I've been refining a 3-slot system for several weeks now. Let me be clear that I'm only working on one system, even though I've gone through various names as I refine it. The current name is MAS, Majority Acceptable Score. Here's my latest definition. Note that I've tweaked the default rule so that it can be said in one sentence. Mathematically it's trickier, but I think it makes some intuitive sense, as explained in the last sentence. *Here’s how MAS works: you can give each candidate 0, 1, or 2. Any candidate that gets a majority of 0’s is eliminated, unless that would eliminate everyone. Of the remaining candidates, highest score wins. * *Blank votes for a candidate are read as 0’s or 1’s; the proportion that count as 0’s is equal to the proportion between the voters that didn't give the candidate in question a 2, and those that gave a 2 to a candidate with a higher explicit score. Basically, that rule assumes that a voter would want to give 0s to they left blank if those candidates were weaker than their favorite, but 1s if those candidates were stronger.* Here's a scenario to illustrate: Candidate 2 votes 1 votes 0 votes Blank votes Explicit score A 30 0 0 70 60 B 25 25 0 50 75 C 42 0 55 3 84 D 8 42 0 50 58 (Note: I think that a scenario like the above, where one candidate got many more explicit 1-votes, would only happen in cases of center squeeze; that is, B's 1-votes probably come primarily from C voters. Thus, B is almost-certainly, but not quite provably, the CW here.) Candidate A has 70 blank votes, and 70 voters who didn't give them a 2. 67 voters gave 2 to a candidate with a higher explicit score (C or B). 67 of A's blank votes count as 0s, leaving 3 1's. A gets a total score of 63, and is eliminated for a majority of 0's. B has 50 blank votes, and 75 voters who didn't give them a 2. 42 voters gave 2 to a candidate with a higher explicit score (C). So 28 of the blank votes count as 0, 22 count as 1; B gets a score of 97. C is eliminated by explicit 0s. D has all their blank votes count as 0 since the number of 2-votes for explicitly stronger candidates is greater than the number who didn't vote for them. They are not quite eliminated. So B wins this scenario. If B had gotten 9 or fewer explicit 1-votes, A would have had a higher explicit score, and after assigning blank votes, A would have won. This default rule does cause the system to technically fail FBC, because giving extra 2-votes to eliminated candidates can change how blank votes are assigned for uneliminated candidates. However, constructing an FBC-violating scenario would be nontrivial; I don't think it would ever happen in practice.

Still working on refining this. Here's version 3.1. I expect the final version to be version 4.0, at which point the earlier versions and numberings will be only a historical curiosity. *Here’s how MAS works: you can give each candidate 0, 1, or 2. Any candidate that gets a majority of 0’s is eliminated, unless that would eliminate everyone. Of the remaining candidates, highest score wins. * *Blank votes for a candidate are read as 0’s or 1’s. The proportion that count as 0’s is equal to the proportion between the voters that didn't give the candidate in question a 2, and those that gave a 2 to a candidate with more 2's. * This default rule gives exactly the result you'd get if blank votes were counted as 0 only for voters who preferred a stronger candidate, under some simple assumptions about which votes come from where: explicit votes of 1 for a given candidate are spread evenly among all voters who didn't give them a 2; explicit votes of 0 come only from voters who preferred a stronger candidate; every voter gives a 2 to exactly one "serious" candidate; and all "nonserious" candidates get fewer 2's than "serious" ones. You need simplifying assumptions like that so that counting can work by simply tallying the votes of each type, without recording how they are combined on each ballot. 