Banzhaf power index

Computer model of the Banzhaf power index from the Wolfram Demonstrations Project

The Banzhaf power index, named after John Banzhaf (originally invented by Lionel Penrose in 1946 and sometimes called Penrose–Banzhaf index; also known as the Banzhaf–Coleman index after James Samuel Coleman), is a power index defined by the probability of changing an outcome of a vote where voting rights are not necessarily equally divided among the voters or shareholders.

To calculate the power of a voter using the Banzhaf index, list all the winning coalitions, then count the critical voters. A critical voter is a voter who, if he changed his vote from yes to no, would cause the measure to fail. A voter's power is measured as the fraction of all swing votes that he could cast. There are some algorithms for calculating the power index, e.g., dynamic programming techniques, enumeration methods and Monte Carlo methods.^[1]

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- WELCOME TO A LESSON ON THE BANZHAF POWER INDEX. IN THIS LESSON, WE'LL DEFINE THE BANZHAF POWER INDEX AND ALSO FIND THE BANZHAF POWER INDEX FOR A WEIGHTED VOTING SYSTEM. THE BANZHAF POWER INDEX WAS ORIGINALLY CREATED IN 1946 BY LIONEL PENROSE, BUT WAS REINTRODUCED IN JOHN BANZHAF IN 1965. THE POWER INDEX IS A NUMERICAL WAY OF LOOKING AT POWER IN A WEIGHTED VOTING SYSTEM. TO CALCULATE THE BANZHAF POWER INDEX, NUMBER ONE, WE WANT TO FIND OR LIST THE WINNING COALITIONS. TWO, IN EACH WINNING COALITION WE WANT TO IDENTIFY THE PLAYERS WHO ARE CRITICAL. THREE, WE WANT TO COUNT HOW MANY TIMES EACH PLAYER IS CRITICAL. AND THEN FINALLY, NUMBER FOUR, WE WANT TO CONVERT THESE COUNTS TO FRACTIONS, DECIMALS, OR PERCENTAGES BY DIVIDING BY THE TOTAL NUMBER OF TIMES ANY PLAYER IS CRITICAL. LET'S LOOK AT AN EXAMPLE. WE WANT TO FIND THE BANZHAF POWER INDEX FOR THE WEIGHTED VOTING SYSTEM GIVEN HERE. SO WE WANT TO BEGIN BY LISTING THE WINNING COALITIONS, BUT TO DO THIS WE'LL LIST ALL THE COALITIONS AND THEN FIND WHICH COALITIONS ARE WINNING COALITIONS. NOTICE HOW IN THIS CASE THOUGH, WE DON'T HAVE TO LIST ONE PLAYER COALITIONS BECAUSE WE DON'T HAVE ANY DICTATORS OR ANY PLAYERS THAT HAVE A WEIGHT THAT'S GREATER THAN OR EQUAL TO THE QUOTA HERE, WHICH IS 16. SO LETS BEGIN BY LISTING THE TWO PLAYER COALITIONS, WHICH WE SEE HERE. WE HAVE PLAYER ONE AND PLAYER TWO, PLAYER ONE AND PLAYER THREE, PLAYER ONE AND PLAYER FOUR, PLAYER TWO AND PLAYER THREE, PLAYER TWO AND PLAYER FOUR, AND PLAYER THREE AND PLAYER FOUR. LET'S FIND THE TOTAL WEIGHT OF THESE COALITIONS TO DETERMINE THE WHICH ARE WINNING COALITIONS. FOR OUR FIRST COALITION THE TOTAL WEIGHT WOULD BE 12 + 6 = 18. SO THIS IS A WINNING COALITION. NEXT, WE HAVE