Discussion:
MinMax Losing Votes (equal-ranking whole) Margins
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C.Benham
2014-04-25 18:40:53 UTC
Permalink
This is my new idea for a Condorcet method that meets Mono-raise and
Chicken Dilemma and is relatively resistant
to Burial strategy.

*Voters rank from the top however many candidates they wish Truncation
and equal-ranking is allowed.

A pairwise matrix is created, giving normal gross scores except that
ballots that explicitly equal rank (not truncate) any two
candidates X and Y give a whole vote to each in that pairwise contest.

Using this information, give each alternative a score that equals the
smallest number of votes it received in a pairwise loss.

Henceforth we are only concerned with the direction of the pairwise
defeats and these individual candidate scores.

Use the Schulze algorithm, weighing each pairwise "defeat" by the
absolute margin of difference between the two candidates'
scores. (Or use Ranked Pairs or River in the same way if you prefer).

Or use the candidate scores for the Margins Sort algorithm.*

This method uses the same type of pairwise matrix as a Schulze (Losing
Votes) variant I suggested earlier. I think this is much
better.

46 A
44 B>C (sincere is B or B>A)
05 C>A
05 C>B

A>B 51-49, B>C 44-10, C>A 54-46. MinMax (Losing Votes) scores:
B49, A46, C10.

The method I suggested earlier elects the buriers' candidate B, but my
new method elects A (the "sincere CW").

The Margins Sort version begins with the MM(LV) order B>A>C, then
notices that the two adjacent candidates with the two most
similar scores are B and A and that A pairwise beats B, so flips that
order and then considers A>B>C and then sees that there for each
pair of adjacent candidates, the one higher in the order pairwise beats
the one lower in the order and so is content and elects the now
highest-ordered candidate.

The other versions look at the pairwise results weighted thus: A > B
(46-49 = -3) B>C (49-10 = 39) C>A (10-46 = -36).

A's pairwise defeat score (negative 36) is by far the lowest so A wins.

Chris Benham
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C.Benham
2014-04-30 19:50:29 UTC
Permalink
My enthusiasm for this method has increased, so I thought I'd add a few
more examples.

46 A>B
44 B>C (sincere is B or B>A)
05 C>A
05 C>B

This is similar to an earlier example, but here the supporters of the
candidate being targeted by the buriers (A)
are all voting B above C instead of truncating. It still elects A.

A>B 51-49, B>C 90-10, C>A 54-46. MinMax (Losing Votes) scores:
B49, A46, C10.

The directions of the defeats and the MMLV scores are all unchanged, so
neither is the winner.

(Winning Votes and Margins both elect the buriers' candidate B. The
scenario was brought up a few years ago by
James Green-Armytage as a prime example where such Condorcet methods are
more vulnerable to strategy than
IRV.)

46 A>C
10 B>A
10 B>C
34 C=B

B>A>C>B.

(Interpreting "strictly ranked below no other candidate" as Top Rating)

Top Ratings: B54 > A46 > C34.

I hope we can agree that if any candidates are top-rated on more than
half the ballots then one of them should win.
Not doing so would be a failure of Forest's Stable Approval Potential
Criterion. Electing A here would be a failure of
Steve Eppley's Non-Drastic Defense criterion, that says that if on more
than half the ballots X is voted both above Y
and below no other candidate (i.e. no lower than equal-top) then Y must
not win.

The point is that MinMax Losing Votes Margins (where ballots that vote
any X and Y equal and above bottom either
have no effect on the X -Y pairwise tallies, just like ballots that
truncate them both, or contribute a half-vote to each)
elects A (just like ordinary Margins).

But the MinMax LosingVotes (equal-ranking whole) Margins method I'm
proposing easily elects B.

C>B 80-54 (with the 34 C=B ballots giving a whole vote to both
sides), B>A 54-46, A>C 56-44.

MMLV(erw) scores: B54 > A46 > C44. Margins Sort gives the order B>A>C.

C>B (44-54 = -10) B>A (54-46 = 8) A>C (46-44 = 8) B's
pairwise defeat is the weakest (lowest defeat score) and so B wins.

