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Re: [EM] Example of Condorcet Missing the Highest Utility Center Candidate

FS
Forest Simmons
Fri, Jun 5, 2015 6:47 PM

Juho,

thanks for your insights.

In your excellent example, candidate D appears to be the IA-MPO winner,
especially if we define implicit approval as "ranked above bottom."

Forest

On Fri, Jun 5, 2015 at 12:03 AM, Juho Laatu juho4880@yahoo.com wrote:

One solution for this problem would be to use some rated method. If every
voter votes according to his travel distances to each candidate center we
could get ideal results. I mean that actually we are discussing here
indirectly whether to use ranked or rated methods.

(In the example one could mandate each voter to vote based on the distance
(in the rated method based election), or calculate the results based on the
addresses of the voters, i.e. without having an election. This way we could
get rid of any possible strategies.  :-) )

When compared to the rated methods, (plain) Condorcet methods focus only
on counting the majorities, not the strength of opinions. In this example
any two 33 voter groups could form a clear majority. They could agree to
vote together to make the center between them the winner. Any sensible
argument against the third 33 voter group could be a sufficient reason (for
the two 33 groups) to do so.

One could say that Condorcet methods aim at electing stable winners. They
try to seek a winner that will not be disliked by some clear majority. (I'm
vague here because the behaviour of different Condorcet methods is somewhat
different.) If you would elect C, there could easily be some majority
alliance that would be interested in trying to change the elected center to
another center. If you elect Mx, the outcome is probably more stable.

One can study this example also from strategic voting point of view. I
guess the given votes are quite stable and there are no obvious strategies
to improve one's (distance based) expected outcome. Condorcet could however
make it possible to strategically make the result worse in the sense that
the voters could reduce their expected outcome in terms of distance but
make the expected outcome better in the sense that the whole society would
benefit of it. I mean that many voters could see that C is obviously in
some sense best for the society, and they could therefore rank C first in
their ballots. No harm done and no risks doing so, if the voters truly
prefer a solution that is good for all, and not just good for them
personally. The point here is that Condorcet does not force them to make
decisions that they consider stupid. If some solution looks stupid to us in
some example, maybe the voters would see that stupidity too, and vote
accordingly (changing their preferences to something more sensible).

Juho

P.S. I'm ok with electing Condorcet losers in some (extreme) scenarios. As
I already said, Condorcet methods tend to seek stable solutions. In the
following example the Condorcet loser (D) is disliked only very mildly in
the pairwise comparisons, while all the others have a strong opposition
against them (in favour of changing them to some other candidate).
17 A>B>D>C
16 A>D>B>C
17 B>C>D>A
16 B>D>C>A
17 C>A>D>B
16 C>D>A>B

On 05 Jun 2015, at 05:11, Forest Simmons fsimmons@pcc.edu wrote:

Suppose that a town with 100 voting citizens has 33 voters residing at
each of the three vertices of an equilateral triangle (with two mile sides,
say), and one voter residing within two hundred yards of the the center of
the triangle.

Proposed sites for the new community center are M1, M2, and M3, at the
respective midpoints of the three sides of the triangle, as well as site C
at the center of the triangle’ a couple of hundred yards from the lone
voter that we just mentioned.

Assuming that voters prefer closer sites over more distant sites the
preferences are

33 M1=M2>C

33 M1=M3>C

33 M2=M3>C

01 C

Note that C is the Condorcet Loser, since each of the M’s beats C pairwise
by almost a two-thirds majority, 66 to 34.

On the other hand, C is the IA winner with 100 percent implicit approval.

Candidate C is also the IA-MPO winner with a score of 100-66, compared
with 66-33 for the other alternatives.

How about average distance of voters from each of the alternatives?

The average distance to alternative C is 0.66 times the square root of
three miles, or about 1.14 miles.

The average distance to any of the M’s is about 1.24 miles.

If the M’s were moved directly away from their midpoint positions to a
position nearly twice as far from the center as the midpoint position, the
preference schedule based on distances would not change, but the average
distance from voter to any of the M’s would go up from about 1.24 to
about 1.52 miles, more than 33 percent farther than the average distance to
C.

Think of it: the center location is about 33 percent better on average.  It
cuts the distance in half for the faction that ends up furthest from the
winning M, and doesn’t give a lopsided solution where 33 voters have to go
twice as far (forever after) as the other 66 voters on the vertices of the
triangle.  It is the geometrical center solution and the approval
solution, but it is the Condorcet Loser.

