[EM] My Recent New Methods Summary with Corrections

[Forest Simmons]

De-cloned Copeland

The Copeland score for candidate X is the difference D(X) defined by ...

Sum(over the set of candidates Y pairwise defeated by X) of f*(Y)
MINUS
Sum(over the candidates Z that pairwise defeat X) of f(Z).

Elect argmax(D(X)).

This is a correction that rewards X for defeating "bad candidates", while
not overly competing with "good candidates."

If makes the method compliant with mono- raise, and reduces the incentive
for compromising.

Next time ... de-cloned Borda ...

El mar., 18 de ene. de 2022 5:57 p. m., Forest Simmons <
forest.simmons21 at gmail.com> escribió:

> Since all of these new methods depend (by default) on the benchmark
> lottery f and its reverse counterpart f*, we begin by defining these
> probability densities.
>
> The context is a set beta of (voted) ranked choice ballots.
>
> For each candidate X, let f(X) be the probability that the random ballot
> favorite candidate is X.  A thought experiment defines this probability: if
> a randomly drawn ballot has more than one contender for favorite,
> additional ballots are drawn sequentially to narrow down to a single
> favorite. The probability that X is the resulting favorite is the value of
> f(X).
>
> Similarly, f*(X) is the random ballot anti-favorite probability for X. if
> a ballot randomly drawn from beta has more than one contender for
> anti-favorite, additional ballots are drawn sequentially to narrow down to
> a single anti-favorite. The probability that X is the resulting
> anti-favorite is the value of f*(X).
>
> Note that if the ballot rankings are all reversed f and f* swap places.
>
> The salient (i.e. sufficient for what follows) properties of f and f* in
> this context are ...
> 1. They are both probability density functions on the set of candidates.
> 2. If X (and only X) is raised on one or more ballots, then ...
>  f(X) does not decrease, nor does f*(X) increase
> AND
> for Y not equal to X, f(Y) does not increase nor does f*(Y) decrease.
> 3. If X (and only X) decreases on one or more ballots, then ...
> f(X) does not increase, nor does f*(X) decrease
> AND
> for Y not equal to X,  f(Y) does not decrease, nor does f*(Y) increase.
> 4. Both f a d f* respect clone sets, which means (in the case of f), if
> candidate X is replaced by a clone set chi, then
> f(X)=Sum(over x in chi) of f(x).
> 5. The procedure that yields f when applied to the ballot set beta, yields
> f* when applied to the reversed ballot set beta*.
> [Condition 5 is not essential, except for the strong reverse symmetry
> property.]
>
> Now that we are all set up, we can define the four methods ... de-cloned
> Kemeny-Young, de-cloned Borda, de-cloned Copeland, and the Loggerheads
> Condorcet Lottery.
>
> I'm going to break here to save what we have so far ...
>
>
>
>
>
>
>
> El lun., 17 de ene. de 2022 1:44 p. m., Forest Simmons <
> forest.simmons21 at gmail.com> escribió:
>
>> I'm changing the name to "Loggerheads" because (1) our two player game is
>> based on players with polar opposite preferences, and (2) it's not quite
>> Condorcet compliant because of its way of handling pairwise ties. However,
>> if we just go back to the customary zero payoff for pairwise ties, the
>> method becomes Condorcet compliant.
>>
>> For now let's leave the pairwise tie payoff open; it may help us
>> distinguish weak CW's from strong ones .... when the tie handling makes a
>> difference, it might be a sign of marginal weakness or instability in the
>> Condorcet winner.
>>
>> Just to pursue this point a little further, the non-zero payoff for the
>> row player when both players choose the same candidate X, is the
>> difference  F(X)-R(X), which is positive only when X has more first place
>> than last place votes. So if X has fewer first than last place votes, then
>> that diagonal payoff entry will be negative, preventing  X from being the
>> sure winner, even if X is the Condorcet candidate ... but not impairing the
>> Condorcet efficiency too much, unless the tie payoff entries are very
>> negative, which would be very unusual.
>>
>> So let's keep open the possibility of non-zero pairwise tie payoffs, but
>> make zero the default payoff for simplicity.
>>
>> Analogously de-cloned Copeland loses its absolute Condorcet efficiency
>> when we allow pairwise ties to count other than zero. So let's go back to
>> the original version there, as default, too:
>>