PLAYER ONE AND PLAYER THREE. THAT WOULD BE 12 + 4 = 16. THIS ALSO MEETS QUOTA. NEXT, WE HAVE PLAYER ONE AND PLAYER FOUR. THAT WOULD BE 12 + 2 = 14. THIS IS NOT A WINNING COALITION. NEXT, WE HAVE PLAYER TWO AND PLAYER THREE. THAT WOULD BE 6 + 4 OR A WEIGHT OF 10, WHICH AGAIN DOES NOT MEET QUOTA. PLAYER TWO AND PLAYER FOUR WOULD HAVE A WEIGHT OF 6 + 2 = 8, WHICH DOES NOT MEET QUOTA. AND THEN FINALLY, PLAYER THREE AND PLAYER FOUR WOULD HAVE A WEIGHT OF 4 + 2 = 6, WHICH DOES NOT MEET QUOTA. SO WE ONLY HAVE TWO-TWO PLAYER WINNING COALITIONS. SO WE'LL COME BACK TO THESE. NOW, LETS FIND THE THREE PLAYER AND FOUR PLAYER WINNING COALITIONS. THERE ARE FOUR POSSIBLE THREE PLAYER COALITIONS AND ONLY ONE FOUR PLAYER COALITION. THE FOUR-THREE PLAYER COALITIONS ARE PLAYERS ONE, TWO, AND THREE THEN PLAYERS ONE, TWO, AND FOUR, THEN PLAYERS ONE, THREE, AND FOUR, AND PLAYERS TWO, THREE AND FOUR. LET'S DETERMINE WHICH OF THESE ARE WINNING COALITIONS. PLAYERS ONE, TWO, AND THREE WOULD HAVE A COMBINED WEIGHT OF 12 + 6 + 4 = 22, WHICH MEETS QUOTA. SO WE HAVE A WINNING COALITION HERE. NEXT, PLAYERS ONE, TWO, AND FOUR, THE WEIGHT WOULD BE 12 + 6 + 2 = 20, WHICH MEETS QUOTA. WE HAVE A WINNING COALITION. NEXT, PLAYERS ONE, THREE, AND FOUR, WE'D HAVE 12 + 4 + 2 = 18. THAT MEETS QUOTA. PLAYERS TWO, THREE, AND FOUR, THAT WOULD BE 6 + 4 + 2 = 12, WHICH DOES NOT MEET QUOTA. AND THE COALITION OF ALL FOUR PLAYERS WOULD BE QUOTA 12 + 6 + 4 + 2 = 24, WHICH OF COURSE DOES MEET QUOTA. IF ANY THREE PLAYER COALITION MEETS QUOTA, OF COURSE THIS FOUR PLAYER COALITION WOULD ALSO HAVE TO MEET QUOTA. SO WE HAVE A TOTAL OF ONE, TWO, THREE, FOUR, FIVE, SIX WINNING COALITIONS, WHICH WE HAVE LISTED HERE. NOW WE'LL GO THROUGH AND DETERMINE WHICH PLAYERS ARE CRITICAL IN EACH WINNING COALITIONS. SO FOR PLAYER ONE AND PLAYER TWO, IF PLAYER ONE LEAVES THE COALITION THE WEIGHT IS 6, WHICH DOESN'T MEET QUOTA, WHICH MEANS PLAYER ONE IS CRITICAL SO UNDERLINING IT IN RED MEANS THAT IT'S CRITICAL. IF PLAYER TWO LEAVES THE COALITION AND WE HAVE A WEIGHT OF 12, WHICH DOES NOT MEET QUOTA. PLAYER TWO IS CRITICAL. NEXT, WE HAVE THE WINNING COALITION OF PLAYER ONE AND PLAYER THREE. IF PLAYER ONE LEAVES, THE REMAINING WEIGHT WOULD BE FOUR, WHICH DOES NOT MEET QUOTA SO PLAYER ONE IS CRITICAL AGAIN. IF PLAYER THREE LEAVES THE WEIGHT WOULD BE 12, WHICH DOES NOT MEET QUOTA SO PLAYER THREE IS CRITICAL. FOR THE COALITION WITH PLAYERS ONE, TWO, AND THREE, IF ONE LEAVES WE'RE LEFT WITH PLAYER TWO AND THREE THAT HAVE A WEIGHT OF 10, WHICH DOES NOT MEET QUOTA. ONCE AGAIN, PLAYER ONE IS CRITICAL. IF PLAYER TWO LEAVES, THE WEIGHT WOULD BE 12 + 4 = 16, WHICH DOES MEET QUOTA SO PLAYER TWO IS NOT CRITICAL. IF PLAYER THREE LEAVES THE REMAINING WEIGHT