35 A
10 A=B
30 B>C
25 C

A>B>C>A. MMLV(erw) scores: A45 > B40 > C25 Margins Sort elects A.

C>A -20, A>B 5, B>C 15. A has by far the weakest defeat and so
wins.

I don't like that both Winning Votes and ordinary Margins both elect B,
who is pairwise beaten and positionally dominated
by A (i.e. A is both more approved and more top-rated) and is also the
least approved candidate.

34 A>B
17 C>A
16 B>C
31 B
02 B>C (sincere is B or B>A)

A is the sincere Condorcet winner (and a sincere Mutual Dominant Third
winner).

A>B>C>A. MMLV(erw) scores: B49 > A34 > C17. The margin between
adjacent candidates B and A (15) is smaller than
that between A and C (17), and A pairwise beats B, so Margins Sort first
flips that order to give A>B>C with no candidate pairwise
beating the next highest in the order, and so elects the candidate
highest in this final order, A.

C>A -17, A>B -15, B>C +32. A has the weakest defeat and so wins.

It looks to me like MMLV(erw)M meets Unburiable Mutual Dominant Third!

(If that is the case, then I was too pessimistic in answering a question
from Kristofer about the mutual compatibility of UMDT,
Condorcet and Mono-raise.)

25 A>B
26 B>C
23 C>A
26 C

C>A>B>C. (Approvals: C75 > B51 > A48. Top Ratings: C49 > B26 >
A25) MMLV(erw) scores: C49 > B26 > A25. Margins Sort settles on C>A>B.

B>C -23, C>A +24, A>B -1. C's defeat is by far the weakest.

I like that MMLV(erw)M elects the prettyist and strongest-looking
candidate, C. Benham/Woodall (like IRV) and Winning Votes all elect B.

Thanks for taking an interest.

Chris Benham
Post by C.Benham
This is my new idea for a Condorcet method that meets Mono-raise and
Chicken Dilemma and is relatively resistant
to Burial strategy.
*Voters rank from the top however many candidates they wish Truncation
and equal-ranking is allowed.
A pairwise matrix is created, giving normal gross scores except that
ballots that explicitly equal rank (not truncate) any two
candidates X and Y give a whole vote to each in that pairwise contest.
Using this information, give each alternative a score that equals the
smallest number of votes it received in a pairwise loss.
Henceforth we are only concerned with the direction of the pairwise
defeats and these individual candidate scores.
Use the Schulze algorithm, weighing each pairwise "defeat" by the
absolute margin of difference between the two candidates'
scores. (Or use Ranked Pairs or River in the same way if you prefer).
Or use the candidate scores for the Margins Sort algorithm.*
This method uses the same type of pairwise matrix as a Schulze
(Losing Votes) variant I suggested earlier. I think this is much
better.
46 A
44 B>C (sincere is B or B>A)
05 C>A
05 C>B
B49, A46, C10.
The method I suggested earlier elects the buriers' candidate B, but my
new method elects A (the "sincere CW").
The Margins Sort version begins with the MM(LV) order B>A>C, then
notices that the two adjacent candidates with the two most
similar scores are B and A and that A pairwise beats B, so flips that
order and then considers A>B>C and then sees that there for each
pair of adjacent candidates, the one higher in the order pairwise
beats the one lower in the order and so is content and elects the now
highest-ordered candidate.
The other versions look at the pairwise results weighted thus: A > B
(46-49 = -3) B>C (49-10 = 39) C>A (10-46 = -36).
A's pairwise defeat score (negative 36) is by far the lowest so A wins.
Chris Benham
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C.Benham
2015-04-18 18:13:29 UTC
Permalink
I've discovered that this method fails both Clone-Winner and Clone-loser.

Say we have this example of the scenario specified in the "Chicken
Dilemma" criterion definition:

34 C
33 A>B
32 B

MMLV(erw) scores: C34, A33, B32. Pairwise C>A>B, so
Margins-Sort is happy with that order and elects C.

Now say we add two clones of C (X and Y) to give:

13 C>Y>X
11 X>C>Y
10 Y>X>C
33 A>B
32 B

C>Y 24-10, Y>X 23-11, X>C 21-13.

