,2,1. For d) we can have 3,2 or 3,2,2. For e) we can have 3,3 or 3,_,2 or _,3,2 or 3,3,2. So we see that a,c,d are all isomorphic.

End of Solutions/Hints to Problem Set 11


Solutions/Hints to Problem Set 12 : The Banzhaf power of a permanent member is C10,4+C10,5+C10,6+C10,7+C10,8+C10,9+C10,10 (call this number P).
The Banzhaf power of a nonpermanent member is C10,3 (Call this number NP).
To compute the total Banzhaf power of this sytem, remember we have 5 permanent member and 10 nonpermanent members, we take 5*P + 10*NP (call this number TBP)
Then we have the Banzhaf index for a permanent member is P/TBP and for a nonpermanent member it's NP/TBP.

End of Solutions/Hints to Problem Set 12




References:
[1]Jonathan K. Hodge and Richard E. Klima.The Mathematics of Voting and Elections: A Hands-On Approach. American Mathematical Society, Providence, R.I., 2005.