[EM] 3 or more choices - Condorcet

[Chris Benham]

Robert,

Can I take it that your "49: A>B A>C" means that 49 voters voted
A>B=C  (meaning exactly the same thing as A>C=B)?  I'll proceed
on the assumption that your answer is yes.

So if 

49: A>B  A>C
48: B>C>A
03: C>A  C>B

really means

49: A>B=C
48: B>C>A
03: C>A=B

you are aware here that there is no Condorcet winner and that Margins
elects B, right?

B>C 48-3 (=45),  C>A 51-49 (=2),  A>B 49-48 (=1).

With only 3 candidates the Condorcet method algorithms you refer to
are all the same as MinMax. 

You wrote:
"again, i will repeat that probably, technically, Schulze is superior to 
Ranked-Pairs.  but it doesn't matter with a Smith Set of 3 candidates or 
less.  Condorcet cycles will be rare.  cycles with more than 3 in the 
Smith set will be rare of the rare."

I agree.

" it's best to get Condorcet of *some* method enacted into law."

 
I agree, if compliance with the Condorcet criterion is so important to you.

"the most realistic path to accomplishing that is *not* to advocate a method 
that cannot be explained to citizen-legislators."

Yes, but it also helps to advocate a method that opponents can't  easily ridicule
with bad examples.

I suggest Smith//Approval or Condorcet//Approval  (interpreting ranking above
equal-bottom as approval).

In this example:

49: A
48: B>C
03: C

those methods elect C, while Margins elects B.

Chris Benham





Robert Bistrow-Johnson wrote (8 Nov 2012):

 

well, i'm not the guy with upper-case letters.  i didn't comment on this 
response to what i said, but looking it over right now, whether people 
vote on a Ranked-Choice ballot as if it were FPP or not, any candidates 
*not* ranked are tied for last place on that ballot.  so

49: A
48: B
03: C

is really
49: A>B  A>C
48: B>A  B>C
03: C>A  C>B


the difference with this:

49: A
48: B>C
03: C

is that C gets a helluva lot more support from B voters than the other 
scenario.

49: A>B  A>C
48: B>C>A
03: C>A  C>B

it's the same old complaint that Rob Ritchie (FairVote) and others make 
against Condorcet (justifying putting all their support behind IRV): 
that (from their POV) Condorcet can elect wishy-washy candidates with 
little primary support.  i (and Condorcet) would say that in the second 
case, C is the best candidate even if he/she got only 3 first-choice 
votes.  might be a nice centrist, no-drama candidate in a polarized 
environment.  in Burlington VT 2009, the 3rd-place finisher from the POV 
or FPP or IRV was the Condorcet winner and nearly everyone i talked with 
would have been much happier with this candidate than with whom actually 
won the IRV or with whom would have won FPP (who suffered a decisive 
defeat in a repeat run in March 2012).  but the margins weren't so wide 
as with this example, candidate C got a lot more than 3% primary support 
votes.

again, i will repeat that probably, technically, Schulze is superior to 
Ranked-Pairs.  but it doesn't matter with a Smith Set of 3 candidates or 
less.  Condorcet cycles will be rare.  cycles with more than 3 in the 
Smith set will be rare of the rare.  it's best to get Condorcet of 
*some* method enacted into law.  the most realistic path to 
accomplishing that is *not* to advocate a method that cannot be 
explained to citizen-legislators.  and i still think that margins is 
better than either winning votes (or the logical complement regarding 
the most losing votes).  margins encompasses *both* winning votes and 
losing votes (the latter with a negative sign, of course).

-- 

r b-j                  rbj at audioimagination.com
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