3.3 A Look at Power

Consider the voting system $open bracket 10: 11, 3, 2 close bracket$ . Notice that in this system, player 1 can reach quota without the support of any other player. When this happens, we say that player 1 is a dictator.

In the voting system $open bracket 8: 6, 3, 2 close bracket$ , no player is a dictator. However, in this system, the quota can only be reached if player 1 is in support of the proposal; player 2 and 3 cannot reach quota without player 1’s support. In this case, player 1 is said to have veto power. Notice that player 1 is not a dictator, since player 1 would still need player 2 or 3’s support to reach quota.

With the system $open bracket 10: 7, 6, 2 close bracket$ , player 3 is said to be a dummy, meaning they have no influence in the outcome. The only way the quota can be met is with the support of both players 1 and 2 (both of which would have veto power here); the vote of player 3 cannot affect the outcome.

To better define power, we need to introduce the idea of a coalition. A coalition is a group of players voting the same way. In the example above, $open brace P sub 1, P sub 2, P sub 4 close brace$ would represent the coalition of players 1, 2 and 4. This coalition has a combined weight of $7 + 6 + 3 = 16$ , which meets quota, so this would be a winning coalition.

A player is said to be critical in a coalition if them leaving the coalition would change it from a winning coalition to a losing coalition. In the coalition $open brace P sub 1, P sub 2, P sub 4 close brace$ , every player is critical. In the coalition $open brace P sub 3, P sub 4, P sub 5 close brace$ , no player is critical, since it wasn’t a winning coalition to begin with. In the coalition $open brace P sub 1, P sub 2, P sub 3, P sub 4, P sub 5 close brace$ , only players 1 and 2 are critical; any other player could leave the coalition and it would still meet quota.

# Math in Society 3.3 -- A Look at Power: dictators, veto power, dummies
# A weighted voting system is written [q: w1, w2, ...] -- quota, then weights.
from itertools import combinations

def analyse(q, w, label=""):
    n, total = len(w), sum(w)
    idx = range(n)
    winning = [c for k in range(1, n + 1) for c in combinations(idx, k)
               if sum(w[i] for i in c) >= q]
    critical = [sum(1 for c in winning
                    if i in c and sum(w[j] for j in c if j != i) < q) for i in idx]
    print(f"{label}[{q}: {', '.join(map(str, w))}]   total weight {total}"
          f"   {len(winning)} of {2**n - 1} coalitions win")
    if q * 2 <= total:
        print("   !! the quota is not more than half the weight: two opposing")
        print("      coalitions could both reach quota, so nothing gets decided")
    if q > total:
        print("   !! the quota exceeds the total weight: nothing can ever pass")
    for i in idx:
        if w[i] >= q:
            tag = "DICTATOR -- reaches quota alone"
        elif critical[i] == 0:
            tag = "DUMMY -- never critical, so never matters"
        elif total - w[i] < q:
            tag = "VETO POWER -- critical in every winning coalition"
        else:
            tag = "ordinary player"
        print(f"   P{i+1}  weight {w[i]:>3}   critical {critical[i]:>3}x   {tag}")
    print()

analyse(16, [7, 6, 3, 3, 2])
analyse(10, [11, 3, 2])
for q in (10, 12, 16):            # EDIT: try other quotas for the same weights
    analyse(q, [10, 5, 3])
analyse(65, [47, 46, 17, 16, 2], "Scottish Parliament 2009  ")

print("Raising the quota on a fixed set of weights walks each player up the same")
print("ladder: dummy -> ordinary -> veto power. A dictator can only appear when one")
print("weight on its own already reaches the quota.")
print("Try it: analyse(q, [4, 4, 4, 1]) for every q from 5 to 13 and find the quota")
print("at which the 1-weight player stops being a dummy.")

Example 2

In the Scottish Parliament in 2009 there were 5 political parties: 47 representatives for the Scottish National Party, 46 for the Labour Party, 17 for the Conservative Party, 16 for the Liberal Democrats, and 2 for the Scottish Green Party. Typically all representatives from a party vote as a block, so the parliament can be treated like the weighted voting system:

$open bracket 65: 47, 46, 17, 16, 2 close bracket$

Show solution

Consider the coalition $open brace P sub 1, P sub 3, P sub 4 close brace$ . No two players alone could meet the quota, so all three players are critical in this coalition.

In the coalition $open brace P sub 1, P sub 3, P sub 4, P sub 5 close brace$ , any player except $P sub 1$ could leave the coalition and it would still meet quota, so only P₁ is critical in this coalition.

Notice that a player with veto power will be critical in every winning coalition, since removing their support would prevent a proposal from passing.

Likewise, a dummy will never be critical, since their support will never change a losing coalition to a winning one.

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.