Question

consider the weighted voting system 9; 7, 4, 1. find the shapley - shubik power distribution of this weighted voting system. list the power for each player as a fraction: p1: p2: p3:

Explanation:

Step1: Calculate total number of permutations

There are 3 players, so the total number of permutations is $n!=3!=3\times2\times 1 = 6$.

Step2: Analyze each permutation

Let the players with weights 7, 4, 1 be $P_1$, $P_2$, $P_3$ respectively.

• Permutation $(P_1,P_2,P_3)$: Cumulative weights are 7, 7 + 4=11, 11+1 = 12. The pivotal player is $P_2$.
• Permutation $(P_1,P_3,P_2)$: Cumulative weights are 7, 7 + 1=8, 8+4 = 12. The pivotal player is $P_2$.
• Permutation $(P_2,P_1,P_3)$: Cumulative weights are 4, 4 + 7=11, 11+1 = 12. The pivotal player is $P_1$.
• Permutation $(P_2,P_3,P_1)$: Cumulative weights are 4, 4 + 1=5, 5+7 = 12. The pivotal player is $P_1$.
• Permutation $(P_3,P_1,P_2)$: Cumulative weights are 1, 1 + 7=8, 8+4 = 12. The pivotal player is $P_1$.
• Permutation $(P_3,P_2,P_1)$: Cumulative weights are 1, 1 + 4=5, 5+7 = 12. The pivotal player is $P_1$.

Step3: Calculate Shapley - Shubik power

The Shapley - Shubik power of a player is the number of times it is a pivotal player divided by the total number of permutations.
$P_1$ is a pivotal player 4 times, so its power is $\frac{4}{6}=\frac{2}{3}$.
$P_2$ is a pivotal player 2 times, so its power is $\frac{2}{6}=\frac{1}{3}$.
$P_3$ is a pivotal player 0 times, so its power is $\frac{0}{6}=0$.

Answer:

$P_1:\frac{2}{3}$
$P_2:\frac{1}{3}$
$P_3:0$