Probabilistic Banzhaf Power¶

Electoral College Case Study¶

Weighted Voting Games¶

Imagine two different voting scenarios. In the first, you are a citizen of Rhode Island voting in a U.S. presidential election. If you were to discover that a citizen of New York had 1.25 as many votes as you did, you would certainly consider this scenario unfair. We have grown to believe that every person's vote should be equal in an electoral process - we will refer to this as the equality principle. In the second scenario, you are a partner in a new company. You own 50% of the shares in the company - the other five investors each own 10% of the shares. If each of the partners had the same number of votes you would certainly consider this scenario unfair. We have also grown to believe that in some circumstances, different people deserve different treatment, and that the difference should be based on some objective criterion (in our example, the number of shares owned). We will refer to this as the equity principle.

The tension between the equality principle and the equity principle is well illustrated in United States history by the debate among the Founding Fathers as to how the legislature should be structured. The advocates of equality (the smaller states) argued that representation in the legislature should be the same for every state; the advocates of equity (the larger states) argued that representation should be based on population. As a result of this argument, we have a bicameral legislature where the equality principle is applied to the Senate and the equity principle is applied to the House of Representatives.

A second important illustration of the conflict between equality and equity is presented to us by the Electoral College. Citizens of the United States expect to have the same number of votes ("one person-one vote") in a presidential election, regardless of the state they live in. But states have different numbers of votes in the Electoral College (number of seats in the House of Representatives plus 2), and because of the "winner takes all rule" in most states, citizens in different states have in effect different amounts of say in a presidential election - notwithstanding the fact that everyone casts one vote.

The focus of this study will be to discuss situations where the criterion for fairness is equity, rather than equality. These situations are common in both the world of politics and the corporate world. Corporations commonly distribute votes to stockholders based on the number of shares owned in the company. Political and private organizations may prefer to weight the votes of particular parties in order to apportion voting power proportional to population, tax base, investment, geographical regions, etc. Governments commonly distribute votes to legislators proportional to the population represented (e.g., county board of supervisors in New York State).

When a voting system is based on the principle of equity, and votes are allocated to the voting parties (called "players") based on a well defined, objective measure (population, tax base, number of shares owned, etc.), the system is called a weighted voting game. (This terminology comes from game theory, the branch of mathematics that gave birth to the mathematical study of equity.) Thus, a weighted voting game consists of players, the players' weights (a numerical assignment of votes to each of the players), and a numerical quota that defines how many votes are needed to succeed in an election. Last, but not least, a weighted voting game is based on the presumption of equity. The question is, how well does a particular weighted voting game succeed in achieving the objective of equity? As we will see, this question is far from trivial. The first step in determining equity is to determine a measure of voting power. The Banzhaf Power Index is used to measure voting power for this analysis.

Historical Context of Banzhaf Power¶

In 1965, a lawyer named John F. Banzhaf published a paper addressing the issue of voting power from a constitutional point of view. The Equal Protection Clause, part of the 14th Amendment of the United States Constitution, provides that no state shall... deny any person within its jurisdiction the equal protection of the laws. Whether or not the Equal Protection Clause pertains to voting rights is arguable; however, a Supreme Court case in 1964 (Reynolds v. Sims) established the one-man-one-vote standard. If each representative on a board has one vote yet each representative does not represent districts with equal populations, individuals from some districts receive more representation than others on the board. This situation does not conform to a standard equal protection under the law or one-man-one-vote principle. The immediate response may be to weight the votes of each representative based on the population of the district they represent. This, as we have demonstrated, is not the best manner to distribute votes if the desire is to give each representative voting power proportional to the populations represented. Banzhaf illustrated that voting power is not proportional to weight, shown in the example below.

The well-known case of the 1964 Nassau County Board of Supervisors brought action to the courts spanning over three decades to ensure the weighted voting game for their Board of Supervisors upheld the one-man-one-vote principle. Instead of weights based on the number of shares held, these exact same weights were based on the populations of the six districts dividing Nassau County, New York. Each municipality was assigned a number of votes based on its population, and a total of 58 votes were needed to pass any resolution. Banzhaf saw a problem with this arrangement. He argued that the apportionment of votes gave a false sense of equitable power to influence legislative decisions when in fact half the districts had no power at all and the three largest districts had all the power. The distribution of votes among six different municipalities of Nassau County, as well as other similar situations in which dummies occurred, inspired Banzhaf to develop a standard measurement of voting power to ensure voters were properly represented under constitutional law.

Table 1

1960 Nassau County Example:

The measure Banzhaf originally proposed was the crude Banzhaf score (shown as # of critical votes above) which simply counted the number of times each player was a critical player for a winning coalition (see example 1.1 in Appendix 1 for more details on critical players). He was mainly concerned with the ratio of power between voters [26]. The popular form of the Banzhaf Power Index calculated today normalizes the Banzhaf score so the Banzhaf Power Distribution always sums to one and represents the distribution of probabilities for each player having a critical vote. The Banzhaf Power Index for the Nassau County Municipalities as shown above is {1/3, 1/3, 1/3, 0, 0, 0}. Although North Hempstead had 21 votes based on its population, it had zero voting power. The Banzhaf Power Index has been the standard voting power index in legal cases as New York's highest court, in a decision in 1967, required counties to use the Banzhaf index as the criterion for measuring the equity of weighted in local legislatures [17].

