<div dir="auto"><div>A note concerning monotonicity testing - even when there is no possible pushover strategy for a voter/group of voters, it doesn't necessarily mean that a given method is monotone.</div><div dir="auto">In these examples, changing ABC to BAC makes C the winner, so the change doesn't make sense from the voter's point of view and will go under the radar of any strategy detector, I guess.</div><div dir="auto"><br></div><div dir="auto">Another thing is that there are some election scenarios which an impartial/spatial simulator might never notice. For the purposes of strategy/critetia testing it might be good to include, for example, a ballot generator that will produce random-size groups of voters, rather that drawing voters one-by-one like a standard impartial generator does.</div><div dir="auto"><br></div><div dir="auto"><br></div><div dir="auto"><br></div><div dir="auto"><div class="gmail_quote" dir="auto"><div dir="ltr" class="gmail_attr">sob., 5 sie 2023, 13:52 użytkownik Kristofer Munsterhjelm <<a href="mailto:km_elmet@t-online.de">km_elmet@t-online.de</a>> napisał:<br></div><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">On 8/5/23 05:49, Kevin Venzke wrote:<br>
> Hi Kristofer,<br>
> <br>
> It wasn't so easy, but regrettably I think I have a monotonicity counter-example:<br>
> <br>
> 408: B>C>A<br>
> 329: A>C>B<br>
> 126: C>A>B<br>
> 91: C>B>A<br>
> 43: A>B>C --> B>A>C<br>
> (total 997)<br>
> <br>
> For the first round, A and B votes both exceed 1/3rd (332.33) and so only C can be<br>
> eliminated.<br>
> The match-up A:B gives B a very slight win of 499 vs 498 for A. C can't score anything.<br>
> Scores: B 499, A 498, C 0.<br>
<br>
I can verify that the scores are B: 499 > A: 498 > C: 0.<br>
<br>
> Now change the 43 to B>A>C, theoretically helping B further.<br>
> First round totals become 329 A, 451 B, 217 C. So it is now allowed to eliminate A.<br>
> Both A and B fare worse against C than against each other and so prefer to score off of<br>
> eliminating C.<br>
> B improves its score to 542 while A's score is reduced to 455.<br>
> However, when A is eliminated, C can score 546 from their matchup with B.<br>
> New scores: C 546, B 542, A 455.<br>
<br>
And I can verify that the scores are C: 546 > B: 542 > A: 455.<br>
<br>
Well done. Well, I would rather have wanted it to be monotone, but it's <br>
better to know the truth! I guess that makes this "very low <br>
nonmonotonicity" rather than monotone - now I know how the IRVists feel <br>
when people complain about nonmonotonicity!<br>
<br>
Here's a minimal example produced by linear programming:<br>
<br>
1: A>B>C<br>
7: A>C>B<br>
8: B>A>C<br>
3: C>A>B<br>
4: C>B>A<br>
<br>
the scores are B: 12 > A: 11 > C: 0, then after changing ABC to BAC the<br>
scores become C: 14 > B: 13 > A: 10.<br>
<br>
Interestingly, for your example, fpA-fpC says that the correct ordering <br>
for the "before" election is C>B>A, whle Carey says B>A>C. My example, <br>
on the other hand, doesn't have this distinction... but it has a <br>
Condorcet cycle both before and after, thus showing that Smith//X won't <br>
solve the problem.<br>
<br>
Despite the example showing that X itself isn't monotone, I'm more <br>
confident now that (properly phrased) DMTBR is compatible with both <br>
monotonicity and Condorcet. Prior to method X, we only had the fpA-fpC <br>
generalizations, IFPP, and IRV; the first were only DMTCBR, while the <br>
latter two were clearly nonmonotone. I was worried that there might be <br>
an impossibility theorem of some kind proving that monotonicity would be <br>
forever out of our grasp for burial-resistant Condorcet methods.<br>
<br>
I can also use method X to find out just what kind of DMTBR should hold, <br>
and then build off that. I think I have another idea that could work, <br>
but it would be so incredibly ugly - basically "IRV with donations".<br>
<br>
Or we could try to find out why X comes so close to monotonicity, since <br>
it's the closest we've got so far. Doing so would require figuring out <br>
why max A>B ("max votes-for") is monotone, and why Smith//method X also <br>
seems to be (nearly) monotone, I think.<br>
<br>
> One thing I noticed is that modifying the quota rule allows you at one extreme to<br>
> implement IRV (i.e. by saying that only the candidate with the fewest votes can be<br>
> eliminated each round) and at the other extreme to implement "max votes-for wins" (by<br>
> imposing no quota requirement at all). While the latter is monotone, it doesn't satisfy<br>
> majority favorite.<br>
<br>
That's right; making the quota more loose (i.e. giving the method more <br>
candidates to choose eliminations from in a given round) doesn't seem to <br>
hurt monotonicity until you go past 1/3, but it does hurt strategy <br>
resistance. Going in the other direction is not strictly possible <br>
because if you're in an n-way tie, every candidate has exactly 1/n of <br>
the first preferences. So you would then need to also allow eliminating <br>
the lowest scorer no matter what. This would make it more like IRV and <br>
thus compromise its monotonicity (further).<br>
<br>
> It's interesting to consider whether any quota rule could at least<br>
> preserve monotonicity and add majority favorite. I'm thinking no, though.<br>
The weakest quota I can think of that will preserve majority is 1/2. <br>
Suppose A is voted first by a majority. Then A can never be eliminated, <br>
so for any other candidate B, it eventually ends up being A vs B, and <br>
since A is a majority favorite, A then wins. However, this is not <br>
strategy resistant; even a constant quota of 1/3 for everything but the <br>
final round (which is what I tried first) destroys strategy resistance.<br>
<br>
Furthermore, as mentioned above, there seems to be a strange <br>
relationship between the quota and the degree of nonmonotonicity - at <br>
least if "Other" is a good indicator. For a three-candidate election, <br>
1/3 is equivalent to "normal" method X, which we now know is (barely) <br>
nonmonotone. However, loosening the quota to 1/2 introduces more <br>
nonmonotonicity; then getting rid of the quota altogether gets us back <br>
into the monotone domain.<br>
<br>
E.g. with fixed quota 1/2, impartial culture, 5 candidates, 97 voters, <br>
7500 elections:<br>
<br>
Burial, no compromise: 218 0.0305793<br>
Compromise, no burial: 1138 0.15963<br>
Burial and compromise: 435 0.0610184<br>
Two-sided: 5297 0.743021<br>
Other coalition strats: 41 0.00575116<br>
==========================================<br>
Manipulable elections: 7129 1<br>
<br>
and with fixed quota 1/3:<br>
<br>
Burial, no compromise: 403 0.0558017<br>
Compromise, no burial: 1544 0.213791<br>
Burial and compromise: 86 0.0119081<br>
Two-sided: 5149 0.71296<br>
Other coalition strats: 0 0<br>
==========================================<br>
Manipulable elections: 7182 0.994461<br>
<br>
-km<br>
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</blockquote></div></div></div>