Example using the Shapley-Shubik Power Index Compute the SSPI for the weighted voting system [6: 4, 3,
2].
Solution:
STEP 1: Make a list of all N! (=3!=6)
sequential coalitions containing N players.
{P1 , P2 , P3}
{P1 , P3 , P2}
{P2 , P1 , P3}
{P2 , P3 , P1}
{P3 , P1 , P2}
{P3 , P2 , P1}
STEP 2: In each of these N! sequential
coalitions, determine the pivotal player.
(There is always exactly one in each of these coalitions for a total of N!.)
(Remember, P1 has 4 votes, P2 has 3 votes, and P3 has
2 votes.
To determine the pivotal Player in each sequential coalition, add votes from left to
right. The Player whose votes cause the coalition's votes to equal or exceed the
quota is the pivotal Player. The pivotal Players are underlined below.
{P1 , P2 , P3} 4 (P1) + 3 (P2) = 7
{P1 , P3 , P2} 4 (P1) + 2 (P3) = 6
{P2 , P1 , P3} 3 (P2) + 4 (P1) = 7
{P2 , P3 , P1} 3 (P2) + 2 (P3) + 4 (P1) = 9
{P3 , P1 , P2} 2 (P3) + 4 (P1) = 6
{P3 , P2 , P1} 2 (P3) + 3 (P2) + 4 (P1) = 9
Here, N! = 3! = 3(2)(1) = 6.
STEP 3: Determine the number of times Player P
is pivotal.
P1 is pivotal 4 times, P2 is pivotal 1 time, P3 is pivotal
1 time.
STEP 4: SSPI(P) is the smaller # (the one from
STEP 3) divided by the larger number (the N! from STEP 2). DO THIS FOR EACH
PLAYER AND YOU HAVE THE SHAPLEY-SHUBIK POWER DISTRIBUTION.
BPI(P1) = 4/6 = 2/3
BPI(P2) = 1/6
BPI(P3) = 1/6
Back |