I think the effect of what we said about ties is usually
the same. Critical to any method, though, would be to record an A=B>C as
having C in the third-place spot.
Things get kind of complicated when you get into 1/2 a 1st
place vote and 1/2 a 2nd place vote. A and B both get the same value, but is
that more or less than one of each compared to the other alternatives? When you
assign values to rankings, you're introducing Borda into the collection process.
I prefer 1,1,3 to be saved as 1,1,3 for A,B,C. 1.5,1.5,3
could have the collection method impact the results.
On 12/14/05, Paul Kislanko <[EMAIL PROTECTED]> wrote:
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of rob brown
Sent: Wednesday, December 14, 2005 5:12 PM
To: Paul Kislanko
Cc: [EMAIL PROTECTED]; Election Methods Mailing List
Subject: Re: [EM] number of possible ranked ballots given N candidatesOur messages overlapped, each trying to find a way to store more data without storing all ballots. :)Just a thought for your endeavor. A way to save "more" information than the pairwise matrix but less than saving counts of each ballot configuration is to save an NxN matrix with column headings being count of #1, #2, ... #N ranks and rows being the Alternatives. Use the rule:If equals are found, assign the next alternative rank (r+1) where where r is the rank the previous alternative listed would've had if there were no equals. I.e. ranks are assigned 1,1,3 for A=B>C.This takes no more space than the pairwise matrix, but contains much more information. See http://football.kislanko.com/2005/bucklincomps.html for an example with 119 alternatives and any number of voters.
Your suggestion seems like something you'd want to store in addtion to the matrix, no? It seems like it would be hard to perform the sort of calculations that the matrix works well for, i.e. picking a candidate in a way that doesn't encourage strategic voting and doesn't punish (or reward) candidates for having other candidates that appeal to the same constituency.
BTW, if it was me I would handle ties differently, for two candidates tied for first, I would probably want to store half a point for first place and half a point for second place for each candidate. In other words treat 2 A=B>C ballots as being functionally identical to 1 A>B>C ballot and 1 B>A>C ballot. But that's just me.
-rob
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