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<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;">Are you referring to how equal-ranking is counted in Borda?</pre>
</blockquote>
Yes, mainly truncated ballots. Do you consider that truncation is
the same thing as ranking equal bottom?<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;">I advocate
that, in these methods’ Borda-count, a ranking give each of its
ranked-candidates a number of points equal to the number of candidates in
the ranking who aren’t ranked over hir.</pre>
</blockquote>
<p>Can you clarify the exact meaning of the phrase "in the
ranking"? Is this a hint that your answer to my last<br>
question is "no"?<br>
<br>
And why is that what you advocate? In some online discussion I
saw it was mentioned (I think by N.Tideman) that<br>
the Baldwin method doesn't meet Condorcet unless the Fractional
(i.e. based on the symmetrically completed ballots)<br>
version of Borda count is used.<br>
<br>
So I've tended to assume that that is the "correct" way of doing
Borda counts.<br>
<br>
Say there are three candidates A,B,C. Say 46 ballots bullet-vote
A. In the fractional version these votes are counted the same as<br>
23 A>B, 23 A>C, giving 92 points to A and 23 points each to
B and C.<br>
</p>
<p>Is that the same as what you advocate, or would you have those
truncating ballots give 2 points to A and zero points to both of B
and C?<br>
Or something else?<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;"><i> So another way of putting it is: "If there is no CW, elect the member of
</i><i> the Smith set with the second-worst score".
</i><i>
</i>
Would you mind telling why that’s another way of putting it?
</pre>
</blockquote>
<br>
By definition the members of the Smith set all pairwise beat all
the non-member candidates, and also have a pairwise defeat at the
hands<br>
of one of the other members (assuming there are three members). <br>
<br>
So the Smith set member with the highest score must be pairwise
beaten by one with a lower score. That much is clear.<br>
<br>
To be honest I may have confused myself again as to further
details. It seems to be possible for the lowest-scored Smith-set
member to win.<br>
<br>
Say the top cycle is (in terms of score order) Middle > High
> Low > Middle.<br>
<br>
Then High is disqualified by being pairwise beaten by Middle and
Middle is disqualified by being pairwise beaten by Low. <br>
</p>
<p>So Low wins.<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;">Remember that sincere top-cycles are vanishingly rare.</pre>
</blockquote>
<br>
What mind-reading technology have you accessed to determine that?
I don't know how you can know that.<br>
<br>
Simple scenarios with lots of truncation and a top cycle look very
plausible to me, so I don't know why I should<br>
accept your assurance that such a thing would be "vanishingly
rare".<br>
<br>
Also it seems to me you are giving yourself a marketing problem:
<br>
<br>
"Ok, we admit that Condorcet methods are very vulnerable to Burial
strategy, so in a effort to combat that we have<br>
to employ this apparently nonsensical, anti-monotonic,
anti-intuitive completion method.<br>
<br>
But don't worry, probably we'll never have to use it."<br>
<br>
Chris B.<br>
</p>
<br>
<blockquote type="cite">
<h1
style="color: rgb(0, 0, 0); font-family: "Times New Roman"; font-style: normal; font-variant-ligatures: normal; font-variant-caps: normal; letter-spacing: normal; orphans: 2; text-align: start; text-indent: 0px; text-transform: none; widows: 2; word-spacing: 0px; -webkit-text-stroke-width: 0px; white-space: normal; text-decoration-thickness: initial; text-decoration-style: initial; text-decoration-color: initial;"> <b
style="color: rgb(0, 0, 0); font-family: "Times New Roman"; font-size: medium; font-style: normal; font-variant-ligatures: normal; font-variant-caps: normal; letter-spacing: normal; orphans: 2; text-align: start; text-indent: 0px; text-transform: none; widows: 2; word-spacing: 0px; -webkit-text-stroke-width: 0px; white-space: normal; text-decoration-thickness: initial; text-decoration-style: initial; text-decoration-color: initial;">Michael
