A weighted voting system is one in
which the preferences of some voters carry more weight than the preferences of other
voters.
(Here we will consider only those situations in which the voting is between 2
alternatives.)
KEY POINT
When a vote deals with only two alternatives, all reasonable voting methods have the
same outcome as "majority rule."
For this reason, our major interest here will not be comparing voting systems but
rather, the concept of POWER: Who has it and how much do they have?
TERMINOLOGY
Motion
Players
Weights
Quota
Dictator
Dummy
Veto Power
Coalition
the weight of a coalition
winning coalitions
losing coalitions
grand coalition
NOTATION
If player P_{1} has w_{1
}votes, player P_{2} has w_{2 }votes,
P_{3} has w_{3 }votes,…, and player
P_{N} has w_{N }votes and the quota is q, then we will write { q : w_{1
}, w_{2 }, w_{3 },…,
w_{N }}.
Example:
The three stockholders in a small company form a Board of Directors to oversee the
company. John (P_{1 }) has 5 votes as the largest
stockholder, Ginny (P_{2 }) has 3 votes, and Ann (P_{3}
) has 2 votes. The quota is 7; that is, it takes 7 or more votes to pass a motion. This
weighted voting system is represented mathematically as { 7: 5, 3, 2 }.
ANOTHER KEY POINT
The weight of a player
is not a good measure of a player’s POWER.
For instance, it is possible for a weighted voting
system to actually reduce to a oneperson, onevote situation in which case all players
have the same POWER even though they don’t all have the same weight. For example, for
the weighted voting system { 5: 4, 3, 2 } none of the players can pass a motion
alone and any two players can join together to pass a motion; so, although the players
have different weights, each player has the same amount of power.
We will discuss two ways of defining POWER.
Banzhaf Power Index
Key Points
 A player whose desertion of a winning coalition turns it
into a losing one is called a critical player.
 A player’s power is proportional to the number of
times the player is critical.
Banzhaf Power Index for Player P: BPI(P)
Computing the BPI for Player P:
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.)
STEP 2: Determine the critical players in each winning
coalition.
STEP 3: Find the number of times all players are critical.
STEP 4: Find the number of times Player P is critical.
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.
Banzhaf Power Index example
Underlying Assumptions:
 Players can enter and leave coalitions freely.
 A player’s power is proportional to the number of
times the player is critical.
ShapleyShubik
Power Index
Key Points
 A sequential coalition is one in which the players
are listed in the order that they entered the coalition.
 A pivotal player is the player in a sequential coalition
who changes the coalition from a losing to a winning one.
 There are N!
sequential coalitions containing all N players.
ShapleyShubik Power Index for Player P: SSPI(P)
Computing the SSPI for Player P:
STEP 1: Make a list of all N! sequential
coalitions containing N players.
STEP 2: In each of these coalitions, determine the pivotal
player.
(There is always exactly one in each of these coalitions.)
STEP 3: Determine the number of times Player P
is pivotal.
STEP 4: SSPI(P) is the smaller # (the one from STEP 3)
divided by the larger number, N! (the one from STEP 1)
DO THIS FOR EACH PLAYER AND YOU HAVE THE SHAPLEYSHUBIK
POWER DISTRIBUTION.
ShapleyShubik Power Index example
Underlying Assumptions:
 Players enter coalitions in a certain order.
 Players are NOT free to leave a coalition once in it.
 A player’s power is proportional to the number of
times the player is pivotal.
Looking at Power from a Banzhaf and ShapleyShubik
Point of View
Example
{4: 3, 2, 2} Which player has the most power?
 from a Banzaf point of view:
Look at all the winning coalitions.
{P_{1}}
{P_{2}}
{P_{3}}
{P_{1}, P_{2}} (winning)
{P_{1}, P_{3}} (winning)
{P_{2}, P_{3}} (winning)
{P_{1}, P_{2}, P_{3}} (winning)
Find the critical players in each.
winning coalition {P_{1}, P_{2}}
{P_{1}, P_{3}}
{P_{2}, P_{3}}
{P_{1}, P_{2}, P_{3}} 
critical players P_{1 }&_{ }P_{2
}P_{1 }&_{ }P_{3
}P_{2 }&_{ }P_{3
}none 
BPI(P) = (# of times player P is critical) / (# times all are critical)
BPI(P_{1 }) = 2/6 = 1/3 ; BPI(P_{2 })
= 2/6 = 1/3 ; BPI(P_{3 }) = 2/6 = 1/3
 from a ShapleyShubik point of view:
Find all sequential coalitions containing all players (N!
of them).
{P_{1}, P_{2}, P_{3}}
{P_{1}, P_{3}, P_{2}}
{P_{2}, P_{1}, P_{3}}
{P_{2}, P_{3}, P_{1}}
{P_{3}, P_{2}, P_{2}}
{P_{3}, P_{2}, P_{1}}
Find the pivotal player in each.
sequential coalition {P_{1}, P_{2}, P_{3}}
{P_{1}, P_{3}, P_{2}}
{P_{2}, P_{1}, P_{3}}
{P_{2}, P_{3}, P_{1}}
{P_{3}, P_{1}, P_{2}}
{P_{3}, P_{2}, P_{1}} 
pivotal player P_{2
}P_{3
}P_{1
}P_{3
}P_{2
}P_{2} 
SSPI(P) = (# of times player P is pivotal) / (# times all
are pivotal)
SSPI(P_{1 }) = 1/6 ; SSPI(P_{2 }) =
3/6 = 1/2 ; SSPI(P_{3}) = 2/6 = 1/3
Example:
{5: 3, 2, 2} Which player has the most power?
 from a Banzaf point of view:
Look at all the winning coalitions.
{P_{1}}
{P_{2}}
{P_{3}}
{P_{1}, P_{2}}(winning)
{P_{1}, P_{3}} (winning)
{P_{2}, P_{3}}
{P_{1}, P_{2}, P_{3}} (winning)
Find the critical players in each.
winning coalitions {P_{1}, P_{2}}
{P_{1}, P_{3}}
{P_{1}, P_{2}, P_{3}} 
critical players P_{1}& P_{2
}P_{1}& P_{3
}P_{1} 
BPI(P) = (# of times player P is
critical) / (# times all are critical)
BPI(P_{1}) = 3/5 ; BPI(P_{2}) = 1/5 ;
BPI(P_{3}) = 1/5
 from a ShapleyShubik point of view:
Find all sequential coalitions containing all players (N!
of them).
{P_{1}, P_{2}, P_{3}}
{P_{1}, P_{3}, P_{2}}
{P_{2}, P_{1}, P_{3}}
{P_{2}, P_{3}, P_{1}}
{P_{3}, P_{1 }, P_{2}}
{P_{3}, P_{2}, P_{1}}
Find the pivotal player in each.
sequential coalitions {P_{1}, P_{2}, P_{3}}
{P_{1}, P_{3}, P_{2}}
{P_{2}, P_{1}, P_{3}}
{P_{2}, P_{3}, P_{1}}
{P_{3}, P_{1}, P_{2}}
{P_{3}, P_{2}, P_{1}} 
pivotal player P_{2
}P_{3
}P_{1
}P_{1
}P_{1
}P_{1} 
SSPI(P) = (# of times player P is
pivotal) / (# times all are pivotal)
SSPI(P_{1}) = 4/6 = 2/3 ; SSPI(P_{2})
= 1/6 ; SSPI(P_{1}) = 1/6
