Compute the BPI for the weighted voting system [6: 4, 3, 2].

Solution:

STEP 1: Determine all WINNING coalitions.
(If you can list just the winning coalitions, then there is no need to list all of the 2^N -1 coalitions.)

Winning Coalitions	Explanation
	No Player can win alone so there are no 1 Player winning coalitions.
{P1 , P2} {P1 , P3}	Players 1 & 2 together have enough votes to win.  Also, Players 1 & 3 together have enough votes to win.  However, Players 2 & 3 do not. So, there are 2 winning coalitions having two players.
{P1 , P2 , P3}	Of course, the coalition containing all 3 Players wins.

 

STEP 2: Determine the critical players in each winning coalition.

(Remember, P1 has 4 votes, P2 has 3 votes, and P3 has 2 votes.
To determine whether or not a Player is a critical Player in a winning coalition, count   the number of votes the coalition has without that particular Player.  If the coalition no longer has enough votes to win, then that Player is critical.)
The critical Players are underlined below.

In {P1 , P2}, both are critical since the coalition loses if either Player leaves.

In {P1 , P3}, both are critical since the coalition loses if either Player leaves.

In {P1 , P2 , P3}, only P1 is critical since the coalition still wins if P2 leaves or if P3 leaves (but not if P1 leaves).

 

STEP 3: Find the number of times all players are critical.

5 (underlined above).

STEP 4: Find the number of times Player P is critical.

(See the underlined Players in STEP 2 above.)

P1 is critical 3 times, P2 is critical 1 time, P3 is critical 1 time.

STEP 5: BPI(P) is the smaller # (the one from STEP 4) divided by the larger number (the one from STEP 3).  DO THIS FOR EACH PLAYER AND YOU HAVE THE BANZHAF POWER DISTRIBUTION.

BPI(P1) = 3/5           BPI(P2) = 1/5           BPI(P3) = 1/5

Back