[EM] Inclusion/Exclusion Counts

[Forest Simmons]

Suppose we start with  Defeat Strength DS(A>B) defined by ...

Sum Top(X) | X does not defeat A
Minus
Sum Top(Y) | B defeats Y,

and add a difference that will confer  reverse symmetry onto the resulting
defeat strength ... something like this ...

Sum Bottom(Z) | B is not defeated by Z
Minus
Sum Bottom(K) | K defeats A

If we put these differences together, and combine the subtrahends to
minimize subtractions, we get a defeat strength expression with reverse
symmetry for the A>B defeat:

Sum Top(X) | X does not defeat A
Plus
Sum Bottom(Z) | B is not defeated by Z
Minus
[Sum Bottom(K) | K defeats A+Sum Top(Y) | B defeats Y]


If A is the CW and B is the CL, this strength is 200 percent, so perhaps we
should divide the whole thing by two.

With all the bottom and top percentage terms involved in one way or
another, this expression should be very decisive.

But in the rare case of a tie or ties, a complete ranking created by random
ballot draws should be an acceptable traditional way of resolving any
ambiguities in the Ranked Pairs finish order.

-Forest


On Mon, Oct 24, 2022, 11:29 AM Forest Simmons <forest.simmons21 at gmail.com>
wrote:

> That is a great insight!
>
> My most recent message in this thread adds a term to the defeat strength
> that achieves reverse symmetry, but it comes at the cost of increasing
> burial incentive.
>
> The term is Bad(B) defined as ...
>
> Sum bottom(Z) | B does not defeat Z
>
> in the context of defeat strength, where B is the pairwise loser of the
> defeat in question. Bottom(Z) is the percentage of bottom ballot positions
> occupied by Z.
>
> So if B is the Condorcet Loser, then Bad(B) is 100%.
>
> Obviously if this term is given equal weight with its reverse symmetry
> counterpart, it will contribute an appreciable burial incentive.
>
> The reverse symmetry counterpart is Good(A) defined by
>
> Sum Top(X) | X does not defeat A
>
> in the context of defeat strength of a defeat where A is the pairwise
> winner.
> Top(X) is the percentage of Too ballot positions occupied by X.
>
> So if A is the CW, then Good(A) is 100percent.
>
> So jt's probably not a good idea to use Bad(B) to create reverse symmetry.
> However, perhaps it could be given non-symmetrical,  infinitesimal weight
> for tie breaking purposes only.
>
> Also, perhaps their is a milder version of Good(A), whose reverse symmetry
> counterpart Bad(B), would have a tolerable burial incentive.
>
> -Forest
>
> On Mon, Oct 24, 2022, 2:43 AM Kristofer Munsterhjelm <km_elmet at t-online.de>
> wrote:
>
>> On 10/23/22 04:55, Forest Simmons wrote:
>>
>> > Critics have always maintained that this idea shows a lack of awareness
>> > of clone dependence. But that judgment assumes that just because there
>> > is a bad way of using those ballots, there can be no good way.
>>
>> Here's a thought that occurred to me, that would explain why the
>> (seemingly out of nowhere) implication that we can't have both reversal
>> symmetry and DMTCBR.
>>
>> First preferences are unaffected by burial, and last preferences are
>> unaffected by compromising. Suppose we had a method that were DMTCBR and
>> reversal symmetric. Then we could freely translate between a method
>> that's very strong against burial and very strong against compromise by
>> just reversing the ballots, since the reversed ballots' first
>> preferences would be last preferences.
>>
>> Thus a method that passes both DMTCBR and rev. sym. would be extremely
>> resistant to both burial and to compromise. But since the favorite
>> betrayal criterion is so hard to pass, we have reason to believe that
>> this is impossible. So no such method can be rev. sym -- which is what
>> we at least see with Condorcet methods!
>>
>> It's thus quite that the implication is stronger: that we can't have all
>> of DMTBR, majority, and reversal symmetry. But the proof is probably a
>> lot harder to find, too.
>>
>> So all of the above implies that when creating a resistant ranked
>> method, we can't both have extreme resistance to burial and compromising
>> - we have to pick one. Fortunately (as James Green-Armytage originally
>> showed), we already get a great deal of compromising resistance from the
>> Condorcet criterion itself (since, for instance, it does the right thing
>> under center squeeze). Thus it's more sensible to choose further burial
>> resistance over further compromise resistance if we can only have one.
>>
>> (Unless we consider maximum compromise resistance absolutely
>> non-negotiable, e.g. Mike O's insistence on the FBC.)
>>
>> ...
>>
>> Finally, it might be useful to see just what the analog of the DMTCBR is
>> for a reversed DMTCBR-compliant method. It's something like...
>>
>> Suppose that more than 1/3 of the ranks some Condorcet loser last. Then
>> nobody who prefers this loser to the current winner can make the loser
>> win by upranking him.
>>
>> -km
>>
>
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