I've been trying to come up with a better method with Bucklin-like virtues
including Later-no-Help, even though
there seems to be no call for any such thing and it results in a method
with a stronger truncation incentive than
I like.

The closest I came fills the bill with 3 candidates, but can probably fail
Majority for Solid Coalitions (aka Mutual Majority)
with more.

Approval seeded by Minimum Gross Score (equal-top rating whole):

*Voters fill out a multi-slot ratings ballot (I suggest as many slots as
there are candidates, up to say 4).
Default rating (truncation) is bottom.

Construct a pairwise matrix in which any ballot that rates/ranks candidate
X and Y equal-top gives a whole vote to both
in the X v Y pairwise comparison. Those that rate at least one of them in
a lower position give a whole vote to one if they
rate that one above the other, otherwise give no vote to either.

We are only concerned with pairwise scores, not defeats or victories or
ties. Select the candidate S whose lowest pairwise
score is higher than any other candidate's lowest pairwise score.

Then interpret all the ballots  that rate S above bottom as approving S
and all other candidates they rate no lower than S,
and all the other ballots as approving all the candidates they rate above
bottom.

Based on that interpretation, elect the most approved candidate.*

I claim that this meets the Favorite Betrayal Criterion, Plurality,
Irrelevant Ballots,  Later-no-Help  (maybe barring some fantastic
scenario with many candidates), Condorcet(Gross), Minimal Defense,
mono-raise, mono-add-top, mono-switch-plump, mono-add-plump,
mono-append, and  3-candidate Majority for Solid Coalitions.

Because I'm sure that it doesn't properly meet Majority for Solid
Coalitions, I don't count this as a complete success.

Without the approval stage and the rule about equal-top rating/ranking, it
is Douglas Woodall's  "MinGS" method  (one of many
he devised for test purposes).

46 A
44 B>C
10 C

C> A 54-46,  A>B 46-44, B>C 44-10.    The method "seeds" A and then elects
C.

Electing A would be a failure of Minimal Defense and electing B would show
a failure of  Later-no-Help.  Unfortunately the election of C
is a failure of Chicken Dilemma (not compatible with the method's
compliance with Plurality and Minimal Defense).

46 A
44 B>C  (sincere might be B or B>A)
05 C>A
05 C>B

C> A 54-46,  A>B 51-49, B>C 44-10.    The method "seeds" A and elects C.

Here it resists Burial strategy better than the MinMax  (Margins) and
(Losing Votes) Condorcet methods, which both elect B.

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

C>B 80-54,  B>A 54-46.  A>C 56-44.  The method seeds B and elects C.

Not electing the only candidate that is top-rated on more than half the
ballots may be an odd look by comparison with Bucklin, but I'm not bothered.
All the candidates are pairwise beaten and the winner is rated above
bottom on 90% of the ballots.

40 A>C
15 B>A
20 B
15 C>A
10 C

A>B 55-35,  A>C 55-25,  C>B 65-35.  The method seeds A and elects A.

There are 100 ballots and A is the Condorcet(Gross) winner (meaning that
in each and all of A's pairwise contests A is
strictly preferred to the other candidate by more than half the voters).
Bucklin elects C.

Chris Benham

Hi Chris,

I like this kind of method where you pick a pivotal candidate and
check if the other
candidates can defeat him, but typically these methods don't satisfy
FBC or mono-raise.

Off the top of my head, what if some voters rank
Favorite=Compromise>...>Worst,
resulting in Favorite being the seeded candidate and the winner being
Worst. Isn't it
possible that if the voters lower Favorite in their rankings that
Favorite will no longer
be the seed, and instead someone else will be the seed? If so, I don't
think there's
a way to promise that Compromise would not go on to win the election.

Regarding mono-raise, while it's obvious that getting raised can't
stop you from being
the seed if you were the seed, what if you weren't the seed but you
won via approval in
the second phase? Getting raised/lowered could change who the seed is,
and it may
be that you can only win in phase 2 when certain candidates are the
seed. In fact,
becoming the seed when you hadn't been the seed could even make you lose.

Kevin

De : C.Benham cbenham@adam.com.au
À : "election-methods@lists.electorama.com"
election-methods@lists.electorama.com
Envoyé le : Mardi 26 mai 2015 1h57
Objet : Approval seeded by MinGS (etrw)

I've been trying to come up with a better method with Bucklin-like
virtues including Later-no-Help, even though
there seems to be no call for any such thing and it results in a method
with a stronger truncation incentive than
I like.

The closest I came fills the bill with 3 candidates, but can probably
fail Majority for Solid Coalitions (aka Mutual Majority)
with more.

Approval seeded by Minimum Gross Score (equal-top rating whole):

*Voters fill out a multi-slot ratings ballot (I suggest as many slots as
there are candidates, up to say 4).
Default rating (truncation) is bottom.

Construct a pairwise matrix in which any ballot that rates/ranks
candidate X and Y equal-top gives a whole vote to both
in the X v Y pairwise comparison. Those that rate at least one of them
in a lower position give a whole vote to one if they
rate that one above the other, otherwise give no vote to either.

