3.4 Calculating Power- Banzhaf Power Index

The Banzhaf power index was originally created in 1946 by Lionel Penrose, but was reintroduced by John Banzhaf in 1965. The power index is a numerical way of looking at power in a weighted voting situation.

# Math in Society 3.4 -- Banzhaf Power Index, coalition by coalition
# List every winning coalition, star the critical players, count the stars.
from itertools import combinations

def banzhaf(q, w, names=None, show=True):
    n = len(w)
    names = names or [f"P{i+1}" for i in range(n)]
    counts = [0] * n
    rows = []
    for k in range(1, n + 1):
        for c in combinations(range(n), k):
            weight = sum(w[i] for i in c)
            if weight < q:
                continue
            crit = {i for i in c if weight - w[i] < q}
            for i in crit:
                counts[i] += 1
            rows.append((c, crit, weight))
    grand = sum(counts)
    print(f"[{q}: {', '.join(map(str, w))}]   {len(rows)} winning coalitions"
          f"   ({grand} critical slots in total)")
    if show:
        for c, crit, weight in rows:
            body = ", ".join(f"*{names[i]}*" if i in crit else names[i] for i in c)
            print(f"   {{{body}}}   weight {weight}")
    for i in range(n):
        print(f"   {names[i]:<20} critical in {counts[i]:>3} of them"
              f"   ->  {counts[i]}/{grand} = {counts[i]/grand:6.1%}"
              + ("   (a dummy)" if counts[i] == 0 else ""))
    print()
    return counts

banzhaf(8, [6, 3, 2])
banzhaf(16, [7, 6, 3, 3, 2])
banzhaf(36, [20, 17, 16, 3])          # EDIT: change the quota or any weight
banzhaf(65, [47, 46, 17, 16, 2], show=False,
        names=["Scottish National", "Labour", "Conservative",
               "Liberal Democrat", "Green"])

print("The Liberal Democrats hold 16 seats and the Greens hold 2, yet their Banzhaf")
print("power is identical: in every coalition where one of them tips the balance,")
print("so does the other. Weight is not power.")
print("Try it: in [36: 20, 17, 16, 3] raise the quota to 41 and re-run. The")
print("3-weight player drops from 8.3% of the power to a dummy, while the other")
print("three all gain veto power -- one number moved, and the whole structure changed.")

Example 1

Find the Banzhaf power index for the voting system $open bracket 8: 6, 3, 2 close bracket$ .

Show solution

We start by listing all winning coalitions. If you aren’t sure how to do this, you can list all coalitions, then eliminate the non-winning coalitions. No player is a dictator, so we’ll only consider two and three player coalitions.

$open brace P sub 1, P sub 2 close brace$ Total weight: 9. Meets quota.

$open brace P sub 1, P sub 3 close brace$ Total weight: 8. Meets quota.

$open brace P sub 2, P sub 3 close brace$ Total weight: 5. Does not meet quota.

$open brace P sub 1, P sub 2, P sub 3 close brace$ Total weight: 11. Meets quota.

Next we determine which players are critical in each winning coalition. In the winning two-player coalitions, both players are critical since no player can meet quota alone. Underlining the critical players to make it easier to count:

$open brace P bar sub 1, P bar sub 2 close brace$

$open brace P bar sub 1, P bar sub 3 close brace$

In the three-person coalition, either $P sub 2$ or $P sub 3$ could leave the coalition and the remaining players could still meet quota, so neither is critical. If $P sub 1$ were to leave, the remaining players could not reach quota, so $P sub 1$ is critical.

$open brace P bar sub 1, P sub 2, P sub 3 close brace$

Altogether, $P sub 1$ is critical 3 times, $P sub 2$ is critical 1 time, and $P sub 3$ is critical 1 time.

Converting to percents:

$P_{1} = 3 / 5 = 60 %$

$P_{2} = 1 / 5 = 20 %$

$P_{3} = 1 / 5 = 20 %$

Example 2

Consider the voting system $open bracket 16: 7, 6, 3, 3, 2 close bracket$ . Find the Banzhaf power index.

Show solution

The winning coalitions are listed below, with the critical players underlined.

$open brace P bar sub 1, P bar sub 2, P bar sub 3 close brace$

$open brace P bar sub 1, P bar sub 2, P bar sub 4 close brace$

$open brace P bar sub 1, P bar sub 2, P sub 3, P sub 4 close brace$

$open brace P bar sub 1, P bar sub 2, P bar sub 3, P sub 5 close brace$

$open brace P bar sub 1, P bar sub 2, P bar sub 4, P sub 5 close brace$

$open brace P bar sub 1, P bar sub 2, P sub 3, P sub 4, P sub 5 close brace$

Counting up times that each player is critical:

$P sub 1 equals 6$

$P sub 2 equals 6$

$P sub 3 equals 2$

$P sub 4 equals 2$

$P sub 5 equals 0$

Total of all: 16

Divide each player’s count by 16 to convert to fractions or percents:

$P_{1} = 6 / 16 = 3 / 8 = 37.5 %$

$P_{2} = 6 / 16 = 3 / 8 = 37.5 %$

$P_{3} = 2 / 16 = 1 / 8 = 12.5 %$

$P_{4} = 2 / 16 = 1 / 8 = 12.5 %$

$P_{5} = 0 / 16 = 0 = 0 %$

The Banzhaf power index measures a player’s ability to influence the outcome of the vote. Notice that player 5 has a power index of 0, indicating that there is no coalition in which they would be critical power and could influence the outcome. This means player 5 is a dummy, as we noted earlier.

