STAR cloneproof variant based on Score Chain Climbing
Here's a proposal for a STAR variant that handles clones:
Top two score winners (A and B), plus the score winner when you exclude the
top score winner ballots (by score weight), that's the clone-proof part.
Call that the exclusive winner, X. In other words, if a ballot gives a
score of 5 out of 10 to the top score winner, remove half that ballot's
weight.
Eliminate any candidates defeated by X. If more than one remains, the
winner is the one who defeats the other.
This follows the logic of Forest Simmons' Score Chain Climbing to resolve
cycles.
The clone problem is that A and B could be clones. Removing A's ballot
contributors finds the non-clone while avoiding pushover incentive. If X
defeats both A and B, it's likely the CW. Otherwise, whichever of A or B is
defeated by X is "weaker" (low probability, but possible in cycles). Using
a lower-scoring candidate as an eliminator reduces burial incentive.
Why do I propose finding X that way instead of by a Hare or Droop quota?
Well, for one thing, it's a summable process. Next, if A has >50% approval,
A and B are probably the two candidates to choose from anyway. If A has
<50% approval, X is being found with something like a Hare quota, moving
more toward Droop as A's approval decreases.
Your thoughts?
Hi Ted,
Ted Stern wrote:
Here's a proposal for a STAR variant that handles clones:
Top two score winners (A and B), plus the score winner when you exclude the
top score winner ballots (by score weight), that's the clone-proof part. Call
that the exclusive winner, X. In other words, if a ballot gives a score of 5
out of 10 to the top score winner, remove half that ballot's weight.
Is it allowed for B and X to be the same candidate? (I'm not sure it's totally
ruled out that even A and X could be the same.)
If a ballot gave A a 10/10, I assume all weight is removed when determining X.
Eliminate any candidates defeated by X. If more than one remains, the winner
is the one who defeats the other.
When you say "if more than one remains" I assume you mean "of A and B." Or
possibly "of A, B, and X."
When you say "the winner is the one who defeats the other," this is a second
round of voting, correct? So in all cases of this second round it will be
between two of {X,A,B}?
This follows the logic of Forest Simmons' Score Chain Climbing to resolve
cycles.
The clone problem is that A and B could be clones. Removing A's ballot
contributors finds the non-clone while avoiding pushover incentive. If X
defeats both A and B, it's likely the CW. Otherwise, whichever of A or B is
defeated by X is "weaker" (low probability, but possible in cycles).
Using a lower-scoring candidate as an eliminator reduces burial incentive.
Not sure I follow. The claim is counter-intuitive because usually a candidate to
be given artificial preferences is a weaker one (lower-scoring) because the
insincere voters believe this candidate can't win, and don't wish for him to win.
For example in your case if either A or B (doesn't matter) believes they will
beat X pairwise then they are able to try to use X to knock the other of A/B
out. Just rate X a 1/10, and the other 0/10. No?
As I say, I think it's not chain climbing generally, but TACC(implicit)
specifically, that has anti-burial value.
Why do I propose finding X that way instead of by a Hare or Droop quota?
Well, for one thing, it's a summable process. Next, if A has >50% approval,
A and B are probably the two candidates to choose from anyway. If A has
<50% approval, X is being found with something like a Hare quota, moving
more toward Droop as A's approval decreases.
Your thoughts?
I think the goal of decloning the election is a bit different from the goal of
selecting the "best" two finalists for a runoff (if this method actually has a
runoff, wasn't sure there). So I'm not so worried if you don't use some more
"proper" decloning method like a quota.
Kevin
Replies inline below:
On Tue, Feb 15, 2022 at 12:09 PM Kevin Venzke stepjak@yahoo.fr wrote:
Hi Ted,
Ted Stern wrote:
Here's a proposal for a STAR variant that handles clones:
Top two score winners (A and B), plus the score winner when you exclude
the
top score winner ballots (by score weight), that's the clone-proof part.
Call
that the exclusive winner, X. In other words, if a ballot gives a score
of 5
out of 10 to the top score winner, remove half that ballot's weight.
