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 [email protected]
"Imagination is more important than knowledge."
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