[EM] I'm liking PAR

[Jameson Quinn]

Still trying to find a PAR-like system which meets FBC. It's not hard to
find a way to eliminate B or C in the scenario I gave earlier (where A is
the center-squeeze CW, and yet B and C both avoid elimination in PAR). But
such "solutions" involve IRV-like and/or Condorcet-like dynamics, and that
means that turkey-raising becomes an issue, so FBC breaks, even though the
scenarios that break it can essentially be made arbitrarily implausible.

So, what's the simplest possible way to fix the given scenario, without
actually fully rescuing FBC? Allow candidate B to voluntarily
self-eliminate before step 3, if there are more than two non-eliminated
candidates after step 2. At that point, all the relevant tallies in the
case where they don't self-eliminate should be public.

This rule could in theory lead to a chicken dilemma between candidates, but
that's very unlikely. In general, either the threat candidate C will be
eliminated, or only one of A or B will be able to beat C. In order for the
chicken dilemma to happen, there'd have to be just the right amount of C
non-rejection from both A and B camps; enough to combine to keep C alive,
but not enough from either side to make the opposite camp not beat C.

Grrrr.... the boundary zone between FBC, chicken dilemma, and
Condorcet/center squeeze is very, very annoying. From pushing as hard as I
have, I suspect there would be an impossibility theorem here, if you
defined my "chicken dilemma non-slippery-slopeness" rigorously enough.

I think that it's possible to make a method which navigates this minefield
in practice. But it's not possible to make one which provably does.

So, between PAR, and PAR-with-voluntary-withdrawals, which do people like
better? Neither meets FBC in theory, though both are close in practice.
PARWVW is closer, but PAR is simpler.


2016-11-02 10:49 GMT-04:00 Jameson Quinn <jameson.quinn at gmail.com>:

> I said earlier that I couldn't think of a realistic scenario where PAR
> fails to choose the CW. I've now thought of one:
>
> 22: A>B
> 4: A>C
> 25: B>A
> 49: C>AB (or C>anything including at least 3% which accept C, and at least
> as many C>A as C>B)
>
> This can be thought of as a variety of center squeeze, with A as the
> center. (Sorry, I know that the convention is to use B as the center, but I
> don't want to rewrite this whole email.)
>
> The B>A voters did not account for the second preference of the A>C
> voters, so, unexpectedly to them, neither B nor C is eliminated. If C had
> been eliminated, A would not have needed the B>A ballots to win; and if B
> had been, the B>A ballots would have rolled over to A. But since neither of
> these things happened, C wins, instead of the CW A.
>
> In order to ensure A can beat C, the B voters would have to
> almost-unanimously top-rate A. But that would mean that B couldn't win if
> the B faction happened to outweigh the A factions.
>
> I find this scenario plausible, but still not very likely. I think that in
> most cases, either B would get less than 25% preferences and be eliminated;
> or, if A and B are both comfortably over 25% preferred, C would get over
> 50% rejection and be eliminated.
>
> I guess that a slightly more-plausible version of the scenario is:
>
>
> 22: A>B
> 4: A>C
> 5: BA
> 20: B>A
> 49: C>AB (or C>anything including at least 3% which accept B, and at least
> as many C>A as C>B)
>
> In this case, C wins, but the 5 BA voters could elect A by voting A>B.
>
> Whoops! That's an FBC violation! Note that it's not a violation if all the
> voters who honestly prefer B>A can strategize as a bloc, but it is a
> violation for any individuals in that group if they know that the rest of
> the group will be using sub-optimal strategy.
>
> So in general, PAR violates FBC in a center-squeeze scenario in situations
> where the Condorcet loser is not majority-rejected. In a situation where
> the honest preferences are roughly as in the second scenario above, there
> are several ways that the CW could still win:
>
>    1. The A voters largely reject B (defensive truncation)
>    2. The B voters almost unanimously prefer A (defensive compromise)
>    3. A few B or AB voters say A>B (defensive betrayal or truncation)
>    4. The C voters almost unanimously reject B (strategically suboptimal
>    over-truncation)
>    5. A few B voters say ∅>AB (defensive, um, I don't know what that is
>    called. "Denormalization"? "Weakening"? Technically, this could be seen as
>    restoring FBC compliance, but that's a stretch. I'd call that "semi-FBC" at
>    best.)
>    6. Combination of 1 and 4 above
>    7. Combination of 3 and 5 above (although since either one requires
>    relatively few voters, it's unlikely that both would be needed.)
>
> In general, I still think that PAR does exceptionally well with naive
> ballots, because I think that cases where the problem arises, but none of
> the above solutions happens naturally, would be rare. But hmmm... failing
> FBC... I recognize that that looks bad.
>
> Is there a way to fix this? I guess you could run PAR's step 3 as
> IRV-style successive eliminations. In that case, B would be eliminated
> first, and the votes would go to A, so C would lose. But.... I suspect it's
> still possible, although massively unlikely in practice, to make a
> pathology where the elimination would go in the wrong order, so that the
> resulting system fails FBC.
>
> Oh... and I guess that solution 5 above doesn't actually give even
> "semi-FBC", because in theory the C voters could counter it by voting
> CB>... in exactly the right proportions (offensive turkey-raising). I don't
> think this would ever work in practice because it requires knowing exactly
> how many of another faction will strategize AND sophisticated
> within-faction vote-management, but it still blows the criterion.
>
> Anybody have any thoughts on any of this?
>
>
> 2016-11-02 9:03 GMT-04:00 Jameson Quinn <jameson.quinn at gmail.com>:
>
>> I made the electowiki page for PAR voting
>> <http://wiki.electorama.com/wiki/Prefer_Accept_Reject_voting>, and I'm
>> liking the system more and more. In the Tennessee voting example, Nashville
>> wins in a strong equilibrium; as far as I can recall, this is the only
>> non-Condorcet system where that's true.
>>
>> Compared to MAS, it loses later-no-help, IIA (though it still has LIIA),
>> and summability (though it has two-pass O(N) summability). It may gain
>> majority voted Condorcet loser, though I don't yet have a proof for that.
>>
>> In practice, I expect that it would be rare to have elections with other
>> than exactly 1 or 2 non-eliminated candidates. In such elections, this
>> system elects the voted pairwise winner of the non-eliminated candidates.
>> In fact, I can't think of a single realistic scenario where the most
>> obvious naive votes don't elect an existing CW, including nonmajority CWs.
>> In particular, it handles both CD and center squeeze using naive votes.
>>
>
>
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