[EM] More on MAS (version 3.0)

Jameson Quinn jameson.quinn at gmail.com
Fri Oct 14 13:34:14 PDT 2016


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 at 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.
>>
>>
>>
>>
>>
>
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