We are only concerned with pairwise scores, not defeats or victories or
ties. Select the candidate S whose lowest pairwise
score is higher than any other candidate's lowest pairwise score.

Then interpret all the ballots  that rate S above bottom as approving S
and all other candidates they rate no lower than S,
and all the other ballots as approving all the candidates they rate
above bottom.

Based on that interpretation, elect the most approved candidate.*

I claim that this meets the Favorite Betrayal Criterion, Plurality,
Irrelevant Ballots,  Later-no-Help  (maybe barring some fantastic
scenario with many candidates), Condorcet(Gross), Minimal Defense,
mono-raise, mono-add-top, mono-switch-plump, mono-add-plump,
mono-append, and  3-candidate Majority for Solid Coalitions.

Because I'm sure that it doesn't properly meet Majority for Solid
Coalitions, I don't count this as a complete success.

Without the approval stage and the rule about equal-top rating/ranking,
it is Douglas Woodall's  "MinGS" method  (one of many
he devised for test purposes).

46 A
44 B>C
10 C

C> A 54-46,  A>B 46-44, B>C 44-10.    The method "seeds" A and then
elects C.

Electing A would be a failure of Minimal Defense and electing B would
show a failure of  Later-no-Help.  Unfortunately the election of C
is a failure of Chicken Dilemma (not compatible with the method's
compliance with Plurality and Minimal Defense).

46 A
44 B>C  (sincere might be B or B>A)
05 C>A
05 C>B

C> A 54-46,  A>B 51-49, B>C 44-10.    The method "seeds" A and elects C.

Here it resists Burial strategy better than the MinMax (Margins) and
(Losing Votes) Condorcet methods, which both elect B.

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

C>B 80-54,  B>A 54-46.  A>C 56-44.  The method seeds B and elects C.

Not electing the only candidate that is top-rated on more than half the
ballots may be an odd look by comparison with Bucklin, but I'm not
bothered.
All the candidates are pairwise beaten and the winner is rated above
bottom on 90% of the ballots.

40 A>C
15 B>A
20 B
15 C>A
10 C

A>B 55-35,  A>C 55-25,  C>B 65-35.  The method seeds A and elects A.

There are 100 ballots and A is the Condorcet(Gross) winner (meaning that
in each and all of A's pairwise contests A is
strictly preferred to the other candidate by more than half the voters).
Bucklin elects C.

Chris Benham

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

46 A
44 B>C
10 C

46 A
44 B>C  (sincere might be B or B>A)
05 C>A
05 C>B

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

40 A>C
15 B>A
20 B
15 C>A
10 C

"It turns out that IA-MPO does satisfy a modified version of
Plurality:  If A is ranked top above C on more ballots than C is
ranked, then C cannot be the IA-MPO winner."

Now I remember the interesting example that shows that IA-MPO can fail
Plurality when equal ranking at top is allowed:

33 A
16 C=A
02 C
16 C=B
33 B

The IA-MPO score for both  A and B is 49-49=0, while the score for C
is 34-33=1, so C wins.

This is a failure of Plurality because A (for example) is top ranked
on 49 ballots, while C is ranked on only34 ballots.

However, any configuration in issue space that could give rise to this
ballot set would be more faithfully reflected in a ballot like

33 A
16 C>A
02 C
16 C>B
33 B

Why would C voters raise A and B to top if they didn't really like
them as well as the Condorcet Winner C?

It could be that (through typical disinformation) voters thought that
C didn't have a chance compared to the two main party candidates A and
B.  They raised their lesser evil compromise candidates to hedge their
bets.

It turns out that IA-MPO does satisfy a modified version of
Plurality:  If A is ranked top above C on more ballots than C is
ranked, then C cannot be the IA-MPO winner.

In any case where this Plurality' would allow C to win while an
ordinary Plurality requirement would preclude C's right to win, the C
voters would (under perfect information conditions) rightly have an
incentive to change each instance of C=A to C>A .

Proof that IA-MPO satisfies this modified Plurality':

First note that the IA winner cannot have a negative IA-MPO score,
because it is ranked on as many (or more) ballots than any other
candidate, including the one that gives it max opposition.

Next note that if A is ranked top above C on more ballots than C is
ranked, then A's pairwise opposition against C is greater than C's IA
score, therefore C's IA-MPO score is negative, and therefore smaller
than the IA-MPO score of the IA, winner, and therefore not maximal.

In a way IA-MPO automatically compensates for voters' hypercautious
raising of compromises to equal top status. This should attract voters
that don't like Approval because they know that (under Approval)
approving their compromise can take the win away from a Condorcet Winner.

For this reason, I suggest that even on two slot approval style
ballots, we use the Approval-MPO score to determine the winner instead
of Approval alone.

Forest

On Mon, Jun 1, 2015 at 7:10 PM, Forest Simmons <fsimmons@pcc.edu
mailto:fsimmons@pcc.edu> wrote:

 Chris,

 it is interesting to me that IA-MMPO (implicit approval minus max
 pairwise opposition) gives the same results as Approval Seeded by
 MinGS (etrw) in the four examples that you offered:


     46 A
     44 B>C
     10 C

  The respective IA-MPO scores for A, B, and C are  46-54, 44-46,
 and 54-46, the only positive one.

