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