Other power indices include Rae (1969), Coleman (1971, 1986), Deegan and Packel (1978), etc [10]. Each voting power index has its own criticism (for more on this see [26]). Coleman defines a pair of measures of voting power; a player can have affirmative power, or power to initiate action, or negative power, or blocking power [33].

Electoral College Application¶

The US Electoral College could also be considered a weighted voting game. Analyzing power for the US Electoral College can be challenging. In some states, the Electoral College does not actually require the Electors to cast their electoral votes for the candidate who wins the popular vote. Although Electors rarely cast dissenting votes, there are still 24 states that do not require Electors to vote for a specific candidate, and in a few recent elections, some Electors have not followed the people's choice (e.g. 1988, 1976, and 1972 [32]). Additionally, Maine and Nebraska may split their electoral college votes between two candidates. For purposes of our analysis, we will assume that Electors cast their all their votes for the winner of the popular vote. We also assume that the probability of any independent candidate to win is zero; i.e., the election is between a Republican and a Democrat. This would make the Electoral College a weighted voting game with a total of 538 votes distributed to 51 players (50 states + DC) with a quota of 270 to win.

The generating function method utilized here was developed by J. Richard Gott and Wes Colley and used in an analysis by Tannenbaum [35]. Each function is explained in more detail in the Appendix. The weights of each of the 51 players in the Electoral College is their respective number of Electoral College votes in 2016.

The difference in Electoral College weight (e.g. 55/538 or 0.1022 for California) and the Banzhaf Power Index (0.1136 for California) can be seen in the bar plot above. Most states have lower BPI voting power than Electoral College Weight. Only California, Florida, New York, and Texas had higher voting power than their respective Electoral College weights.

The electoral college actually performs fairly well when compared to the Banzhaf Power Index. California was the only state with a relatively large difference. With 55 electoral votes, California is indeed a critical player in many coalitions. Though as we are all aware, California is not a very important player before presidential elections. It is often ignored by campaigns because it has an extremely low probability of voting Republican. The outcome is all but known. Note that the expected outcome of the Electoral college votes is different than the percent in the state who voted for a candidate. The average percent of votes for a Democratic candidate over the last six elections is 57%, but we would not assign a 57% probability of California's Electoral College votes to go to a Democrat. It would likely be much higher.

One issue with the Banzhaf Power Index (as well as many of the other power indices) lies in the assumption that all voting parties are assigned an equal probability of voting for or against a measure. The Banzhaf Power Index does not incorporate the voting party's preference.

Probabilistic Power Distribution for the Electoral College¶

The analysis below uses a probabilistic voting power function. In order to use a probabilistic method, we must assume probabilities for each of the 51 players in the Electoral College (the 50 states plus the District of Columbia).

There are various methods of determining probabilities for a state to vote for a particular candidate. Political forecasting is a large field. Many forecasts make use of polling, data on what matters to voters, demographics, etc. Prediction markets, such as the Iowa Electronic Market or PredictWise, can also be a form of measuring the probability a state voting for candidate. In a prior analysis, data from PredictWise was utilized for the probabilities of the 51 EC voting parties; however, polls and prediction markets are not necessarily reliable (as witnessed in 2016). Hummel and Rothschild [38] do not use voter intentions in their forecasting model, rather economic and political indicators available months before the election to analyze state-level data. This so-called Fundamental Model results in much lower prediction error. For this analysis, we use the Fundamental Probability for each voting party to vote for a Republican [39] and the Electoral College weights from 2016.

The graph above shows the probabilities of each state voting for a Republican presidential candidate. Using this probability model, swing states are identified as Nevada, Colorado, Iowa, Wisconsin, Ohio, Pennsylvania, New Hampshire, Virginia, and Florida. This is similar to FiveThirtyEight identification of swing states [37], though it does not include North Carolina and Minnesota.

Next, we take the 2016 Electoral College weights and the probabilities of casting Electoral College votes for a Republican into the probabilistic generating function. We calculate the probability of the voting party to be a positive critical player (vote Republican and a Republican is elected) and a negative critical player (vote Democrat and a Republican is not elected). The probabilistic Banzhaf Power Index is as follows (see Appendix 2 for more details on the calculation):

2.5 Conclusion¶

When we examine voters' preferences as well as the weights of their votes, all voters cannot be equal, no matter how hard we try to instill the principle of one-person-one-vote in our voting systems. The differences of each voter - weight, preference, residence, etc. - give a difference in their respective voting powers. The question then becomes, how can we measure the differences in power? By looking at the Electoral College where all voters have probabilities of 50% of voting for a Republican or Democratic candidate, we see this a priori consideration determines the faithfulness of the voting system to the distribution electoral votes for each state. Using the Banzhaf Index as a measure of voting power, the Electoral College performs fairly well.

The second part of the analysis takes Electoral College voting parties' preferences as probabilities and measures the notion of voting power by measuring the probability for each player to be a critical player. We combined the probabilities for players to swing the outcome of a 'yes' coalition with the probabilities for them to swing the outcome of a 'no' coalition, thereby playing with Coleman's ideas of measuring both affirmative and negative power to calculate criticality of players for probabilistic weighted voting games. With generating functions, we can easily calculate the voting powers for practically any voting scenario, even one as large as the Electoral College. We might have predicted that the swing states have slightly more probabilistic Banzhaf power; however, the dramatic difference between the intended weight from the Electoral College votes and voting power for California, Texas, and New York was unforeseen. Texas and New York both had approximately 1% less voting power than intended and California had almost 5% less voting power. Note that each of those states is considered a solid "red" or "blue" state.

