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First Steve's comment is wrong as shown below: A > B > C.<br>
<blockquote type="cite">
<pre wrap="">33: A > B | C
31: B > C | A
33: C | A > B
3: B | A > C
C is eliminated with 33 votes as support.
B is eliminated with 34 votes as support.
A is last eliminated but receives no rallying voters and finishes with 33
votes as support.
</pre>
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<blockquote type="cite">
<pre wrap="">B wins.</pre>
</blockquote>
Second, as written before, scores or supports matter, not meaningless
winners which could not get elected with SPPA in the end...<br>
<br>
S.Rouillon<br>
<br>
Steve Eppley a écrit :
<blockquote cite="mid47654F70.9030209@alumni.caltech.edu" type="cite">
<pre wrap="">Hi,
Assuming I'm correctly understanding a voting method Stéphane Rouillon
used in a recent message (excerpted below), which he called "Repetitive
Condorcet (Ranked Pairs(Winning Votes)) elimination," it is
unnecessarily complicated because it chooses the same winner as Ranked
Pairs(Winning Votes), which of course is simpler.
Ranked Pairs(Winning Votes), also known as MAM, satisfies H Peyton
Young's criterion Local Independence of Irrelevant Alternatives (LIIA).
One implication of LIIA is that elimination of the last-ranked
candidate(s) does not change the ranking of the remaining candidates.
By the way, a different criterion has been masquerading as LIIA in
Wikipedia. Peyton Young defined the real LIIA in his 1994 book Equity
In Theory And Practice (if not earlier).
--Steve
--------------------------------------
Stéphane Rouillon wrote:
-snip-
</pre>
<blockquote type="cite">
<pre wrap="">Let's try a counter-example:
3 candidates A, B, C and 100 voters.
Ballots:
35: A > B > C
33: B > C > A
32: C > A > B
Repetitive Condorcet (Ranked Pairs(winning votes) ) elimination would
produce
at round 1:
68: B > C
67: A > B
Thus ranking A > B > C
C is eliminated.
at round 2:
67: A > B is the ranking
B is eliminated
at round 3:
A wins.
</pre>
</blockquote>
<pre wrap=""><!---->-snip-
----
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