Forest,

 "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."

 Your presumption about my motive is wrong. I did it that way
 because (perhaps because of lack of sleep) that was the only way
 that occurred to me.
 I don't like 2-slot ballots and if they are used I can't take
 seriously the idea that anything other than Approval should be
 used to determine the winner.

 Also I wasn't suggesting or contemplating using the symmetric
 completion at the top to modify IA-MPO, rather I was just
 suggesting using it to test
 whether or not the result is in compliance with the Plurality
 criterion.

 Unfortunately your second example shows that even the newly
 modified version of IA-MPO (that works on the ballots
 symmetrically completed at
 the top) miserably fails Plurality.

 Chris Benham



 On 6/5/2015 8:15 AM, Forest Simmons wrote:

 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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 Checked by AVG - www.avg.com <http://www.avg.com>
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 06/04/15

Forest, Sorry you're right, I somehow miscounted. However the method certainly fails "Pairwise Plurality", and I can't see what the justification (in terms of criterion compliances) is for the bad failure of Condorcet Loser. Chris Benham On 6/6/2015 4:06 AM, Forest Simmons wrote: > Chris, > > in the second example the symmetrically completed ballots are > > 27 A > 11 A>C > 11 C>A > 02 C > 11 C>B > 11 B>C > 27 B > > Candidate C, the IA-MPO winner, is ranked on 46 ballots. Neither A nor > B is top ranked on more than 38 ballots. So it seems to me that > Plurality is not violated. > > In summary, IA-MPO does not violate the strong version of Plurality on > the symmetrically completed ballots, and does not violate the > (original) weaker version of Plurality (what I called Plurality') when > applied to the original ballots. I think it is too early to throw it out. > > Forest > > Forest > > On Fri, Jun 5, 2015 at 9:46 AM, C.Benham <cbenham@adam.com.au > <mailto:cbenham@adam.com.au>> wrote: > > Forest, > >> "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." > > Your presumption about my motive is wrong. I did it that way > because (perhaps because of lack of sleep) that was the only way > that occurred to me. > I don't like 2-slot ballots and if they are used I can't take > seriously the idea that anything other than Approval should be > used to determine the winner. > > Also I wasn't suggesting or contemplating using the symmetric > completion at the top to modify IA-MPO, rather I was just > suggesting using it to test > whether or not the result is in compliance with the Plurality > criterion. > > Unfortunately your second example shows that even the newly > modified version of IA-MPO (that works on the ballots > symmetrically completed at > the top) miserably fails Plurality. > > Chris Benham > > > > On 6/5/2015 8:15 AM, Forest Simmons wrote: >> >> >> 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! >> >> No virus found in this message. >> Checked by AVG - www.avg.com <http://www.avg.com> >> Version: 2015.0.5961 / Virus Database: 4355/9941 - Release Date: >> 06/04/15 >> > > > No virus found in this message. > Checked by AVG - www.avg.com <http://www.avg.com> > Version: 2015.0.5961 / Virus Database: 4355/9950 - Release Date: 06/05/15 >