2016-10-13 6:47 GMT-04:00 Jameson Quinn <jameson.quinn@gmail.com>: > I've been refining a 3-slot system for several weeks now. Let me be clear > that I'm only working on one system, even though I've gone through various > names as I refine it. The current name is MAS, Majority Acceptable Score. > Here's my latest definition. Note that I've tweaked the default rule so > that it can be said in one sentence. Mathematically it's trickier, but I > think it makes some intuitive sense, as explained in the last sentence. > > *Here’s how MAS works: you can give each candidate 0, 1, or 2. Any > candidate that gets a majority of 0’s is eliminated, unless that would > eliminate everyone. Of the remaining candidates, highest score wins. * > > *Blank votes for a candidate are read as 0’s or 1’s; the proportion that > count as 0’s is equal to the proportion between the voters that didn't give > the candidate in question a 2, and those that gave a 2 to a candidate with > a higher explicit score. Basically, that rule assumes that a voter would > want to give 0s to they left blank if those candidates were weaker than > their favorite, but 1s if those candidates were stronger.* > > Here's a scenario to illustrate: > > Candidate > > 2 votes > > 1 votes > > 0 votes > > Blank votes > > Explicit score > > A > > 30 > > 0 > > 0 > > 70 > > 60 > > B > > 25 > > 25 > > 0 > > 50 > > 75 > > C > > 42 > > 0 > > 55 > > 3 > > 84 > > D > > 8 > > 42 > > 0 > > 50 > > 58 > > (Note: I think that a scenario like the above, where one candidate got > many more explicit 1-votes, would only happen in cases of center squeeze; > that is, B's 1-votes probably come primarily from C voters. Thus, B is > almost-certainly, but not quite provably, the CW here.) > > Candidate A has 70 blank votes, and 70 voters who didn't give them a 2. 67 > voters gave 2 to a candidate with a higher explicit score (C or B). 67 of > A's blank votes count as 0s, leaving 3 1's. A gets a total score of 63, and > is eliminated for a majority of 0's. > > B has 50 blank votes, and 75 voters who didn't give them a 2. 42 voters > gave 2 to a candidate with a higher explicit score (C). So 28 of the blank > votes count as 0, 22 count as 1; B gets a score of 97. > > C is eliminated by explicit 0s. D has all their blank votes count as 0 > since the number of 2-votes for explicitly stronger candidates is greater > than the number who didn't vote for them. They are not quite eliminated. > > So B wins this scenario. If B had gotten 9 or fewer explicit 1-votes, A > would have had a higher explicit score, and after assigning blank votes, A > would have won. > > This default rule does cause the system to technically fail FBC, because > giving extra 2-votes to eliminated candidates can change how blank votes > are assigned for uneliminated candidates. However, constructing an > FBC-violating scenario would be nontrivial; I don't think it would ever > happen in practice. > > > > >

Still working on refining this. Here's version 3.2. I expect the final version to be version 4.0, at which point the earlier versions and numberings will be only a historical curiosity. *Here’s how MAS works: you can give each candidate 0, 1, or 2 points. * *Then,** any candidate that gets a majority of 0’s is eliminated**, unless that would eliminate everyone.** Of the remaining candidates, highest score wins. * *Blank votes for a candidate count as 1 point in the same percentage as that candidate gets of 2-point votes. Otherwise, they count as 0.* Here's some examples of how that default rule works out: 2-votes 1-votes 0-votes Blanks Total 0-votes Score A 30 0 0 70 0+70*.7=49 81 B 25 19 0 52 0+52*.75=39 83 C 40 0 40 20 40+20*.6=52 (88) D 25 7 0 68 0+68*.75=51 (74) As you can see, it takes just under 30% support to save a candidate from elimination if nobody explicitly downvotes them; and at around that level, it takes just under 4 explicit 1-votes to make up for a deficit in explicit 2-votes. Since this rule looks at only one candidate at a time, it's easy to implement, and it easily passes FBC and participation. It fails consistency, but only in Simpson's-paradox-like situations, in which arguably consistency