Voting Theory: Borda count


– WELCOME TO A LESSON ON THE
BORDA COUNT METHOD OF VOTING. IN THIS LESSON WE’LL DEFINE
THE BORDA COUNT METHOD, THEN DETERMINE WINNERS
OF ELECTIONS USING THE BORDA COUNT METHOD. BORDA COUNT IS A VOTING METHOD
NAMED AFTER JEAN CHARLES DE BORDA, WHO
DEVELOPED THE SYSTEM IN 1770. IN THIS METHOD POINTS ARE
ASSIGNED TO CANDIDATES BASED UPON THEIR RANKING,
WHERE CANDIDATES RECEIVE 1 POINT FOR LAST, 2 POINTS FOR SECOND
TO THE LAST, AND SO ON. SO FOR EXAMPLE, IF THERE WERE
FOUR CHOICES, FIRST CHOICE VOTES
WOULD BE WORTH 4 POINTS, SECOND CHOICE VOTES
WORTH 3 POINTS, THIRD CHOICE VOTES
WORTH 2 POINTS, AND FINALLY FOURTH CHOICE VOTES
WOULD BE WORTH 1 POINT. THEN THE POINTS OF THE BALLOTS
ARE TOTALED, AND THE CANDIDATE
WITH THE LARGEST POINT TOTAL IS THE WINNER. LET’S LOOK AT AN EXAMPLE. HERE’S THE PREFERENCE TABLE
FOR A CAMPING CLUB THAT IS DECIDING WHERE TO GO ON
THEIR NEXT TRIP, WHERE “A” REPRESENTS THE ARCHES,
G REPRESENTS THE GRAND CANYON, Y REPRESENTS YOSEMITE,
AND Z=ZION. NOTICE HOW WE HAVE FOUR CHOICES
HERE, AND THEREFORE FIRST PLACE, OR FIRST CHOICE,
WOULD RECEIVE 4 POINTS, SECOND PLACE WOULD RECEIVE 3
POINTS, THIRD PLACE 2 POINTS, AND FOURTH PLACE 1 POINT. WHICH MEANS EVERY VOTE
IN THIS FIRST ROW WOULD BE WORTH 4 POINTS, EVERY VOTE IN THE SECOND CHOICE
ROW IS WORTH 3 POINTS, AND SO ON. SO TO FIND THE TOTAL POINTS, WE’LL SET THIS UP USING A LARGER
TABLE, AS WE SEE HERE. AGAIN, ALL OF THESE FIRST PLACE
VOTES HERE ARE WORTH 4 POINTS EACH FOR EACH CANDIDATE,
OR IN THIS CASE EACH LOCATION. SO LOOKING AT THIS FIRST ROW
FOR A MOMENT, NOTICE HOW 12 VOTERS VOTED
YOSEMITE FIRST. SO YOSEMITE RECEIVES 12 x 4
OR 48 POINTS. LOOKING AT THE SECOND COLUMN
HERE, NOTICE HOW FIVE VOTERS VOTED
THE GRAND CANYON FIRST, AND THEREFORE THE GRAND CANYON
RECEIVED 5 x 4 OR 20 POINTS. THE SECOND CHOICE ROW HERE IS
WORTH 3 POINTS FOR EVERY VOTE. THE THIRD CHOICE ROW
IS WORTH 2 POINTS, AND THE FOURTH CHOICE ROW
IS WORTH 1 POINT. SO YOU MAY WANT TO PAUSE
THE VIDEO NOW JUST TO VERIFY
THAT I DID CALCULATE THESE POINTS CORRECTLY. LET’S GO AHEAD AND FIND THE
TOTAL FOR EACH CHOICE. LET’S FIRST FIND THE POINTS
FOR THE ARCHES, SO WE’LL LOOK FOR ALL THE “A”s. SO IN THE FIRST ROW WE HAVE
AN “A” HERE AND HERE, THE SECOND ROW WE HAVE AN “A”
HERE, THE THIRD ROW WE HAVE AN “A”
HERE AND HERE, AND THE FOURTH ROW WE HAVE AN
“A” HERE. SO “A” RECEIVES 36 + 24 + 12 +
10 + 20 + 12 POINTS, OR 114 POINTS. NEXT WE’LL FIND THE POINTS FOR
THE GRAND CANYON, SO WE’LL FIND ALL OF THE Gs. FIRST ROW WE HAVE A G HERE,
SECOND ROW WE HAVE A G HERE, THIRD ROW WE HAVE A G HERE,
HERE, AND HERE, AND THE FOURTH ROW WE HAVE A G
HERE. SO THE GRAND CANYON RECEIVES 20 + 30 + 24 + 8 + 18 + 6,
OR 106 POINTS. NEXT, FOR YOSEMITE WE HAVE A Y
HERE IN THE FIRST ROW, SECOND ROW WE HAVE A Y HERE AND
HERE, THIRD ROW WE HAVE A Y HERE, AND FOURTH ROW WE HAVE A Y
HERE AND HERE. SO YOSEMITE RECEIVES 48 + 15 +
