First of all, I call it pairwise-count, because "Condorcet" properly
applies to Condorcet's own proposals for solving circular ties.
IRV-completed pairwise-count has been proposed many times. Here's an
example:
3 candidates: A, B, & C.
Sincere rankings:
40: ABC
25: BAC
35: CBA
Voted ranking
Alex Small wrote:
> ...
> I'm curious if any work has been done comparing the Condorcet efficiencies
> of Approval and IRV. It's been a few months since I looked at Brams and
> Fishburn, and I don't have a copy handy, so I don't know if they compared
> the two. When my copy arrives (ordered it
Donald has claimed that using the Condorcet criterion is inherently
prejudicial, since only methods in the Condorcet family can comply.
However, there's also a concept called "Condorcet Efficiency": Quantify
the likelihood that a given non-Condorcet method will elect the Condorcet
winner. We ca
Adam-
I like your geometric proof. However, I'm rusty on geometry. For a given
triangle, will the perpendicular bisectors of its 3 sides always meet at a
single point?
Also, when working in issue space, I assume you're measuring the distance
between two points (x1, y1) and (x2, y2) as |x1-x2|
Alex wrote:
> If all candidates fit on a one-dimensional spectrum then
> this is certainly true. The person whose preference
> order is Bush>Nader>Buchanan>Gore needs to start reading
> some newspapers. The person whose preference order is
> Buchanan>Gore>Nader>Bush probably lives in Flor
