[EM] FWD: Borda Count by Paul Dumais
Bart Ingles
bartman at netgate.net
Sat Apr 17 23:48:16 PDT 1999
Here is a reply to Mr. Dumais's request.
--Bart
The following are examples of failures in the Borda Count voting
method. The first example shows a failure mode common (in some form) to
most if not all popular election methods that use ranked ballots, and
the second shows a failure that appears to be peculiar to Borda.
The examples show voters' suitability ratings of the candidates on a
scale of 0-100. I am assuming that each voter has the right to give a
maximum score (100) to his/her first choice, and minimum score (0) to
his last choice, and that the voter has the right to place middle
candidates at any position along that scale. I use average ratings to
score the overall suitability of a candidate.
While voters' sincere ratings are probably not measurable, I believe it
is a mistake to pretend they do not exist. Even though we may not be
able to use ratings directly in an actual election, we can set up
examples based on sets of voter ratings and see how various election
methods behave.
-----------------------------------------------------
In the first example, 45% of the voters prefer A over all other
candidates. 15% prefer B, and 40% prefer C. The full ratings are as
follows:
EXAMPLE 1: Voters' private suitability ratings
Rating:
100 80 60 40 20 0
----------------------------------
45 A B C
15 B C A
40 C B A
---
100 votes total
Average sincere ratings:
Candidate A = (45% x 100) = 45.0 points
Candidate B = (15% x 100) + (85% x 10) = 23.5 points
Candidate C = (15% x 90) + (40% x 100) = 53.5 points
Borda Count:
Candidate A = 90 points
Candidate B = 115 points
Candidate C = 95 points
Borda picks B as the winner based on rankings, although B has only half
the rating of the other two candidates.
Note that under Approval Voting, the voters could vote as follows:
45 A
15 BC
40 C
C would probably win, with around 55 votes, versus 45 for A and 15 for
B. The A voters could prevent C from winning by voting for both A and
B, but are unlikely to for two reasons First, A has enough votes that
they may think that A has a chance of winning. Second, since they think
that B is a poor candidate, it will be unlikely that many of them will
be willing to vote for B the same as for their favorite.
----------------------------------------------------
As bad as the Borda results were in the last example, it gets even
worse. In the following example, B is considered totally unsuitable by
all of the A and C supporters.
EXAMPLE 2:
Rating:
100 80 60 40 20 0
----------------------------------
45 A C B
15 B C A
40 C A B
---
100 votes total
Average sincere ratings:
Candidate A = (45% x 100) + (40% x 50) = 65.0 points
Candidate B = (15% x 100) = 15.0 points
Candidate C = (45% x 50) + (15% x 20) + (40% x 100) = 65.5 points
Under Borda C should get 140, A should get 130, and B should get 30,
right? The only problem is, since B is perceived as a weak candidate,
the A supporters are more concerned with defeating their major opponent,
C. The A voters rank insincerely as:
45 A B C
15 B C A
40 C A B
---
100 votes total
Now A has 130, C has only 95, and B has 75. What should the C voters
do? If they adopt the same strategy by voting CBA, they can knock out
A, but will be worse off with last choice B winning (as in the Borda
results for Example 1). They may have to do so, though, in order to
discourage the other side from attempting the strategy again in the
future.
Granted that such severe Mutual Assured Destruction tactics involving
every voter are unlikely. However, it is possible for smaller numbers
of the A and C supporters to engage in the same tactic. The sensible
strategy would be to try to defeat the major opponent, while being
careful not to give the election away to a "dummy" opponent. The result
would be a tendency to equalize the scores of all three candidates,
resulting in a nearly random outcome.
Note again that under Approval voting, voters could vote:
45 A
15 BC
40 C
Here, the A and C voters have no incentive at all to vote for a second
candidate. The B voters know that they have no chance of winning, but
can get a moderate improvement in outcome by including a vote for C
along with B.
-----------------------------------------------------------
While a couple of examples alone don't prove that Approval is a better
method than Borda (or other ranked methods), they do seem to poke holes
in some of the arguments to the contrary. How good a method is at
distinguishing between to nearly equal candidates is immaterial when the
method is also capable of letting a very bad candidate defeat both of
them.
Donald E Davison wrote:
>
> -----------Forwarded letter -------------
> Date: Fri, 16 Apr 1999 10:22:44 -0600
> From: Paul Dumais <paul at amc.ab.ca>
> Reply-To: paul at amc.ab.ca
> Organization: AMC
> MIME-Version: 1.0
> To: Donald E Davison <donald at mich.com>
> Subject: Re: Salva Voting - multi-seat example
>
> Hi,
>
> In response to this, I'd like to discuss Borda Count. Sorry if this is
> somewhat off topic. I think Borda count is clearly the best choice
> (clear to me at least ;-)). The best way I to convince others of this is
> by many examples and comparissons with other voting methods. Looking at
> Borda count, I have never found an example of an "innapropriate" result
> or a result which was more "appropriate" via another method. There has
> been several mentions of borda count having it's second and subsequent
> choices working against your first. I feel this is appropriate. Your
> second choice is an inication of compromise. Where a race is close
> between several candidates, often the first choice count gives close to
> a tie (two or three way). Second and subsequent choices here are all
> important. Borda count is similar to STV and condorcet, but more fare.
> Condorcet is like stv except the "lowest" candidate does not have his
> votes transferred. All votes are transferred to all combinations of
> pairs. The choice of the "lowest" candidate using the first choices is
> unfare and not necessary. Borda count is similar to condorcet in that
> one vote is given to a candidate for each paring it wins. Condorcet
> breaks up the votes ito pairing though, which wastes votes. I will
> provide in-detail examples in later posts to clarify this.
> A description of how Borda count should work: Each voter ranks his
> candidates ie ABC. He need not rank them all. He could only give one
> choice if he wanted. Suppose there are 5 candidates, a vote of ABC would
> give A - 4
> B - 3
> C - 2
> D - 0.5
> E - 0.5
>
> Notice that D and E split the remaining points since they were not
> ranked. This point system is equivalent to a round robin tourney where a
> point is given for every match one. Since ech candidate has 4 matches,
> ech candidate gets 0-4 points. D and E tie, so 0.5. All ties get 0.5. So
> if someone votes AD, the points would be:
> A-4
> D-3
> C-1
> D-1
> E-1
>
> Each candidate is given an appropriate number of points for each vote.
> After the points are added for each voter, the ranking of candidates is
> easily made. The top vote-getters get the available seats - simply and
> fairly. Borda count is better than Salva voting, not all the votes are
> used equally in any method which transfers votes. The "lower candidates"
> get there votes transferred. The choice of "lower candidates" is unfair.
> The first choice of the voters might reveal a very close race, but there
> may be overwhelming support for a candidate when looking at second
> choices. Yet, for single seat elections, the Salva method might transfer
> the votes of those who chose this popular candidate as their first
> choice. Transferring votes will always be unfair, because any fair
> measure of the "lowest candidate(s)" will also be a fair method of
> choosing the "highest" candidates, which eliminates the need to transfer
> votes in the first place. Borda count is such a method.
> I can construct examples where borda count is superior to other
> methods. I have not found any examples where other methods are superior
> to borda count. Perhaps someone can help.
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