>> The (default) De-Cloned Copeland Score of candidate X is ...
>>
>> The Sum (over all candidates Y pairwise defeated by X) of F(Y)
>> Minus
>> The Sum (over all Z that pairwise defeat X) of R(Z)
>>
>> Now continuing on with "Loggerheads" .... since the two players have
>> polar opposite preferences, it seems that their optimal strategies must be
>> maximally resistant to manipulation... your optimal defensive strategy
>> against your most antagonistic enemy should hold up against lesser foes, as
>> well!
>>
>> At least that is my basic heuristic for this method.
>>
>> The first Condorcet Lottery method that we learned about, nearly two
>> decades ago, disappointly turned out to be non-monotonic, as did the more
>> advanced Rivest method that incorporated pairwise defeat scores into the
>> payoff matrix.
>>
>> It seems that the problem was the same basic problem we faced when trying
>> to preserve monotonicity while de-cloning Kemeny-Young, Borda, and Copeland.
>>
>> Our recent (last week) breakthrough in that context is the impetus for
>> this Loggerhead method.
>>
>> One way of looking at the breakthrough is this: making a clear
>> distinction between passive lack of approval and active disapproval allows
>> us to de-couple mono-raising of one candidate from lowering (mono or
>> otherwise) of another candidate.
>>
>> In our original unsuccessful versions we did not distinguish the role of
>> F from the role of R.  There we just used "lack of F" as a proxy for R.
>>
>> Fixing that crucial defect not only made monotonicity possible, but also,
>> as a pleasant surprise,  made possible the strong reverse symmetry enjoyed
>> by all of these new methods.
>>
>> Some people resist lotteries as legitimate election methods, but if, as
>> we have been assured by our RCV friends the 440 real life elections they
>> analyzed all enjoyed Condorcet Winners, irrespective of employing a
>> non-Condorcet compliant method ... almost all of these lotteries will be
>> zero entropy lotteries ... the possibility of chance serving only as a
>> deterrent to insincere rankings.
>>
>> And suppose that a sincere rock, paper, scissors cycle should exist....
>> it is comforting to know that the support of the winning lottery is always
>> a subset of the Dutta Set, a kind of special subset of the better known
>> Banks, Landau, and Smith sets.
>>
>> It has often been suggested that in the absence of a sincere CW, the best
>> thing might be to choose randomly from the Smith Set.  Well, that's
>> precisely what this Loggerheads method does ... and with probabilities
>> calculated to make sincere voting optimal.
>>
>> We'll continue when I get some more free time.
>>
>> In the mean time, somebody in contact with James Green-Armytage could
>> help by passing this message along to him ... I seem to remember him
>> expressing interest in the Rivest Lottery recently. It would be nice to get
>> him, and others with a game theoretic bent, thinking along these lines.
>>
>> Forest
>>
>> El lun., 17 de ene. de 2022 12:17 a. m., Forest Simmons <
>> forest.simmons21 at gmail.com> escribió:
>>
>>> I would like to propose this Rivest-like two-player, zero-sum game
>>> related to the de-cloned versions of Kemeny-Young, Borda, and Copeland that
>>> I recently posted.
>>>
>>> For each candidate k, let F(k) be the random ballot Favorite probability
>>> of candidate k, and let R(k) be the random ballot favorite of candidate k
>>> on the Reversed ballots.
>>>
>>> Let P be the payoff matrix for the row player defined as follows:
>>>
>>> P(i, j) is F(j) if candidate i pairwise defeats j.
>>> P(i, j) is -R(i) if candidate i is pairwise defeated by j.
>>> P(i,  j) is F(j)-R(i) if candidates i and j are pairwise tied, including
>>> the case of i=j.
>>>
>>> Remember the game is zero sum, so the column player's payoff is the
>>> opposite of the row player's payoff.
>>>
>>> In general optimal strategies for the players are stochastic mixtures of
>>> the respective pure deterministic strategies, i.e. they are Lotteries.
>>>
>>> Let L and L* be the respective optimal lotteries for the respective row
>>> and column players.
>>>
>>> L(k) and L*(k) are the probabilities with which the respective players
>>> should bet on row or column k.
>>>
>>> For the un-reversed ballots, the method winner is chosen by L.
>>>
>>> For the reversed ballots the winner is chosen by L*.
>>>
>>> That's the method ... more commentary next time....
>>>
>>> Forest
>>>
>>
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