Ossipoff</b><span
style="color: rgb(0, 0, 0); font-family: "Times New Roman"; font-size: medium; 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; white-space: normal; background-color: rgb(255, 255, 255); text-decoration-thickness: initial; text-decoration-style: initial; text-decoration-color: initial; display: inline !important; float: none;"><span> </span></span><a
href="mailto:election-methods%40lists.electorama.com?Subject=Re%3A%20%5BEM%5D%20Duncan%20Proposal%20Draft&In-Reply-To=%3CCAOKDY5D-CUNWajrZxSi8VauLY8FAys8vnt6SvX%3D7KiQFeEpeJw%40mail.gmail.com%3E"
title="[EM] Duncan Proposal Draft"
style="font-family: "Times New Roman"; font-size: medium; 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; white-space: normal;">email9648742
at gmail.com</a></h1>
<i
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Oct 21 12:38:58 PDT 2023</i><span
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<p
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</p>
<hr
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<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 Sat, Oct 21, 2023 at 11:05 C.Benham <<a
href="http://lists.electorama.com/listinfo.cgi/election-methods-electorama.com">cbenham at adam.com.au</a>> wrote:
"Am I right in assuming that the Borda counts are based on the symmetrically
completed ballots?"
Are you referring to how equal-ranking is counted in Borda? I advocate
that, in these methods’ Borda-count, a ranking give each of its
ranked-candidates a number of points equal to the number of candidates in
the ranking who aren’t ranked over hir.
><i> Duncan Definition:
</i>><i>
</i>><i> In the vast majority of the cases ... those in which the pairwise counts
</i>><i> of the ballots unambiguously identify the candidate that pairbeats each of
</i>><i> the others ... elect that candidate.
</i>><i>
</i>><i> Otherwise, elect the highest score candidate that pairbeats every
</i>><i> candidate with lower score.
</i>><i>
</i>><i>
</i>><i> So another way of putting it is: "If there is no CW, elect the member of
</i>><i> the Smith set with the second-worst score".
</i>><i>
</i>
Would you mind telling why that’s another way of putting it?
><i>
</i>><i> To put it bluntly, that is bound to have monotonicity problems and doesn't
</i>><i> fly philosophically.
</i>><i>
</i>
How doesn’t it fly philosophically?
><i>
</i>><i>
</i>><i> Trying to deter or frustrate order-reversal Burial strategy is fine, but
</i>><i> the algorithm should "appear fair" and be able to be justified when
</i>><i> we assume that all the votes are sincere (or even just all equally likely
</i>><i> to be sincere).
</i>><i>
</i>
Remember that sincere top-cycles are vanishingly rare. That’s why
Sequential-Pairwise’s Pareto failure isn’t important, & it’s why
MinMax(wv)’s Condorcet Loser failure isn’t important.
It matters much more what happens in a strategic cycle. Does the method
reward or penalize burial? …or neither?
Duncan tends to often do neither, because it will likely disqualify both
Bus & BF. (I defined those usages when I defined CTE.
I prefer CTE to Duncan, because it more often penalizes, instead of merely
not rewarding.
But I remind you that these methods are intended to deter
probabilisticallly, but aren’t claimed to penalize burial in every possible
example.
><i>
</i>><i>
</i>
><i>
</i>><i> That breaks (at least one version of) "Double Defeat". B is pairwise
</i>><i> beaten by a candidate with a higher "score".
</i>><i>
</i>><i>
</i>><i> Chris B.
</i>><i>
</i>><i>
</i>><i> On 14/10/2023 4:43 am, Michael Ossipoff wrote:
</i>><i>
</i>><i> Yes, I like Duncan because burying the CW in an attempt to help your
</i>><i> favorite won’t help hir when it causes hir disqualification, as it probably
</i>><i> will.
</i>><i>
</i>><i> …& Duncan is remarkably briefly-defined, needing only a very slight
</i>><i> modification of Black’s method.
</i>></pre>
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