I finally had a chance to look this over. I think there are some good
ideas here.
The first matrix (of which you have shown the sub diagonal elements) is
the symmetric matrix whose entry in the i_th row and J_th column is the
number of voters who approved both candidate i and candidate j. If the
On Thu, 6 Mar 2003, Forest Simmons wrote:
>
> (8) We could combine your method (2) with method (6), by finding the max
> of each row in your (transpose of pairwise) matrix and then subtracting
> the min of the corresponding column. The candidate corresponding to the
> minimum
On Thu, 6 Mar 2003, [iso-8859-1] Venzke Kevin wrote:
>
> Looking at some of your proposals for picking the
> winner:
These were not really intended as proposals so much as examples to
establish a context for understanding your method relative to commonly
understood methods.
>
> >(1) Take the
Has anybody ever proposed minimizing the maximum opposition rather than
minimizing the maximum defeat?
I know that theoretically this could elect the Condorcet Loser, but it
seems very unlikely that it would do so.
It seems to me that if equality were allowed in the rankings, then this
method wou
I must be missing something. Could you give an example in which the
approval winner is not the winner of the method?
Forest
On Mon, 10 Mar 2003, Steve Eppley wrote:
> On 10 Mar 2003 at 11:36, Kevin Venzke wrote:
> > My recent "MinMax" message concluded with a
> > half-hearted attempt at a syste
Forest
On Tue, 11 Mar 2003, Steve Eppley wrote:
> On 12 Mar 2003 at 0:33, Kevin Venzke wrote:
> > Forest Simmons wrote:
> > >I must be missing something. Could you give an
> > >example in which the approval winner is not the
> > >winner of the method?
>
On Sun, 9 Mar 2003, Olli Salmi wrote:
> Forest,
>
> I've found this method of yours fascinating.
>
> At 18:45 +0200 6.2.2002, Forest Simmons wrote:
> >> One option might be to decay a voter's ballot based on the position of the
> >> elected candi
On Thu, 13 Mar 2003, [iso-8859-1] Kevin Venzke wrote:
> Forest,
>
> This is an interesting idea.
>
> I was trying to do some examples with it, but I'm not
> sure how to create the four matrices. These are the
> ballots I was trying to use:
>
> 12: A at 10 (fill 8 and 2 circles)
> 11: B at 7 (fi
See typo correction below:
On Thu, 13 Mar 2003, Forest Simmons wrote:
>
> If there are fewer than five candidates there is no need to have four bit
> ballots. With three candidates two bits are sufficient, so the method
> would be the same as the one Venzke described.
Should be &
Dear Steve,
what I once called "Approval Seeded Bubble Sort" but would now (for
increased respectibility) call the "Local Kemenization of the Approval
Order" is a member of the family of methods that satisfy your conditions
(1) and (2).
We initialize with the approval order and then, starting at
Here's another idea along these lines in a different direction.
The margins versus winning votes controversy seems to be a question of
priorities of majorities, loosely speaking, at least in the context of
Ranked Pairs.
Suppose that (infinitely patient) voters voted directly on the order of
impor
t Favorite cannot beat.
You would have to be pretty fickle to abandon Favorite on account
of some poll pretending to be accurate enough to predict all of that.
Forest
On Fri, 7 Mar 2003, Steve Eppley wrote:
> On 7 Mar 2003 at 10:42, Forest Simmons wrote:
>
> > Has anybody ever p
I had a simpler example in mind:
Three teams in a round robin tournament compile the following scores:
A beats B four to three, B beats C two to one, and A beats C two to one.
A is the Condorcet winner while C is not only the Condorcet loser, but
also the method winner under this variant of MinM
On Fri, 14 Mar 2003, [iso-8859-1] Kevin Venzke wrote:
> I suspect that this system (applying a Condorcet
> method to the ballots) is identical to Borda with a
> fixed number of ranks. I'd bet that voters would use
> Approval strategy (give only 15's and 0's).