One question that arises when we discuss the calculation of probabilistic voting power is, could these calculations be used for power gains? California residents have seen that being a clearly Democratic state prevents them from being able to manipulate presidential politics in their favor. Swing states tend to have more political power and more probabilistic voting power than states with the same number of Electoral Votes but with a clear party preference. In general, the player(s) with positions closer to the middle ground have relatively more power than if they were completely politically aligned because they become decisive of the outcome of the vote. Even if a definite political preference suggests a lack of power and is apparent to voters, there is no way to control the likelihood of residents voting in favor of or against any particular election. The only way these calculations could possibly be used for power gain is if one person could control the perceived likelihood of their vote. For example, they could strategically put out the message that they are undecided. If players could 'fake' preferences, the computations of probabilistic voting power would become very complicated.

Topics for further research include developing other generating functions to compute probabilistic power measurements, such as the Shapley-Shubik Index. The Shapley Shubik Index could be insightful in that considers the sequence of voting - by the time votes are counted on the west coast on election night the election is often already called. Another interesting topic would be to determine which preferences maximize voting power of any particular player under various voting environments; i.e., weighted voting systems where either the preferences or the weights of the players vary. If we set the probabilities of certain swing states lean towards voting for a Democratic candidate, the Banzhaf power for California significantly changes. Another challenge would be to determine the a priori measurements of power by finding a sort of 'inverse function'. In other words, what weights should be assigned to the players to achieve a specific distribution of the power? We know that there are a finite number of power distributions for a particular number of players, but as the number of players gets larger, the number of power distributions grows exponentially. How many possible distributions of power are there? Tolle writes that there are five distributions for a four-player game [5]. For a five-player game, Cutler and Rolle determine approximately 300 feasible power distributions [27]. This is a finite problem but computationally complex. The analysis of a probabilistic weighted voting game can become quickly complex; however, many open-ended questions remain.

Appendix 1 Banzhaf Power Index¶

Example 1.1¶

Imagine a corporation with 6 shareholders (the players): Anderson and Brad own 31 shares each, Connie owns 28 shares, Don owns 21 shares, and Eryn and Frank own 2 shares each for a total of 115 shares. Let's consider the most intuitive approach for achieving equity: assign to each shareholder one vote per share. This would give Anderson and Brad 31 votes each, Connie 28 votes, Don 21 votes, and Eryn and Frank 2 votes each. The number of votes needed to pass a motion (the quota) is 58, a strict majority of the 115 votes owned by all the players.

Suppose Don decides to vote 'yes'. Who else can he recruit to join him in a coalition that will win? If any two of the first three partners, Anderson, Brad, or Connie, join Don to vote 'yes', the motion passes. But then, whether or not Don votes 'yes' does not matter anymore; the vote would pass with or without Don's vote. Thus, a necessary and sufficient condition for a motion to pass is that any two of the 'big three' (Anderson, Brad, Connie) vote 'yes'. This implies that Don, Eryn, and Frank have no power to influence the outcome of vote whereas Anderson, Brad, and Connie hold all the power (the details of this analysis are provided in sections 1.2 and 1.3). Quantitatively, Anderson, Brad, and Connie hold one-third of the power each and Don, Eryn, and Frank hold zero power.

The key observation of the proceding analysis is that the distribution of real power differs dramatically from the intended weight distribution that represents the intention of equity.

Table 2

Weight distribution versus power in Example 1.1

Don has the largest discrepancy between his relative weight and power. He has 18% of the votes, but zero percent power.

Example 1.1 illustrates how the voting power can be entirely different from the goal originally intended, even in elections with few voters and only two voting options. Imagine how quickly complications arise with many voters and many choices. Understanding how to calibrate the assignment of weights to reflect voting power is key to creating a weighted voting game faithful (with the least variance) to the intent of its creators.

1.2 Notation and Terminology for Weighted Voting Games¶

The mathematical structure of weighted voting games can be used to represent any election with yes-no votes, or motions. The idea of motions can be extended to any two options (e.g., a resolution that either passes or fails, or an election between exactly two candidates). Real-life examples of weighted voting games can be found everywhere in life, from the local school board to the US Electoral College or the European Union's Council of Ministers. The problem with weighted voting is that rarely are the weights of the individual players proportional to the desired player's influence on the outcome of the decision. The discrepancy between weights and power illustrated in Example 1.1 is the rule rather than the exception. Despite the importance of measuring inadequacies in such weighted voting systems, relatively little research and energy has been spent to further the understanding of the effects of weighted voting on a representative's power.

John F. Banzhaf, a lawyer, was one of a few among mathematicians, political scientists and economists to create a power index to measure power in weighted voting systems. Section 1.3 provides an overview of the Banzhaf Index; some historical background on measuring voting power for weighted voting games is provided in sections 1.4 and 1.5. This section will provide a basic introduction to the notation and terminology for weighted voting games (most of which conform to those in [14]).

The subject of weighted voting games derives much of its terminology from mathematical game theory. The voters in Example 1.1 (shareholders Anderson, Brad, Connie, Don, Eryn, and Frank) can be easily represented generically as players 1 through 6, or $P_1, P_2, P_3$, ...$P_6$. In general, we let n denote the number of players in any weighted voting system and $P_1,...P_n$ denote players 1 through n. The weights for players 1 through $n$ are the number of votes controlled by each player and are represented by {$w_1, w_2,...w_n$}, respectively.

