Question

consider the weighted voting system 8: 8, 4, 2, 1
find the banzhaf power distribution of this weighted voting system.
list the power for each player as a fraction:
$p_1$:
$p_2$:
$p_3$:
$p_4$:

Explanation:

Step1: Define winning coalitions

A coalition is a subset of players. A winning coalition is one whose total weight is at least the quota. Here, the quota is 8 and the weights of players are 8, 4, 2, 1 for $P_1$, $P_2$, $P_3$, $P_4$ respectively. The winning coalitions are: $\{P_1\}$, $\{P_1,P_2\}$, $\{P_1,P_3\}$, $\{P_1,P_4\}$, $\{P_1,P_2,P_3\}$, $\{P_1,P_2,P_4\}$, $\{P_1,P_3,P_4\}$, $\{P_1,P_2,P_3,P_4\}$, $\{P_2,P_3,P_4\}$.

Step2: Determine pivotal players

A pivotal player in a coalition is one whose addition makes the coalition winning.

• For $\{P_1\}$, $P_1$ is pivotal.
• For $\{P_1,P_2\}$, $P_1$ is pivotal (since adding $P_1$ makes it winning).
• For $\{P_1,P_3\}$, $P_1$ is pivotal.
• For $\{P_1,P_4\}$, $P_1$ is pivotal.
• For $\{P_1,P_2,P_3\}$, $P_1$ is pivotal.
• For $\{P_1,P_2,P_4\}$, $P_1$ is pivotal.
• For $\{P_1,P_3,P_4\}$, $P_1$ is pivotal.
• For $\{P_1,P_2,P_3,P_4\}$, $P_1$ is pivotal.
• For $\{P_2,P_3,P_4\}$, $P_2$ is pivotal (as $4 + 2+1=7$ without $P_2$ and with $P_2$ it is $4 + 2+1 + 4=11$).

The number of times $P_1$ is pivotal is 8, the number of times $P_2$ is pivotal is 1, the number of times $P_3$ is pivotal is 0, and the number of times $P_4$ is pivotal is 0.

Step3: Calculate Banzhaf power

The total number of pivotal - player occurrences is $8 + 1+0 + 0=9$.
The Banzhaf power of $P_1$ is $\frac{8}{9}$, of $P_2$ is $\frac{1}{9}$, of $P_3$ is $\frac{0}{9}=0$, and of $P_4$ is $\frac{0}{9}=0$.

Answer:

$P_1:\frac{8}{9}$
$P_2:\frac{1}{9}$
$P_3:0$
$P_4:0$