27 + 12 + 4 + 10, OR 116 POINTS. AND THEN FINALLY ZION IS HERE
IN THE FIRST ROW AND HERE, SECOND ROW WE HAVE A Z
HERE AND HERE. THERE ARE NO Zs IN THE THIRD
ROW, BUT THE FOURTH ROW WE HAVE A Z
HERE AND HERE. SO ZION RECEIVES 16 + 40 + 36 +
18 + 5 + 9 POINTS, WHICH IS 124 POINTS. AND SINCE ZION
HAS THE MOST POINTS, UNDER THE BORDA COUNT METHOD
ZION IS THE WINNER. NOW LET’S TALK ABOUT THE FLAW, OR WHAT’S WRONG WITH THE BORDA
COUNT METHOD. ONE POTENTIAL FLAW OF BORDA
COUNT IS A CANDIDATE COULD RECEIVE A MAJORITY OF THE FIRST CHOICE
VOTES, AND STILL LOSE THE ELECTION. TO ILLUSTRATE THIS, LET’S TAKE
A LOOK AT ANOTHER EXAMPLE. LET’S CONSIDER THE FOLLOWING
PREFERENCE SCHEDULE. FIRST, NOTICE IN THIS ELECTION
THERE ARE A TOTAL OF 20 VOTES, AND NOTICE HOW CANDIDATE “A”
RECEIVES 9 + 2, OR 11 OF THE 20 VOTES,
WHICH WOULD BE 55%, OR A MAJORITY WIN. NOTICE CANDIDATE B RECEIVES
8 FIRST CHOICE VOTES, OR 40% OF THE FIRST CHOICE
VOTES, AND CANDIDATE C ONLY RECEIVES 1,
OR 5% OF THE FIRST CHOICE VOTES. SO “A” WOULD WIN BY MAJORITY. BUT NOW LET’S FIND THE WINNER
OF THIS ELECTION BY USING THE BORDA COUNT METHOD. FIRST, SINCE THERE ARE THREE
CANDIDATES, FIRST CHOICE VOTES
ARE WORTH 3 POINTS, SECOND CHOICE VOTES
ARE WORTH 2 POINTS, AND THIRD CHOICE VOTES
ARE WORTH 1 POINT. I DIDN’T SHOW AS MUCH WORK HERE, BUT NOTICE HOW LOOKING AT THIS
FIRST COLUMN 9 VOTERS VOTED “A” FIRST, SO “A” RECEIVES 9 x 3,
OR 27 POINTS. IN THIS COLUMN 2 VOTERS VOTED
“A” FIRST, SO “A” RECEIVES ANOTHER 2 x 3,
OR 6 POINTS. HERE B IS VOTED FIRST BY 8
VOTERS, SO B RECEIVES 8 x 3,
OR 24 POINTS, AND SO ON. EVERY VOTE IN THE SECOND ROW
IS WORTH 2 POINTS, SO B RECEIVES 18 POINTS HERE,
SINCE 9 x 2=18, AND SO ON. SO NOW LET’S FIND THE TOTAL
NUMBER OF POINTS FOR EACH CANDIDATE. “A” RECEIVES 27 + 6 + 8 + 1
OR 42 POINTS. REMEMBER “A” WAS THE MAJORITY
WINNER. BUT NOW IF WE TAKE A LOOK
AT CANDIDATE B, B RECEIVES 24 + 18 + 2 + 2, WHICH NOTICE HOW IT COMES
TO 46 POINTS. NOTICE HOW CANDIDATE B
HAS MORE POINTS THAN “A”, EVEN THOUGH “A” IS THE MAJORITY
WINNER. AND THEN FINALLY CANDIDATE C
RECEIVES 3 + 4 + 16 + 9, OR 32 POINTS. SO NOTICE HOW, AGAIN,
B WOULD WIN USING BORDA COUNT, WHICH BRINGS UP ANOTHER FAIRNESS
CRITERION, THE MAJORITY CRITERION. IF A CHOICE HAS A MAJORITY
OF FIRST PLACE VOTES, THAT CHOICE
SHOULD BE THE WINNER. BORDA COUNT IS SOMETIMES
DESCRIBED AS A CONSENSUS
BASED VOTING SYSTEM, SINCE IT CAN SOMETIMES CHOOSE
A MORE BROADLY ACCEPTABLE OPTION OVER ONE WITH MAJORITY SUPPORT. THIS IS A DIFFERENT APPROACH
THAN PLURALITY AND INSTANT RUNOFF VOTING THAT
FOCUS ON FIRST CHOICE VOTES. BORDA COUNT CONSIDERS
VOTERS ENTIRE RANKING TO DETERMINE THE OUTCOME. BECAUSE OF THIS CONSENSUS
BEHAVIOR, BORDA COUNT, OR SOME VARIATION OF IT, IS COMMONLY USED IN SELECTING
SPORTS AWARDS. VARIATIONS ARE USED TO DETERMINE THE MOST VALUABLE PLAYER
IN BASEBALL, TO RANK TEAMS IN THE NCAA, AND ALSO TO AWARD THE HEISMAN
TROPHY. I HOPE YOU FOUND THIS HELPFUL.  