Election-Methods mailing list - see http://electorama.com/em for list info

Juho, thanks for your insights. In your excellent example, candidate D appears to be the IA-MPO winner, especially if we define implicit approval as "ranked above bottom." Forest On Fri, Jun 5, 2015 at 12:03 AM, Juho Laatu <juho4880@yahoo.com> wrote: > One solution for this problem would be to use some rated method. If every > voter votes according to his travel distances to each candidate center we > could get ideal results. I mean that actually we are discussing here > indirectly whether to use ranked or rated methods. > > (In the example one could mandate each voter to vote based on the distance > (in the rated method based election), or calculate the results based on the > addresses of the voters, i.e. without having an election. This way we could > get rid of any possible strategies. :-) ) > > When compared to the rated methods, (plain) Condorcet methods focus only > on counting the majorities, not the strength of opinions. In this example > any two 33 voter groups could form a clear majority. They could agree to > vote together to make the center between them the winner. Any sensible > argument against the third 33 voter group could be a sufficient reason (for > the two 33 groups) to do so. > > One could say that Condorcet methods aim at electing stable winners. They > try to seek a winner that will not be disliked by some clear majority. (I'm > vague here because the behaviour of different Condorcet methods is somewhat > different.) If you would elect C, there could easily be some majority > alliance that would be interested in trying to change the elected center to > another center. If you elect Mx, the outcome is probably more stable. > > One can study this example also from strategic voting point of view. I > guess the given votes are quite stable and there are no obvious strategies > to improve one's (distance based) expected outcome. Condorcet could however > make it possible to strategically make the result worse in the sense that > the voters could reduce their expected outcome in terms of distance but > make the expected outcome better in the sense that the whole society would > benefit of it. I mean that many voters could see that C is obviously in > some sense best for the society, and they could therefore rank C first in > their ballots. No harm done and no risks doing so, if the voters truly > prefer a solution that is good for all, and not just good for them > personally. The point here is that Condorcet does not force them to make > decisions that they consider stupid. If some solution looks stupid to us in > some example, maybe the voters would see that stupidity too, and vote > accordingly (changing their preferences to something more sensible). > > Juho > > P.S. I'm ok with electing Condorcet losers in some (extreme) scenarios. As > I already said, Condorcet methods tend to seek stable solutions. In the > following example the Condorcet loser (D) is disliked only very mildly in > the pairwise comparisons, while all the others have a strong opposition > against them (in favour of changing them to some other candidate). > 17 A>B>D>C > 16 A>D>B>C > 17 B>C>D>A > 16 B>D>C>A > 17 C>A>D>B > 16 C>D>A>B > > > > On 05 Jun 2015, at 05:11, Forest Simmons <fsimmons@pcc.edu> wrote: > > > Suppose that a town with 100 voting citizens has 33 voters residing at > each of the three vertices of an equilateral triangle (with two mile sides, > say), and one voter residing within two hundred yards of the the center of > the triangle. > > > Proposed sites for the new community center are M1, M2, and M3, at the > respective midpoints of the three sides of the triangle, as well as site C > at the center of the triangle’ a couple of hundred yards from the lone > voter that we just mentioned. > > > Assuming that voters prefer closer sites over more distant sites the > preferences are > > > 33 M1=M2>C > > 33 M1=M3>C > > 33 M2=M3>C > > 01 C > > Note that C is the Condorcet Loser, since each of the M’s beats C pairwise > by almost a two-thirds majority, 66 to 34. > > > On the other hand, C is the IA winner with 100 percent implicit approval. > > > Candidate C is also the IA-MPO winner with a score of 100-66, compared > with 66-33 for the other alternatives. > > > How about average distance of voters from each of the alternatives? > > The average distance to alternative C is 0.66 times the square root of > three miles, or about 1.14 miles. > > > The average distance to any of the M’s is about 1.24 miles. > > > If the M’s were moved directly away from their midpoint positions to a > position nearly twice as far from the center as the midpoint position, the > preference schedule based on distances would not change, but the average > distance from voter to any of the M’s would go up from about 1.24 to > about 1.52 miles, more than 33 percent farther than the average distance to > C. > > > Think of it: the center location is about 33 percent better on average. It > cuts the distance in half for the faction that ends up furthest from the > winning M, and doesn’t give a lopsided solution where 33 voters have to go > twice as far (forever after) as the other 66 voters on the vertices of the > triangle. It is the geometrical center solution and the approval > solution, but it is the Condorcet Loser. > ---- > Election-Methods mailing list - see http://electorama.com/em for list info > > >