Section 2.4 Try it Now Solutions
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Try it Now 2.1.7 If we represent the players as \(M_1,\) \(M_2,\) \(M_3,\) \(M_4,\) \(U_1,\) \(U_2,\) \(U_3\text{,}\) then we may be tempted to set up a system like [4: 1, 1, 1, 1, 1, 1, 1]. While this system would meet the first requirement that four members must support a proposal for it to pass, this does not satisfy the requirement that at least one member of the union must support it.
To accomplish that, we might try increasing the voting weight of the union members: [5: 1, 1, 1, 1, 2, 2, 2]. The quota was set at 5 so that the four management members alone would not be able to reach quota without one of the union members. Unfortunately, now the three union members can reach quota alone. To fix this, three management members need to have more weight than two union members.
After trying several other guesses, we land on the system [13: 3, 3, 3, 3, 4, 4, 4]. Here, the four management members have combined weight of 12, so cannot reach quota. Likewise, the three union members have combined weight of 12, so cannot reach quota alone. But, as required, any group of 4 members that includes at least one union member will reach the quota of 13. For example, three management members and one union member have combined weight of 3+3+3+4=13, and reach quota.
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Try it Now 2.2.8 In the voting system \([q : 10, 5, 3]\text{,}\) if the quota is 10, then player 1 is a dictator since they can reach quota without the support of the other players. This makes the other two players automatically dummies.
If the quota is 12, then player 1 is necessary to reach quota, so has veto power. Since at this point either player 2 or player 3 would allow player 1 to reach quota, neither player is a dummy, so they are regular players (not dictators, no veto power, and not a dummy).
If the quota is 16, then both players 1 and 2 are necessary to reach quota, so they both have veto power. Player 3’s support is not necessary, so player 3 is a dummy.
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Try it Now 2.3.8 The voting system tells us that the quota is 36, that Player 1 has 20 votes (or equivalently, has a weight of 20), Player 2 has 17 votes, Player 3 has 16 votes, and Player 4 has 3 votes.A coalition is any group of one or more players. What we’re looking for is winning coalitions - coalitions whose combined votes (weights) add to up to the quota or more. So the coalition \(\{P3, P4\}\) is not a winning coalition because the combined weight is 16+3=19, which is below the quota.
So we look at each possible combination of players and identify the winning ones:
\(\{P1, P2\}\) (weight: 37) \(\{P1, P3\}\) (weight: 36) \(\{P1, P2, P3\}\) (weight: 53) \(\{P1, P2, P4\}\) (weight: 40) \(\{P1, P3, P4\}\) (weight: 39) \(\{P1, P2, P3, P4\}\) (weight: 56 \(\{P2, P3, P4\}\) (weight: 36)
Now, in each coalition, we need to identify which players are critical. A player is critical if the coalition would no longer reach quota without that person. So, in the coalition \(\{P1, P2\}\text{,}\) both players are necessary to reach quota, so both are critical. However in the coalition \(\{P1, P2, P3\}\text{,}\) we can see from the earlier two coalitions that either P2 or P3 could leave the coalition and it would still reach quota. But if P1 left, it would not reach quota, so P1 is the only player critical in this coalition. We evaluate the rest of the coalitions similarly, giving us this (underlining the critical players) \(\{P1, P2\}\) \(\{P1, P3\}\) \(\{P1, P2, P3\}\) \(\{P1, P2,P4\}\) \(\{P1, P3, P4\}\) \(\{P1, P2, P3, P4\}\) \(\{P2, P3, P4\}\)
Next we count how many times each player is critical:P1: 5 timesP2: 3 timesP3: 3 timesP4: 1 time In total, there were 5+3+3+1 = 12 times anyone was critical, so we take our counts and turn them into fractions, giving us our Banzhaf power:P1: 5/12P2: 3/12 = 1/4P3: 3/12 = 1/4P4: 1/12
[36: 20, 17, 15].