[EM] My Recent New Methods Summary with Corrections

[Forest Simmons]

We start with Kemeny-Young:

The cost of replacing a rank order R with a new rank order R' is the sum of
the costs of all of the (non-gratuitous) transpositions (or half
transpositions when equal rankings and truncations are allowed) necessary
to effect the change.

The cost of convertng X<Y to X>Y is given by f*(X)f(Y), while the cost of
converting X>Y to X<Y is f*(Y)f(X).

Think of the original order as a ballot order and the final order as a
"social order." The cost represents the disappointment of the voter when
her will is contradicted.

More generally, if H is a set of equal ranked candidates, and K is another
such set, then the cost of raising H from below K to above K is

Sum(over h in H, and k in K)of f*(h)f(k)

Furthermore, half of that cost is from raising H to the level of K, and the
other half is from finishing the move.

Let c(R) be the total cost of converting (one by one) each of the ballots
to the order R.

The winning social order is argmin(c(R)).

Next is de-cloned Copeland ...

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 and 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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