[EM] Does IRV elect "majority winners?"
Kristofer Munsterhjelm
km-elmet at broadpark.no
Thu Jan 8 02:51:36 PST 2009
Dave Ketchum wrote:
> On Re: [EM] Does IRV elect "majority winners?" Kristofer Munsterhjelm
> wrote:
>> Dave Ketchum wrote:
>>
>>> Condorcet certainly costs more for the system than Plurality. Costs
>>> bullet-voters nothing - provides a service to whichever voters like
>>> to do more than bullet vote.
>>> Actually can be a service to candidates. Clinton and Obama had
>>> to try to kill their competitor's campaign for the Democrat
>>> nomination they could not share. A similar race in Condorcet would
>>> let them both get nominated and have a more civilized fight as to
>>> which should be ranked higher than the other on the ballot.
>>
>>
>> If people tend to bullet-vote, it may be the case that elections in
>> general suffer from vote-splitting - simply because if C splits into
>> C1 and C2, people either bullet-vote C1 or C2.
>
> In the Condorcet election described above, Clinton and Obama would have
> been ENCOURAGING voters to vote for both of them - that the voters
> should consider both C>O and O>C according to which they preferred.
>
> That is consistent with bullet voting in the many elections where that
> is appropriate.
Probably so, yes.
>> I would assume that if one does A = B > Y and A is eliminated, then
>> the ballot becomes B > Y next - "the ballots are transformed as if the
>> candidate in question never ran".
>
> Note that your words imply that A and B are each counted when neither
> has yet been eliminated - an advantage over the voter having to vote
> either A>B or B>A.
>
Yes, though if it's fractional, each vote only counts half.
>> Range reduces to Approval if enough people use strategy. I think that
>> any version of cardinal ratings should either be DSV or have some sort
>> of Condorcet analysis (like CWP does, or perhaps not that far). Those
>> are my opinions, though, and others (like Abd) may disagree.
>
> Agreed that voters can CHOOSE to express the same restricted thoughts in
> Range as offered by Approval - but Range includes abilities beyond
> Approval's limited ability (just as Condorcet voters can express
> Plurality's limited thought).
The point is that it's an advantage to the voters to express their
thoughts in this manner. It's kind of like if Condorcet weighted votes
by 1 / (number of ranks specified) so that it made sense to bullet-vote.
>>> I am still trying to promote series thought as to need for a majority
>>> for other than Plurality or Approval.
>>
>>
>> A worst-case point of view might be to consider the groups maximally
>> different. That is, nobody who voted A > B also voted A > C. From that
>> point of view, and a strict interpretation of "majority", one would
>> have to have the weakest victory be one of a majority - that is, for
>> the candidate X so that the magnitude of the win of A against X is
>> least, A must beat X by a majority.
>
> Let me offer bullet voting in Condorcet:
> 32 A
> 33 B
> 34 C
>
> C wins because, with 99 voters, C's 34 makes it CW.
That's true. I was talking about Condorcet majorities, though, and none
of those would have such a majority.
In general, if you have a voting method and everybody bullet-votes, then
you pretty much have to reduce to Plurality, since there's no other
information available.
>> This encompasses the standard majority setting where a majority votes
>> A > [everybody else]. It's not equal to it, as one may see from this
>> example:
>>
>> A > B > C > D
>> C > B > D > A
>> D > B > C > A
>> A > B > D > C
>>
>> B has a majority against C, D, and A.
>
> Huh? 2 A>B vs 2 B>A?
Oops, I calculated based on 50% as majority. It should be relatively
easy to fix though.
A > B > C > D
C > B > D > A
D > B > C > A
B > D > A > C
Now 3 B>A, 1 A>B. Also, 1 C>B, 1 D>B, so no other has 50%+1 over all others.
>> It's also a worst-case point of view because it errs "safe" in the
>> case of truncation - truncation so that A is not ranked on the ballot
>> means that no victory for A above some other candidate will be counted
>> for that ballot.
>
> Each ranked candidate counts toward beating each unranked candidate, as
> well as toward beating each lower ranked candidate!
Of course. What I meant was, if you have
1: B > A > C
2: B
the 2: B is the same as 2: B > A = C, which means that it doesn't
contribute to A's victory count. "Truncation so that A is not ranked on
the ballot means that no victory for A above some other candidate will
be counted for that ballot".
>> I don't know much about the cost of optical scanning machines, but
>> presumably getting one with 8 or 10 sensors shouldn't be that more
>> expensive than one with 3. They wouldn't have to be specialized,
>> either, since optical scanning is used for other things than counting
>> ballots.
>
> Eight sensors may be affordable. How much space on the paper ballot is
> required for the eight targets to be sensed - for you do this for each
> candidate for each race.
Something like http://www.indybay.org/uploads/2007/10/22/ballot2007.gif
seems quite reasonable, yet that has 25 ranks - so I don't foresee 8
being a problem.
>> The ideal solution as far as granularity is concerned would be to have
>> a machine that does OCR, and where voters just write a number in a box
>> next to the candidate (1 for first place, 20 for twentieth). That
>> would be quite a bit more expensive, though, and would also need some
>> sort of fallback... or just manual counting.
>
> Tempting - needs more thought.
I'd imagine it would be somewhat like this: You have an OCR mechanism
that's been tested on numbers written in various styles. It either
returns "this is [number]" or "I don't know what it is", with very
stringent requirements on the number being correct if it gives the first
output. If it returns "I don't know what it is", send the ballot to a
manual fallback to check what number it really is.
This won't work if the OCR's so bad it returns "I don't know what it is"
more often than it returns an actual result, or if the manufacturers
don't bother to follow the accuracy requirements and states don't bother
to verify them (Diebold all over again). Also, I'd prefer the counting
function to be isolated in hardware so that all it gets is the number
(so that there's no way it can cheat).
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