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Comments

  1. Is there, by chance, a method of the Borda count that includes a "0" ranking for choices that voters just cannot abide? I think that would be an excellent addition.

    Like in this video example – What if I reeeally wanted to go to the Yosemite so…

    I rank Yosemite first, giving it 4 points.

    I think the Grand Canyon would also be cool so I give that my 3 points.

    I don't know anything about Arches, really, so since it's a 3rd choice, would need to rank it with 2 pts. (rather than skipping a rank to give it a 1, I would think. (Now that I think of it, I didn't consider how to deal with potential "skips" and that may be an issue with voting. Will have to think about that one.)

    But I can't stand the idea of going to Zion so I give it a "0" in order to purposefully denote my stance against the idea and lower the point count for the choice.

    Is there a method that considers that?

    Further, can one make a "0" rank or "non-vote" for more than one option? Say I want to vote "0" for Zion and Arches…

    That being said, would it be better to offer a ranking system of top-down, bottom-up to distinguish the ones you like best to least and also the ones you really don't like most to least? Would that require a "null" voting rank to put a buffer between the two options "preferred" and "not-preferred". Maybe it should be a 10 pt. down through 1 pt. ranking (unless there are more voting option; perhaps double the count or the votes?)
    Or would a negative point ranking system work for the "non-preferred" candidates? If so, would (or could) those count against the other candidates overall positive votes?

    Not sure about that last one. At least with governmental elections, anyway. Sounds like it could be used as a weapon, so to speak. It's already convoluted. And I definitely fear the potential for fraud & tampering.
    So, perhaps it would be best to stick to zeroes. But that means that the entire candidate list from all parties (including 3rd parties, etc.) would, of course, have to be ranked.

    Sorry… Got lost in thinking "out loud" so to speak. : )

  2. Thank you SO much for making his video!!!! I totally understand this topic now! Right on time for my test thursday. Thank you so much!

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