[EM] Strong Reverse Symmetry Clone Proof Borda

[Forest Simmons]

To give credit where credit is due, I want to thank Richard Lung for
getting us thinking along the inclusion/exclusion axis corresponding to
near-first/near-last ballot ranks ... That contributed to my confidence in
the lottery/anti-lottery idea as a means of de-cloning Borda, Copeland. and
Kemeny-Young, which has panned out so beautifully.

Thanks, Richard!!!

El vie., 14 de ene. de 2022 8:18 a. m., Forest Simmons <
forest.simmons21 at gmail.com> escribió:

> Two equivalent ways of decloning Borda ...  both relying on a combination
> of the benchmark and anti-benchmark lotteries BML and ABML.
>
> First method:
>
> Ballot B contributes to candidate X's score the sum of BML(j) [summed over
> candidates j ranked below X] minus the sum of ABML(k) [summed over
> candidates k ranked above X].
>
> Think of it this way ... the importance of candidate X to the voter of
> ballot B is the estimated strength of the candidates she likes less than X,
> minus the estimated weakness of the candidates she likes better:
>
> [If the ones you like better are weak and the ones you dislike in
> comparison are strong, you should support X.]
>
> [Also this is a natural DSV kind of way to specify an approval cutoff]
>
> Method 2.
>
> The score for X is the difference in the BML expectation of X's pairwise
> matrix row and the ABML expectation of X's pairwise matrix column.
>
> The former is a source of pride for X ...the weighted average pairwise
> scores of X over the candidates, where the weights are estimates of the
> strengths of those candidates.
>
> The latter expectation is a source of chagrin for X ... the weighted
> average of the scores against X by the other candidates, where the weights
> are estimates of the weakness of those candidates... if you are soundly
> trounced by a weak candidate, you should feel chagrin, if not shame!
>
> This gives another idea for a kind of MinMax/MaxMin method:
>
> Let M(X) be the Max value of the product of ABML(k)*P(k,X), where P(k, X)
> is the percentage of ballots on which k is ranked ahead of X.
>
> Similarly, let m(X) be the min value of P(X, j)*BML(j), where P(X,j) is
> the percentage of ballots ranking X above j.
>
> Elect argmin(M(X)-m(X)).
>
> [Maybe this is another key to fpC-fpA]
>
> Note the strong symmetry of this method, also.
>
> All in all, I'm pretty excited about the usefulness of the anti-benchmark
> lottery in conjunction with the benchmark lottery.
>
> If BML(X) is an estimate of X's viability, then ABML(X) is an estimate of
> X's weakness.
>
> Of course, the BML is not the only clone-free lottery ...Jobst's MaxParC,
> the Martin Harper Lottery, the Random Approval Lottery, the Nash Lottery
> and its "Ultimate Lottery" generalization ...all of these and others have
> their corresponding anti-lottery counterparts ... just reverse the ballots
> to get them.
>
> Why didn't we think of this before?  It was staring us right in the face!
>
> What other obvious ideas are still invisible to us?
>
> -Forest
>
>
> El vie., 14 de ene. de 2022 6:24 a. m., Forest Simmons <
> forest.simmons21 at gmail.com> escribió:
>
>> Kristofer,
>>
>> I have a feeling you might like this method:
>>
>> For simplicity assume complete rankings.
>>
>> For each candidate X, let fp(X) and lp(X), respectively, be the total
>> number of first place votes and last place votes, respectively, of the
>> candidates that are pairwise defeated by X, and pairwise defeat X,
>> respectively.
>>
>> Elect argmax(fp(X)-lp(X))
>>
>> What do you think?
>>
>> -Forest
>>
>>
>>
>>
>> El vie., 14 de ene. de 2022 2:11 a. m., Kristofer Munsterhjelm <
>> km_elmet at t-online.de> escribió:
>>
>>> On 14.01.2022 11:01, Kristofer Munsterhjelm wrote:
>>> > On 14.01.2022 04:05, Forest Simmons wrote:
>>> >> I've always admired the basic idea of Kemeny-Young, namely a cost
>>> metric
>>> >> for conversion of one ranking into another ... simply the number of
>>> >> transpositions (elementary swaps) required.
>>> >>
>>> >> The deal-breaker draw-back of the Kemeny cost metric is that it is
>>> clone
>>> >> dependent ... if it were independent of clones the " cost" of
>>> >> transposing AB to BA would be the same as the net cost of transposing
>>> >> the rank order of respective clones of A and B:
>>> >>
>>> >> the order a1a2a3b1b2b3
>>> >>  to the order b3b2b1a3a2a1
>>> >
>>> > Of course, I have to restate here that Ranked Pairs is such a method:
>>> > the metric is just leximax. (ordering a is better than ordering b if
>>> the
>>> > greatest pairwise victory consistent with a is higher than the greatest
>>> > pairwise victory consistent with b, with tiebreaks for second greatest,
>>> > third greatest, etc.) And it's cloneproof.
>>>
>>> Here's a thought: let parameterized p-Kemeny be: take the lp norm of the
>>> vector of all pairwise victories consistent with the output ordering.
>>> Break ties by gradually lowering p, i.e. l(p-epsilon), further ties by
>>> l(p-2 epsilon), etc.
>>>
>>> Then Kemeny is l_1 and RP is the limit as p approaches infinity, l_inf.
>>> I think? Are any other values of p interesting?
>>>
>>> -km
>>>
>>
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