Self Check Answers

1. The sum of the weights is 15; so the quota must not be greater than 15. The quota cannot be less than half the sum of the weights, so the quota is at least 8. The possible values for the quota are 8, 9, 10, 11, 12, 13, 14, and 15.
2. The dictator’s voting weight must be greater than or equal to . If there is a dictator, it has to be the weight-8 voter. The quota must be at least 8, by the result of Self Check 1. Thus if the quota is 8, the first voter is dictator. If the quota is more than 8, no voter is dictator.
3. The dummy would be the weight-1 voter. If is an even number, this voter will make no difference. If some of the other voters get together to pass a motion, and their weights add up to a number less than the quota, will be even, so at least 2 more votes will be needed to reach the quota. If the quota is 8, then the weight-8 voter is dictator, so the other three voters are dummies. If the quota is 12, the weight-2 voter is also a dummy. For or 14, the weight-1 voter is the only dummy. On the other hand, if the quota is odd—9, 11, 13, or 15—there are no dummy voters.
4. The weight-8 voter has veto power for any value of the quota from 8 to 15, because if she votes “no,” there are at most 7 votes in favor. If the quota is 12 or more (up to 15), the weight-4 voter also has veto power; if he votes “no,” there are at most 11 votes in favor. If the quota is 14 or 15, the weight-2 voter has veto power, because if she votes “no,” there are at most 13 votes in favor. Thus, if the quota is 12 or 13, exactly two voters, those with weights 8 and 4, have veto power. For , both of the other voters are dummies; for , no voter is a dummy.
5. In the voting permutation Allen, Cao, Betty, the first member, Allen, has 5 votes, less than the quota. Cao and Allen together have 6 votes, which is equal to the quota, so these two voters can pass a measure. Hence Cao is the pivot.
6. The quota must be more than either voter’s weight, because there is no dictator. Therefore, both voters are needed to pass a motion; so in both permutations of the voters, the second voter is pivotal. Thus, each voter is pivotal in one permutation, and each has a Shapley- Shubik index equal to .
7. We have seen that with five voters there are 120 permutations. With six voters, there are permutations. Finally, with seven voters, there are permutations.
8. Four permutations would acquire new pivots: would be the new pivot in and (the pivots had been and , respectively); and would be the new pivot in and (again, the pivots had been and , respectively). Thus, and each lost two permutations, and and each gained two. Now is pivot in 12 permutations, while ,, and are each pivot in four. The Shapley–Shubik index of is now , while ,, and have Shapley–Shubik indices equal to .
9. Alice will be the pivot in any permutation with between 30 and 59 weight-1 voters coming before her— in other words, if she is in position 31 up to 60. If we separate the permutations into 90 groups, according to Alice’s position, she is the pivot in every permutation in 30 of the groups and not the pivot in the remaining 60. Her Shapley–Shubik power index is therefore . The other 89 members of the committee have equal shares of the remaining of the power. Each weight-1 voter’s Shapley–Shubik power index is .
10. Bill is one vote short of being a dictator. He will be pivot in any permutation when he is in position 2, 3, or 4. He will not be pivot in position 1, for then the pivot will be whoever is in position 2, and he will not be pivot in position 5, for then the pivot will be the occupant of position 4. Therefore, he is pivot in of the permutations. Alice’s power is the same as a weight-1 voter’s, so she and the three weight-1 voters share the remaining of the power. Each has a Shapley–Shubik index equal to .
11. No state was a critical voter. It is only necessary to see if California, which had the largest number of electoral votes in the winning coalition, was not a critical voter. The winning coalition had 332 electoral votes, 62 more than the quota. California had 55 votes, so the Democratic ticket would still win by a margin of 7 without California.

12. The weight-4 voter is a dictator and is thus a critical voter in every winning coalition. Each winning coalition consists of the dictator and a subset of the three weight-1 voters. There are subsets of a three-element set—hence there are 8 winning coalitions. The dictator’s Banzhaf power index is therefore 8; the other voters are dummies, and their Banzhaf power indices are all 0.

13. No. Suppose that voter V is a critical voter in the grand coalition, and let be any coalition. If V does not belong to , then is a subset of the grand coalition with V removed, which we know is a losing coalition. Therefore, all winning coalitions must include V, and hence V has veto power.
14. Yes. In Example 12, we saw that {Allen, Betty} was a minimal winning coalition. It has 2 extra votes. The extra votes of a minimal winning coalition must be less than the voting weight of each member of the coalition.
15. You could get by removing from , and by removing from .

19. 		Approximation			Approximation
    2	50.0%		10	24.61%
    4	37.5%		20	17.62%
    6	31.2%		100	7.96%

20. Suppose X is a voter who belongs to no minimal winning coalition. If X happens to belong to some winning coalition W, then because W is winning, it contains a minimal winning coalition М. We know X does not belong to М, so the coalition formed by removing X from W would still be a winning coalition. This means X has no opportunity to cast a critical vote in any winning coalition: X is a dummy.
21. No, and do not overlap.
22. Let X be a voter who belongs to all of the minimal winning coalitions. If С is a coalition that X doesn’t belong to, then С contains no minimal winning coalition and is thus a losing coalition. Therefore the vote of X is necessary to win; that is, X has veto power.
23. There are four minimal winning coalitions; each consists of the dean and three of the other four members.