WOULD BE 12 + 6 = 18, WHICH DOES MEET QUOTA. THEREFORE, PLAYER THREE IS NOT CRITICAL. NEXT, WE HAVE PLAYERS ONE, TWO, AND FOUR. IF PLAYER ONE LEAVES WE'RE LEFT WITH PLAYER TWO AND FOUR. THEY WOULD HAVE A WEIGHT OF 6 + 2 = 8. PLAYER ONE IS CRITICAL. IF PLAYER TWO LEAVES WE HAVE PLAYER ONE AND PLAYER FOUR THAT HAVE A COMBINED WEIGHT OF 14, WHICH DOES NOT MEET QUOTA AND THEREFORE PLAYER TWO IS CRITICAL. IF PLAYER FOUR LEAVES THE REMAINING WEIGHT WOULD BE 12 + 6 = 18, WHICH DOES MEET QUOTA. PLAYER FOUR IS NOT CRITICAL. MOVING ALONG TO THE WINNING COALITION OF PLAYERS ONE, THREE, AND FOUR. IF ONE LEAVES, PLAYERS THREE AND HAVE A WEIGHT OF SIX, WHICH DOES NOT MEET QUOTA. PLAYER ONE IS CRITICAL. IF PLAYER THREE LEAVES WE'RE LEFT WITH PLAYER ONE AND PLAYER FOUR WITH A COMBINED WEIGHT OF 14, DOES NOT MEET QUOTA. THEREFORE, PLAYER THREE IS CRITICAL. IF PLAYER FOUR LEAVES WE HAVE PLAYER ONE AND THREE THAT HAVE A COMBINED WEIGHT OF 16. NOTICE IT'S 12 + 4, WHICH DOES STILL MEET QUOTA SO PLAYER FOUR IS NOT CRITICAL. AND THEN FOR THE LAST WINNING COALITION THAT CONTAINS ALL FOUR PLAYERS, IF PLAYER ONE LEAVES WE'RE LEFT WITH THE WEIGHT OF 6 + 4 + 2 = 12. THAT DOES NOT MEET QUOTA. PLAYER ONE IS CRITICAL. IF PLAYER TWO LEAVES WE'RE LEFT WITH THE WEIGHT OF 12 + 4 + 2 = 18, WHICH DOES MEET QUOTA. PLAYER TWO IS NOT CRITICAL. IF PLAYER THREE LEAVES, WE'RE LEFT WITH THE WEIGHT OF 12 + 6 + 2 = 20, WHICH DOES MEET QUOTA. PLAYER THREE IS NOT CRITICAL. AND FINALLY PLAYER FOUR, IF PLAYER FOUR LEAVES WE HAVE A WEIGHT OF 12 + 6 + 4, WHICH DOES MEET QUOTA. PLAYER FOUR IS NOT CRITICAL. NOW TO FIND THE BANZHAF POWER INDEX WE'LL FIRST COUNT THE NUMBER OF TIMES EACH PLAYER IS CRITICAL. SO PLAYER ONE IS CRITICAL ONE, TWO, THREE, FOUR, FIVE, SIX TIMES. PLAYER TWO IS CRITICAL ONE, TWO TIMES. PLAYER THREE IS CRITICAL ONE, TWO TIMES. AND PLAYER FOUR IS CRITICAL ZERO TIMES. NOW WE'LL FIND THE TOTAL NUMBER OF TIMES ANY PLAYER IS CRITICAL. 6 + 2 + 2 + 0 = 10 AND THEREFORE, TO FIND THE BANZHAF POWER INDEX FOR PLAYER ONE WOULD BE 6 DIVIDED BY 10, OR 3/5, WHICH WOULD BE 0.6, OR 60%. FOR PLAYER TWO THE POWER INDEX WOULD BE 2 DIVIDED BY 10 OR 1/5, WHICH IS EQUAL TO 20% AND THE SAME THING FOR PLAYER THREE. 2 DIVIDED BY 10, OR 1/5, WHICH IS EQUAL TO 20%. PLAYER FOUR WOULD JUST BE ZERO OR 0%. LOOKING AT THESE NUMBERS NOTICE HOW THIS TELLS US THAT PLAYER THREE HAS THREE TIMES THE VOTING POWER OF PLAYER TWO AND PLAYER THREE, BUT ALSO NOTICE THAT PLAYER TWO AND PLAYER THREE HAVE THE SAME POWER INDEX, WHICH MEANS THEY HAVE THE SAME VOTING POWER EVEN THOUGH THEIR WEIGHT IN THE VOTING SYSTEM IS DIFFERENT. ALSO, RECOGNIZE THAT P SUB ONE HAS VETO POWER AND P SUB FOUR IS A DUMMY PLAYER. I HOPE YOU FOUND THIS HELPFUL.

