Question

find the banzhaf power distribution of the weighted voting system 25: 16, 11, 9, 4. give each players power as a fraction or decimal value. $p_1 = $ $p_2 = $ $p_3 = $ $p_4 = $

Explanation:

Step1: Define winning - losing coalitions

A weighted - voting system is given as $[q: w_1,w_2,w_3,w_4]=[25:16,11,9,4]$. A coalition is a subset of the set of players $\{P_1,P_2,P_3,P_4\}$. A coalition is winning if the sum of the weights of its members is at least $q = 25$, and losing otherwise.

Step2: List all possible coalitions and their weights

The number of possible coalitions of 4 players is $2^4=16$.

• $\varnothing$ (weight = 0, losing)
• $\{P_1\}$ (weight = 16, losing)
• $\{P_2\}$ (weight = 11, losing)
• $\{P_3\}$ (weight = 9, losing)
• $\{P_4\}$ (weight = 4, losing)
• $\{P_1,P_2\}$ (weight = $16 + 11=27$, winning)
• $\{P_1,P_3\}$ (weight = $16+9 = 25$, winning)
• $\{P_1,P_4\}$ (weight = $16 + 4=20$, losing)
• $\{P_2,P_3\}$ (weight = $11 + 9=20$, losing)
• $\{P_2,P_4\}$ (weight = $11 + 4=15$, losing)
• $\{P_3,P_4\}$ (weight = $9 + 4=13$, losing)
• $\{P_1,P_2,P_3\}$ (weight = $16+11 + 9=36$, winning)
• $\{P_1,P_2,P_4\}$ (weight = $16+11 + 4=31$, winning)
• $\{P_1,P_3,P_4\}$ (weight = $16+9 + 4=29$, winning)
• $\{P_2,P_3,P_4\}$ (weight = $11+9 + 4=24$, losing)
• $\{P_1,P_2,P_3,P_4\}$ (weight = $16+11 + 9+4=40$, winning)

Step3: Determine pivotal players

A player is pivotal in a winning coalition if the coalition without that player is losing.

• In $\{P_1,P_2\}$, $P_1$ is pivotal (since $\{P_2\}$ is losing).
• In $\{P_1,P_3\}$, $P_1$ is pivotal (since $\{P_3\}$ is losing).
• In $\{P_1,P_2,P_3\}$, $P_1$ is pivotal (since $\{P_2,P_3\}$ is losing), $P_2$ is not pivotal, $P_3$ is not pivotal.
• In $\{P_1,P_2,P_4\}$, $P_1$ is pivotal (since $\{P_2,P_4\}$ is losing), $P_2$ is not pivotal, $P_4$ is not pivotal.
• In $\{P_1,P_3,P_4\}$, $P_1$ is pivotal (since $\{P_3,P_4\}$ is losing), $P_3$ is not pivotal, $P_4$ is not pivotal.
• In $\{P_1,P_2,P_3,P_4\}$, $P_1$ is not pivotal.

The Banzhaf power index of a player $P_i$ is the number of times $P_i$ is a pivotal player.

• For $P_1$: $P_1$ is pivotal in $\{P_1,P_2\}$, $\{P_1,P_3\}$, $\{P_1,P_2,P_3\}$, $\{P_1,P_2,P_4\}$, $\{P_1,P_3,P_4\}$, so $BI_1 = 5$.
• For $P_2$: $P_2$ is pivotal in no winning - coalitions, so $BI_2 = 0$.
• For $P_3$: $P_3$ is pivotal in no winning - coalitions, so $BI_3 = 0$.
• For $P_4$: $P_4$ is pivotal in no winning - coalitions, so $BI_4 = 0$.

The total number of pivotal - player occurrences is $5+0 + 0+0=5$.

Step4: Calculate Banzhaf power distribution

$P_1=\frac{5}{5}=1$
$P_2=\frac{0}{5}=0$
$P_3=\frac{0}{5}=0$
$P_4=\frac{0}{5}=0$

Answer:

$P_1 = 1$
$P_2 = 0$
$P_3 = 0$
$P_4 = 0$