Definition 1.2.1 The quota is the minimum number of votes needed to pass a motion.

A weighted voting game with quota $q$ and weights {$w_1, w_2,...w_n$} is denoted as [ $q$; $w_1, w_2,...w_n$]; e.g., [58; 31, 31, 28, 21, 2, 2] for Example 1.1. The weights are usually listed in decreasing order. We will use this convention throughout this paper.

In Example 1.1, there are a total of 115 votes distributed to the six players. Imagine if the quota were changed from 58 to 50 (or less than one-half of the total votes). If In Example 1.1, if $P_1$ and $P_2$ voted yes and the remaining players voted 'no', the 'yes' votes would add to 62 and the 'no' votes would add to 53; therefore both 'yes' and 'no' would have enough votes to pass. In order to avoid this situation, we must place restrictions on the quota.

Remark 1.2.2 The quota must be over 50% and less than or equal to 100% of the total number of votes.

Definition 1.2.3 The term coalition is used to describe any set of players who vote the same way.

If a player votes for a motion and the motion passes, that player is affirmatively successful. Similarly, if a player votes against a motion and the motion fails, that player is negatively successful.

In Example 1.1, if $P_1$ and $P_2$ voted 'yes'and $P_3, P_4, P_5, and P_6$ voted 'no', two coalitions {$P_1, P_2$} and {$P_3, P_4, P_5, P_6$} would divide the players. The first coalition {$P_1, P_2$} would be the winning coalition with a total weight of 62 votes (greater than the quota of 58), and the second, {$P_3, P_4, P_5, P_6$}, a losing coalition with a total weight of 53 (less than the quota of 58). A single player is a winning coalition if and only if that player is a dictator; i.e., that player has a number of votes larger than or equal to the quota. A coalition of all players is always a winning coalition and is known as the grand coalition.

Definition 1.2.4 A player is considered to be a critical player in a winning coalition if the winning coalition would become a losing coalition without that particular player's vote. A player is considered an affirmative critical player if the player is critical to a coalition that votes 'yes' and a negative critical player if the player is critical to a coalition that votes 'no'. For an individual voter, P_i, to be able to cast a critical vote, the sum of the other votes in the coalition must be greater than or equal to the quota with P_i's vote and less than the quota without P_i's vote. In other words, the winning coalitions becomes a losing coalition without P_i's vote so that P_i is critical to the outcome.

Remark 1.2.5 P_i is a critical player in a coalition if $T \ge q$ and $T-w_i < q$ (where $T$ is the total weight of the coalition, $w_i$ is the weight of player i, and $q$ is the quota).

For indices that measure voting power using possible critical votes per player, the more coalitions for which player i has a critical vote, the more voting power that player is considered to have. A player's power increases with greater criticality.

A weighted voting game can have only one dictator, and according to the convention of listing the players in decreasing order of their weights, that player will always be player 1. If players have no influence on the outcome of the vote, they are called dummies and are always listed last. Don, Eryn, and Frank are dummies in Example 1.1; i.e., they have no voting power.

Definition 1.2.6 A player has veto power if no coalition has the ability to win without having that player in the coalition.

Example 1.1 (continued) Let's examine Example 1.1 in terms of winning coalitions and critical players. The winning coalitions for [58; 31, 31, 28, 21, 2, 2] are as follows (with critical players underlined):

Table 3

Winning coalitions (with critical players) for Example 1.1

Note that any coalition with A, B, and C together is a winning coalition, but when the three are combined, none of them are critical players; therefore, these combinations are not listed in the above Table 3. We summarize the number of possible coalitions for which each voter has a critical vote in the Table 4 below. Number of critical votes per player in Example 1.1

Table 4

A, B, and C are all critical exactly 16 times, whereas D, E, and F are never critical; thus, the voting power distribution based on the number of times each player is critical would be {16,16,16, 0, 0, 0}. If we normalize the probability as a percent (or probability) we get {16/48, 16/48, 16/48, 0/48, 0/48, 0/48} or {1/3, 1/3, 1/3, 0, 0, 0}. This is the Banzhaf Power Distribution for players A, B, C, D, E, and F.

1.3 Banzhaf Power Index Definition¶

A step-by-step procedure for calculating the Banzhaf Power Index is provided below:

• Start by listing all coalitions (all possible nonempty subsets of {P₁, P₂,..., P_n}

• Determine which coalitions are winning coalitions.

• In the winning coalitions, determine which players are critical players.

• Count how many times each player $P_i$ is critical; call this number $B_i$.

Let $\beta_{i}$ = $\frac{B_{i}}{\sum_{j = i}^{n} B_{j}}$

Then, $\beta_{i}$ is the Banzhaf Power Index of $P_i$ and the distribution {$\beta_{1}$, $\beta_{2}$,..., $\beta_{n}$} is the Banzhaf Power Distribution.

1.4 Banzhaf Power Generating Function¶

Calculating Banzhaf Power by manually listing coalitions can become very cumbersome; even in an example with 10 players, the number of subsets is $2^{10}-1$=1023 coalitions. Generating all of these is not an easy chore; then determining which are winning presents yet another tedious computation, and then identifying the critical players in each adds another layer of complexity. Instead of calculating the Banzhaf Power Index manually each time, we can use an efficient generating function.