Examples

Voting game

Simple voting game

A simple voting game, taken from Game Theory and Strategy by Philip D. Straffin:^[2]

[6; 4, 3, 2, 1]

The numbers in the brackets mean a measure requires 6 votes to pass, and voter A can cast four votes, B three votes, C two, and D one. The winning groups, with underlined swing voters, are as follows:

AB, AC, ABC, ABD, ACD, BCD, ABCD

There are 12 total swing votes, so by the Banzhaf index, power is divided thus:

A = 5/12, B = 3/12, C = 3/12, D = 1/12

U.S. Electoral College

Consider the United States Electoral College. Each state has different levels of voting power. There are a total of 538 electoral votes. A majority vote is 270 votes. The Banzhaf power index would be a mathematical representation of how likely a single state would be able to swing the vote. A state such as California, which is allocated 55 electoral votes, would be more likely to swing the vote than a state such as Montana, which has 3 electoral votes.

Assume the United States is having a presidential election between a Republican (R) and a Democrat (D). For simplicity, suppose that only three states are participating: California (55 electoral votes), Texas (38 electoral votes), and New York (29 electoral votes).

The possible outcomes of the election are:

California (55)	Texas (38)	New York (29)	R votes	D votes	States that could swing the vote
R	R	R	122	0	none
R	R	D	93	29	California (D would win 84–38), Texas (D would win 67–55)
R	D	R	84	38	California (D would win 93–29), New York (D would win 67–55)
R	D	D	55	67	Texas (R would win 93–29), New York (R would win 84–38)
D	R	R	67	55	Texas (D would win 93–29), New York (D would win 84–38)
D	R	D	38	84	California (R would win 93–29), New York (R would win 67–55)
D	D	R	29	93	California (R would win 84–38), Texas (R would win 67–55)
D	D	D	0	122	none

The Banzhaf power index of a state is the proportion of the possible outcomes in which that state could swing the election. In this example, all three states have the same index: 4/12 or 1/3.

However, if New York is replaced by Georgia, with only 16 electoral votes, the situation changes dramatically.

California (55)	Texas (38)	Georgia (16)	R votes	D votes	States that could swing the vote
R	R	R	109	0	California (D would win 55–54)
R	R	D	93	16	California (D would win 71–38)
R	D	R	71	38	California (D would win 93–16)
R	D	D	55	54	California (D would win 109–0)
D	R	R	54	55	California (R would win 109–0)
D	R	D	38	71	California (R would win 93–16)
D	D	R	16	93	California (R would win 71–38)
D	D	D	0	109	California (R would win 55–54)

In this example, the Banzhaf index gives California 1 and the other states 0, since California alone has more than half the votes.

History

What is known today as the Banzhaf power index was originally introduced by Lionel Penrose in 1946^[3] and went largely forgotten.^[4] It was reinvented by John F. Banzhaf III in 1965,^[5] but it had to be reinvented once more by James Samuel Coleman in 1971^[6] before it became part of the mainstream literature.

Banzhaf wanted to prove objectively that the Nassau County board's voting system was unfair. As given in Game Theory and Strategy, votes were allocated as follows:^[2]

Hempstead #1: 9
Hempstead #2: 9
North Hempstead: 7
Oyster Bay: 3
Glen Cove: 1
Long Beach: 1

This is 30 total votes, and a simple majority of 16 votes was required for a measure to pass.^[a]

In Banzhaf's notation, [Hempstead #1, Hempstead #2, North Hempstead, Oyster Bay, Glen Cove, Long Beach] are A-F in [16; 9, 9, 7, 3, 1, 1]

There are 32 winning coalitions, and 48 swing votes:

AB AC BC ABC ABD ABE ABF ACD ACE ACF BCD BCE BCF ABCD ABCE ABCF ABDE ABDF ABEF ACDE ACDF ACEF BCDE BCDF BCEF ABCDE ABCDF ABCEF ABDEF ACDEF BCDEF ABCDEF

The Banzhaf index gives these values:

