# [EM] number of possible ranked ballots given N candidates

Paul Kislanko kislanko at airmail.net
Wed Dec 14 15:24:11 PST 2005

```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.

_____

From: rjbrown at gmail.com [mailto:rjbrown at gmail.com] On Behalf Of rob brown
Sent: Wednesday, December 14, 2005 5:12 PM
To: Paul Kislanko
Cc: honky98 at aggienetwork.com; Election Methods Mailing List
Subject: Re: [EM] number of possible ranked ballots given N candidates

On 12/14/05, Paul Kislanko <kislanko at airmail.net> wrote:

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
<http://football.kislanko.com/2005/bucklincomps.html>  for an example with
119 alternatives and any number of voters.

Our messages overlapped, each trying to find a way to store more data
without storing all ballots. :)

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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