[EM] number of possible ranked ballots given N candidates
rob brown
rob at karmatics.com
Wed Dec 14 13:50:24 PST 2005
On 12/14/05, Rob LeGrand <honky1998 at yahoo.com> wrote:
>
> 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
Although i might not want to count a=b=c, which I would sorta consider "not
voting"...
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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