2.8: Borda Count
Borda Count is another voting method, named for Jean-Charles de Borda, who developed the system in 1770.
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.
A 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.
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 3 rd place, 2 points for 2 nd place, and 3 points for 1 st 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.