In the example profiles below 100 = P+Q+R, and 50>P>Q>R>0. One consequence of these constraints is that in all three profiles below the cycle A>B>C>A will obtain. I am interested in simple methods that always ... (1) elect candidate A given the following profile: P: A Q: B>>C R: C, and (2) elect candidate C given P: A Q: B>C>> R: C, and (3) elect candidate B given P: A Q: B>>C (or B>C) R: C>>B. (or C>B) I have two such methods in mind, and I'll tell you one of them below, but I don't want to prejudice your creative efforts with too many ideas. Here's the rationale for the requirements: Condition (1) is needed so that when the sincere preferences are P: A Q: B>C R: C>B, the B faction (by merely disapproving C without truncation) can defend itself against a "chicken" attack (truncation of B) from the C faction. Condition (3) is needed so that when the C faction realizes that the game of Chicken is not going to work for them, the sincere CW is elected. Condition (2) is needed so that when sincere preferences are P: A>C Q: B>C R: C>A, then the C faction (by proactively truncating A) can defend the CW against the A faction's potential truncation attack. Like I said, I have a couple of fairly simple methods in mind. The most obvious one is Smith\\Approval where the voters have control over their own approval cutoffs (as opposed to implicit approval) with default approval as top rank only.The other method I have in mind is not quite as simple, but it has the added advantage of satisfying the FBC, while almost always electing from Smith.
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In the example profiles below 100 = P+Q+R, and 50>P>Q>R>0. One consequence of these constraints is that in all three profiles below the cycle A>B>C>A will obtain.
Forest, In your profile (3), isn't B simply the Condorcet winner? (And so there is no "cycle A>B>C>A") You presumably have in mind that the ballot will allow voters indicate an approval threshold in their rankings. In that case one method fills the bill is good old Approval Sorted Margins: https://wiki.electorama.com/wiki/Approval_Sorted_Margins I think that method is somewhat better
at resisting Burial than Smith//Approval(explicit), which in this
April 2002 EM post by Adam Tarr is called
"Approval-Completed Condorcet": http://lists.electorama.com/pipermail/election-methods-electorama.com//2002-April/073341.html The following are the sincere preferences of my example electorate: 49: Bush>Gore>Nader 12: Gore>Bush>Nader 12: Gore>Nader>Bush 27: Nader>Gore>Bush Say that some of the Gore>Bush>Nader voters were extremely non-strategic and decided to approve both Bush and Gore. So the votes now look like: 49: Bush>>Nader>Gore 6: Gore>Bush>>Nader 6: Gore>>Bush>Nader 6: Gore>>Nader>Bush 6: Gore>Nader>>Bush 27: Nader>Gore>>Bush Now, Bush wins the approval runoff 55-51-33. This is where ACC's favorite betrayal scenario comes in. Since Bush wins the approval vote, the only way the majority can guarantee a Gore win is to make Gore the initial Condorcet winner, which requires that the Nader camp vote Gore in first place. Where Smith//explicit Approval fails, Approval Sorted Margins easily elects the sincere Condorcet winner. Gore's approval score is 51 and Nader's is 33. Both adjacent pairs (B-N and N-B) are pairwise out of order. The gap between 55 and 51 is (much) smaller than that between 51 and 33, so we flip the order of that pair to give the final order N>B>G which has no adjacent pair out of order pairwise. Also giving the same result would be to use Approval(explicit) Margins as the measure of defeat strength in a traditional Condorcet method like Schulze or Ranked Pairs or Smith//MinMax. Chris Benham On 31/05/2019 8:03 am, Forest Simmons wrote:
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On 31/05/2019 00.33, Forest Simmons wrote:
> In the example profiles below 100 = P+Q+R, and 50>P>Q>R>0. One > consequence of these constraints is that in all three profiles below the > cycle A>B>C>A will obtain. > > I am interested in simple methods that always ... > > (1) elect candidate A given the following profile: [snip] You might be able to do something with my (three-candidate) fpA-fpC method, since it elects A in the situation where: P: A Q: B>C R: C since you have an ABCA cycle. In a 3-cycle, fpA-fpC lets each candidate's score be the number of first preferences for that candidate, minus the number of first preferences for whoever beats him pairwise. Highest score wins. Thus the scores become: A: fpA - fpC = P - R B: fpB - fpA = Q - P C: fpC - fpB = R - Q Since P > Q > R, P - R > 0, but Q-P and R-Q < 0, so A wins. Extending this method to four candidates so it meets Smith and still both resists strategy and passes mono-raise is hard, and is one of the things I'm working on (on and off) at the moment. ---- Election-Methods mailing list - see https://electorama.com/em for list info |