Example 3

Revisiting the Scottish Parliament, with voting system $open bracket 65: 47, 46, 17, 16, 2 close bracket$ , the winning coalitions are listed, with the critical players underlined.

Show solution

$open brace P bar sub 1, P bar sub 2 close brace; open brace P bar sub 1, P bar sub 2, P sub 3 close brace, open brace P bar sub 1, P bar sub 2, P sub 4 close brace; open brace P bar sub 1, P bar sub 2, P sub 5 close brace, open brace P bar sub 1, P bar sub 3, P bar sub 4 close brace; open brace P bar sub 1, P bar sub 3, P bar sub 5 close brace, open brace P bar sub 1, P bar sub 4, P bar sub 5 close brace; open brace P bar sub 2, P bar sub 3, P bar sub 4 close brace, open brace P bar sub 2, P bar sub 3, P bar sub 5 close brace; open brace P sub 1, P sub 2, P sub 3, P sub 4 close brace, open brace P sub 1, P sub 2, P sub 3, P sub 5 close brace; open brace P bar sub 1, P sub 2, P sub 4, P sub 5 close brace, open brace P bar sub 1, P sub 3, P sub 4, P sub 5 close brace; open brace P bar sub 2, P bar sub 3, P sub 4, P sub 5 close brace$

Counting up times that each player is critical:

District	Times critical	Power index
$P sub 1 (Scottish National Party)$	9	$9 / 27 equals 33.3 %$
$P sub 2 (Labour Party)$	7	$7 / 27 equals 25.9 %$
$P sub 3 (Conservative Party)$	5	$5 / 27 equals 18.5 %$
$P sub 4 (Liberal Democrats Party)$	3	$3 / 27 equals 11.1 %$
$P sub 5 (Scottish Green Party)$	3	$3 / 27 equals 11.1 %$

Interestingly, even though the Liberal Democrats party has only one less representative than the Conservative Party, and 14 more than the Scottish Green Party, their Banzhaf power index is the same as the Scottish Green Party’s. In parliamentary governments, forming coalitions is an essential part of getting results, and a party’s ability to help a coalition reach quota defines its influence.

Your Turn

Try it Now 3

Find the Banzhaf power index for the weighted voting system $open bracket 36: 20, 17, 16, 3 close bracket$ .

Answer

The voting system tells us that the quota is 36, that Player 1 has 20 votes (or equivalently, has a weight of 20), Player 2 has 17 votes, Player 3 has 16 votes, and Player 4 has 3 votes.

A coalition is any group of one or more players. What we're looking for is winning coalitions - coalitions whose combined votes (weights) add to up to the quota or more. So the coalition $open brace P 3, P 4 close brace$ is not a winning coalition because the combined weight is $16 + 3 = 19$ , which is below the quota.

So we look at each possible combination of players and identify the winning ones:

$open brace P 1, P 2 close brace open parenthesis weight: 37 close parenthesis, open brace P 1, P 3 close brace (weight: 36 close parenthesis; open brace P 1, P 2, P 3 close brace (weight: 53 close parenthesis, open brace P 1, P 2, P 4 close brace (weight: 40 close parenthesis; open brace P 1, P 3, P 4 close brace (weight: 39 close parenthesis, open brace P 1, P 2, P 3, P 4 close brace (weight: 56 close parenthesis; open brace P 2, P 3, P 4 close brace open parenthesis weight: 36 close parenthesis$

Example 4

Banzhaf used this index to argue that the weighted voting system used in the Nassau County Board of Supervisors in New York was unfair. The county was divided up into 6 districts, each getting voting weight proportional to the population in the district, as shown below. Calculate the power index for each district.

# Math in Society 3.4 -- Nassau County: the case Banzhaf actually argued
# Six districts were given voting weight proportional to population. Banzhaf's
# point was that proportional WEIGHT does not produce proportional POWER.
from itertools import combinations

districts = [("Hempstead #1", 31), ("Hempstead #2", 31), ("Oyster Bay", 28),
             ("North Hempstead", 21), ("Long Beach", 2), ("Glen Cove", 2)]
QUOTA = 58        # simple majority of the 115 total. EDIT: try 77 (a two-thirds rule)

names = [d for d, _ in districts]
w = [x for _, x in districts]
n, total = len(w), sum(w)

counts = [0] * n
winning = 0
for k in range(1, n + 1):
    for c in combinations(range(n), k):
        weight = sum(w[i] for i in c)
        if weight < QUOTA:
            continue
        winning += 1
        for i in c:
            if weight - w[i] < QUOTA:
                counts[i] += 1
grand = sum(counts)

