Again, very interesting.Forgive me if this is again reinventing, but I can certainly think of ways that might eliminate cycling.The problem seems to be that there are still "cusp points" that need to be smoothed out. Averaging the totals helped, but it wasn't enough. When one candidate's total pa
Rob Brown wrote:
> 1) start with a much lower cutoff. Say 10 or 20. Or, if the
ballots
> are simply ranked, start by giving a "yes" to all but the
bottom-most
> candidates.
No, the same B->A->C->B cycle obtains given the votes
A B C D
33: 100 70 30 0
16: 10 100 70 0
17: 0
On 12/9/05, Warren Smith <[EMAIL PROTECTED]> wrote:
this whole idea, including the fact that it can lead to perpetual cycling,and other versions of this idea (this is related to "declared strategy voting")have already been discussed on the RV bulletin board.
I think the perpetual cycling problem is
Fascinating, thanks!So do you think it still won't find a condorcet winner if it is modified as I suggested:1) start with a much lower cutoff. Say 10 or 20. Or, if the ballots are simply ranked, start by giving a "yes" to all but the bottom-most candidates.
2) use an average of all previous totals
you (Venzke) really should send you criticism to Brams directly.
He is indeed (far as I know) unaware of the EM list and the many fine
developments that have sprung from it. Including your naming system
with the // and so on 9which seems to me quite excellent although I'm not
sure I fully underst
Rob Brown wrote:
> What I suggest is that, prior to tabulating, each Range ballot be
> processed into an Approval ballot. But, it should be done taking
into
> account optimum strategy, with knowledge of how others are voting.
You
> could call this a "Nash equilibrium seeking" system, as it keeps
this whole idea, including the fact that it can lead to perpetual cycling,
and other versions of this idea (this is related to "declared strategy voting")
have already been discussed on the RV bulletin board.
My suggestion is to join that bulletin board insetad of reinventing the wheel.
wds
e
On 12/8/05, Scott Ritchie <[EMAIL PROTECTED]> wrote:
Then make it 5 A>B>C, 4 B>C>A, and 3 C>A>B, and watch the same thinghappen.I looked into this a bit, and see what's going on and think it is easily fixable.My approach would be to stick with the conceptual point of view that each voter has one so
On 12/8/05, Scott Ritchie <[EMAIL PROTECTED]
> wrote:
On Thu, 2005-12-08 at 18:40 -0800, rob brown wrote:> 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>BThen make it 5 A>B>C, 4 B>C>A, and 3 C>A>B, and watch th
