-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED]
> ] On Behalf Of Kevin Venzke
> Sent: Sunday, November 07, 2004 5:20 PM
> To: [EMAIL PROTECTED]
> Subject: RE: [EM] More easily hand-counting three-slot Condorcet
>
> Paul,
>
> --- Paul Kislanko <[EMAIL PROTECT
Dear Jobst,
I apologize if you feel that I haven't taken your ideas seriously. I
haven't been reading EM very much until recently. I will spend some
time reading past messages.
I did understand how your first proposal for measuring defeat strength
was a special case of the "grand compromise," t
Paul,
--- Paul Kislanko <[EMAIL PROTECTED]> a écrit :
> Well, you're wrong, although as usual you guys are really arrogant.
I don't see how I was wrong, except that you forgot to mention that you don't
care what I'm talking about.
Regarding your several paragraphs:
> I was objecting to phras
Dear Kevin!
Let us recall: you suggest to use only three slots and to interpret
candidates in slots 1 and 2 as approved of and candidates in slot 3 as
not approved of. Then you came up with the topic of how to measure
defeat strength best without having to count all winning votes, and
suggested to
m to yourself. You never
advance a discussion with ad hominems.
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED]
> ] On Behalf Of Kevin Venzke
> Sent: Sunday, November 07, 2004 4:21 PM
> To: [EMAIL PROTECTED]
> Subject: [Norton AntiSpam] RE
Paul,
--- Paul Kislanko <[EMAIL PROTECTED]> a écrit :
> Kevin Venzke wrote:
> > My intended strategy is to discard "noise"
> > candidates. For example, again:
> >
> > 9 A>B>C
> > 8 B>C>A
> > 7 C>A>B
> >
> > I don't want to see B win here. Nine A voters will kick themselves
> > if B wins. T
Kevin Venzke wrote:
> My intended strategy is to discard "noise"
> candidates. For example, again:
>
> 9 A>B>C
> 8 B>C>A
> 7 C>A>B
>
> I don't want to see B win here. Nine A voters will kick themselves
> if B wins. The best resolution seems to be to pretend the (clearly)
> weakest candidate,
Dear Jobst,
--- Jobst Heitzig <[EMAIL PROTECTED]> a écrit :
> > (The scenario I have in mind is: Approval ranking: A>B>C. Pairwise:
> > B>A>C>B. B wins. If B's approval is simply reduced somehow, with no
> > other changes: Approval ranking: A>C>B. Now A wins.)
>
> Isn't that very strange? We
Dear Kevin!
you wrote:
> Hmm. I note that if A is going to be elected despite being defeated
> pairwise by B, this must mean that A has greater approval than B, and
> that A's wins will be locked first.
This is what happens in your version, right. And at first it seems to be
intuitive to requi
Jobst,
Thanks for taking the time to look at this idea.
--- Jobst Heitzig <[EMAIL PROTECTED]> a écrit :
> But the vector
> (0,1,1,0,0,1,0,0,0)
> is linearly independent from
> (1,1,1,1,1,1,1,1,1), [...]
Interesting, this is precisely what I ended up doing to convince
myself that the magnit
Dear Kevin!
you wrote:
> If there are three ranks, and R1 is the number of candidates
> placed in the first rank, R2 for second, and R3 the number of
> unmarked candidates, then using the above method the manual
> counter will have to mark
>
> (R1*R2) + (R1*R3) + (R2*R3) tallies.
...
> For a give
Hello all,
I accidentally came up with an easier way to hand-count
three-slot Condorcet. By "easier" I mean fewer tally marks
(potentially by far) per ballot.
The intuitive way to count Condorcet methods is to take a
ballot, and mark a vote in a matrix at X,Y for every candidate
X ranked above
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