is actually a bad idea. It passes a weakened form of Frohnemayer balance: > This default rule gives exactly the result you'd get if blank votes were > counted as 0 only for voters who preferred a stronger candidate, under some > simple assumptions about which votes come from where: explicit votes of 1 > for a given candidate are spread evenly among all voters who didn't give > them a 2; explicit votes of 0 come only from voters who preferred a > stronger candidate; every voter gives a 2 to exactly one "serious" > candidate; and all "nonserious" candidates get fewer 2's than "serious" ones. > You need simplifying assumptions like that so that counting can work by > simply tallying the votes of each type, without recording how they are > combined on each ballot. > > > 2016-10-13 6:47 GMT-04:00 Jameson Quinn <jameson.quinn@gmail.com>: > >> I've been refining a 3-slot system for several weeks now. Let me be clear >> that I'm only working on one system, even though I've gone through various >> names as I refine it. The current name is MAS, Majority Acceptable Score. >> Here's my latest definition. Note that I've tweaked the default rule so >> that it can be said in one sentence. Mathematically it's trickier, but I >> think it makes some intuitive sense, as explained in the last sentence. >> >> *Here’s how MAS works: you can give each candidate 0, 1, or 2. Any >> candidate that gets a majority of 0’s is eliminated, unless that would >> eliminate everyone. Of the remaining candidates, highest score wins. * >> >> *Blank votes for a candidate are read as 0’s or 1’s; the proportion that >> count as 0’s is equal to the proportion between the voters that didn't give >> the candidate in question a 2, and those that gave a 2 to a candidate with >> a higher explicit score. Basically, that rule assumes that a voter would >> want to give 0s to they left blank if those candidates were weaker than >> their favorite, but 1s if those candidates were stronger.* >> >> Here's a scenario to illustrate: >> >> Candidate >> >> 2 votes >> >> 1 votes >> >> 0 votes >> >> Blank votes >> >> Explicit score >> >> A >> >> 30 >> >> 0 >> >> 0 >> >> 70 >> >> 60 >> >> B >> >> 25 >> >> 25 >> >> 0 >> >> 50 >> >> 75 >> >> C >> >> 42 >> >> 0 >> >> 55 >> >> 3 >> >> 84 >> >> D >> >> 8 >> >> 42 >> >> 0 >> >> 50 >> >> 58 >> >> (Note: I think that a scenario like the above, where one candidate got >> many more explicit 1-votes, would only happen in cases of center squeeze; >> that is, B's 1-votes probably come primarily from C voters. Thus, B is >> almost-certainly, but not quite provably, the CW here.) >> >> Candidate A has 70 blank votes, and 70 voters who didn't give them a 2. >> 67 voters gave 2 to a candidate with a higher explicit score (C or B). 67 >> of A's blank votes count as 0s, leaving 3 1's. A gets a total score of 63, >> and is eliminated for a majority of 0's. >> >> B has 50 blank votes, and 75 voters who didn't give them a 2. 42 voters >> gave 2 to a candidate with a higher explicit score (C). So 28 of the blank >> votes count as 0, 22 count as 1; B gets a score of 97. >> >> C is eliminated by explicit 0s. D has all their blank votes count as 0 >> since the number of 2-votes for explicitly stronger candidates is greater >> than the number who didn't vote for them. They are not quite eliminated. >> >> So B wins this scenario. If B had gotten 9 or fewer explicit 1-votes, A >> would have had a higher explicit score, and after assigning blank votes, A >> would have won. >> >> This default rule does cause the system to technically fail FBC, because >> giving extra 2-votes to eliminated candidates can change how blank votes >> are assigned for uneliminated candidates. However, constructing an >> FBC-violating scenario would be nontrivial; I don't think it would ever >> happen in practice. >> >> >> >> >> >

The table I sent had mistakes in the B row. Here it is, fixed: 2-votes 1-votes 0-votes Blanks Total 0-votes Score A 30 0 0 70 0+70*.7=49 81 B 25 19 0 56 0+56*.75=42 83 C 40 0 40 20 40+20*.6=52 (88) D 25 7 0 68 0+68*.75=51 (74)