I derived those formulas a couple of years ago, and
verified them by hand through 5 alternatives. It was a major headache, and it
would not surprise me if I made a mistake.
Rating all candidates equally is the same as "not voting"
only when you apply a counting-side rule. I don't mind ballots being discarded
because they don't affect the outcome, but in approval those ballots would
contribute to all the candidates' percentages, so IN GENERAL, on the collection
side it is better to not make such assumptions.
(In my idealized model of stored ballots A=B=C>D=E=F is
how you'd record approval-style ballots).
I don't have a philosophical objection to using the
pairwise matrix on the counting side of the system at all. I just think when
comparing different Condorcet methods it would be helpful to have the original
voters' ballots, because when we get into hypotheticals like "if x B>A voters
had changed to B>C>A then..." we'd have the data on how likely that might
be.
On 12/14/05, Rob LeGrand <[EMAIL PROTECTED]> wrote:
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of rob brown
Sent: Wednesday, December 14, 2005 3:50 PM
To: [EMAIL PROTECTED]
Cc: Election Methods Mailing List
Subject: Re: [EM] number of possible ranked ballots given N candidatesAlthough i might not want to count a=b=c, which I would sorta consider "not voting"...Paul Kislanko wrote:
> The number of full ranked ballots is just the number of permutations of
> N alternatives = N!
>
> If equal ranknigs are allowed, it's N! + 2^N - 1
I assume you mean that there are N! ways to arrange the candidates
strictly, and then N - 1 spaces in between that could hold either a > or
a =, giving N! * 2^(N - 1). But I think that's an overestimate because
it would count both A=B>C and B=A>C. Did you mean something different?
I did a quick check with 3 candidates, and his formula appeared to give the correct answer (note that I eliminated redundant ones like your example):
a=b=c
a>b>c
a>b=c
a=b>c
a>c>b
a=c>b
b>a>c
b>a=c
b>c>a
b=c>a
c>a>b
c>a=b
c>b>a
which is 13, which agrees with http://www.google.com/search?q=3%21%2B2%5E3-1
Paul Kislanko wrote:
And double yes to I'm glad someone's working on a way to model the different methods from a universal ballot format. It should make it easier to see the differences between Condorcet-based methods.Cool. While I may not be 100% aligned with your philosophical objection to the matrix as an intermediate step, I am certainly interested in exploring some methods which don't use it. (I'm not trying to model the exisiting condorcet methods this way, but looking at other, new[?] methods that don't use the matrix)
Unfortunately with 10 candidates, there are about 4 million possible unique ballots, so this non-lossy compression scheme may be less helpful than I hoped
rob
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