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<blockquote type="cite">"for Benham, what if we count fractional
(for equal rank top) as you suggest when doing the IRV
eliminations, but check at each step for a pairwise beats all
candidate in the usual way?"</blockquote>
<br>
Forest, That wouldn't be too bad. But it seems to me that it
would make Pushover strategizing less risky (than not allowing
above-bottom equal-rankings) and since the method<br>
fails FBC I didn't see sufficient justification for the extra
complexity.<br>
<br>
<blockquote type="cite">"In your example below, since B beats A
pairwise 31 to zero and B beats C 65 to 35, no IRV elimination
step is required, so how equal rank top is counted in this
example does not seem to matter."</blockquote>
<br>
Yes, Benham has less of a Pushover vulnerability problem than
IRV. I don't understand your other question. Benham only checks
for a single CW. "Symmetrically completing" pairwise<br>
contests can't make any difference to that.<br>
<br>
A better example of my suggested way of getting extra "pushover
resistance":<br>
<br>
04: A=C (sincere is A or A>B)<br>
41 A<br>
28 B>A<br>
27 C>B<br>
<br>
B>A 55-45, A>C 69-27, C>B 31-28. Top Preferences
(erf): A43 > C29 > B28<br>
<br>
B is the sincere CW (and so also the sincere Benham winner) and
the sincere IRV winner. Benham and IRV elect A.<br>
<br>
My suggested variant looks at the order of candidates according to
their TP(erf) scores and on seeing that A is higher in that order
than C assigns the whole value of A=C ballots to A<br>
(and none of it to C) to give A45 > B28 > C27 (purely for
the purpose of the IRV component and not the pairwise component in
Benham) and then eliminates C and elects B.<br>
<br>
Chris Benham<br>
<br>
<br>
On 5/21/2014 9:12 AM, Forest Simmons wrote:<br>
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<div>Chris,<br>
<br>
</div>
for Benham, what if we count fractional (for equal rank
top) as you suggest when doing the IRV eliminations, but
check at each step for a pairwise beats all candidate in
the usual way?<br>
<br>
</div>
In your example below, since B beats A pairwise 31 to zero
and B beats C 65 to 35, no IRV elimination step is required,
so how equal rank top is counted in this example does not
seem to matter.<br>
<br>
</div>
Or is there some reason for doing a "symmetric completion" of
equal rankings for the pairwise contests as well?<br>
<br>
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Forest<br>
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<blockquote class="gmail_quote" style="margin:0
0 0 .8ex;border-left:1px #ccc
solid;padding-left:1ex">
<br>
Date: Wed, 21 May 2014 03:53:38 +0930<br>
From: "C.Benham" <<a moz-do-not-send="true"
href="mailto:cbenham@adam.com.au">cbenham@adam.com.au</a>><br>
To: <a moz-do-not-send="true"
href="mailto:election-methods@lists.electorama.com">election-methods@lists.electorama.com</a><br>
Subject: Re: [EM] New Criterion<br>
Message-ID: <<a moz-do-not-send="true"
href="mailto:537B9DAA.4030406@adam.com.au">537B9DAA.4030406@adam.com.au</a>><br>
Content-Type: text/plain;
charset="iso-8859-1"; Format="flowed"<br>
<br>
Forest,<br>
<br>
I've been meaning to remind you: for IRV and
Benham (and Woodall and<br>
similar) I'm strongly opposed to allowing
voters to do any equal-ranking<br>
apart from truncating because it makes
Push-over strategizing much less<br>
risky and more likely to succeed.<br>
<br>
Two versions of ER-IRV have been discussed,
one where an A=B ballot<br>
gives a "whole vote" to each and one where it
gives half a vote to each,<br>
i.e,<br>
ER-IRV(whole) and ER-IRV(fractional). The
problem I referred to is<br>
much worse for the former and so I consider
the latter to less bad.<br>
<br>
But if we insist on allowing above-bottom
equal-ranking and don't mind a<br>
lot of extra complexity, I have this
suggestion:<br>
<br>
*Before each elimination, order the candidates
according to their<br>
ER-IRV(fractional), (so that among continuing
candidates a ballot that<br>
equal-top<br>
ranks n candidates give 1/n of a vote to
each).<br>
<br>
Then assign each of the ballots that equal-top
rank more than one<br>
candidate to whichever of them is highest in
that order.<br>
<br>
Then eliminate the candidate with the fewest
ballots assigned to hir.*<br>
<br>
34 A=B<br>
31 B<br>
35 C<br>
<br>
So in this example of Forest's, to create the
initial order the 34 A=B<br>
ballots give half a vote each to A and B, to
give the scores<br>
B (31+17=48) > C35 > A17.<br>
<br>
B is above A in this order, so all of the A=B
ballots are assigned to B.<br>
This gives the scores B65 > C35 > A0. A
has the lowest score so<br>
A is eliminated and B wins.<br>
<br>
Chris Benham<br>
<br>
<br>
<br>
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