<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=us-ascii">
<style type="text/css" style="display:none;"> P {margin-top:0;margin-bottom:0;} </style>
</head>
<body dir="ltr">
<div style="font-family: Calibri, Arial, Helvetica, sans-serif; font-size: 12pt; color: rgb(0, 0, 0);" class="elementToProof">
I may not understand the method, but consider the following with candidates X, Y, Z:</div>
<div style="font-family: Calibri, Arial, Helvetica, sans-serif; font-size: 12pt; color: rgb(0, 0, 0);" class="elementToProof">
<br>
</div>
<div style="font-family: Calibri, Arial, Helvetica, sans-serif; font-size: 12pt; color: rgb(0, 0, 0);" class="elementToProof">
51 X<Y<Z</div>
<div style="font-family: Calibri, Arial, Helvetica, sans-serif; font-size: 12pt; color: rgb(0, 0, 0);" class="elementToProof">
49 Y>Z>X</div>
<div style="font-family: Calibri, Arial, Helvetica, sans-serif; font-size: 12pt; color: rgb(0, 0, 0);" class="elementToProof">
<br>
</div>
<div style="font-family: Calibri, Arial, Helvetica, sans-serif; font-size: 12pt; color: rgb(0, 0, 0);" class="elementToProof">
Then X is the CW, and X is ranked over or pairwise defeats a'(B) only on those ballots B in the group of 51. Thus N(X)=51.</div>
<div style="font-family: Calibri, Arial, Helvetica, sans-serif; font-size: 12pt; color: rgb(0, 0, 0);" class="elementToProof">
<br>
</div>
<div style="font-family: Calibri, Arial, Helvetica, sans-serif; font-size: 12pt; color: rgb(0, 0, 0);" class="elementToProof">
Meanwhile, Y is ranked over a'(B) on all ballots, so N(Y)=100.</div>
<div style="font-family: Calibri, Arial, Helvetica, sans-serif; font-size: 12pt; color: rgb(0, 0, 0);" class="elementToProof">
<br>
</div>
<div style="font-family: Calibri, Arial, Helvetica, sans-serif; font-size: 12pt; color: rgb(0, 0, 0);" class="elementToProof">
If I've understood the method correctly, it is not Condorcet efficient.</div>
<div style="font-family: Calibri, Arial, Helvetica, sans-serif; font-size: 12pt; color: rgb(0, 0, 0);" class="elementToProof">
<br>
</div>
<div style="font-family: Calibri, Arial, Helvetica, sans-serif; font-size: 12pt; color: rgb(0, 0, 0);" class="elementToProof">
Jim Faran<br>
</div>
<div id="appendonsend"></div>
<hr style="display:inline-block;width:98%" tabindex="-1">
<div id="divRplyFwdMsg" dir="ltr"><font face="Calibri, sans-serif" style="font-size:11pt" color="#000000"><b>From:</b> Election-Methods <election-methods-bounces@lists.electorama.com> on behalf of Forest Simmons <forest.simmons21@gmail.com><br>
<b>Sent:</b> Wednesday, April 6, 2022 2:57 AM<br>
<b>To:</b> EM <Election-methods@lists.electorama.com>; Kristofer Munsterhjelm <km_elmet@t-online.de><br>
<b>Subject:</b> [EM] Decloned Copeland Borda Hybrid</font>
<div> </div>
</div>
<div>
<div dir="auto">On each ballot B identify the anti-favorite a'(B) as the candidate a' among those bottom listed on ballot B that is bottom listed the most on the rest of the ballots.
<div dir="auto"><br>
</div>
<div dir="auto">[A candidate is bottom listed on a ballot if it is ranked over no candidate on that ballot]<br>
<div dir="auto"><br>
</div>
<div dir="auto">Elect the candidate X that, on the most ballots B, either is ranked ahead of or pairwise defeats a'(B).</div>
<div dir="auto"><br>
</div>
<div dir="auto">In other words, elect argmax N(X), where N(X) is the cardinality of the set of ballots</div>
<div dir="auto">{B | X is ranked ahead of or pairwise defeats a'(B) (or both)}.</div>
<div dir="auto"><br>
</div>
<div dir="auto">The key to monotonicity is that if some candidate X is uniquely raised on some ballot B, then for any other candidate Y, the difference N(X)-N(Y) does not decrease.</div>
<div dir="auto"><br>
</div>
<div dir="auto">The only non-trivial case is where X starts in the a'(B) position and swaps positions with some Z that was ranked above it. </div>
<div dir="auto"><br>
</div>
<div dir="auto">This definitely adds a point to N(X) because X is newly ranked ahead of a'(B). This will also add a point to Y if Z is pairwise defeated by Y.</div>
<div dir="auto"><br>
</div>
<div dir="auto">In any case, N(X)-N(Y) does not decrease.</div>
<div dir="auto"><br>
</div>
<div dir="auto"> Does this method satisfy the Condorcet Criterion?</div>
<div dir="auto"><br>
</div>
<div dir="auto">Well, if X pairwise defeats all other candidates, then N(X) is equal to the total number of ballots minus the number of ballots on which X is in the a' position, which must be fewer than half of the ballots, otherwise X would be the Condorcet
loser.</div>
<div dir="auto"><br>
</div>
<div dir="auto">Is it possible for N(Y) to be greater than this N(X) while losing pairwise to the CW X?</div>
<div dir="auto"><br>
</div>
<div dir="auto">Perhaps ... if so, then this method is not a stand alone Condorcet method.</div>
<div dir="auto"><br>
</div>
<div dir="auto">It would be a valuable task for someone to test its Condorcet efficiency experimentally.</div>
<div dir="auto"><br>
</div>
<div dir="auto">Thanks!</div>
<div dir="auto"><br>
</div>
<div dir="auto">-Forest</div>
<div dir="auto"><br>
</div>
<div dir="auto"><br>
</div>
</div>
</div>
</div>
</body>
</html>