<html><body><div style="color:#000; background-color:#fff; font-family:times new roman, new york, times, serif;font-size:12pt"><div style="RIGHT: auto"><SPAN style="RIGHT: auto">Forest,<BR></SPAN></div>
<div style="RIGHT: auto"><SPAN style="RIGHT: auto">"I think in general that if the approval scores are at all valid I would go for the enhanced DMC winner over <BR>any of the chain building methods we have considered. I think other considerations over-ride the <BR>importance of being uncovered."</SPAN></div>
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<div style="RIGHT: auto"><SPAN style="RIGHT: auto">I agree. I think the chain building method in comparison seems a bit arbitrary and less philosophically justified.</SPAN></div>
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<div style="RIGHT: auto"><SPAN style="RIGHT: auto">Also the method has a fairly straight-forward description that doesn't need to mention "Smith set" or "the Condorcet winner".<VAR id=yui-ie-cursor></VAR></SPAN></div>
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<div style="RIGHT: auto"><SPAN style="RIGHT: auto">So of these similar methods (that include Smith//Approval and all elect the same winner if the Smith set contains 3 members or 1 member), </SPAN></div>
<div style="RIGHT: auto"><SPAN style="RIGHT: auto">I think this is my favourite. </SPAN></div>
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<div style="RIGHT: auto"><SPAN style="RIGHT: auto">Maybe it could use a new name? :)<BR></SPAN></div>
<div style="RIGHT: auto"><SPAN style="RIGHT: auto">Chris</div></SPAN>
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<DIV style="BORDER-BOTTOM: #ccc 1px solid; BORDER-LEFT: #ccc 1px solid; PADDING-BOTTOM: 0px; LINE-HEIGHT: 0; MARGIN: 5px 0px; PADDING-LEFT: 0px; PADDING-RIGHT: 0px; HEIGHT: 0px; FONT-SIZE: 0px; BORDER-TOP: #ccc 1px solid; BORDER-RIGHT: #ccc 1px solid; PADDING-TOP: 0px" class=hr contentEditable=false readonly="true"></DIV><B><SPAN style="FONT-WEIGHT: bold">From:</SPAN></B> "fsimmons@pcc.edu" <fsimmons@pcc.edu><BR><B><SPAN style="FONT-WEIGHT: bold">To:</SPAN></B> C.Benham <cbenhamau@yahoo.com.au><BR><B><SPAN style="FONT-WEIGHT: bold">Cc:</SPAN></B> election-methods-electorama.com@electorama.com<BR><B><SPAN style="FONT-WEIGHT: bold">Sent:</SPAN></B> Monday, 12 September 2011 8:50 AM<BR><B><SPAN style="FONT-WEIGHT: bold">Subject:</SPAN></B> Re: Enhanced DMC<BR></FONT><BR>Very good Chris. <BR><BR>I tried to build a believable profile of ballots that would yield the approval order and defeats of this <BR>example without success, but I am sure
that it is not impossible.<BR><BR>I think in general that if the approval scores are at all valid I would go for the enhanced DMC winner over <BR>any of the chain building methods we have considered. I think other considerations over-ride the <BR>importance of being uncovered.<BR><BR>----- Original Message -----<BR>From: "C.Benham" <BR>Date: Sunday, September 11, 2011 10:08 am<BR>Subject: Enhanced DMC<BR>To: <A href="mailto:election-methods-electorama.com@electorama.com" ymailto="mailto:election-methods-electorama.com@electorama.com">election-methods-electorama.com@electorama.com</A><BR>Cc: Forest W Simmons <BR><BR>> Forest Simmons wrote (15 Aug 2011):<BR>> <BR>> >Here's a possible scenario:<BR>> ><BR>> >Suppose that approval order is alphabetical from most approval <BR>> to least A, B, C, D.<BR>> ><BR>> >Suppose further that pairwise defeats are as follows:<BR>> ><BR>> >C>A>D>B>A
together with B>C>D .<BR>> ><BR>> >Then the set P = {A, B} is the set of candidates neither of <BR>> which is pairwise<BR>> >beaten by anybody with greater approval.<BR>> ><BR>> >Since the approval winner A is not covered by B, it is not <BR>> covered by any<BR>> >member of P, so the enhanced version of DMC elects A.<BR>> ><BR>> >But A is covered by C so it cannot be elected by any of the <BR>> chain building<BR>> >methods that elect only from the uncovered set.<BR>> ><BR>> <BR>> Forest,<BR>> <BR>> Is the "Approval Chain-Building" method the same as simply <BR>> electing the <BR>> most approved uncovered candidate?<BR>> <BR>> I surmise that the set of candidates not pairwise beaten by a <BR>> more <BR>> approved candidate (your set "P", what I've<BR>> been referring to as the "Definite Majority set") and the <BR>> Uncovered set <BR>> don't