---------- Forwarded message --------- From: Forest Simmons <[hidden email]> Date: Sat, Jun 1, 2019 at 12:01 PM Subject: Re: [EM] What are some simple methods that accomplish the following conditions? To: Kristofer Munsterhjelm <[hidden email]> Great! This is the kind of creativity that will continue to keep this EM list alive and relevant. On Sat, Jun 1, 2019 at 8:13 AM Kristofer Munsterhjelm <[hidden email]> wrote: On 31/05/2019 00.33, Forest Simmons wrote: ---- Election-Methods mailing list - see https://electorama.com/em for list info |
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Hi Forest,
I had two ideas. Idea 1: 1. If there is a CW using all rankings, elect the CW. 2. Otherwise flatten/discard all disapproved rankings. 3. Use any method that would elect C in scenario 2. (Approval, Bucklin, MinMax(WV).) So scenario 1 has no CW. The disapproved C>A rankings are dropped. A wins any method. In scenario 2 there is no CW but nothing is dropped, so use a method that picks C. In both versions of scenario 3 there is a CW, B. If step 3 is Approval then of course step 2 is unnecessary. In place of step 1 you could find and apply the majority-strength solid coalitions (using all rankings) to disqualify A, instead of acting based on B being a CW. I'm not sure if there's another elegant way to identify the majority coalition. Idea 2: 1. Using all rankings, find the strength of everyone's worst WV defeat. (A CW scores 0.) 2. Say that candidate X has a "double beatpath" to Y if X has a standard beatpath to Y regardless of whether the disapproved rankings are counted. (I don't know if it needs to be the *same* beatpath, but it shouldn't come into play with these scenarios.) 3. Disqualify from winning any candidate who is not in the Schwartz set calculated using double beatpaths. In other words, for every candidate Y where there exists a candidate X such that X has a double beatpath to Y and Y does not have a double beatpath to X, then Y is disqualified. 4. Elect the remaining candidate with the mildest WV defeat calculated earlier. So in scenario 1, A always has a beatpath to the other candidates, no matter whether disapproved rankings are counted. The other candidates only have a beatpath to A when the C>A win exists. So A has a double beatpath to B and C, and they have no path back. This leaves A as the only candidate not disqualified. In scenario 2, the defeat scores from weakest to strongest are B>C, A>B, C>A. Every candidate has a beatpath to every other candidate no matter whether the (nonexistent) disapproved rankings are counted. So no candidate is disqualified. C has the best defeat score and wins. In scenario 3, the first version: B has no losses. C's loss to B is weaker than both of A's losses. B beats C pairwise no matter what, so B has a double beatpath to C. However C has no such beatpath to A, nor has A one to B, nor has B one to A. The resulting Schwartz set disqualifies only C. (C needs to return B's double beatpath but can't, and neither A nor B has a double beatpath to the other.) Between A and B, B's score (as CW) is 0, so he wins. Scenario 3, second version: B again has no losses, and also has double beatpaths to both of A and C, neither of whom have double beatpaths back. So A and C are disqualified and B wins. I must note that this is actually a Condorcet method, since a CW could never get disqualified and would always have the best worst defeat. That observation would simplify the explanation of scenario 3. I needed the defeat strength rule because I had no way to give the win to B over A in scenario 3 version 1. But I guess if it's a Condorcet rule in any case, we can just add that as a rule, and greatly simplify it to the point where it's going to look very much like idea 1. I guess all my ideas lead me to the same place with this question. Oh well, I think the ideas are interesting enough to post. Kevin >Le jeudi 30 mai 2019 à 17:32:42 UTC−5, Forest Simmons <[hidden email]> a écrit : > >In the example profiles below 100 = P+Q+R, and 50>P>Q>R>0. One consequence of these constraints is that in all three profiles below the cycle >A>B>C>A will obtain. > >I am interested in simple methods that always ... > >(1) elect candidate A given the following profile: > >P: A >Q: B>>C >R: C, >and >(2) elect candidate C given >P: A >Q: B>C>> >R: C, >and >(3) elect candidate B given > >P: A >Q: B>>C (or B>C) >R: C>>B. (or C>B) > >I have two such methods in mind, and I'll tell you one of them below, but I don't want to prejudice your creative efforts with too many ideas. > >Here's the rationale for the requirements: > >Condition (1) is needed so that when