[EM] Did someone not hear what I said about Approval vs Condorcet?

Mon May 27 01:42:16 PDT 2024

No, but some humans are. Count-fraud has been demonstrated to have
occurred, in 2000 & 2004.

On Mon, May 27, 2024 at 01:39 Chris Benham <cbenhamau at yahoo.com.au> wrote:

> Are your computers really necessarily so untrustworthy?
>
> I think counting Hare would be somewhat easier. Probably very weak
> candidates could be eliminated in batches (while ensuring the correct
> one-at-a-time elimination result). One possible kludgey solution could be
> to apply the Condorcet method to the Hare last X candidates.  X could be
> say 7.
>
> Something I've been meaning to point out: the Approval Sorted Margins
> method I promoted in the poll but the other voters here either ignored or
> claimed they couldn't understand  (and then failed to give it even majority
> approval) has the big practical advantage over
> other Condorcet methods that it normally doesn't use (and therefore
> doesn't have to know) the full pairwise matrix.  The more candidates there
> are I should say the greater would this advantage be.
>
> This is because (as you all recall) it only needs to check if adjacent
> pairs of candidates in the Approval-score order are in "pairwise harmony"
> with that order.  Probably most of the time most of them will be.
>
> Chris B.
>
> On 27/05/2024 4:28 pm, Michael Ossipoff wrote:
>
> …or of course you could, for the voting, just ask each voter to write on
> hir (big) ballot, which member of each of the 18090/2 candidate-pairs s/he
> prefers to the other. S/he might be in the voting/booth for a while…
>
> On Sun, May 26, 2024 at 23:51 Michael Ossipoff <email9648742 at gmail.com>
> wrote:
>
>> Suppose that there are 300 million voters. Round 18090 off to 20000x
>>
>> 3E8 X 2E4 = 6E12
>>
>> 6 trillion.
>>
>> That’s roughly the number of miles in a light-year.
>>
>> Whatever the number of voters, nationally, is, suppose that every one of
>> them participated in the handcount.
>>
>> With the work divided among them all, each has “only” an amount of work
>> equal to that of determining which member of each one of 18090/2 pairs of
>> candidates is ranked over the other on a ballot.
>>
>> 135 candidates in each ranking. On the average, s/he’d only have to look
>> at half of the ranking’s candidates to find each of the 2 members of each
>> pair.
>>
>> …but s/he’d have to do it for both. So the order-determination for each
>> candidate-pair, on the average will require looking at 135 candidates.
>>
>> Suppose that s/he can skim over 10 of them in a second, when doing those
>> searches.
>>
>> Then it would take (135)(18090/2)/10 seconds. That’s about 1.4 days. But
>> say she only does it for 8 hours per day (with no breaks). Now it’s more
>> like 4.2 days.
>>
>> But supervision is the whole point. If s/he’s working alone, she can say
>> that the pairwise vote-totals are whatever s/he wants them to be.
>>
>> So in reality, it would be counting *teams* each wit representatives of
>> several parties. Say (optimistically) there are 10 parties.
>>
>> Now it will take 42 days. But the parties don’t really have equal numbers
>> of members. It’s going to take longer.
>>
>> …&, realistically, they aren’t going to scan 10 ranked candidates every
>> second for 8 hours with no breaks.
>>
>> It would obviously take months. Might it not, in fact, be measured in
>> years…with every one of the nation’s voters participating in that handcount?
>>
>>
>>
>> On Sun, May 26, 2024 at 22:11 Michael Ossipoff <email9648742 at gmail.com>
>> wrote:
>>
>>> …& of course multiply that 18090 by the number of ballots, to get an
>>> idea of what’s involved in the 18090 order-determinations to be done on
>>> each ballot, & recorded, &  then summed, to obtain each of the 18090
>>> pairwise vote-totals.
>>>
>>> …each of which then must be carried or transmitted to where the central
>>> count is done.
>>>
>>> On Sun, May 26, 2024 at 22:03 Michael Ossipoff <email9648742 at gmail.com>
>>> wrote:
>>>
>>>>
>>>>
>>>> On Sun, May 26, 2024 at 12:56 Chris Benham <cbenhamau at yahoo.com.au>
>>>> wrote:
>>>>
>>>>>
>>>>> In 2003  there was California  gubernatorial election with 135
>>>>> candidates.
