[EM] Alright, next try. Range voting fix, version 2.

[rob brown]

Well, your example is not only a Condorcet cycle, but a pure 3-way tie in
condorcet terms.

It is effectively:
A>B>C
B>C>A
C>A>B

So, no condorcet method could resolve it either as anything but an
out-and-out tie. (right?)  I would not expect this to do so either -- at
least not in its simplest implementation.  In fact, I'm pretty happy that it
indeed met my prediction of behaving similarly to condorcet methods, in that
it gets stumped on the same data sets. :)

Note that the method as I first described it basically ignores the rating
values and relies only on the ordering (except for the initial condition).
However, I mentioned at the end that it was possible that it could be more
stable, as well as use the ratings more effectively, if -- on each round --
we more gradually moved each ballot toward the "optimum strategy", by simply
adjusting the cutoff point toward that optimum strategy value.  This way you
are less likely to get ties, and more likely for it to stabilize.  (I
suppose I should follow that with "I think")

-rob

On 12/8/05, Scott Ritchie <scott at open-vote.org> wrote:
>
> On Thu, 2005-12-08 at 17:10 -0800, rob brown wrote:
> > I am going to guess a few things:
> >
> > 1) that it will be extremely rare that it does not find a Nash
> > equilibrium in fewer than 20 rounds
>
> Here's an example where it will never resolve:
>
> Voter 1:
> A: 100
> B: 40
> C: 0
>
> Voter 2:
> A: 45
> B: 0
> C: 100
>
> Voter 3:
> A: 0
> B: 100
> C: 60
>
> A simple Condorcet cycle.
>
> Let's say it starts at 50 as the approval margin.  The total is then:
> A: 1
> B: 1
> C: 2
>
> So, C will win, but Voter 3 can get a better option with a lower margin.
> So we then downrate his vote...similarly voter 1 can get a better option
> by lowering his margin, so in the next round of tabulating we get:
> A: 2
> B: 2
> C: 1
>
> This one won't resolve.  Ever.
>
> > 2) that the results of this will be remarkably similar to the results
> > obtained by some of the better Condorcet methods
>
> Might want to make sure it's deterministic first.
>
> Thanks,
> Scott Ritchie
>
>
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