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Mathematics LibreTexts

2.8: Borda Count

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    36232
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    Borda Count is another voting method, named for Jean-Charles de Borda, who developed the system in 1770.

    Borda Count

    In this method, points are assigned to candidates based on their ranking; 1 point for last choice, 2 points for second-to-last choice, and so on. The point values for all ballots are totaled, and the candidate with the largest point total is the winner.

    Example 8

    Map showing the locations of 4 cities: Olympia, Tacoma, Puyallup, and SeattleA group of mathematicians are getting together for a conference. The members are coming from four cities: Seattle, Tacoma, Puyallup, and Olympia. Their approximate locations on a map are shown to the right.

    The votes for where to hold the conference were:

    \(\begin{array}{|l|l|l|l|l|}
    \hline & 51 & 25 & 10 & 14 \\
    \hline 1^{\text {st }} \text { choice } & \text { Seattle } & \text { Tacoma } & \text { Puyallup } & \text { Olympia } \\
    \hline 2^{\text {nd }} \text { choice } & \text { Tacoma } & \text { Puyallup } & \text { Tacoma } & \text { Tacoma } \\
    \hline 3^{\text {rd }} \text { choice } & \text { Olympia } & \text { Olympia } & \text { Olympia } & \text { Puyallup } \\
    \hline 4^{\text {th }} \text { choice } & \text { Puyallup } & \text { Seattle } & \text { Seattle } & \text { Seattle } \\
    \hline
    \end{array}\)

    Solution

    In each of the 51 ballots ranking Seattle first, Puyallup will be given 1 point, Olympia 2 points, Tacoma 3 points, and Seattle 4 points. Multiplying the points per vote times the number of votes allows us to calculate points awarded:

    \(\begin{array}{|l|l|l|l|l|}
    \hline & 51 & 25 & 10 & 14 \\
    \hline 1^{\text {st choice }} & \text { Seattle } & \text { Tacoma } & \text { Puyallup } & \text { Olympia } \\
    4 \text { points } & 4 \cdot 51=204 & 4 \cdot 25= 100 & 4 \cdot 10=40 & 4 \cdot 14=56 \\
    \hline 2^{\text {nd choice }} & \text { Tacoma } & \text { Puyallup } & \text { Tacoma } & \text { Tacoma } \\
    3 \text { points } & 3 \cdot 51=153 & 3 \cdot 25=75 & 3 \cdot 10=30 & 3 \cdot 14=42 \\
    \hline 3^{\text {rd }} \text { choice } & \text { Olympia } & \text { Olympia } & \text { Olympia } & \text { Puyallup } \\
    2 \text { points } & 2 \cdot 51=102 & 2 \cdot 25=50 & 2 \cdot 10=20 & 2 \cdot 14=28 \\
    \hline 4^{\text {th }} \text { choice } & \text { Puyallup } & \text { Seattle } & \text { Seattle } & \text { Seattle } \\
    1 \text { point } & 1 \cdot 51=51 & 1 \cdot 25=25 & 1 \cdot 10=10 & 1 \cdot 14=14 \\
    \hline
    \end{array}\)

    Adding up the points:

    • Seattle: \(204 + 25 + 10 + 14 = 253\) points
    • Tacoma: \(153 + 100 + 30 + 42 = 325\) points
    • Puyallup: \(51 + 75 + 40 + 28 = 194\) points
    • Olympia: \(102 + 50 + 20 + 56 = 228\) points

    Under the Borda Count method, Tacoma is the winner of this vote.

    Try it Now 4

    Consider again the election from Try it Now 1. Find the winner using Borda Count. Since we have some incomplete preference ballots, for simplicity, give every unranked candidate 1 point, the points they would normally get for last place.

    \(\begin{array}{|l|l|l|l|l|l|l|l|}
    \hline & 44 & 14 & 20 & 70 & 22 & 80 & 39 \\
    \hline 1^{\text {st }} \text { choice } & \mathrm{G} & \mathrm{G} & \mathrm{G} & \mathrm{M} & \mathrm{M} & \mathrm{B} & \mathrm{B} \\
    \hline 2^{\text {nd }} \text { choice } & \mathrm{M} & \mathrm{B} & & \mathrm{G} & \mathrm{B} & \mathrm{M} & \\
    \hline 3^{\text {rd }} \text { choice } & \mathrm{B} & \mathrm{M} & & \mathrm{B} & \mathrm{G} & \mathrm{G} & \\
    \hline
    \end{array}\)

    Answer

    Using Borda Count:

    We give 1 point for 3rd place, 2 points for 2nd place, and 3 points for 1st place.

    \(\begin{array}{|l|l|l|l|l|l|l|l|}
    \hline & 44 & 14 & 20 & 70 & 22 & 80 & 39 \\
    \hline 1^{\text {st }} \text { choice } & \mathrm{G} & \mathrm{G} & \mathrm{G} & \mathrm{M} & \mathrm{M} & \mathrm{B} & \mathrm{B} \\
    & 132 \mathrm{pt} & 42 \mathrm{pt} & 60 \mathrm{pt} & 210 \mathrm{pt} & 66 \mathrm{pt} & 240 \mathrm{pt} & 117 \mathrm{pt} \\
    \hline 2^{\text {nd }} \text { choice } & \mathrm{M} & \mathrm{B} & & \mathrm{G} & \mathrm{B} & \mathrm{M} & \\
    & 88 \mathrm{pt} & 28 \mathrm{pt} & & 140 \mathrm{pt} & 44 \mathrm{pt} & 160 \mathrm{pt} & \\
    \hline 3^{\text {rd }} \text { choice } & \mathrm{B} & \mathrm{M} & \mathrm{M} 20 \mathrm{pt} & \mathrm{B} & \mathrm{G} & \mathrm{G} & \mathrm{M} 39 \mathrm{pt} \\
    & 44 \mathrm{pt} & 14 \mathrm{pt} & \mathrm{B} 20 \mathrm{pt} & 70 \mathrm{pt} & 22 \mathrm{pt} & 80 \mathrm{pt} & \mathrm{G} 39 \mathrm{pt} \\
    \hline
    \end{array}\)

    G: \(132+42+60+140+22+80+39 = 515\) pts

    M: \(88+14+20+210+66+160+39 = 597\) pts

    B: \(44+28+20+70+44+240+117 = 563\) pts

    McCarthy (M) would be the winner using Borda Count.


    2.8: Borda Count 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.