print(f"[{QUOTA}: {', '.join(map(str, w))}]   total weight {total}"
      f"   {winning} winning coalitions\n")
print(f"{'district':<18}{'weight':>7}{'weight share':>14}{'critical':>10}{'Banzhaf power':>15}")
for i in range(n):
    note = "   <-- DUMMY" if counts[i] == 0 else ""
    print(f"{names[i]:<18}{w[i]:>7}{w[i]/total:>14.1%}{counts[i]:>10}"
          f"{counts[i]/grand:>15.1%}{note}")
print(f"{'total':<18}{total:>7}{'100.0%':>14}{grand:>10}{'100.0%':>15}")

dummies = [i for i in range(n) if counts[i] == 0]
if dummies:
    smallest = min(len(c) for k in range(1, n + 1) for c in combinations(range(n), k)
                   if sum(w[i] for i in c) >= QUOTA)
    print(f"\n{len(dummies)} of the {n} districts are DUMMIES: "
          + ", ".join(names[i] for i in dummies) + ".")
    print(f"The smallest coalition that reaches quota has {smallest} members, and the")
    print("big districts can always assemble one without help, so a dummy's vote can")
    print("never turn a losing coalition into a winning one.")
gap = max(range(n), key=lambda i: w[i] / total - counts[i] / grand)
print(f"\nWidest gap: {names[gap]} holds {w[gap]/total:.1%} of the weight"
      f" and {counts[gap]/grand:.1%} of the power.")

print("\nTry it: set QUOTA = 77 (two thirds of 115) and re-run. Now no two districts")
print("can pass anything alone; North Hempstead stops being a dummy and the four")
print("large districts end up with exactly equal power, 25% each.")
print("Try it: QUOTA = 59, 60, ... and watch for the first quota that gives")
print("Long Beach and Glen Cove any power at all.")

District	Weight
Hempstead #1	31
Hempstead #2	31
Oyster Bay	28
North Hempstead	21
Long Beach	2
Glen Cove	2

Show solution

Translated into a weighted voting system, assuming a simple majority is needed for a proposal to pass:

$open bracket 58: 31, 31, 28, 21, 2, 2 close bracket$

Listing the winning coalitions and marking critical players:

$open brace H 1 bar, H 2 bar close brace, open brace H 1 bar, O B bar, N H close brace, open brace H 2 bar, O B bar, N H, L B close brace; open brace H 1 bar, O B bar close brace, open brace H 1 bar, O B bar, L B close brace, open brace H 2 bar, O B bar, N H, G C close brace; open brace H 2 bar, O B bar close brace, open brace H 1 bar, O B bar, G C close brace, open brace H 2 bar, O B bar, L B, G C close brace; open brace H 1 bar, H 2 bar, N H close brace, open brace H 1 bar, O B bar, N H, L B close brace, open brace H 2 bar, O B bar, N H, L B, G C close brace; open brace H 1 bar, H 2 bar, L B close brace, open brace H 1 bar, O B, N H, G C close brace, open brace H 1, H 2, O B close brace; open brace H 1 bar, H 2 bar, G C close brace, open brace H 1 bar, O B bar, L B, G C close brace, open brace H 1, H 2, O B, N H close brace; open brace H 1 bar, H 2 bar, N H, L B close brace, open brace H 1 bar, O B bar, N H, L B. G C close brace, open brace H 1, H 2, O B, L B close brace; open brace H 1 bar, H 2 bar, N H, G C close brace, open brace H 2 bar, O B bar, N H close brace, open brace H 1, H 2, O B, G C close brace; open brace H 1 bar, H 2 bar, L B, G C close brace, open brace H 2 bar, O B bar, L B close brace, open brace H 1, H 2, O B, N H, L B close brace; open brace H 1 bar, H 2 bar, N H, L B, G C close brace, open brace H 2 bar, O B bar, G C close brace, open brace H 1, H 2, O B, N H, G C close brace; open brace H 1, H 2, O B, N H, L B, G C close brace$

There are a lot of them! Counting up how many times each player is critical,

District	Times critical	Power index
Hempstead #1	16	$16 / 48 = 1 / 3 = 33 %$
Hempstead #2	16	$16 / 48 = 1 / 3 = 33 %$
Oyster Bay	16	$16 / 48 = 1 / 3 = 33 %$
North Hempstead	0	$0 / 48 equals 0 %$
Long Beach	0	$0 / 48 equals 0 %$
Glen Cove	0	$0 / 48 equals 0 %$

It turns out that the three smaller districts are dummies. Any winning coalition requires two of the larger districts.

The weighted voting system that Americans are most familiar with is the Electoral College system used to elect the President. In the Electoral College, states are given a number of votes equal to the number of their congressional representatives (house + senate). Most states give all their electoral votes to the candidate that wins a majority in their state, turning the Electoral College into a weighted voting system, in which the states are the players. As I’m sure you can imagine, there are billions of possible winning coalitions, so the power index for the Electoral College has to be computed by a computer using approximation techniques.

Adapted from Math in Society by David Lippman, hosted on LibreTexts (math.libretexts.org) and licensed under CC BY-SA 3.0. Changes were made. License: CC-BY-SA-3.0.