C/X/Y >A 34-33, A>B 33-32, B>C/X/Y 65-34

MMLV(erw) scores: A33, B32, C13, X11, Y10.

Pairwise X>C and Y>X. The MMLV(erw) score margin between the latter pair
is the lowest, so Margins-Sort gives the order A>B>C>Y>X
and so elects A. To not fail Clone-Winner the winner has to be one of
C, X, Y.

Say instead of cloning the winner we have X and Y be clones of A:

34 C
12 A>Y>X>B
11 X>A>Y>B
10 Y>X>A>B
32 B

A>Y 23-10, Y>X 22-11, X>A 21-12.

C>A/X/Y 34-33, A/X/Y>B 33-32, B>C 65-34.

MMLV(erw) scores: C34, B32, A12, X11, Y10

This is pairwise out of order, and Margins-Sort breaks the score-margins
tie between X>A and Y>X by demoting the lower-ranked candidate,
so it goes C>B>A>Y>X, then B>C>A>Y>X and then is content and elects B.
To meet Clone-Loser the winner has to remain C.

For these failures to occur there had to be a sub-cycle inside the
Smith-set, which I think would be extremely unlikely in any real election.

I think the problem can be fixed. I suggest replacing MMLV(erw)
Margins-Sort with *MMLV(erw) Margins-Sort Elimination*. We order the
candidates
as before (with MMLV(erw) Margins-Sort) but instead of simply electing
the candidate highest in that order, we eliminate (drop from the
ballots) the
lowest-ordered candidate, recalculate the MMLV(erw) scores, and repeat
until one candidate remains (or giving the same result and saving time,
until
3 candidates remain and then elect the highest-ordered candidate).

In all the example elections I gave in earlier posts on MMLV(erw)
Margins , this method gives the same winners.

To take the clone-winner example I gave, MMLV(erw) Margins-Sort gives
the order A>B>C>Y>X. After eliminating X, C's smallest number of votes
in a pairwise
defeat (C's MMLV(erw) score) changes from 13 back to 34. Then
Margins-Sort gives the order C>A>B>Y, and then we eliminate Y and C wins.

In the clone-loser example I gave MMLV(erw) Margins-Sort gives the order
B>C>A>Y>X. After eliminating X, A's MMLV(erw) score changes from 12
back to 33.
Then Margins-Sort gives the order C>A>Y>B and then we eliminate B and
C wins.

Chris Benham
Post by C.Benham
This is my new idea for a Condorcet method that meets Mono-raise and
Chicken Dilemma and is relatively resistant
to Burial strategy.
*Voters rank from the top however many candidates they wish Truncation
and equal-ranking is allowed.
A pairwise matrix is created, giving normal gross scores except that
ballots that explicitly equal rank (not truncate) any two
candidates X and Y give a whole vote to each in that pairwise contest.
Using this information, give each alternative a score that equals the
smallest number of votes it received in a pairwise loss.
Henceforth we are only concerned with the direction of the pairwise
defeats and these individual candidate scores.
Use the Schulze algorithm, weighing each pairwise "defeat" by the
absolute margin of difference between the two candidates'
scores. (Or use Ranked Pairs or River in the same way if you prefer).
Or use the candidate scores for the Margins Sort algorithm.*
This method uses the same type of pairwise matrix as a Schulze
(Losing Votes) variant I suggested earlier. I think this is much
better.
46 A
44 B>C (sincere is B or B>A)
05 C>A
05 C>B
B49, A46, C10.
The method I suggested earlier elects the buriers' candidate B, but my
new method elects A (the "sincere CW").
The Margins Sort version begins with the MM(LV) order B>A>C, then
notices that the two adjacent candidates with the two most
similar scores are B and A and that A pairwise beats B, so flips that
order and then considers A>B>C and then sees that there for each
pair of adjacent candidates, the one higher in the order pairwise
beats the one lower in the order and so is content and elects the now
highest-ordered candidate.
The other versions look at the pairwise results weighted thus: A > B
(46-49 = -3) B>C (49-10 = 39) C>A (10-46 = -36).
A's pairwise defeat score (negative 36) is by far the lowest so A wins.
Chris Benham
----
Election-Methods mailing list - see http://electorama.com/em for list info
----
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