necessarily overlap.<BR>> <BR>> If forced to choose between electing from the Uncovered set and <BR>> electing <BR>> from the "DM" set, I tend towards<BR>> the latter.<BR>> <BR>> Since Smith//Approval always elects from the DM set, and your <BR>> suggested <BR>> "enhanced DMC" (elect the most<BR>> approved member of the DM set that isn't covered by another <BR>> member) <BR>> doesn't necessarily elect from the Uncovered set;<BR>> there doesn't seem to be any obvious philosophical case that <BR>> enhanced <BR>> DMC is better than Smith//Approval.<BR>> <BR>> (Also I would say that an election where those two methods <BR>> produce <BR>> different winners would be fantastically unlikely.)<BR>> <BR>> A lot of Condorcet methods are promoted as being able to give <BR>> the <BR>> winner just from the information contained in the<BR>> gross pairwise matrix. I think that the same is true of
these <BR>> methods <BR>> if we take a candidate X's highest gross pairwise<BR>> score as X's approval score. Can you see any problem with that?<BR>> <BR>> <BR>> Chris Benham<BR>> <BR>> <BR>> <BR>> <BR>> ----- Original Message -----<BR>> From:<BR>> Date: Friday, August 12, 2011 3:12 pm<BR>> Subject: Enhanced DMC<BR>> To: election-methods at lists.electorama.com,<BR>> <BR>> > > From: "C.Benham"<BR>> > > To: election-methods-electorama.com at electorama.com<BR>> > > Subject: [EM] Enhanced DMC<BR>> ><BR>> > > Forest,<BR>> > > The "D" in DMC used to stand for *Definite*.<BR>> ><BR>> > Yeah, that's what we finally settled on.<BR>> ><BR>> > ><BR>> > > I like (and I think I'm happy to endorse) this Condorcet method<BR>> > > idea, and consider it to be clearly better than regular DMC<BR>> > ><BR>> > >
Could this method give a different winner from the ("Approval<BR>> > > Chain Building" ?) method you mentioned in the "C//A" thread <BR>> (on 11<BR>> > > June 2011)?<BR>> ><BR>> > Yes, I'll give an example when I get more time. But for all <BR>> practical <BR>> > purposes they both pick the highest approval Smith candidate.<BR>> <BR>> <BR>> <BR>> Here's a possible scenario:<BR>> <BR>> Suppose that approval order is alphabetical from most approval <BR>> to least <BR>> A, B, C, D.<BR>> <BR>> Suppose further that pairwise defeats are as follows:<BR>> <BR>> C>A>D>B>A together with B>C>D .<BR>> <BR>> Then the set P = {A, B} is the set of candidates neither of <BR>> which is <BR>> pairwise<BR>> beaten by anybody with greater approval.<BR>> <BR>> Since the approval winner A is not covered by B, it is not <BR>> covered by any<BR>> member of
P, so the enhanced version of DMC elects A.<BR>> <BR>> But A is covered by C so it cannot be elected by any of the <BR>> chain building<BR>> methods that elect only from the uncovered set.<BR>> <BR>> <BR>> Forest Simmons wrote (12 June 2011):<BR>> <BR>> > I think the following complete description is simpler than anything<BR>> > possible for ranked pairs:<BR>> ><BR>> > 1. Next to each candidate name are the bubbles (4) (2) (1). The<BR>> > voter rates a candidate on a scale from<BR>> > zero to seven by darkening the bubbles of the digits that add <BR>> up to<BR>> > the desired rating.<BR>> ><BR>> > 2. We say that candidate Y beats candidate Z pairwise iff Y <BR>> is rated<BR>> > above Z on more ballots than not.<BR>> ><BR>> > 3. We say that candidate Y covers candidate X iff Y pairwise beats<BR>> > every candidate that X pairwise<BR>> > beats
or ties.<BR>> ><BR>> > [Note that this definition implies that if Y covers X, then Y <BR>> beats X<BR>> > pairwise, since X ties X pairwise.]<BR>> ><BR>> > Motivational comment: If a method winner X is covered, then the<BR>> > supporters of the candidate Y that<BR>> > covers X have a strong argument that Y should have won instead.<BR>> ><BR>> > Now that we have the basic concepts that we need, and <BR>> assuming that<BR>> > the ballots have been marked<BR>> > and collected, here's the method of picking the winner:<BR>> ><BR>> > 4. Initialize the variable X with (the name of) the <BR>> candidate that<BR>> > has a positive rating on the greatest<BR>> > number of ballots. Consider X to be the current champion.<BR>> ><BR>> > 5. While X is covered, of all the candidates that cover X, <BR>> choose the<BR>> > one that has the greatest number
of<BR>> > positive ratings to become the new champion X.<BR>> ><BR>> > 6. Elect the final champion X.<BR>> ><BR>> > 7. If in step 4 or 5 two candidates are tied for the number of<BR>> > positive ratings, give preference (among the<BR>> > tied) to the one that has the greatest number of ratings <BR>> above level<BR>> > one. If still tied, give preference<BR>> > (among the tied) to the one with the greatest number of <BR>> ratings above<BR>> > the level two. Etc.<BR>> ><BR>> > Can anybody do a simpler description of any other Clone Independent<BR>> > Condorcet method?<BR>> <BR>> <BR>> <BR>> <BR><BR><BR></DIV></DIV></div></body></html>