To find the Shapley-Shubik power distribution for the voting arrangement [tex]$[15: 12,7,3,1]$[/tex], we follow these detailed steps:
1. Understand the Setup:
- We have a quota of 15.
- There are 4 players with weights 12, 7, 3, and 1 respectively.
2. Permutations of Players:
- Consider all possible permutations of the 4 players. With 4 players, there are [tex]$4! = 24$[/tex] permutations.
3. Calculate Shapley-Shubik Index:
- For each permutation, determine the pivotal player. The pivotal player is the one who, when added to the coalition, causes the total weight to meet or exceed the quota.
- Track how many times each player is the pivotal player across all permutations.
4. Pivotal Players:
- For each permutation, sum up the weights until the quota of 15 is reached or exceeded. Identify the player who contributes to reaching or exceeding the quota.
- Count how often each player is pivotal in these 24 permutations.
- So, let's label our players as P1 (with weight 12), P2 (with weight 7), P3 (with weight 3), and P4 (with weight 1).
- Let's consider permutations one at a time and count the pivotal player for each permutation.
After compiling all this, we've found:
[tex]\[
\sigma_1 = \frac{16}{24} = \frac{2}{3},
\sigma_2 = \frac{4}{24} = \frac{1}{6},
\sigma_3 = \frac{4}{24} = \frac{1}{6},
\sigma_4 = \frac{0}{24} = 0.
\][/tex]
5. Result:
- Player 1 (P1): [tex]\( \sigma_1 = \frac{2}{3} \)[/tex]
- Player 2 (P2): [tex]\( \sigma_2 = \frac{1}{6} \)[/tex]
- Player 3 (P3): [tex]\( \sigma_3 = \frac{1}{6} \)[/tex]
- Player 4 (P4): [tex]\( \sigma_4 = 0 \)[/tex]
So, the Shapley-Shubik power distribution for the voting arrangement [tex]$[15: 12,7,3,1]$[/tex] is:
[tex]\[
\sigma_1 = \frac{2}{3}, \quad \sigma_2 = \frac{1}{6}, \quad \sigma_3 = \frac{1}{6}, \quad \sigma_4 = 0.
\][/tex]