A generating function is a formal power series in which coefficients provide information about a formal sequence <a_n> [29]. The ordinary generating function of a sequence <a_n> is:

$G(a_n, x)= {\sum_{n = 0}^{\infty} a_n x^n}$

Tannenbaum (1997) wrote a generating function to calculate the Banzhaf Power Index in Mathematica [2]. The main ideas used in his algorithm is the following: each time all of the conditions (i), (ii), and (iii) below are satisfied, player i is critical.

(i)  Player i is not a member of coalition A.  
(ii) Coalition A is a losing coalition (weight<quota).  
(iii) Adding the weight of player i makes A a winning  coalition. 

The Banzhaf Power Index is traditionally calculated by listing the winning coalitions and counting how many times player i is critical. In the generating function we are looking at losing coalitions which will win with player i's vote. The generating function finds all coalitions by finding the product of (1+$x^{w_{1}}$)(1+$x^{w_{2}}$) ... (1+$x^{w_{n}}$)

The resulting expanded polynomial allows us to easily calculate how many coalitions exist of each weight: the coefficient of $x^{t}$ gives the number of coalitions with weight t.

Take, for example, the weighted voting system [4; 3,2,1]. The generating function to calculate the sums of the weights of all the subsets is:

(1+$x^{3}$)(1+$x^{2}$)(1+$x$) = 1 + $x$ + $x^{2}$ + $2x^{3}$ + $x^{4}$ + $x^{5}$ + $x^{6}$

Each x-term in the polynomial represents a possible coalition with a total weight equal to its power. For the purposes of calculating Banzhaf power, we are only interested in the winning coalitions. The winning coalitions are represented by the x-terms with power $q$ or greater. In this example, $x^{4}$, $x^{5}$, and $x^{6}$ represent winning coalitions {$P_1, P_3$}, {$P_1, P_2$}, and {$P_1, , P_2, P_3$}, respectively.

If we remove $P_i$ from the list of weights, calculate all coalitions with weights between $q-w_i$ (the quota minus the weight of $P_i$) and $q-1$, we identify all the coalitions which lose without the weight of $P_i$ but win with $P_i$, thus calculating the number of times player i is critical.

The generating function for the coalitions without player i is $MinusPlayer$. The $MinusPlayer$ function is the same as the $genFn$ function except that the calculation excludes player i from any coalition. Below is an example of the generating function for [4;3,2,1], without player 1 (0 in Python).

1 + $x$ + $x^{2}$ + $x^{3}$

Total Banzhaf Power Index of player i is obtained by adding the coefficients of $x^j$, in $MinusPlayer$, where q-w_i $\le$ j $\le$ q-1 and j is between 0 and [Total Weights]-1. The Banzhaf power distribution is the normalized distribution or Banzhaf Power Indexes.

As shown in section 1.2, the resulting Banzhaf Power Distribution from the generating function for example 1.1 is {1/3,1/3,1/3,0,0,0}:

Appendix 2 Probabilistic Power Measures¶

2.1 Background on Probabilistic Power Measures¶

In 1971, following a New York's court decision requiring counties to use the Banzhaf index, Justice Harlan repudiated the Banzhaf index as a measure of voter power because of its sensitivity to variations in assumptions [12]. Harlan stated,

This model uses as a measure of voting power the probability that a given voter will cast a tie-breaking ballot in an election. Two further assumptions are made: first, that the voting habits of all members of the electorate are alike; and second that each voter is equally likely to vote for either candidate before him...However, Professor Banzhaf's model also reveals that minor variations in assumptions can lead to major variations in results. For instance, if the temper of the electors are changed by one-half of one percent [if the probability for a voter to select a certain candidate changes from 50% to 50.5%], each individuals voting power is reduced by a factor of 1,000,000.

Harlan makes a good point in that the Banzhaf index, as traditionally calculated, does not accurately measure the voting power when the voter's preferences are taken into account. His dissenting opinion is analyzed in detail and vindicated in Fair Apportionment and the Banzhaf Index[19]. Groffman supports Harlan's mathematical claim that a change in one half of one percent in the voter's preference can cause such a drastic change in the calculated Banzhaf voting power.

Of course, the importance of measuring voting power goes beyond the U.S. courts and beyond the one-man-one-vote issue. Voting blocs in parliaments and Congress can effectively involve weighted voting games when representatives vote along party lines. When the parties' preferences are well-known, sometimes all it takes is one player to reverse the outcome of the vote.

One of the main drawbacks of the Banzhaf Power Index (as well as many of the other power indices) lies in the assumption that all voting parties are assigned an equal probability of voting for or against any given measure. As Justice Harlan points out, the situation models the equivalent of every player flipping a coin to decide which way to vote. This in no way represents reality. In presidential elections, for example, analysts and news reporters often peg certain states' electoral votes to specific presidential candidates. Would it be wise to put money on California electing a Republican president and Texas a Democrat? Likewise, a country's position on a proposition in the European Union, World Trade Organization, or United Nations is often well-known. Using a probability of one-half doesn't always make sense. Banzhaf explicitly states, 'No assumptions are made as to the relative likelihood of any combination....' [17]. Felsenthal and Machover protest that without such an assumption, we cannot justify assigning equal probability to all coalitions [26]. The problem with incorporating each voter's preference in a voting power measurement lies both in the difficulty of measuring the probability of the player's vote and in calculating the voting power of the player once the probability is measured. We will focus solely on the measurement of the voting power once the probability of the voter to vote 'yes' on a motion within the immediate future is provided by external sources.

