3-valued Booleans inside rules, passing Condorcet (Re: [EM] "More often" (was: IRV and Condorcet operating identically)

[Craig Carey]

At 03\03\03 10:31 +0100 Monday, Markus Schulze wrote:
 >Dear Craig,
 >
 >fortunately, "incompleteness" is not an issue since nobody
 >suggests to use a Condorcet method without a tie-breaker.
 >

Perhaps you did not read this.
It limits your freedom to alter the interpretation of the
word Condorcet. A tiebreaker would not be considered by a good
rule since no peeking into the internals (but an exception would
easily occur when the output is unpredictable).


------------------------------------------------------------------------
    Date: Sun, 02 Mar 2003 09:58:47 +1300
    To: election-methods-list at eskimo.com
    From: Craig Carey <research at ijs.co.nz>
    Subject: 3-valued Booleans inside rules, passing Condorcet (Re: [EM]
             "More often" (was: IRV and Condorcet operating identically)
    In-Reply-To: <3E60C869.E657420C at sympatico.ca>
    References: <3E600A0A.1C7E7822 at alumni.tu-berlin.de>

At 03\03\02 09:58 +1300 Sunday, Craig Carey wrote:
 >At 03\03\01 09:49 -0500 Saturday, Stephane Rouillon wrote:
 >...
 > >what is the participation criterion?
 > >
 > >Steph
 > >
 > >Markus Schulze a écrit :
 > >
 > >> > FBC is the only criteria that favors Approval
 > >> > over Condorcet.
 > >>
 > >> Condorcet violates the participation criterion.
 > >> Approval Voting meets the participation criterion.
 > >>
 >
 >And the missing third sentence is: both statements are so
 >unimportant as to be best ignored.
 >
 >There is no clue there on the weightiness of the claims. I am
 >sure that they are not important. Certainly that view can locked
 >down if the definition of the "criterion" is not available.
 >
 >To the extent possible, please regard the following comments being
 >about an election having only the papers (AB), (B), and (C). For
 >that election, the Condorcet method has an undefined region of
 >quite a big size.
 >
 >It is controversial to create a weak rule and see that it passes
 >some methods and not others. Instead a plausible rule that fails
 >all the methods that need to pass can be used, but it is length
 >(or bigness) of the worst failure is measured. It looked
------------------------------------------------------------------------

The 3 paper limitation continues until the end of this thread.

What was the argument that you are replying to?.

That above  is the first message in this thread according to its
Subject header information.

AB  a
B.  b
C.  c

a+b+c=1

B beaten by A = (b<a)   % passes through (C) vertex
C beaten by B = (c<a+b) = (1/2<c)
A beaten by C = (a<c)   % passes through (B) vertex

Failure region = (b<a)(c<a+b)(a<c)  [or could indifferently use "<="]

The formula does not contain a "tiebreaker".


Recall that I sent to you (5 minutes after the date of the message
I reply to) some lengthy good arguments indicating that the Condorcet
method is not a variant of itself.  The mailing list did not receive
a copy.

Did Mr Schulze try to alter the meaning of the word "Condorcet" ?.





Craig Carey



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