On Jan 8, 2007, at 19:38 , Chris Benham wrote:
>
> Juho,
>
> 26: A>B
> 25: C>A
> 49: B>C (sincere is B>A or B)
>
>
> Juho wrote:
>
>> But I'll however mention some random observations that the
>> example that you used made me think.
>> - One could also claim that these votes are a result of st
Juho,
26: A>B
25: C>A
49: B>C (sincere is B>A or B)
Juho wrote:
>But I'll however mention some random observations that the example
>that you used made me think.
>- One could also claim that these votes are a result of strategic
>voting but in another way than what you described. Instead
Ok, the method that I proposed is not as defensive against burial as
the original one. My target was just to make the method better with
sincere votes (not to seek the ultimate most strategy resistant
method). I'll come back with this method and also some other variants
when I find some mor
Simmons, Forest wrote:
>Here's a version that is both clone proof and monotonic:
>
>The winner is the alternative A with the smallest number of ballots on which
>alternatives that beat A pairwise are ranked in first place. [shared first
>place slots are counted fractionally]
>
>That's it.
>
Simmons, Forest wrote:
>In view of comments and suggestions from Chris, Warren, Markus, and all (for
>which I thank you all warmly) I would like to suggest that this attempt at
>clone proofing Copeland be used in a three slot setting.
>
>I'll restate it for the record:
>
>For each candidate
In view of comments and suggestions from Chris, Warren, Markus, and all (for
which I thank you all warmly) I would like to suggest that this attempt at
clone proofing Copeland be used in a three slot setting.
I'll restate it for the record:
For each candidate X let p(X) be the probability tha
If I understand this, it's a Condorcet cycle resolution system based
purely on who was 1st pick on each ballot. So, count up the virtual
round robin matrix, and count 1st place votes separately for later if
needed.
It's incomplete. A Condorcet method can elect someone no one put in
for fir
Juho wrote:
>How about "the smallest number of ballots on which some alternative
>that beats A pairwise is ranked higher than A"?
>
>Juho
>
>
No, that would have nothing like the same strength or resistance to Burial.
26: A>B
25: C>A
49: B>C (sincere is B>A or B)
The Simmons method narrow
[EMAIL PROTECTED] wrote:
Chris Benham wrote:
I'm happy with its performance in this old example:
101: A
001: B>A
101: C>B
It easily elects A. Schulze (like the other Winning Votes "defeat
dropper" methods) elects B.
It meets my "No Zero-Information Strategy" criterion, which m
Chris Benham wrote:
> I'm happy with its performance in this old example:
>
> 101: A
> 001: B>A
> 101: C>B
>
> It easily elects A. Schulze (like the other Winning Votes "defeat
dropper" methods) elects B.
>
> It meets my "No Zero-Information Strategy" criterion, which means that
the voter with no
How about "the smallest number of ballots on which some alternative
that beats A pairwise is ranked higher than A"?
Juho
On Dec 31, 2006, at 3:52 , Simmons, Forest wrote:
> Here's a version that is both clone proof and monotonic:
>
> The winner is the alternative A with the smallest number o
Dear Forest Simmons,
you wrote (30 Dec 2006):
> Here's a version that is both clone proof and monotonic:
>
> The winner is the alternative A with the smallest number
> of ballots on which alternatives that beat A pairwise are
> ranked in first place. [shared first place slots are counted
> fracti
Simmons, Forest wrote:
Here's a version that is both clone proof and monotonic:
The winner is the alternative A with the smallest number of ballots on which
alternatives that beat A pairwise are ranked in first place. [shared first
place slots are counted fractionally]
That's it.
This met
Here's a version that is both clone proof and monotonic:
The winner is the alternative A with the smallest number of ballots on which
alternatives that beat A pairwise are ranked in first place. [shared first
place slots are counted fractionally]
That's it.
This method satisfies the Smith C
Here's an idea for clone proofing Copeland:
1. First (as in Copeland) compute the pairwise win/lose matrix, which has a
+1, -1, or zero in row i column j according as alternative i beats, loses
to, or ties with candidate j in the (i, j) pairwise comparison.
2. Then (unlike Copeland) mul
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