Let's turn back to Example 1.1. We know from analysis in Part 1 that the Banzhaf index for the shareholders is {1/3, 1/3, 1/3, 0, 0, 0}. Suppose the shareholders have to make a decision, such as whether or not to hire a specific CEO for the company. The above calculation assumes that each voter has a likelihood of exactly 50 percent to vote 'yes' (or 'no') for the hire. If we now recognize the probabilistic voting behaviors of the players, we are allowed to reflect voters' preferences and voting behavior by summarizing them in probabilistic terms.

Example 2.1 If we knew ahead of the vote that Anderson was certain he wanted to hire the candidate, Brad was completely opposed to hiring him, and Connie and the rest of the shareholders didn't have a preference either way, the 'affirmative' probabilities would change to {1,0,0.5,0.5,0.5,0.5}, where Anderson was 100% sure of voting 'yes' and Brad was 100% sure of voting 'no' (or had a 0% probability of voting 'yes'). Intuitively, Connie holds the deciding vote. Does Connie now have more voting power than Anderson or Brad? In sections 2.2 and 2.3 will explore a method for computing Banzhaf Power that incorporates the probabilities of each player voting for or against a motion.

2.2 Probabilistic Power Notation and Terminology¶

In their article Assessment of Voting Situations: The Probabilistic Foundations, Laurelle and Valenciano laid out a theory for measuring probabilistic voting power based on the concepts of 'criticality' (or decisiveness) and 'success' [10]. The probabilistic power measures that follow make extensive use of the ideas, notation, and terminology from this article.

The traditional Banzhaf power index measures the probability of each voter to be a critical player in a 'yes' coalition, given that they have a 50% probability of voting 'yes'. In Part 1 we calculated the traditional Banzhaf power for the weighted voting game [58; 31, 31, 28, 21, 2, 2]. We now want to recalculate the Banzhaf power for [58; 31, 31, 28, 21, 2, 2] where the respective probabilities of the players are {1,0,0.5,0.5,0.5,0.5}. Using our knowledge that p_A=1 we know that Anderson will be in every 'yes' coalition and that with p_B=0, Brad never be in a 'yes' coalition. Connie doesn't have a known preference; at p_C=0.5, she has 50% probability to vote 'yes' and so do D, E, and F. Although the last three voters have no Banzhaf power, we still acknowledge their voting preferences.

Remark 2.2.1 In general, the probability for player i to vote 'yes' is denoted p_i. The 'probability vector' (p₁, p₂,..., p_n) represents the probabilities of players 1 through n, respectively, voting 'yes'.

The weights still determine which coalitions win; however, using the players' probabilities to vote 'yes', we can now incorporate the relative likelihood of each of the winning coalitions.

Remark 2.2.2 In Part 1, we denote a weighted voting system with weights {w₁, w₂,..., w_n} for players {1, 2, 3...,n} and a quota of q as [q; w₁, w₂,..., w_n]. Let $V$ be a general probabilistic voting system consisting of [q; w₁, w₂,..., w_n] together with a probability vector (p₁, p₂,..., p_n) and denoted by

$V$ = [q; w₁, w₂,..., w_n; p₁, p₂,..., p_n]

2.2.1 General Criticality¶

The ability to change the outcome of a winning coalition to a losing coalition is the basic notion behind the traditional measurement of voting power in the Banzhaf index. Our intention here is to show how to generalize this concept to a general probabilistic voting system. If we include the player's ability to change the outcome of a winning $\overline{S}$ ('no') coalition to a losing coalition, we have a player's general criticality (what Valenciano and Laurelle refer to as a player's decisiveness [10]). A player's general criticality is the potential for that player's vote to be critical in both an affirmative and negative sense. Whether or not a player is a critical player still primarily depends on the allotted weights of each player and the quota. With a probabilistic weighted voting system, $V$, the probability of a player to be a critical player now incorporates the relative likelihood that each coalition will form. The new winning coalitions $S \in W$ with critical players for Example 2.1 with probabilities {1,0,0.5,0.5,0.5,0.5} have zero probability that they will include any S coalitions with Brad.

Table 5

Winning 'yes' coalitions (with critical players) for Example 2.1

This example provides a clear distribution of probabilities for 'yes' coalitions with critical players. The coalitions in red are impossible because they include Brad, who will not vote 'yes' (p = 0). If we changed Brad's probability of voting 'yes' to less than 0%, for example if p_B=0.05, the 16 winning S coalitions with Brad in red would be possible, no matter how slight. Brad has no power for an affirmative vote; however, he does have power for a negative vote.

Remark 2.2.5 If the probability for player i to vote 'yes' is denoted p_i, the probability for player i to vote 'no' is 1-p_i. So {1-p₁, 1-p₂,...,1-p_n} represents probabilities of players 1 through n, respectively, voting 'no'. The probability of forming coalition S, denoted by p_S is given by

p_S = $\prod_{i \in S} p_{i}$$\prod_{j \in S} (1-p_{i})$

This is called the probabilistic polynomial (see Straffin [33]).

Definition 2.2.6 An affirmative critical player, is a player whose desertion of a winning coalition (successful affirmative coalition) turns it into a losing coalition.

Let $\varphi^{+}(S) = {\varphi^{+}}_{i}$ denote the probability that i is an affirmative critical player. Then, $\varphi^{+}$ is the sum of all probabilities in S (where S: $i \in S$, $S \in W$, the winning coalition set and $S-i$ is not in the winning coalition set. Note that when p_i=0.5 for all i an affirmative critical player is simply a critical player in the traditional Banzhaf Power Index computation.

