<div dir="ltr"><div><div>Chris,<br><br></div>I agree, just give points to the equal top candidates.<br><br></div>Or if more extensive rankings are wanted for the purpose of finding Smith, use standard ordinal ballots with an extra check box beside each candidate name. Check the box iff you want to contribute a point to that candidate in the point count should she make it into Smith.<br><div><br></div></div><div class="gmail_extra"><br><div class="gmail_quote">On Sun, Oct 9, 2016 at 8:18 PM, C.Benham <span dir="ltr"><<a href="mailto:cbenham@adam.com.au" target="_blank">cbenham@adam.com.au</a>></span> wrote:<br><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">
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<div>On 10/10/2016 7:43 AM, Forest Simmons
wrote:<br>
<blockquote type="cite">I want to propose a new Condorcet Method
below, but first a simple three slot method inspired by
Jameson's MAS, but one that truly satisfies the Chicken Defense
Criterion:<br>
<br>
Ballots are scores or ratings on a scale of zero to three.<br>
<br>
The Smith candidate with the highest average score wins.</blockquote>
<br>
Forest,<br>
<br>
What is the "Chicken Defense Criterion"? <br>
<br>
Your suggested 3-slot Smith//Score doesn't meet the Chicken
Dilemma criterion.<br>
<br>
<a href="http://wiki.electorama.com/wiki/Chicken_Dilemma_Criterion" target="_blank">http://wiki.electorama.com/<wbr>wiki/Chicken_Dilemma_Criterion</a><br>
<br>
33: A>B<br>
32: B<br>
34: C<br>
<br>
A>B>C>A , all candidates in the Smith set. 0-1-2
Scores: B 97 > C 68 > A 66.<br>
<br>
B easily wins, but the Chicken Dilemma criterion specifies that B
must not win.<br>
<br>
Chris Benham<br>
<br>
<br>
<br>
</div>
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<div>Do I remember correctly
that MAM is just Ranked
Pairs with a better tie
breaker?<br>
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C: MAM ("Maximum Affirmed Majorities") is Ranked Pairs (Winning
Votes) with a specific random-ballot based tie-breaker.<br>
<br>
<a href="http://wiki.electorama.com/wiki/Maximize_Affirmed_Majorities" target="_blank">http://wiki.electorama.com/<wbr>wiki/Maximize_Affirmed_<wbr>Majorities</a><br>
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It seems like MAM and Ranked
Pairs put in as many defeats
as possible without creating
cycles, but like Juho says do
we really need all of those
defeats to decide the winner?
<br>
<br>
River recognizes that we just
need one defeat for each
non-winner. It seems to me
that the one defeat for each
non-winner should be as strong
as possible, but (having
already been eliminated) their
other defeats don't matter.<br>
<br>
</div>
I want to propose a new
Condorcet Method below, but
first a simple three slot method
inspired by Jameson's MAS, but
one that truly satisfies the
Chicken Defense Criterion:<br>
<br>
</div>
Ballots are scores or ratings on a
scale of zero to three.<br>
<br>
</div>
The Smith candidate with the highest
average score wins.<br>
<br>
</div>
In other words, the method is
Smith//Score (or is it Smith\\Score
?).with three slot ballots.<br>
<br>
</div>
Example:<br>
<br>
</div>
49 C<br>
</div>
27 A>B<br>
</div>
24 B (sincere is B>A)<br>
<br>
</div>
With sincere votes A is elected as the only
member of Smith.<br>
<br>
</div>
Under the B faction defection C wins as the Smith
candidate with the highest Score.<br>
<br>
</div>
Now, how do we adapt this to general rankings? We
assume that equal top rankings and equal bottom or
multiple truncations are allowed.<br>
<br>
</div>
For each ballot on which a candidate is ranked above
bottom but below top that candidte receives one
point. For each ballot on which the candidate is
ranked top or equal top that candidate receives two
points. <br>
<br>
</div>
The Smith candidate with the greatest number of points
wins.<br>
<br>
</div>
[End of definition] <br>
<br>
</div>
Note that the method does satisfy CD unlike
Smith//ImplicitApproval. Jameson's idea of three slot
scores makes it work.<br>
<br>
</div>
How does it do on burial?<br>
<br>
</div>
Forest<br>
</div>
<br>
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