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Mathematics LibreTexts

3.2: Beginnings

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    We’ll begin with some basic vocabulary for weighted voting systems.

    Vocabulary for Weighted Voting

    Each individual or entity casting a vote is called a player in the election. They’re often notated as \(\bf{P_1, P_2, P_3, ... , P_N}\) where \(N\) is the total number of voters.

    Each player is given a weight, which usually represents how many votes they get.

    The quota is the minimum weight needed for the votes or weight needed for the proposal to be approved.

    A weighted voting system will often be represented in a shorthand form:

    \([q: w_1, w_2, w_3, ... , w_N]\)

    In this form, \(q\) is the quota, \(w_1\) is the weight for player 1, and so on.

    Example 1

    In a small company, there are 4 shareholders. Mr. Smith has a 30% ownership stake in the company, Mr. Garcia has a 25% stake, Mrs. Hughes has a 25% stake, and Mrs. Lee has a 20% stake. They are trying to decide whether to open a new location. The company by-laws state that more than 50% of the ownership has to approve any decision like this. This could be represented by the weighted voting system:

    \([51: 30, 25, 25, 20]\)


    Here we have treated the percentage ownership as votes, so Mr. Smith gets the equivalent of 30 votes, having a 30% ownership stake. Since more than 50% is required to approve the decision, the quota is 51, the smallest whole number over 50.

    In order to have a meaningful weighted voting system, it is necessary to put some limits on the quota.

    Limits on the Quota

    The quota must be more than \(\frac{1}{2}\) the total number of votes.

    The quota can’t be larger than the total number of votes.

    Why? Consider the voting system \( [q; 3, 2, 1]\)

    Here there are 6 total votes. If the quota was set at only 3, then player 1 could vote yes, players 2 and 3 could vote no, and both would reach quota, which doesn’t lead to a decision being made. In order for only one decision to reach quota at a time, the quota must be at least half the total number of votes. If the quota was set to 7, then no group of voters could ever reach quota, and no decision can be made, so it doesn’t make sense for the quota to be larger than the total number of voters.

    Try it Now 1

    In a committee there are four representatives from the management and three representatives from the workers’ union. For a proposal to pass, four of the members must support it, including at least one member of the union. Find a voting system that can represent this situation.


    If we represent the players as \(M_{1}, M_{2}, M_{3}, M_{4}, U_{I}, U_{2}, U_{3}\) then we may be tempted to set up a system like \([4: 1, 1, 1, 1, 1, 1, 1]\). While this system would meet the first requirement that four members must support a proposal for it to pass, this does not satisfy the requirement that at least one member of the union must support it.

    3.2: Beginnings is shared under a CC BY-SA 3.0 license and was authored, remixed, and/or curated by David Lippman via source content that was edited to conform to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

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