[EM] IRV ballot pile count (proof of closed form)

[robert bristow-johnson]

On Feb 4, 2010, at 7:51 PM, Kathy Dopp wrote:

> The general formula for the number of possible rankings (for strict
> ordering, without allowing equal rankings) for N candidates when
> partial rankings are allowed and voters may rank up to R candidates
> (N=R if voters are allowed to rank all candidates) on a ballot is
> given on p. 6 of this doc:
>
> http://electionmathematics.org/ucvAnalysis/US/RCV-IRV/ 
> InstantRunoffVotingFlaws.pdf

the only issue, Kathy, is whether the lower limit is i=0 or i=1.  you  
have to defend your use of i=0 for the case illustrated below (using  
the rules in Burlington VT and Cambridge MA).

> In the US, R=3 in most IRV elections.


in Burlington it was 5 in both 2006 and 2009.  N was also 5 (not  
counting any write-in).

but Kathy, suppose N=R=3 and it's the "regular-old" IRV rules that do  
not require any minimum number of candidates ranked and do not allow  
ties.  to be clear, i need to also point out that only *relative*  
ranking is salient (at least in Burlington).  if a voter only ranks  
two candidates and mistakenly marks the ballot 1 and 3, the IRV  
tabulation software will close up the gaps and treat that precisely  
as if it was marked 1 and 2.

now, given those parameters, are you telling us that the 9 tallies  
shown on Warren's page:
  http://rangevoting.org/Burlington.html (i have very similar  
numbers, no more different than 4), are not sufficient to apply the  
IRV rules and resolve the election?

     1332  M>K>W
      767  M>W>K
      455  M
     2043  K>M>W
      371  K>W>M
      568  K
     1513  W>M>K
      495  W>K>M
     1289  W

is there any reason those 9 tallies could not have been summed from  
subtotals coming from all 7 wards of Burlington?

please tell us why those 9 piles are not enough, given the parameters  
stated above?


--

r b-j                  rbj at audioimagination.com

"Imagination is more important than knowledge."