On Jan 16, 2010, at 3:00 PM, Kathy Dopp wrote:
On Sat, Jan 16, 2010 at 2:30 PM, robert bristow-johnson
<[email protected]> wrote:
I was talking about IRV voting. Where do you get "9" piles from?
it's 3!/0! + 3!/2! = 6 + 3
OK. If you prefer to write the formula that way, you're still
incorrect.
It is 3!/0! + 3!/1! + 3!/2! = 15 unique vote combinations in IRV, but
that is also *not* the same as the number of piles you'll need to sort
into to count IRV, which is less. I haven't and don't plan to, figure
out that formula but do know that the answer is less than 9 for three
candidates when counting IRV manually, so I am still uncertain what
your 9 relates to.
(9 would be the number of Condorcet tallies for 3 candidates,
no, that would be 6. for N candidates, i think there would be
Condorcet can always be counted by an n x n matrix where n is the
number of candidates. However you are correct that the diagonal has no
entries so 9 - 3 agrees with your six.
However, your fundamental formula below is incorrect for Condorcet and
for IRV and will not give correct answers for Condorcet except maybe
in the case of two and three candidates (your formula is also overly
complex and easily simplified but does not seem to apply to anything.
For instance, storing all the ballot choices for Condorcet can be done
for four candidates, as always, in a 4 x 4 matix with 4 diagonal
entries blank or used to store other useful items such as the number
of ballots cast, number of spoiled ballots, or whatever.
Again, I suggest you sit down and actually try to count some sample
ballots in either Condorcet or IRV. That would help anyone to go from
the theory to practice.
N-1
SUM{ N!/n! } - N!/1!
n=0
... piles if only relative ranking is salient.
Your formula would be correct for the number of tallies for IRV if you
delete the second expression that you subtract, but is not correct for
anything to do with Condorcet in general.
the second term of the summation (in the case of N=3, it's the
number of
permutations of ranking 2 candidates out of a pool of 3), counts a
superfluous permutation because when only one candidate is
unranked, it's
equivalent to ranking him last. but we have to account for the
case where 2
or more candidates are unranked (and tied for last).
Don't know what you're talking about.
consider Burlington 2009 with the inconsequential candidates Simpson
and "write-in" eliminated and very real (but otherwise last)
candidate Dan Smith eliminated. that least Wright, Montroll, and
Kiss. with only those three left, these are the pile counts of the
only salient permutations of marked ballots:
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
now, Kathy, ask yourself why there are no piles marked just M>K or
M>W or K>M or K>W or W>M or W>K? (those are the 6 piles you want to
enumerate in your 15.)
All your formulas are incorrect.
and, since you don't understand your opponent's argument, then your
evaluation of it is authoritative.
<sigh>
--
r b-j [email protected]
"Imagination is more important than knowledge."
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