the sincere preferences are > >P: A >Q: B>C >R: C>B, >the B faction (by merely disapproving C without truncation) can defend itself against a "chicken" attack (truncation of B) from the C faction. > >Condition (3) is needed so that when the C faction realizes that the game of Chicken is not going to work for them, the sincere CW is elected. > >Condition (2) is needed so that when sincere preferences are > >P: A>C >Q: B>C >R: C>A, >then the C faction (by proactively truncating A) can defend the CW against the A faction's potential truncation attack. > >Like I said, I have a couple of fairly simple methods in mind. The most obvious one is Smith\\Approval where the voters have >control over their own approval cutoffs (as opposed to implicit approval) with default approval as top rank only. The other >method I have in mind is not quite as >simple, but it has the added advantage of satisfying the FBC, while almost always electing from Smith. ---- Election-Methods mailing list - see https://electorama.com/em for list info ---- Election-Methods mailing list - see https://electorama.com/em for list info |
Kevin, Chris, and Kristofer: Great ideas!. DMC (explicit approval version) also comes to mind as fairly simple to describe: Remove candidates from the bottom of the approval list until there is a ballot pairwise beats-all candidate among the remaining candidates (to elect). And analogous to Chris's equivalence between Approval Sorted Margins and Condorcet(approval margins), we have the equivalence between DMC and Condorcet(winning approval). Of course we are talking explicit approval every time we say approval in this context. We know that none of these Condorcet efficient methods satisfies the FBC, but how about MDDA with explicit approval instead of its implicit approval default? This has the problem that Chris has pointed out in MJ and other Bucklin versions where addition of irrelevant ballots can change a majority of ballots into a minority. So here's my idea: do MDDA with explicit approval AND symmetric completion of all of the ballots except for the Top Rank. By excusing the top rank from symmetric completion, we preserve the FBC, but we lose the Condorcet Criterion. [This shows how close we can get to both the CC and the FBC, if we do a full symmetric completion including the top rank, then this version of MDDA satisfies the CC but not the FBC. We can toggle back and forth. Suppose that we balance on the edge (of symmetric completion in the top rank or not) with asset voting so the proxies could finesse this difference. The unsophisticated voters could just vote their favorites, etc. Would this come even closer to satisfying both the CC and the FBC? In other words mightn't this expedient externalize the strategizing enough to satisfy both FBC and CC for all practical purposes?] I think that the FBC is more important in this context, so suggest that we exempt at least the top rank from symmetric completion if not all of the approved ranks. Now what about the irrelevant ballots problem? I think that symmetric completion, if only for the truncated candidates, would suffice. If X and Y are both truncated on N new ballots then N/2 of them count for X>Y and N/2 of them count for Y>X so a majority defeat between X and Y is preserved. Note that symmetric completion among only the truncated candidates is the same as "half power truncation." This brings me to another question. Suppose that voters knew that candidate B became a plurality loser because of insincere truncation. Would they feel so bad it B won anyway? This leads to a weak form of Plurality: If a candidate fails Plurality, even after the ballots have been symmetrically completed, then that candidate must not be elected. Would this weak form be adequate? Thanks for your great ideas and interest in these questions! Forest On Sat, Jun 1, 2019 at 12:49 PM Kevin Venzke <[hidden email]> wrote: Hi Forest, ---- Election-Methods mailing list - see https://electorama.com/em for list info |
just one other tweak to MDDA: Kevin's official version of MDDA says that in the case of two of more candidates surviving the disqualification test, in that case the most approved candidate should be elected. My tweak is to elect from among the undisqualified candidates the one furthest from being disqualified; no need to switch horses in this case. On Sat, Jun 1, 2019 at 2:41 PM Forest Simmons <[hidden email]> wrote:
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So Forest, are you saying one would still use approval in the case that *all* candidates were disqualified? I'd guess that wouldn't be that common. I'm not totally sure what "furthest from being disqualified" means. If it means according to pairwise opposition, then this method will be very similar to MMPO. If it refers to actual pairwise defeats then it probably won't satisfy FBC. Kevin