>>>>
>>>>
>>>> That’s 18090 pairwise vote-totals to determine at each precinct from
>>>> the rankings, by examining each ranking to determine which member of each
>>>> possible candidate-pair is ranked over the other on that ballot.
>>>>
>>>> …& 18080 pairwise vote-totals for the precincts to sum, store, &
>>>> transmit or carry to the central count location.
>>>>
>>>> …& verify in an audit.
>>>>
>>>>
>>>>
>>>>>
>>>>>
>>>>> https://en.wikipedia.org/wiki/2003_California_gubernatorial_recall_election#Results
>>>>>
>>>>> Chris B.
>>>>>
>>>>> On 26/05/2024 9:29 pm, Kristofer Munsterhjelm wrote:
>>>>> > On 2024-05-26 07:28, Michael Ossipoff wrote:
>>>>> >> Someone keeps repeating that the voters shouldn’t have to vote
>>>>> >> strategically. He wants the method to do it all for us, after we
>>>>> >> merely state our sincere-rankings.
>>>>> >>
>>>>> >> That’s of course a common attitude:
>>>>> >>
>>>>> >> …wanting a high-tech, computation-intensive,computer-dependent
>>>>> system
>>>>> >> to do it all for us, taking all the actual choosing responsibility
>>>>> >> off of us.…sheltering & isolating us from the choice.
>>>>> >
>>>>> > I would prefer that you do not attribute opinions that the
>>>>> proponents
>>>>> > have not expressed. Nowhere have I said that Condorcet "[isolates]
>>>>> us
>>>>> > from the choice" we make.
>>>>> >
>>>>> > What you call "sheltering" and "isolating", I see as the method
>>>>> taking
>>>>> > proper responsibility - proper responsibility to turn the voters'
>>>>> > unambiguous honest opinions into an outcome without dumping the
>>>>> > algorithmic calculation upon te voter themselves. See my "kick the
>>>>> can
>>>>> > down the road" post for more info.
>>>>> >
>>>>> > Now, I could make a caricature of Approval itself. Perhaps something
>>>>> > about a calculator that just says "IDK, do the base conversion
>>>>> > yourself, I only accept input numbers in factoradic". But
>>>>> caricatures
>>>>> > only make people angry. Let's not stoop to them, shall we?
>>>>> >
>>>>> >> I’ll ask this for the 3^rd time:
>>>>> >>
>>>>> >> …
>>>>> >>
>>>>> >> How would like you to handount-audit a Condorcet count for a
>>>>> >> many-candidate national presidential election?
>>>>> >
>>>>> > I think your question assumes something that won't hold. If you have
>>>>> a
>>>>> > 25-candidate presidential election, you've already lost, because
>>>>> > nobody is going to rank 25 candidates, irrespective of whether the
>>>>> > method is Condorcet, IRV, or Borda.
>>>>> >
>>>>> > I'm not familiar with minor parties in the US. Has there ever been a
>>>>> > 25-candidate presidential election?
>>>>> >
>>>>> > I don't think I can comment beyond that: I don't know enough about
>>>>> > poll workers. I'll leave that to someone with experience. (Although
>>>>> > roughly calculating: suppose 6 candidates like in Burlington. That's
>>>>> > 30 pairs. Five times the work if counting a particular preference is
>>>>> > as before. So you'd either need 5x the workers, or five times the
>>>>> > time, or some combination of the two.)
>>>>> >
>>>>> > On an aside, though, I would say that I generally wouldn't want to
>>>>> get
>>>>> > computers anywhere near election counting. However, if you
>>>>> absolutely
>>>>> > have to have them, there are ways of making sure they don't cheat:
>>>>> > formal verification. You could also create special-purpose tools
>>>>> that
>>>>> > say, only turn ranks into matrices and nothing else: there's no
>>>>> reason
>>>>> > (apart from programmer convenience) why an election tool should be a
>>>>> > general purpose computer that you could hide all sorts of
>>>>> shenanigans in.
