[EM] Just to let you know...

robert bristow-johnson rbj at audioimagination.com
Fri Jan 13 21:21:28 PST 2023

Kristofer is close.  Here is my spin on it.

Markus, I think that, even if BTR-IRV is the method that ultimately decides a state-wide RCV election, it makes sense for precincts to publish pairwise defeat subtotals (the defeat matrix, but I have to say that I think that the presentation format as a matrix is kinda dumb, even if that's the practice I see in voting methods lit).  If N is the number of candidates that's N(N-1) tallies to publish.  Now, already with RCV elections, we publish the tallies of first-choice votes.  So if you publish both, then the number of tallies reported at each precinct is N(N-1)+N = N^2.  Five candidates, that's 25 numbers.  If it's Hare RCV, it would have to be 205 numbers, clearly impractical to be printed, posted, and summed.

Now, in the U.S., we have had something more than 500 RCV elections that have been analyzed by FairVote and others, and of those elections, all but three have elected the Condorcet winner.  Two of the three RCV elections had a Condorcet winner and Hare failed to elect that candidate.  One of the three (2021 Minneapolis Ward 2) had no Condorcet winner.  So for 99.8% of the time, there is a CW and it doesn't matter what Condorcet method we use, these N(N-1) tallies suffice to tell us who wins.  Perhaps, for the 0.2% of the time there is no CW, we will be forced to learn of the outcome of the election from the seat of government because summing the pairwise defeat values will show there is a cycle.

But both Hare IRV and BTR elect someone whether there is a CW or not.  Now, suppose the Smith set is 3 candidates.  Rock-Paper-Scissors.  And let's say that the semifinal round has these 3 candidates.  Without loss of generality, let's label "Rock" as the candidate with the most votes in the semifinal round.

The order of candidate votes (not pairwise defeat) could be either Rock>Paper>Scissors or Rock>Scissors>Paper.  With Hare RCV, in the first case Paper wins and in the second case Rock wins.  But with BTR, Paper and Scissors have their runoff, Scissors wins, advances to the final round, and gets defeated by Rock.

So, for that 0.2% of elections that have a cycle, how often do you think that the cycle will be any more complex than Rock-Paper-Scissors?  With BTR-IRV, we know that the result will be the same as Condorcet-Plurality even *more* often than the 99.8%  Only in the super-rare case that there is a cycle and it's nastier than a Smith triplet will those N^2 tallies published at the precincts be insufficient to tell us who won the election.

If the election has a CW and the CW is elected, then if any loser is removed, the winner remains the same.  Then there is no spoiler and precinct subtotals suffice to tell us who wins on election night.  If there is no CW, this will happen only rarely (ca. 0.2%), and if it's Condorcet-plurality, the N^2 tallies still suffice to tell us who wins on election night.  If there is no CW and it's BTR-IRV, the N^2 tallies suffice to tell us who wins on election night if the Smith set is 3 candidates.  In the super-rare case that there is a Smith set bigger than 3, then we have to go to the seat of government and learn the outcome of the election in (given the track record of Alaska, Maine, or NYC) about 2 weeks and we do not have the redundant double-checking of the outcome that Precinct Summability would otherwise afford.

At least, that's how I'm gonna sell the product.

I'm not enamored by BTR.  I think Schulze or Tideman based on margins would be better.  But it's really hard to write self-contained legislation describing either procedures.  And if the language is too complicated, it's hard to sell to legislators who are suspicious of changing our voting method anyway.  Eric Maskin suggested to me that simple Condorcet with a completion method if no CW might be the best language.  So I also wrote templates for Condorcet-plurality, because it's simple.  The issue is, how do we tell the public the rationale for how the winner was selected in the case of no CW?

Condorcet-plurality is pretty simple to explain.  BTR is a little harder but, in comparison to IRV, not too bad to explain.  Ranked-pairs is harder yet, and my stab at legislative language is much wordier.  And I can't write legislative language for Beatpath.  Markus, might you post a concise and understandable description in common English usage that can be written into law?  I dunno how to do it.

So that's my spin on it.


> On 01/13/2023 4:16 PM EST Kristofer Munsterhjelm <km_elmet at t-online.de> wrote:
> On 13.01.2023 20:38, Markus Schulze wrote:
> > Dear Robert,
> > 
> > in your paper, you criticize IRV for violating
> > precinct summability and then you propose
> > bottom-two-runoff as an alternative. This
> > doesn't make much sense.
> The impression I got from reading Robert's paper is that he goes:
> "Here are some properties that Condorcet methods have that IRV doesn't. 
> Among these are summability. IRV advocates don't seem to care much about 
> summability, so I'll play by their game; I'll propose a variant of IRV 
> that patches its greatest flaw."
> I.e. that BTR is a way to fix IRV, but with the implication that better 
> methods exist if the voters want summability. He doesn't go into what 
> those methods are because it's beside the point.
> I agree that it could be more clearly stated, though. And as a Condorcet 
> nitpicker, I would probably also say that the paper could mention the 
> theoretical existence of Condorcet cycles in passing, so that the 
> IRVists can't say "hey, the property he desires, that if A beats B 
> pairwise then B can't be elected... it sometimes is impossible to pass! 
> So avert your eyes from this propaganda and support IRV instead!".
> -km
> ----
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r b-j . _ . _ . _ . _ rbj at audioimagination.com

"Imagination is more important than knowledge."


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