# [EM] MinMax Losing Votes (equal-ranking whole) Margins

Sat Apr 18 11:13:29 PDT 2015

```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

On 4/26/2014 4:10 AM, C.Benham wrote:
> 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
>
> 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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>

```