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<p><br>
Ted,<br>
<br>
You asked me a while back to explain how Margins Sorted Top
Ratings "fails Independence from Clones".<br>
<br>
The answer is that wearing that badge I think it technically
doesn't, because if Ratings ballots are used <br>
I think a (say) pair of candidates have to be given the same
rating on every ballot to qualify "clones".<br>
<br>
But if we instead call it "Margins Sorted FPP(Whole)" using
ranking ballots, then it can fail Clone-Winner<br>
as in this example:<br>
</p>
<p>34 C<br>
33 A>B<br>
32 B<br>
<br>
The FPP order is C>A>B and C pairwise beats A and A pairwise
beats B (and that is all we need to know)<br>
so C wins.<br>
<br>
Now say we add a clone of C, X.<br>
<br>
18 C>X<br>
16 X>C<br>
33 A>B<br>
32 B<br>
<br>
Now the FPP order is A>B>C>X and as before no adjacent
pair is pairwise out of order, i.e A (still) pairwise<br>
beats B and B pairwise C and C pairwise beats X, so cloning C
changes the winner from C to A.<br>
<br>
I find this the more natural and useful way of looking at it.<br>
<br>
Chris Benham<br>
<br>
<br>
<br>
<br>
<blockquote type="cite">
<pre
style="white-space: pre-wrap; color: rgb(0, 0, 0); font-style: normal; font-variant-ligatures: normal; font-variant-caps: normal; font-weight: 400; letter-spacing: normal; orphans: 2; text-align: start; text-indent: 0px; text-transform: none; widows: 2; word-spacing: 0px; -webkit-text-stroke-width: 0px; text-decoration-thickness: initial; text-decoration-style: initial; text-decoration-color: initial;">On Mon, Jan 29, 2024 at 10:55 AM C.Benham <<a
href="http://lists.electorama.com/listinfo.cgi/election-methods-electorama.com">cbenham at adam.com.au</a>> wrote:
><i> (This is a re-send with an error corrected)
</i>><i>
</i>><i> I think Margins Sorted Top Ratings would be a good relatively burial
</i>><i> resistant public proposal.
</i>><i>
</i>><i> * Voters rank from the top however many candidates they wish.
</i>><i> Equal-ranking allowed.
</i>><i>
</i>><i> Give each candidate a score according to the number of ballots on which
</i>><i> they are ranked below
</i>><i> no others.
</i>><i>
</i>><i> Line them up in that order, highest to lowest.
</i>><i>
</i>><i> Check to see if all the candidates above bottom in this order pairwise
</i>><i> beat the candidate immediately
</i>><i> below them.
</i>><i>
</i>><i> If they do then elect the candidate highest in the order.
</i>><i>
</i>><i> If not begin with the pair that is pairwise out of order by the lowest
</i>><i> margin and swap them.
</i>><i> (if there is an exact tie in the size of the margin then swap the
</i>><i> tied-margin pair lowest in the order).
</i>><i>
</i>><i> Repeat until no pair of adjacent candidates are pairwise out of order
</i>><i> and then elect the highest-ordered
</i>><i> candidate. *
</i>><i>
</i>><i> This could also use ratings ballots.
</i>><i>
</i>><i> This meets Condorcet, but can be at least be explained (if not sold)
</i>><i> without reference to Condorcet or Smith.
</i>><i>
</i>><i> It would be as monotonic as it is possible for a Condorcet method to be.
</i>><i>
</i>><i> For the sake of simplicity (and elegance) it has some short-comings.
</i>><i> When there is a top cycle, voters who
</i>><i> didn't top-rate (rank below no other candidates) any of the candidates
</i>><i> in the Smith set are disadvantaged by comparison
</i>><i> those that did. It would also fail Clone-Independence.
</i>><i>
</i>
Could you explain how Top-rated Sorted Margins fails Independence from
Clones? I'm not seeing how that would happen, unless you have unusual
restrictions on top-rating</pre>
</blockquote>
<br>
</p>
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