Let's turn our attention back to the probabilistic version of Example 2.1 where the weighted voting game is $V$[58; 31,31,28,21,2,2; 1,0,0.5,0.5,0.5,0.5], so that the probability of Anderson being an affirmative critical player is:

=p_AB+p_ABD+p_ABE+p_ABF+p_ABDF+p_ABDE+p_ABEF+p_ABDEF+p_AC+p_ACD+p_ACE+p_ACF+p_ACDF+p_ACDE+p_ACEF+p_ACDEF

= $(1)(0)(0.5)^{4}+...+(1)(0)(0.5)^4+(1)(1-0)(0.5)^{4}+... +(1)(1-0)(0.5)^{4}$

=$8(0)+ 8(0.5)^{4}$

=$0.5$

Anderson has a 0.5 probability of being an affirmative critical player. The first eight S coalitions have zero probability of occurring since they include Brad. Our calculations are further simplified because the last four players have 0.5 probability of voting yes; therefore p_j=1-p_j, for j = C,D,E,F.

The probabilities for each of the shareholders to be an affirmative critical player are {0.5,0,0.5,0,0,0}. If we measure voting power solely using the probability to be an affirmative critical player, as we do in the traditional Banzhaf Power Index, why should Brad have zero power when he decides to vote 'no' whereas Anderson has one-half the power because he decided to vote 'yes'? Clearly, calculating the probability of being an affirmative critical player doesn't provide enough information about the player's preferences to portray an accurate voting power distribution. We also need to look at a player's veto power (see Definition 1.2.6).

From Brad's perspective, success is a negative outcome, not a positive outcome. Brad does not want to swing the vote in a positive direction (i.e., be an affirmative critical player); Brad would like to swing the vote in the 'no' direction. To balance the calculation of the criticality of a player, we must account for the player's veto power, or negative power, as well as that player's affirmative power. The goal is to combine the probability of being an affirmative critical player with the probability of being a negative critical player to obtain the probability of being a general critical player as a measure of probabilistic voting power.

Remark 2.2.7 The positive probabilistic voting system we use to calculate the probabilities of players to be affirmative critical players is $V^{+}$ = [q; w₁,w₂,...w_n; p₁, p₂,...,p_n] and the negative probabilistic voting system in which we calculate the probabilities of players to be negative critical players is $V^{-}$ = [w_N-q+1; w₁,w₂,...w_n; 1-p₁, 1-p₂,..., 1-p_n] , where W_N= $\sum_{i = 1}^{n} w_{i}$

Definition 2.2.8 A negative critical player, is a player whose desertion of a successful negative coalition turns it into a losing coalition.

Let $\varphi^{-}(S)$ = ${\varphi^{-}}_{i}$ denote the probability that i is an negative critical player. Then, $\varphi^{-}$ is the sum of all probabilities in $\overline{S}$ (where $\overline{S}$: $i \in \overline{S}$, $\overline{S} \in W$, the winning coalition set, and $\overline{S}-i$ is not in the winning coalition set).

We calculate the probability to be a negative critical player the same way we calculated the probability to be a positive critical player, except that we use the probabilities for 'no' coalitions and the quota for a winning negative outcome, given by $\overline{q}$ = $\sum_{i = 1}^{n} w_{i}-q+1$

Table 6

Winning 'no' coalitions (with critical players) for Example 2.1

Returning to Example 2.1, we go through the same tedious calculations for calculating the negative critical player as we did for the affirmative critical player to demonstrate the chore of manually calculating the voting power ratio for a probabilistic weighted voting game with relatively few players and simple probabilities.
The probability of Brad being a negative critical player is as follows:

=p_AB+p_ABD+p_ABE+p_ABF+p_ABDF+p_ABDE+p_ABEF+p_ABDEF+p_BC+p_BCD+p_BCE+p_BCF+p_BCDF+p_BCDE+p_BCEF+p_BCDEF

= $(1-1)(1-0)(0.5)^{4}+...+8(1)(1-0)(0.5)^{4}$

=$8(0)+ 8(0.5)^{4}$

=$0.5$

The probability distribution for each of the shareholders to be a negative critical player is {0,0.5,0.5,0,0,0}. It makes sense that A has zero ability to cause a negative outcome since he is 100% certain he will vote 'yes'.

Having just shown an example where the probabilities for players to be affirmative and negative players differ, we will demonstrate that if the voter's do not have preferences - the distribution of probabilities for the players to vote 'yes' or 'no' is {0.5, 0.5,..., 0.5} - the probabilities for players to be affirmative and negative players will be the same. Next, we will combine the positive and negative criticality to provide a single measure of probabilistic voting power.

Definition 2.2.9 Player i is a general critical player if it is an affirmative critical player or a negative critical player.

Letting $\varphi_{i}(V)$ = $\varphi^{+}$ + $\varphi^{-}$

The probabilities that each player in Example 2.1 are {0.5,0,0.5,0,0,0}+{0,0.5,0.5,0,0,0} = {0.5,0.5,1,0,0,0}

Here we can see that Connie has more power than the other players because she can swing the vote in either direction. The independent voter has more power, which is intuitively reasonable.

