Consider the weighted voting system [q: 8,7,6,2,1,1] . What is the smallest value that the quota q can take so that P 1, P 2 and P 3 are the only players to have veto power?
A) 20
B) 21
C) 22
D) 23
E) None of the above.

Question

Consider the weighted voting system  [q: 8,7,6,2,1,1] . What is the smallest value that the quota  q  can take so that  P 1, P 2  and  P 3  are the only players to have veto power?
A) 20
B) 21
C) 22
D) 23
E) None of the above.

Quizplus · Accepted Answer

For P1, P2, and P3 to be the only players with veto power, the quota must be set such that any two of them can meet the quota, but none of the other players (P4, P5, P6) can contribute enough to meet the quota without at least one of P1, P2, or P3. The weights are 7, 6, and 2 for P4, P5, and P6, respectively, totaling 15 when combined. To ensure P1, P2, and P3 are the only ones with veto power, the quota must be more than 15 (the total of the other players) but not more than the smallest combination of P1, P2, and P3, which is 8+7 (P1 and P2) = 15. To ensure they have veto power, the quota must be set to the smallest value that requires at least one of them to form a winning coalition, which is 15 + 1 = 16. However, since the correct answer must be among the provided options and my initial calculation seems to have misunderstood the question's requirement, let's correct the approach based on the options given.Given the weights, for P1, P2, and P3 to be the only ones with veto power, the quota must be such that any coalition without at least one of them cannot reach the quota. The total weight of P4, P5, and P6 is 4 (2+1+1), meaning any quota above 14 ensures that at least one of P1, P2, or P3 is needed for a winning coalition. However, to ensure P1, P2, and P3 are the only ones with veto power, the quota must be set just above the sum of the weights of any two of the three smallest players (P4, P5, P6), which is not correct since we're considering veto power, not just forming coalitions.The correct approach is to ensure that the quota is such that if any one of P1, P2, or P3 is removed, the remaining players cannot meet or exceed the quota. The sum of the weights of the three largest players (P1, P2, P3) is 21. For them to be the only ones with veto power, the quota must be set to 21, because if it's any lower, other combinations without one of them could potentially meet the quota, and if it's any higher, it would require their unanimous agreement, which doesn't isolate their individual veto power. Therefore, the smallest quota that ensures P1, P2, and P3 are the only ones with veto power is 21, as it is the smallest number that requires the participation of at least one of them to form a winning coalition, ensuring their individual veto power.

Consider the Weighted Voting System [Q: 8,7,6,2,1,1]