Le samedi 1 juin 2019 à 17:32:04 UTC−5, Forest Simmons <[hidden email]> a écrit :
just one other tweak to MDDA: Kevin's official version of MDDA says that in the case of two of more candidates surviving the disqualification test, in that case the most approved candidate should be elected. My tweak is to elect from among the undisqualified candidates the one furthest from being disqualified; no need to switch horses in this case. On Sat, Jun 1, 2019 at 2:41 PM Forest Simmons <[hidden email]> wrote:
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Forest (and interested others), On 2/06/2019 8:01 am, Forest Simmons
wrote:
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Chris, Yes, we did, and I still do believe that Approval Sorted Margins is a superior method to DMC. In fact, Approval sorted Margins and Covering Enhanced Approval are my two favorite Condorcet compliant methods. However, the order for ease of explaining and selling to the public seems to be the opposite of my preference order in the case of these three methods. I could be wrong on this point. BTW, I like the recent tweaks you are making to IBIFA. I don't consider Asset Voting a settled deterministic method, since the are no agreed upon constraints about how the candidates are allowed to use their "assets." Here's a version that I would support: A base election method is established well before voting takes place. Voters submit as much ranking or rating information (relevant to the base method) as they like, and if not sure where to place the approval cutoff (for example) they can indicate both an upper and lower limit for the approval cutoff, and (by some code) request their favorite to choose where in that interval to mark the cutoff. At the other extreme of voter trust of favorite, voters may simply indicate that they would like their ballot to be an exact replica of their favorite's. In any case, the voter specifies how much discretionary "power of attorney" so to speak gets delegated to their indicated proxy. Once the candidate proxies have made their requested amendments to the ballots, then the election is decided by those ballots in the previously agreed upon base method. If the voter has n candidates tied for favorite and cannot make up her mind among them, then she can submit the entire set of n favorites, in which case each of those favorites gets a copy of her ballot to amend,, and each of those amended ballots gets a weight of 1/n in the election. Can you think of anything closer to Strong FBC compliance than this candidate proxy enhancement of any good base method? What voters want more than anything else is Strong FBC compliance. They know that Approval does not satisfy the Strong FBC, and they (wrongly) believe that IRV does satisfy it. That's the main reason they think IRV is better than Approval. The other reason they prefer IRV is that they don't feel confident about deciding on approval cutoff's. Candidate Proxy Ballot Enhancement (CPBE?) takes care of their approval cutoff indecision, and gives them (the closest possible thing to) true compliance with the Strong FBC. I would suggest Approval Sorted Margins for the base method, since that method (by adjusting approval cutoffs) takes care of Chicken Defense and other basic defensive strategy requirements. Voters submit rankings or ratings with upper and lower limits for the approval cutoff (when in doubt), etc.(as indicated above). Forest On Sun, Jun 2, 2019 at 3:34 PM C.Benham <[hidden email]> wrote:
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Earlier in response to this I suggested some Condorcet methods.
Here is a non-Condorcet method On 31/05/2019 8:03 am, Forest Simmons
wrote:
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A simple, easy to understand solution! On Wed, Jun 26, 2019 at 10:10 PM C.Benham <[hidden email]> wrote:
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Forest, I like the idea of an Approval-Sorted-Margins based method. I think it is simple enough to explain and handles cycles in an transparent fashion. You, Chris and I had some private conversations about this some years back. I would prefer an explicit approval cutoff. I think the following version is easy to understand:
What is Strong FBC, and why doesn't Approval meet it? Chris's "tweaked IRV" seems to be a version of DMC/ASM with elimination. Does that elimination change approval cutoffs and other pairwise counts? If not, it could be summable. If it does, and is not summable and doesn't satisfy either FBC or CC, why do you like it? Ted On Thu, Jun 27, 2019 at 2:55 PM Forest Simmons <[hidden email]> wrote:
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