FS
Forest Simmons
Mon, Jan 17, 2022 9:44 PM
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@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
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@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
>
FS
Forest Simmons
Mon, Jan 17, 2022 10:42 PM
One subtle detail that needs emphasis is that informally we say that F(X)
and R(X) are (respectively) the numbers or percentages of ballots that
respectively rank X first and last. But that is only precisely true in the
case of complete rankings. Otherwise go back to the definitions of random
ballot Favorite for F, and the Reverse of that for R, which could be called
anti-favorite.
Suppose you draw a ballot which leaves five candidates unranked. Which of
these is the anti-favorite? (assuming none of them is so designated) In
other words, how does this affect the calculation of their R values?
Answer: count them fractionally... each counts as one fifth of a last place.
The main purpose of F and R is to distinguish strong support and strong
rejection, while the secondary, but still crucial purpose is to ensure
clone independence. It is mostly this secondary aspect that requires the
precise fractional count.
El lun., 17 de ene. de 2022 1:44 p. m., Forest Simmons <
forest.simmons21@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@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
One subtle detail that needs emphasis is that informally we say that F(X)
and R(X) are (respectively) the numbers or percentages of ballots that
respectively rank X first and last. But that is only precisely true in the
case of complete rankings. Otherwise go back to the definitions of random
ballot Favorite for F, and the Reverse of that for R, which could be called
anti-favorite.
Suppose you draw a ballot which leaves five candidates unranked. Which of
these is the anti-favorite? (assuming none of them is so designated) In
other words, how does this affect the calculation of their R values?
Answer: count them fractionally... each counts as one fifth of a last place.
The main purpose of F and R is to distinguish strong support and strong
rejection, while the secondary, but still crucial purpose is to ensure
clone independence. It is mostly this secondary aspect that requires the
precise fractional count.
El lun., 17 de ene. de 2022 1:44 p. m., Forest Simmons <
forest.simmons21@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@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
>>
>
FS
Forest Simmons
Wed, Jan 19, 2022 1:57 AM
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 ...
- They are both probability density functions on the set of candidates.
- 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.
- 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.
- 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).
- 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@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@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
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@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@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
>>
>
FS
Forest Simmons
Wed, Jan 19, 2022 3:01 AM
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@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 ...
- They are both probability density functions on the set of candidates.
- 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.
- 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.
- 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).
- 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@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@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
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@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@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@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
>>>
>>
FS
Forest Simmons
Wed, Jan 19, 2022 3:19 AM
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@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 ...
- They are both probability density functions on the set of candidates.
- 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.
- 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.
- 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).
- 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@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@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
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@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@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@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
>>>
>>
FS
Forest Simmons
Wed, Jan 19, 2022 3:53 AM
Decloned-Borda
Think of Borda as Copeland applied at the ballot level with pairwise defeat
replaced by "out ranking" locally:
Ballot B contributes the following difference d(X) to X's Borda sum:
Sum(over candidates Y out ranked by X on ballot B) of f*(Y)
MINUS
Sum(over candidates Z that out-rank X on ballot B) of f(Y)
Let D(X) be Sum(over B in beta) of d(X).
Elect argmax(D(X)).
Here's the Pairwise matrix formulation of the exact same method:
Let P be the matrix whose j_th entry in its i_th row is the number of
ballots on which candidate i outranks candidate j.
Let D(X) be the difference given by ...
The f* expectation of the X_th row of P
MINUS
The f expectation of the X_th column of P.
This D(X) value is the same as the sum of the d(X) defined above.
Remark ... since we don't expect this method to be Condorcet Compliant, it
might be better to use the inclusive form of P ... P(i, j) is the number of
ballots on which j does not outrank i.
Next ... our Loggerheads Condorcet Lottery ...
El mar., 18 de ene. de 2022 7:19 p. m., Forest Simmons <
forest.simmons21@gmail.com> escribió:
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@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 ...
