>robert bristow-johnson
(please ignore my horrible editing in webmail.)
>hi,
>this is my first post to this list. i subscribed to it a while ago after some
>internet searching on
>issues regarding multi-candidate elections which i did after our recent
>mayoral race in Burlington
>Vermont (dun
On Oct 13, 2009, at 9:01 PM, robert bristow-johnson wrote:
On Oct 13, 2009, at 1:58 AM, Juho wrote:
Welcome to the list!
thanks.
On Oct 13, 2009, at 7:48 AM, robert bristow-johnson wrote:
it is also important to have a deterministic and monotonic measure
of voter support that is under
On Oct 13, 2009, at 1:58 AM, Juho wrote:
Welcome to the list!
thanks.
On Oct 13, 2009, at 7:48 AM, robert bristow-johnson wrote:
it is also important to have a deterministic and monotonic measure
of voter support that is understandable to the less scholarly
I have often promoted the me
Welcome to the list!
On Oct 13, 2009, at 7:48 AM, robert bristow-johnson wrote:
it is also important to have a deterministic and monotonic measure
of voter support that is understandable to the less scholarly
I have often promoted the measure of least additional votes required
to become a
hi,
this is my first post to this list. i subscribed to it a while ago
after some internet searching on issues regarding multi-candidate
elections which i did after our recent mayoral race in Burlington
Vermont (dunno if you heard about it or not).
On Oct 12, 2009, at 2:22 PM, Dave Ket
To look for a better sway to resolve cycles is worthy. However, I
still do not see the gain in what you offer, considering the expense.
Each of the members of a cycle would be winner if only one of them ran.
When they are close to a tie it matters little which wins.
When far away the
The additional complexity is to break cycles and come up with a complete
preference order. In one of the examples below,
5: A>B>C
3: B>C>A
4: C>A>B
the cycle was resolved as A>B>C.
I just wanted to show that it picked the Condorcet winner in the absence
of cycles. Like other Condorcet methods
What is the point to all this? Looks much more complex than I sketch
below, but I do not see value in the extra effort:
For 10 candidates, fill a 10x10 matrix - for A vs B need an entry
counting A>B and one for B>A.
With luck about 9 comparisons will identify winner since each
identifies
This method always selects the Condorcet winner if there are an odd
number of voters with no ties or circular ambiguities. *With respect to
a particular candidate,* the other candidates must form a single
uniquely-preferred rank order, with a single peak in ranking score at
the (composite) medi
Dear Michael,
very interesting, I don't think I saw anything like this before.
When trying do evaluate a new method, I always try to check very simple
criteria first, like neutrality and anonymity (obviously fulfilled
here), Pareto efficiency, monotonicity, etc. Concerning the latter two,
I was n
As usual with such posts, there is a good chance someone has come up
with the same (or very similar) method, but I thought it had interesting
properties, and was wondering what glaring voting paradoxes it had. In
addition, the number of possible orders is overwhelming if there are a
large numbe
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