>
> Given a Borda ballot of A>B>C>D>E
corrections:
"one" changed to "TWO", "two" changed to "THREE" in the last sentence
below:
On Fri, 14 Mar 2003, Forest Simmons wrote:
> I had a simpler example in mind:
>
> Three teams in a round robin tournament compile the following scores:
There is a folk theorem about the impossibility of the Strong Favorite
Betrayal Criterion. In other words, any method that doesn't allow
equality in the top position of the ballot will give incentive to betray
favorite in some situations.
On the other hand some methods that allow equal ranking in
Small wrote:
> Forest Simmons said:
> > For example, MinMax(pairwise opposition) satisfies the FBC when equality
> > at the top is allowed. This method chooses as winner the candidate
> > whose maximum pairwise opposition is minimal.
> ...
> > As near as I can tell this
On Sun, 9 Mar 2003, [iso-8859-1] Kevin Venzke wrote:
> Looking at Schulze's example of:
> 20 ABCD
> 20 BCAD
> 20 CABD
> 13 DABC
> 13 DBCA
> 13 DCAB
>
> The scores are 66 for ABC and 60 for D, so D winning
> is of course not best. But I think the problem is
> that an ordering low in th
Olli,
Here's another idea for PR that makes use of Borda style ballots:
Suppose that we have a family F of subsets of candidates, and that we want
to see which member of the family would better represent the voters
according to the information contained on their ranked preference ballots.
First
Is there some non-deterministic resolution of Condorcet that would
completely discourage insincere ranking?
Neither random ballot nor random candidate does the job, but how about
some probabilities based on how close a candidate is to being a CW, or
something like that?
If we found a non-determin
On Thu, 20 Mar 2003, [iso-8859-1] Kevin Venzke wrote:
> --- Forest Simmons <[EMAIL PROTECTED]> a écrit :
> > > 20 ABCD
> > > 20 BCAD
> > > 20 CABD
> > > 13 DABC
> > > 13 DBCA
> > > 13 DCAB
> > >
> > then
Here's a strategy free method that will work in certain situations:
The main requirement is that there has to be an incumbent or status quo
option.
The voters indicate which candidates they prefer over the status quo. If
the status quo is the Condorcet Winner (i.e. no alternative is preferred
ov
tatus quo can't win.
>
> Are you sure it's strategy-free?
>
>
> Forest Simmons said:
> > Here's a strategy free method that will work in certain situations:
> >
> > The main requirement is that there has to be an incumbent or status
> > quo option
This is the first day of April, but I'm really serious about the subject
line.
If I am not mistaken, it is indeed possible to satisfy the strong FBC with
the right kind of ballot, though it seems impossible with any ballot type
that has been proposed previously.
The ballot has two parts. The fir
Good example!
I passed over this one in my quest for the Strong FBC because, as you
point out, optimal strategy sometimes requires not approving favorite,
which most folks would consider as betrayal of favorite.
But it does have its appeal as an honest way of eliciting sincere
rankings.
It could
On Wed, 2 Apr 2003, Alex Small wrote in part:
> One interesting thing about Approval Runoff is that strategically it
> satisfies the Majority Criterion: When there is a candidate whom a
> majority of the voters consider their first choice then he should win.
> With adequate polling data, the maj
On Thu, 3 Apr 2003, Chris Benham wrote in part:
>
> I see..with your method a voter would have no reason not to approve
> Favourite vs. Lesser Evil.
> I think that it should include a default "bullet-vote" feature : that if
> a voter just ranks one candidate 1, then that voter is automatically
>
t pair
locked in would be A>B.
I suppose another solution would be to simply force compliance with Pareto
at the outset by removing all candidates that are strictly dominated by
some other candidate.
Would this destroy compliance with the FBC?
Forest
On Tue, 1 Apr 2003, Forest Simmons wrote
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