>>>>> >
>>>>> >> I’ve discussed that at length in previous posts, & it probably
>>>>> isn’t
>>>>> >> necessary to again post about ways of choosing how to vote in
>>>>> Approval.
>>>>> >>
>>>>> >> But, just summarize:It’s easy.Whichever of the various ways you
>>>>> >> prefer to use, for choosing whom to approve, it’s easy.…& no, it
>>>>> >> doesn’t require knowing your objectively-optimal vote.
>>>>> >
>>>>> >>
>>>>> >> I’ve many times pointed out that Approval’s Myerson-Weber
>>>>> equilibrium
>>>>> >> is the voter-median.
>>>>> >>
>>>>> >> i.e. Approval soon homes in on where the Condorcet-Winner is.
>>>>> >
>>>>> > Not necessarily. See the following paper:
>>>>> > https://arxiv.org/abs/2001.04216v2
>>>>> >
>>>>> > According to the authors, under their model, if every voter follows
>>>>> a
>>>>> > particular thresholding rule, then iterative approval generally
>>>>> > arrives at the Condorcet winner (but not always). However, if voters
>>>>> > are left to choose any thresholding rule they want (as you propose),
>>>>> > then anything is possible. The arrangement might even elect a
>>>>> > Condorcet loser, and the outcome may be slow to converge or who wins
>>>>> > may depend on how many polls you hold.
>>>>> >
>>>>> > I'm reminded of a quote about distributed algorithms that I read
>>>>> > somewhere: "It's really easy to design distributed algorithms that
>>>>> > suffer from deadlock, network floods or widely unpredictable and
>>>>> > bizarre oscillations". That's in the context of computer science -
>>>>> > deadlocks might not be applicable to election methods. But it does
>>>>> > justify a starting position of skepticisim when considering schemes
>>>>> > that offload more of the work to the voters by turning a one-shot
>>>>> > method into a dynamical system.
>>>>> >
>>>>> >> It seems to me that, in every one of EM’s polls, including the
>>>>> recent
>>>>> >> one, Approval chose the CW.
>>>>> >>
>>>>> >> Have we forgotten that?
>>>>> >
>>>>> > I haven't, nor have I forgotten that Approval wasn't actually the
>>>>> poll
>>>>> > winner.
>>>>> >
>>>>> > But let's take this reasoning at face value. I don't have the other
>>>>> > polls' ballot data available at the moment, so let's consider the
>>>>> > latest one and pick... say, Borda.
>>>>> >
>>>>> > As one can see by going to
>>>>> > https://munsterhjelm.no/km/rbvote/calc.html, pasting in the data,
>>>>> and
>>>>> > clicking Borda, Ranked Pairs is also the Borda winner.
>>>>> >
>>>>> > But I don't think I'm going to start advocating for Borda.
>>>>> >
>>>>> >
>>>>> > To not be accused of tu quoque, let me clarify the point. As
>>>>> > Burlington shows, an election method needs to handle the hard cases,
>>>>> > not just the easy ones, or there may be a considerable uproar when
>>>>> the
>>>>> > method is faced with a hard case and then stumbles. (In addition,
>>>>> > people trying to make sure a stumble doesn't happen may start to do
>>>>> > mass compromising, further entrenching two-party rule.)
>>>>> >
>>>>> > So it's quite possible that our polls are easy cases. But like
>>>>> > FairVote claiming that IRV gets the Condorcet winner more than 90%
>>>>> of
>>>>> > the time, that doesn't by itself tell us much, because the failures
>>>>> > have such a strong impact.
>>>>> >
>>>>> > I illustrate the example above by picking a method we know to be bad
>>>>> > (Borda being so extremely easy to fool with cloning and burial), and
>>>>> > showing that the poll result comes out right. If a bad method can
>>>>> get
>>>>> > a good result, then "getting a good result" is less useful than it
>>>>> > might appear at first glance.
>>>>> >
>>>>> > -km
>>>>> > ----
>>>>> > Election-Methods mailing list - see https://electorama.com/em for
>>>>> list
>>>>> > info
>>>>>
>>>>
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