On Feb 4, 2010, at 8:29 PM, Kathy Dopp wrote:
On Thu, Feb 4, 2010 at 8:18 PM, robert bristow-johnson
<[email protected]> wrote:
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).
False Robert. Starting the index at 0 or 1 is completely irrelevant.
so you're saying that
N-1 N-1
SUM{ N!/i! } = SUM{ N!/i! } ?
i=0 i=1
that the N!/0! term is equal to zero?
Any formula is easily adjusted to either initial index.
i ain't talking about any substitution of dummy variable, i, and
changing the limits.
You are obviously not a mathematician.
i guess not. just a Neanderthal electrical engineer who does signal
processing algs for a living.
in Burlington it was 5 in both 2006 and 2009. N was also 5 (not
counting
any write-in).
My formula gives the general case for R equals anything, as I said.
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.
So? What's your point?
the point is that a ballot marked with 1 and 3 goes on the same pile
as a "properly marked" ballot marked with the same two candidates as
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?
Obviously you did not read my email. I'll read and respond to yours
after you've tried to read and understand my points. Otherwise I am
not wasting my time responding to you.
no need to, but...
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?
... you're running away from the salient question. are those 9 piles
good enough to resolve the IRV election with 3 candidates or not?
was salient information lost when the M>K pile was combined with the
M>K>W pile, enough that could cause the IRV election (with the rules
above) to be decided differently?
--
r b-j [email protected]
"Imagination is more important than knowledge."
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