[EM] number of possible ranked ballots given N candidates

[Paul Kislanko]

Rob LeGrand is correct. You can't derive the pairwise matrix solely from a
matrix that only records the counts for each rank. 

I think it is possible, though, to use the combination of pairwise matrix
and the counts-by-rank matrix together to retrieve the the original ballot
preferences. 

> -----Original Message-----
> From: election-methods-bounces at electorama.com 
> [mailto:election-methods-bounces at electorama.com] On Behalf Of 
> Rob LeGrand
> Sent: Wednesday, December 14, 2005 5:37 PM
> To: Election Methods Mailing List
> Subject: Re: [EM] number of possible ranked ballots given N candidates
> 
> Paul Kislanko wrote:
> > Yes. The pairwise matrix plus this "how many voters ranked 
> A at rank R"
> > could be combined to give more information. In fact, the pairwise
> > matrix can be derived as easily from the "how many voters 
> ranked A at
> > rank R", but not vice-versa.
> 
> I don't believe that's true.  The two ballot sets
> 
> 3:A>B>C
> 2:B>C>A
> 2:C>A>B
> 
> and
> 
> 2:A>B>C
> 1:A>C>B
> 1:B>A>C
> 1:B>C>A
> 1:C>A>B
> 1:C>B>A
> 
> result in different pairwise matrices (no Condorcet winner in 
> the first,
> A is CW in the second), but they both have the same rank profile:
> 
> A is first on 3 ballots, second on 2 ballots, third on 2 ballots
> B is first on 2 ballots, second on 3 ballots, third on 2 ballots
> C is first on 2 ballots, second on 2 ballots, third on 3 ballots
> 
> --
> Rob LeGrand, psephologist
> rob at approvalvoting.org
> Citizens for Approval Voting
> http://www.approvalvoting.org/
> 
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