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Here's an email regarding PAR:

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.

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