Hempstead #1 = 16/48
Hempstead #2 = 16/48
North Hempstead = 16/48
Oyster Bay = 0/48
Glen Cove = 0/48
Long Beach = 0/48

Banzhaf argued that a voting arrangement that gives 0% of the power to 16% of the population is unfair.^[b]

Today,^[when?] the Banzhaf power index is an accepted way to measure voting power, along with the alternative Shapley–Shubik power index. Both measures have been applied to the analysis of voting in the Council of the European Union.^[7]

However, Banzhaf's analysis has been critiqued as treating votes like coin-flips, and an empirical model of voting rather than a random voting model as used by Banzhaf brings different results.^[8]

Notes

^ Banzhaf did not understand how voting in Nassau County actually worked. Initially 24 votes were apportioned to Hempstead, resulting in 36 total votes. Hempstead was then limited to half of the total, or 18, or 9 for each supervisor. The six eliminated votes were not voted, and the majority required to pass a measure remained at 19.
^ Many sources claim that Banzhaf sued (and won). In the original Nassau County litigation, Franklin v. Mandeville 57 Misc.2d 1072 (1968), a New York court ruled that voters in Hempstead were denied equal protection equal because while the town had a majority of the population, they did not have a majority of the weighted vote. Weighted voting would be litigated in Nassau County for the next 25 years, until it was eliminated.

References

Bibliography

Banzhaf, John F. (1965). "Weighted Voting Doesn't Work: A Mathematical Analysis". Rutgers Law Review. 19 (2): 317–343. ISSN 0036-0465.
Coleman, James S. (1971). "Control of Collectives and the Power of a Collectivity to Act". In Lieberman, Bernhardt (ed.). Social Choice. New York: Gordon and Breach. pp. 192–225.
Felsenthal, Dan S.; Machover, Moshé (1998). The Measurement of Voting Power Theory and Practice, Problems and Paradoxes. Cheltenham, England: Edward Elgar.
Felsenthal, Dan S.; Machover, Moshé (2004). "A Priori Voting Power: What is it All About?" (PDF). Political Studies Review. 2 (1): 1–23. doi:10.1111/j.1478-9299.2004.00001.x. ISSN 1478-9302. S2CID 145284470.
Gelman, Andrew; Katz, Jonathan; Tuerlinckx, Francis (2002). "The Mathematics and Statistics of Voting Power". Statistical Science. 17 (4): 420–435. doi:10.1214/ss/1049993201. ISSN 0883-4237.
Lehrer, Ehud (1988). "An Axiomatization of the Banzhaf Value" (PDF). International Journal of Game Theory. 17 (2): 89–99. CiteSeerX 10.1.1.362.9991. doi:10.1007/BF01254541. ISSN 0020-7276. S2CID 189830513. Retrieved 30 August 2017.
Matsui, Tomomi; Matsui, Yasuko (2000). "A Survey of Algorithms for Calculating Power Indices of Weighted Majority Games" (PDF). Journal of the Operations Research Society of Japan. 43 (1): 71–86. doi:10.15807/jorsj.43.71. ISSN 0453-4514. Retrieved 30 August 2017.
Penrose, Lionel (1946). "The Elementary Statistics of Majority Voting". Journal of the Royal Statistical Society. 109 (1): 53–57. doi:10.2307/2981392. ISSN 0964-1998. JSTOR 2981392.
Straffin, Philip D. (1993). Game Theory and Strategy. New Mathematical Library. Vol. 36. Washington: Mathematical Association of America.
Varela, Diego; Prado-Dominguez, Javier (2012). "Negotiating the Lisbon Treaty: Redistribution, Efficiency and Power Indices". Czech Economic Review. 6 (2): 107–124. ISSN 1802-4696. Retrieved 30 August 2017.

External links

Online Power Index Calculator (by Tomomi Matsui)
Banzhaf Power Index Includes power index estimates for the 1990s U.S. Electoral College.
Voting Power Perl calculator for the Penrose index.
Computer Algorithms for Voting Power Analysis Web-based algorithms for voting power analysis
Power Index Calculator Computes various indices for (multiple) weighted voting games online. Includes some examples.
Computing Banzhaf power index and Shapley–Shubik power index with Python and R (by Frank Huettner)
Banzhaf Power Index at the Wolfram Demonstrations Project

This page was last edited on 4 April 2024, at 23:19

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