- They are both probability density functions on the set of candidates.
- 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.
- 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.
- 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).
- 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@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@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
Decloned-Borda
Think of Borda as Copeland applied at the ballot level with pairwise defeat
replaced by "out ranking" locally:
Ballot B contributes the following difference d(X) to X's Borda sum:
Sum(over candidates Y out ranked by X on ballot B) of f*(Y)
MINUS
Sum(over candidates Z that out-rank X on ballot B) of f(Y)
Let D(X) be Sum(over B in beta) of d(X).
Elect argmax(D(X)).
Here's the Pairwise matrix formulation of the exact same method:
Let P be the matrix whose j_th entry in its i_th row is the number of
ballots on which candidate i outranks candidate j.
Let D(X) be the difference given by ...
The f* expectation of the X_th row of P
MINUS
The f expectation of the X_th column of P.
This D(X) value is the same as the sum of the d(X) defined above.
Remark ... since we don't expect this method to be Condorcet Compliant, it
might be better to use the inclusive form of P ... P(i, j) is the number of
ballots on which j does not outrank i.
Next ... our Loggerheads Condorcet Lottery ...
El mar., 18 de ene. de 2022 7:19 p. m., Forest Simmons <
forest.simmons21@gmail.com> escribió:
> 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@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@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@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
>>>>
>>>
FS
Forest Simmons
Wed, Jan 19, 2022 4:32 AM
Finally, Loggerheads:
Let RowPayoff be the matrix whose (i, j) entry is f*(j), if candidate i
pairwise defeats j, Else -f(i) if j defeats i pairwise, Else 0 otherwise.
The column player payoffs are the opposite of the row players ... to make
the game zero sum.
The winner is determined by the optimum mixed strategy lottery of the row
player.
Note that I corrected the PayOff matrix to reflect the natural policy of
making the reward/bounty proportional to the evil of the bad guy.
That pretty much does it for the summary.
Comments and questions are welcome!
I'll answer to the best of my ability... but I'm no game theory expert!
-Forest
El mar., 18 de ene. de 2022 5:57 p. m., Forest Simmons <
forest.simmons21@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 ...
- They are both probability density functions on the set of candidates.
- 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.
- 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.
- 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).
- 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@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@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
Finally, Loggerheads:
Let RowPayoff be the matrix whose (i, j) entry is f*(j), if candidate i
pairwise defeats j, Else -f(i) if j defeats i pairwise, Else 0 otherwise.
The column player payoffs are the opposite of the row players ... to make
the game zero sum.
The winner is determined by the optimum mixed strategy lottery of the row
player.
Note that I corrected the PayOff matrix to reflect the natural policy of
making the reward/bounty proportional to the evil of the bad guy.
That pretty much does it for the summary.
Comments and questions are welcome!
I'll answer to the best of my ability... but I'm no game theory expert!
-Forest
El mar., 18 de ene. de 2022 5:57 p. m., Forest Simmons <
forest.simmons21@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@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@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
>>>
>>
FS
Forest Simmons
Wed, Jan 19, 2022 8:45 PM
As a bonus, let's see how easily the first of the following two
formulations of de-cloned Borda can be parlayed into a monotonic DSV
(Designated Strategy Voting) Approval method. See the comment inserted
inline below *********
El mar., 18 de ene. de 2022 7:53 p. m., Forest Simmons <
forest.simmons21@gmail.com> escribió:
Decloned-Borda
Think of Borda as Copeland applied at the ballot level with pairwise
defeat replaced by "out ranking" locally:
Ballot B contributes the following difference d(X) to X's Borda sum:
Sum(over candidates Y out ranked by X on ballot B) of f*(Y)
MINUS
Sum(over candidates Z that out-rank X on ballot B) of f(Y)
Let D(X) be Sum(over B in beta) of d(X).
Elect argmax(D(X)).
