<div dir="ltr"><br><div class="gmail_extra"><br><br><div class="gmail_quote">On Mon, Apr 21, 2014 at 2:39 PM, C.Benham <span dir="ltr"><<a href="mailto:cbenham@adam.com.au" target="_blank">cbenham@adam.com.au</a>></span> wrote:<br>
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<div>Forest,<div class=""><br>
<br>
<blockquote type="cite">48 C<br>
27 A>B<br>
25 B</blockquote>
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</div><div class=""><blockquote type="cite">Borda, TACC, and IRV based methods like
Woodall and Benham elect C.<br>
<br>
But Borda is clone dependent, and the IRV style elimination
based methods fail monotonicity. So TACC is a leading contender
if we really take the Chicken Dilemma seriously.</blockquote>
<br></div>
Benham and Woodall are a lot more resistant to Burial than TACC
(and other Condorcet methods that meet mono-raise, aka
monotonicity) because they meet <br>
"Unburiable Mutual Dominant Third" that says that if a set S of
candidates are all voted together on top of more than a third of
the ballots (i.e. on more than a third of the ballots no other
candidates are voted between or above them) and all the members of
S pairwise beat all the non-member candidates, then the winner X
must come from S (so far the MDT criterion); and if some of the
ballots that vote any Y above X are altered by further lowering X
then if the winner is changed it can't be to Y.<br>
<br>
34 A>B<br>
17 C>A<br>
16 B>C<br>
33 B<br>
<br>
A>B 51-49, A>C 34-33, B>C 83-17<br>
<br>
A is the DMT winner, but if 2 of the 33 B ballots are changed to
B>C then TACC (like some better methods) elects B.<br>
<br>
Also, for what it's worth, Benham and Woodall have no zero-info.
truncation incentive, and little informed truncation incentive.
IRV has none.<br>
<br>
Whenever there are 3-candidates in a cycle, TACC always elects
the candidate that pairwise beats the least approved candidate
(which strikes me as weird and arbitrary).<br>
If the approval order is A>B>C and the pairwise results go
A>B>C>A, then TACC will elect B.<br>
<br>
That is a violation of a criterion (I like) that says that the
winner can't be pairwise-beaten by a more approved candidate. (You
may have had a name for it).<br></div></div></blockquote><div><br></div><div>We talked about this before, and I agreed with you on that. But that was before I started taking the Chicken Dilemma seriously. <br><br>
</div><div>Now look at this cycle A > B > C > A with the approval order A>B>C.<br><br></div><div>On this information alone we might say that A is strong because it doubly beats B, both in approval and pairwise. However, when we realize that <br>
A is pairwise beaten by the weakest candidate C, and is on the losing end of a short beat path from B, some of the wind is taken from her sails.<br></div><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">
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<br>
A much better method that meets Chicken Dilemma, Condorcet and
Mono-raise is Approval Margins. We determined that it is
equivalent to suggestion of yours, "Approval Margins Sort".<br>
From early (probably March) 2005, does this ring a bell?<br></div></div></blockquote><div><br></div><div>Yes, it rings a loud bell. I'll have to revisit it in the context of the chicken dilemma. <br></div><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">
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<br>
<blockquote type="cite">
<pre>I wonder if the following Approval Margins Sort
</pre>
<pre>(AMS) is equivalent to your Approval Margins method:
1. List the alternatives in order of approval with
highest approval at the top of the list.
2. While any adjacent pair of alternatives is out of
order pairwise ... among all such pairs swap the
members of the pair that differ the least in approval.
This method is clone independent and monotonic, and
yields a social order that reverses exactly when the
ballots are reversed. If AMS and AM are the same, it
might be useful to have this alternative description.
</pre>
<pre>AMS is monotonic in a strong sense: if ballots are
changed so as to increase alternative X's approval or
to give X a victory that it didn't have before, while
leaving all of the other approvals and pairwise
defeats the same, then X cannot move down in the
social order produced by this AMS method. In other
words, AMS is monotonic with respect to the entire
social order it produces.
</pre>
<pre>After one example it is pretty obvious that AM and
AMS are equivalent when there are only three
alternatives, since they both yield the CW when there
is one, and they both preserve the approval order if
the only upward defeat arrow is from the bottom to the
top, and they both reverse the closest approval margin
pair, otherwise.
</pre>
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<br>
Chris Benham<div><div class="h5"><br>
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On 4/20/2014 8:58 AM, Forest Simmons wrote:<br>
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<div>Let's consider the following
ballot set:<br>
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48 C<br>
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27 A>B<br>
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25 B<br>
<br>
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Plurality says that A cannot win, because C
has more top votes than A is ranked.<br>
<br>
</div>
The Chicken Dilemma Criterion says that B
cannot win, because there is a possibility
that the B faction is defecting from a true
preference of B>A.<br>
<br>
</div>
That leaves C as the only acceptable winner for
this ballot set. <br>
<br>
</div>
How do various methods stack up on this ballot
set?<br>
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MinMaxPairwiseOpposition (MMPO) elects A.<br>
<br>
</div>
Condorcet(wv) and Condorcet(margins) both elect B.<br>
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</div>
Implicit Approval elects B.<br>
<br>
</div>
Borda, TACC, and IRV based methods like Woodall and Benham
elect C.<br>
<br>
</div>
But Borda is clone dependent, and the IRV style elimination
based methods fail monotonicity. So TACC is a leading
contender if we really take the Chicken Dilemma seriously.<br>
<br>
</div>
But what if the ballot set is sincere? <br>
<br>
</div>
<div>In that case it seems like B should be the winner.<br>
<br>
</div>
<div>The problem is that standard election methods have no way
of detecting ballot sincerity.<br>
<br>
</div>
<div>Therefore there is no strategy free method for covering
both scenarios.<br>
<br>
</div>
<div>The question now becomes which requires more drastic
strategy?<br>
<br>
</div>
<div>Note that if the above ballot set represents sincere
preferences, the A faction can help elect B by voting A=B,
which requires no order reversal.<br>
<br>
</div>
<div>This move will work in any of the above mentioned methods!<br>
<br>
</div>
<div>Now suppose that the true preferences were<br>
<br>
</div>
<div>48 C>B<br>
</div>
<div>27 A>B<br>
</div>
<div>25 B>C<br>
<br>
</div>
<div>Then B is the Condorcet Winner.<br>
<br>
</div>
<div>But under both Condorcet wv and margins, there is a burial
incentive for the C>B faction to change votes to 48 C>A
. If B takes no defensive action, this burial strategy will
succeed.<br>
<br>
</div>
<div>Under TACC that burial strategy will backfire by electing
electing A.<br>
<br>
</div>
<div>However, if the C>B faction may still be tempted to take
the a more sincere approach of merely truncating B. When the
B faction responds to this threat by truncating C, we are led
back to the original ballot set given above:<br>
<br>
48 C<br>
27 A>B<br>
25 B<br>
</div>
<div><br>
</div>
<div>Which must have C as the winner if we want to honor
Plurality and CD. <br>
<br>
</div>
<div>But then the A>B faction has a strong incentive to raise
B to 27 A=B.<br>
<br>
</div>
<div>This solves the problem without any order reversals.<br>
<br>
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<div>Forest<br>
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</div></div><pre>----
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