On Jan 21, 2010, at 4:26 PM, Abd ul-Rahman Lomax wrote:
But ... it raises some security issues. And with central counting
there are other issues. This is a red herring, because we are
talking about precinct summability, and when the number of
candidates is very small, precinct summability isn't relevant,
because the raw ballot data may be transmitted.
no, the problem is that the raw ballot data may be the only practical
information to transmitt if the number of candidates is *large*, not
very small. when the number of candidates is very small, then it
makes sense to transmit the tallies for piles because the number of
piles, which are precinct summable, is manageable.
So, back to the real question: is precinct summability an important
practical criterion to be applied to voting systems?
i would ask instead if precinct summability is important for
security? i believe that it is. and i believe that it is perfectly
practical when the number of *credible* candidates is small. doesn't
matter what the voting system is. IRV, or whatever.
for 3 candidates, that number is 9.
Okay, three candidates, A, B, C, the ballot possibilities are, to
be complete, much more than 9. I'll assume that write-ins are
illegal and void the ballot. Some of the possibilities are legally
equivalent to others, and in actual IRV ballot imaging, they are
collapsed and reported the same, to the displeasure of voting
security people who do want to know the "error rate," which
includes overvoting and exact overvoting patterns. So-called ballot
images are not, generally. They are processed data reducing a
ballot to legally equivalent votes. The reduced set is this:
A
B
C
A>B
A>C
B>A
B>C
C>A
C>B
Note that this assumes a 2-rank ballot.
no, it can be a 3-rank ballot where the voter declines to rate their
last choice. "3rd choice" is left unmarked.
It also assumes that majority vote isn't important.
bullshit. it (the number of consequential ballot permutations) has
nothing to do with it (whether or not majority vote is important).
If it's important, as it would be in an IRV election under Robert's
Rules, we have some more possibilities. They are all the three-rank
permutations.
A>B>C
A>C>B
B>A>C
B>C>A
C>A>B
C>B>A
Each of these is equivalent, for the purposes of finding a
plurality winner, to a two-candidate combination.
it's equivalent for the purposes of IRV or Condorcet or *any* method
that relies solely on the relative rank of candidates. those 6
markings are equivalent to the corresponding 6 above.
if you or
Kathy say it's 15, then you're wrong (and it's your slip that's
showing).
Well, I won't speak about Kathy, but in terms of practical
elections in the U.S., she's right. You did not state enough
information to establish your reduced count, ...
yes i did state enough information. may i remind you? i said that
there is *no* consequential difference in these two marked ballots
(in the case of N=3). there is no consequential difference between a
ballot marked A>B to one marked A>B>C . there is no election
scenario, whether it's IRV, Condorcet, Borda or any other method
using ranked ballots that will count those two ballots differently.
there is no need to separate the A>B and A>B>C into two piles.
for N candidates, the number of piles necessary, P(N) is
N-1
P(N) = SUM{ N!/n! }
n=1
not
N-1
P(N) = SUM{ N!/n! }
n=0
which is appears to be the formula you and Kathy continue to insist
is correct.
Which it is under some conditions and yours is correct under some
conditions. I assume. I haven't checked them because it's more work
than I can put in now.
want me to spell it out. it's a simple application of combinatorial
analysis, what is the first chapter of my introductory probability
textbook (of a course i took more than 3 decades ago). you're doing
it already for the specific case of 3 candidates A, B, and C. if you
want to look it up, look for language that says something like: "how
many unique ways can a group of n items be selected from a pool of N
items when the order of selection is relevant?" and the answer to
that is N!/n! . but there is one more fact that you need to toss
in. and that fact is that all candidates unmarked or unranked are
tied for last place. if there is only one candidate left unmarked,
we know how all N candidates are ranked, including the unmarked
candidate.
<everything else between is deleted without comment>
A vote of A>B>C, is that the same as A>B? Robert assumes, yes. But
what about write-ins? A>B>C is equivalent to A>B>C>W.
that's not 3 candidates. that's four. you just changed the
premise. that's an official logical fallacy. a form of "straw man".
if the write-ins are insignificant (usually the case) we can sweep
them all into a single insignificant candidate and we have 4
candidates and 40 piles. but we'll see that even if all write-ins
(with different names) are lumped into a single person, that the
number of ballots with candidate W marked higher than last, will have
lower totals than any counted with W marked last. but that is an
*assumption* that the aggregate write-ins are insignificant (i've
voted for 32 years and have never seen a governmental election that
was otherwise). then we're back to 3 credible candidates and 9 piles
that need be reported for public consumption (so we can keep them
election officials honest).
not that i am a defender of IRV. but, you haven't laid a hand on it
regarding "precinct summability".
I didn't use my hand. I used my head.
and your head fails.
so, before pointing out that someone's slip is showing, it might be
safer to adjust where one's own fig leaf is hanging.
I threw the fig leaf away years ago. What you see is what you get.
Take it or leave it.
i am picking and choosing. lest i get sucked into "argumentum
verbosium" and i've wised up to that technique.
--
r b-j [email protected]
"Imagination is more important than knowledge."
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