Question

consider the weighted voting system 12: 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 group of players. A winning coalition is one whose total weight is at least the quota. Here, the quota $q = 12$ and the weights of players are $w_1=8$, $w_2 = 4$, $w_3=2$, $w_4 = 1$. The winning coalitions are: $\{P_1,P_2\}$ (weight $8 + 4=12$), $\{P_1,P_3\}$ (weight $8+2 = 10<12$ is not winning), $\{P_1,P_4\}$ (weight $8 + 1=9<12$ is not winning), $\{P_2,P_3\}$ (weight $4+2 = 6<12$ is not winning), $\{P_2,P_4\}$ (weight $4 + 1=5<12$ is not winning), $\{P_3,P_4\}$ (weight $2+1 = 3<12$ is not winning), $\{P_1,P_2,P_3\}$ (weight $8 + 4+2=14$), $\{P_1,P_2,P_4\}$ (weight $8 + 4+1=13$), $\{P_1,P_3,P_4\}$ (weight $8+2 + 1=11<12$ is not winning), $\{P_2,P_3,P_4\}$ (weight $4+2 + 1=7<12$ is not winning), $\{P_1,P_2,P_3,P_4\}$ (weight $8 + 4+2+1=15$).

Step2: Determine pivotal players

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

• For $\{P_1,P_2\}$: $P_1$ is pivotal (if we add $P_1$ first, the partial - coalition $\{P_1\}$ has weight $8<12$, adding $P_2$ makes it winning), $P_2$ is also pivotal (if we add $P_2$ first, weight is $4<12$, adding $P_1$ makes it winning).
• For $\{P_1,P_2,P_3\}$: $P_3$ is pivotal (since $\{P_1,P_2\}$ has weight $12$ and adding $P_3$ is the last step to form this coalition).
• For $\{P_1,P_2,P_4\}$: $P_4$ is pivotal (since $\{P_1,P_2\}$ has weight $12$ and adding $P_4$ is the last step to form this coalition).
• For $\{P_1,P_2,P_3,P_4\}$: $P_3$ and $P_4$ are non - pivotal (because $\{P_1,P_2\}$ is already winning).

The total number of times a player is pivotal:

• $P_1$ is pivotal $2$ times.
• $P_2$ is pivotal $2$ times.
• $P_3$ is pivotal $2$ times.
• $P_4$ is pivotal $2$ times.

The total number of pivotal player occurrences is $2 + 2+2 + 2=8$.

Step3: Calculate Banzhaf power

The Banzhaf power of a player is the number of times it is pivotal divided by the total number of pivotal player occurrences.

• For $P_1$: $\frac{2}{8}=\frac{1}{4}$.
• For $P_2$: $\frac{2}{8}=\frac{1}{4}$.
• For $P_3$: $\frac{2}{8}=\frac{1}{4}$.
• For $P_4$: $\frac{2}{8}=\frac{1}{4}$.

Answer:

$P_1:\frac{1}{4}$
$P_2:\frac{1}{4}$
$P_3:\frac{1}{4}$
$P_4:\frac{1}{4}$