[EM] Inclusion/Exclusion Counts

[Forest Simmons]

Jobst pointed out clone dependence of a related method I see the need to
fix this one, too.

Let Good(A) be the "goodness" of A,  (judging from the superficial first
place lottery probabilities) defined as ...

Sum {fp(X)|A is not defeated by X},

where fo(X) is the percentage of first place ballot positions occupied by X.

Likewise, let Bad(B) be the superficial "badness" of B, defined as ..

Sum {bot(Z)|Z is not defeated by B},

where bot(Z) is the percentage of bottom ballot positions occupied by Z.

The pairwise defeat A>B is what you would expect when B is bad and A is
good. The worse B, and the better A, the bigger numerically both Bad(B) and
Good(A). So their sum Bad(B)+Good(A) is a natural measure of defeat
strength that gives equal importance to top and bottom ballot positions.

Presently (meaning soon) we'll test it on a few examples to see if it lives
up to expectations!

Forest

On Sat, Oct 22, 2022, 7:55 PM Forest Simmons <forest.simmons21 at gmail.com>
wrote:

> Richard Lung has been patiently teaching and reminding us that the
> information from ballot last choices is just as valuable as the top choice
> information ... if only we incorporate it wisely ... as suggested by
> Richard and other proponents of Binomial STV, for eexample.
>
> For decades we have seen an echo of this wisdom in the form of ballots
> that allow one vote for, and one vote against.
>
> 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.
>
> Just as "vote one" plurality ballots are perfectly adequate for the
> benchmark lottery, the vote  "one against" ballot is perfectly adequate for
> the anti-favorite lottery.
>
> We have seen already that these two lotteries have equally important roles
> in de-cloning the Kendall-tau metric for making a clone free version of
> Kemeny-Young.
>
> Here's another application of this idea whose time has come:
>
> Gauge pairwise defeat strength of Y by X as the product  f(X)*f'(Y), where
> f and f' give the benchmark and anti-benchmark lottery probabilities,
> respectively.
>
> Condorcet with this defeat strength gauge is the method we should be
> testing for dishonest strategy resistance, and for Voter Satisfaction
> Efficiency!
>
> It could be the best all around single winner deterministic Universal
> Domain method ever!
>
> Thanks to Richard for your patience in keeping us on the right track!
>
> And thanks to Kevin and Kristofer for keeping us honest by experimental
> testing of our ideas ... two versatile scientists with both analytical and
> experimental creativity and insight.
>
> And Richard ... a well read renaissance man of wide ranging historical
> social and hard science intellectual interest ... with amazing intuition!
>
> -Forest
>
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