********* For DSV Approval, just replace this D(X), by A(X) defined as
Sum(over B in beta) of sign(d(X)),
and then elect argmax(A(X)).
This DSV Approval is a Universal Domain version of Approval, like Implicit
Approval, but in contrast to IA, at its best in the context of complete
rankings. Kristofer and Daniel have been testing other UD methods (and some
non-UD methods like ordinary Approval) for manipulability. It seems to me
that this compromise resistant UD method would compare favorably with any
other UD method, such as IRV, Benham, Ranked Pairs, MinMax, and even
Implicit Approval Chain Climbing (IACC) ...perhaps not as burial resistant
as IACC, but more compromise resistant ... if my intuition is correct.
Here's the Pairwise matrix formulation of the exact same method:
Let P be the matrix whose j_th entry in its i_th row is the number of
ballots on which candidate i outranks candidate j.
Let D(X) be the difference given by ...
The f* expectation of the X_th row of P
MINUS
The f expectation of the X_th column of P.
This D(X) value is the same as the sum of the d(X) defined above.
Remark ... since we don't expect this method to be Condorcet Compliant, it
might be better to use the inclusive form of P ... P(i, j) is the number of
ballots on which j does not outrank i.
Next ... our Loggerheads Condorcet Lottery ...
El mar., 18 de ene. de 2022 7:19 p. m., Forest Simmons <
forest.simmons21@gmail.com> escribió:
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@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 ...
- They are both probability density functions on the set of candidates.
- 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.
- 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.
- 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).
- 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@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@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
As a bonus, let's see how easily the first of the following two
formulations of de-cloned Borda can be parlayed into a monotonic DSV
(Designated Strategy Voting) Approval method. See the comment inserted
inline below *********
El mar., 18 de ene. de 2022 7:53 p. m., Forest Simmons <
forest.simmons21@gmail.com> escribió:
> Decloned-Borda
>
> Think of Borda as Copeland applied at the ballot level with pairwise
> defeat replaced by "out ranking" locally:
>
> Ballot B contributes the following difference d(X) to X's Borda sum:
>
> Sum(over candidates Y out ranked by X on ballot B) of f*(Y)
>
> MINUS
>
> Sum(over candidates Z that out-rank X on ballot B) of f(Y)
>
> Let D(X) be Sum(over B in beta) of d(X).
>
> Elect argmax(D(X)).
>
********* For DSV Approval, just replace this D(X), by A(X) defined as
Sum(over B in beta) of sign(d(X)),
and then elect argmax(A(X)).
This DSV Approval is a Universal Domain version of Approval, like Implicit
Approval, but in contrast to IA, at its best in the context of complete
rankings. Kristofer and Daniel have been testing other UD methods (and some
non-UD methods like ordinary Approval) for manipulability. It seems to me
that this compromise resistant UD method would compare favorably with any
other UD method, such as IRV, Benham, Ranked Pairs, MinMax, and even
Implicit Approval Chain Climbing (IACC) ...perhaps not as burial resistant
as IACC, but more compromise resistant ... if my intuition is correct.
>
> Here's the Pairwise matrix formulation of the exact same method:
>
> Let P be the matrix whose j_th entry in its i_th row is the number of
> ballots on which candidate i outranks candidate j.
>
> Let D(X) be the difference given by ...
> The f* expectation of the X_th row of P
> MINUS
> The f expectation of the X_th column of P.
>
> This D(X) value is the same as the sum of the d(X) defined above.
>
> Remark ... since we don't expect this method to be Condorcet Compliant, it
> might be better to use the inclusive form of P ... P(i, j) is the number of
> ballots on which j does not outrank i.
>
> Next ... our Loggerheads Condorcet Lottery ...
>
> El mar., 18 de ene. de 2022 7:19 p. m., Forest Simmons <
> forest.simmons21@gmail.com> escribió:
>
>> 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@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@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@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
>>>>>
>>>>