To get a traditional power measure we normalize these probabilities. Some argue that normalizing the probability distribution causes distortions (e.g., [10]); however, in order to compare the values with other power indices, normalizing is critical. Once the probabilities are normalized, we should not combine the normalized values with other probability distributions; rather, if we normalize, we will normalize after combining distributions. If we normalize the probability distribution that each player is a general critical player we get the probabilistic Banzhaf Power Index {0.5/2,0.5/2,1/2,0/2,0/2,0/2} = {0.25,0.25,0.5,0,0,0}. This normalized distribution shows us that Anderson and Brad have one-quarter of the voting power and Connie has one-half.

Definition 2.2.10 Player i has affirmative success if i votes 'yes' and S wins.

Definition 2.2.11 Player i has negative success if i votes 'no' and S loses.

The probability distribution that each player is affirmatively successful in Example 2.1 is: {0.5,0,0.5,0.25,0.25,0.25}. If we performed the same calculations to compute the probabilities for each player to be negatively successful (using $V^-$) we would get the probability distribution of negative success {0,0.5,0.5,0.25,0.25,0.25}.

2.3 Using Probabilistic Generating Functions¶

As with the traditional Banzhaf Power calculation, we can use a generating function to calculate probabilistic voting power. Tannenbaum describes the method used for calculating affirmative critical power in his unpublished paper entitled Probabilistic Models of Banzhaf Power [31]. In addition to affirmative critical power, these methods are extended in this section to calculate negative critical power, general criticality, success, and luck.

2.3.1 Criticality¶

Finally we get to use technology to aid us in calculating probabilistic voting power. We essentially use the same method as we used for calculating Banzhaf Power in the first part, except, here we take into account the distribution of probabilities representing the players' preferences. Tannenbaum describes a generating function of a general weighted voting game, $V$ which calculates the probabilities and weights of all coalitions by finding the product of

($(1-p_{1})+p_{1}x^{w_{1}})$$((1-p_{2})+p_{2}x^{w_{2}})$...$((1-p_{n})+p_{n}x^{w_{n}})$

where $p_i$ is the probability to vote 'yes', and $w_i$ is the weight of player i.

The generating function probGenFn generates the probabilities for each coalition of different weights. The coefficient of $x^r$ represents the probability of the specific winning coalition with weight r, where $r \ge q$.

Affirmative Critical Player

As in the prior section, we start by calculating the probability of each player to be an affirmative critical player. Player i is an affirmative critical player in coalition S if i votes 'yes' and S could not win without i, or the total weight of coalition S-i is between q-s_i and q-1.

The steps to calculating the probability of each player in Example 2.1 to be an affirmative critical player are as follows:

• Calculate the probabilistic generating function for $V^{+}$=[58; 31,31,28,21,2,2; 1,0,0.5,0.5,0.5,0.5] which is the expansion of ($0+1x^{31}$)($1+0x^{31}$)($0.5+0.5x^{28}$)($0.5+0.5x^{21}$)($0.5+0.5x^{2}$)($0.5+0.5x^2$). The coefficient of $x^{31}$ in the expanded function represents the probability of 0.0625 that a coalition with a total weight of 31 occurs. In this particular example, the only probable coalition with weight 31 would be S={A}. The probability of Anderson voting 'yes' and all other voters 'no' is: p_S=$(1)(1-0)(1-0.5)^4$ = $0.0625$ Below, we see how this number comes up as the coefficient of $x^{31}$ in the probabilistic generating function.

• Define the generating function for probabilities of all coalitions without player i as probMinusPlayer.

Below, probMinusPlayer calculates the expansion of the probabilistic case of Example 2.1 without player 1 (0 in Python), excluding Anderson's vote.

Note that without Anderson, the probability of a coalition of weight 31 occurring is zero (i.e., an x-term with exponent 31 doesn't exist in the probMinusPlayer expanded function for Anderson since Brad won't vote yes).

• Calculate the probability that player i will be an affirmative critical player, or PhiPos. The probability that player i is an affirmative critical player is the sum of the coefficients of $x^j$, where q-w_i $\le$ j $\le$ q-1 in ProbMinusPlayer for i multiplied by p_i, the probability that player i will vote 'yes'. This gives us the sum of the probabilities of each S occurring and winning only with i.

• We repeat this for all the players and create a list of PhiPos for all players called PhiPosList.

Negative Critical Player

The steps to calculating the probability that each player is a negative critical player are essentially the same as for an affirmative critical player, except instead of $V^{+}$ we use $V^{+}$ =[58; 31,31,28,21,2,2; 0,1,0.5,0.5,0.5,0.5]. Player i is a negative critical player in $\overline{S}$, if without i voting "no", the negative votes in $\overline{S}$ do not win.

We define the probability that player i will be a negative critical player as PhiNeg. The probability that player i is a negative critical player is the sum of the coefficients of $x^r$ in probMinusPlayer, where $\overline{q}-w_i \le r \le \overline{q}-1$, where $\overline{q}$ = $(\sum w_{i}) - q + 1$, multiplied by the probability of player i to vote 'no'. This gives us the probability of $\overline{S}$ occurring and winning only with i.

Again, we create a list of PhiNeg for each player called PhiNegList. The PhiNegList is the probability distribution for each of the shareholders being a negative critical player.

General Critical Player

The probability of i being a general critical player is the sum of the probabilities of i being an affirmative critical player and a negative critical player. We define the list of such probabilities as PhiList. We can normalize PhiList into a power index PhiNorm.

Brad does in fact end up with a quarter of the voting power, along with Anderson, since Anderson has affirmative power and Brad has negative power. Connie has the most power, as the swing voter.

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