At 08:49 PM 12/8/2005, Scott Ritchie wrote:
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.
Three voters, each with a different first place preference among
three candidates. In FPTP,
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: ABC BCA
CABThen make it 5 ABC, 4 BCA, and 3 CAB, and watch the same
On 12/8/05, Scott Ritchie [EMAIL PROTECTED] wrote:
Then make it 5 ABC, 4 BCA, and 3 CAB, 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 software
I'm gonna take another stab at a method that uses the UI of Range Voting, but tabulates it in a way that makes more sense. I'll admit, I actually like the fact that Range Voting collects very rich information about voter preferences, moreso than ranked ballots. (Of course this is no good if people
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
Well, your example is not only a Condorcet cycle, but a pure 3-way tie in condorcet terms.It is effectively:ABCBCACABSo, 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.
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:
ABC
BCA
CAB
Then make it 5 ABC, 4 BCA, and 3 CAB, and watch the same thing
happen.
So, no condorcet method could resolve it