     46 A
     44 B>C  (sincere might be B or B>A)
     05 C>A
     05 C>B

 The respective IA-MPO scores are  51-54, 49-51, and 54-46, again
 the only positive one.

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

 The respective IA-MPO scores are  56-54, 54-46, and 90-56. C wins
 again.


     40 A>C
     15 B>A
     20 B
     15 C>A
     10 C

 The respective scores are 70-35, 35-65, and 65-55.  This time A
 wins with an IA-MPO score of 35 compared to C's 10.

 This IA-MPO method does satisfy the FBC, but is not chicken
 proof.  However, small defensive moves can thwart chicken threats.

 It seems like I might have suggested IA-MPO before, but we were
 trying for something fancier at the time.

 Forest

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...

Forest, I'm not sure that this isn't the same as the normal  Plurality
criterion.  The reference to "first preference" in the Plurality criterion
definition I think refers to exclusive first preference.

(I gather that Woodall's criteria are only about strict rankings from the
top, which may or may not be truncated,) I suppose it could and should be
extended to applying to ballots
that are symmetrically "completed" only at the top. Doing that to your
example gives:

41 A
18 C
41 B

Electing C on these ballots is insane and I don't see how electing C on
the original ballots (where some of the votes are given half to one
candidate and half to another) is
really any more justified.

Yes, this convinces me that the Plurality criterion should definitely be
applied to to the ballots symmetrically completed at the top and that we
can without regret
kiss  IA-MPO  goodbye.

Another version of the criterion is "Pairwise Plurality"  (suggested a
while ago by Kevin or me): If candidate X's lowest pairwise score is higher
than candidate Y's highest
pairwise score, then Y must not be elected".

I like this. Both IA-MPO and  SMD,TR fail it, as in the two examples.

"Symmetrical completion normally would replace  16 A=C with 8 A>C and
8 C>A .  I understand why you didn't do it that way:  you didn't want
to go outside the category of two slot ballots.  But just because the
voters have to vote two slot ballots doesn't mean that we are
prohibited from using a counting method that creates auxiliary data
structures like matrices or three slot rankings."

On Wed, Jun 3, 2015 at 7:43 PM, C.Benham <cbenham@adam.com.au
mailto:cbenham@adam.com.au> wrote:

 ...

 Forest, I'm not sure that this isn't the same as the normal 
 Plurality criterion.  The reference to "first preference" in the
 Plurality criterion definition I think refers to exclusive first
 preference.

 (I gather that Woodall's criteria are only about strict rankings
 from the top, which may or may not be truncated,) I suppose it
 could and should be extended to applying to ballots
 that are symmetrically "completed" only at the top. Doing that to
 your example gives:

 41 A
 18 C
 41 B

 Electing C on these ballots is insane and I don't see how electing
 C on the original ballots (where some of the votes are given half
 to one candidate and half to another) is
 really any more justified.

 Yes, this convinces me that the Plurality criterion should
 definitely be applied to to the ballots symmetrically completed at
 the top and that we can without regret
 kiss  IA-MPO  goodbye.

Symmetrical completion normally would replace  16 A=C with 8 A>C and
8 C>A .  I understand why you didn't do it that way:  you didn't want
to go outside the category of two slot ballots.  But just because the
voters have to vote two slot ballots doesn't mean that we are
prohibited from using a counting method that creates auxiliary data
structures like matrices or three slot rankings.

If we did this (I think more appropriate) kind of symmetric
completion, the working ballots would become

33 A
08 A>C
08 C>A
02 C
08 C>B
08 B>C
33 B
The resulting respective IA-MPO scores for A, B, and C would become
49-49, 49-49, and 34-41, so this version of IA-MPO with a front end of
symmetric completion at the top would give a tie to A and B, the only
candidates with a non-negative score.

Let's try it on

27 A
22 A=C
02 C
22 B=C
27 B

Candidates A and B are tied for Approval Winner with 49 approvals each
against 46 for C, making C the ballot Condorcet Loser.

Let's do the natural symmetric completion to see the likely sincere
ballots that would be voted if equal ranking at top were not allowed
(nor practically required,as in Approval):

27 A
11 A>C
11 C>A
02 C
11 C>B
11 B>C
27 B

The respective IA-MPO scores for A, B, and C are 49-49, 49-49, and
46-38, the only positive difference. So C wins.  Note that C is still
the ballot Condorcet Loser.

Whether or not we like this result probably reflects how much we
prefer a centrist over an extremist, all else being equal.

 Another version of the criterion is "Pairwise Plurality" 
 (suggested a while ago by Kevin or me): If candidate X's lowest
 pairwise score is higher than candidate Y's highest
 pairwise score, then Y must not be elected".

 I like this. Both IA-MPO and  SMD,TR fail it, as in the two examples.

Nice idea!

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