[EM] number of possible ranked ballots given N candidates

[Paul Kislanko]

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.


  _____  

From: election-methods-bounces at electorama.com
[mailto:election-methods-bounces at electorama.com] On Behalf Of rob brown
Sent: Wednesday, December 14, 2005 3:50 PM
To: honky98 at aggienetwork.com
Cc: Election Methods Mailing List
Subject: Re: [EM] number of possible ranked ballots given N candidates


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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