Is it allowed for B and X to be the same candidate? (I'm not sure it's
totally
ruled out that even A and X could be the same.)
In the absence of clones, X could very easily be B.
If a ballot gave A a 10/10, I assume all weight is removed when
determining X.
Exactly. A ballot voting for A gets A-score/max-score removed.
Eliminate any candidates defeated by X. If more than one remains, the
winner
is the one who defeats the other.
When you say "if more than one remains" I assume you mean "of A and B." Or
possibly "of A, B, and X."
Assuming A, B and X are distinct:
If X defeats both A and B, only X remains and X wins.
If X is defeated by A, but defeats B, B is removed, and A wins.
If X is defeated by B, but defeats A, A is removed and B wins.
When you say "the winner is the one who defeats the other," this is a
second
round of voting, correct? So in all cases of this second round it will be
between two of {X,A,B}?
See above
This follows the logic of Forest Simmons' Score Chain Climbing to resolve
cycles.
The clone problem is that A and B could be clones. Removing A's ballot
contributors finds the non-clone while avoiding pushover incentive. If X
defeats both A and B, it's likely the CW. Otherwise, whichever of A or B
is
defeated by X is "weaker" (low probability, but possible in cycles).
Using a lower-scoring candidate as an eliminator reduces burial
incentive.
Not sure I follow. The claim is counter-intuitive because usually a
candidate to
be given artificial preferences is a weaker one (lower-scoring) because the
insincere voters believe this candidate can't win, and don't wish for him
to win.
For example in your case if either A or B (doesn't matter) believes they
will
beat X pairwise then they are able to try to use X to knock the other of
A/B
out. Just rate X a 1/10, and the other 0/10. No?
As I say, I think it's not chain climbing generally, but TACC(implicit)
specifically, that has anti-burial value.
I'd love to see an example of what you're suggesting. Can you think of
one? I don't necessarily need my proposal to win, I'm just looking for the
best way to cloneproof STAR while still selecting the best candidates.
Why do I propose finding X that way instead of by a Hare or Droop quota?
Well, for one thing, it's a summable process. Next, if A has >50%
approval,
A and B are probably the two candidates to choose from anyway. If A has
<50% approval, X is being found with something like a Hare quota, moving
more toward Droop as A's approval decreases.
Your thoughts?
I think the goal of decloning the election is a bit different from the
goal of
selecting the "best" two finalists for a runoff (if this method actually
has a
runoff, wasn't sure there). So I'm not so worried if you don't use some
more
"proper" decloning method like a quota.
Kevin
One more comment, below.
On Tue, Feb 15, 2022, 12:51 Ted Stern dodecatheon@gmail.com wrote:
Replies inline below:
On Tue, Feb 15, 2022 at 12:09 PM Kevin Venzke stepjak@yahoo.fr wrote:
Hi Ted,
Ted Stern wrote:
Here's a proposal for a STAR variant that handles clones:
Top two score winners (A and B), plus the score winner when you exclude
the
top score winner ballots (by score weight), that's the clone-proof
part. Call
that the exclusive winner, X. In other words, if a ballot gives a score
of 5
out of 10 to the top score winner, remove half that ballot's weight.
Is it allowed for B and X to be the same candidate? (I'm not sure it's
totally
ruled out that even A and X could be the same.)
In the absence of clones, X could very easily be B.
If a ballot gave A a 10/10, I assume all weight is removed when
determining X.
Exactly. A ballot voting for A gets A-score/max-score removed.
Eliminate any candidates defeated by X. If more than one remains, the
winner
is the one who defeats the other.
When you say "if more than one remains" I assume you mean "of A and B." Or
possibly "of A, B, and X."
Assuming A, B and X are distinct:
If X defeats both A and B, only X remains and X wins.
If X is defeated by A, but defeats B, B is removed, and A wins.
If X is defeated by B, but defeats A, A is removed and B wins.
And if both A and B defeat X, you get regular STAR, and the winner of A vs.
B wins.
When you say "the winner is the one who defeats the other," this is a
second
round of voting, correct? So in all cases of this second round it will be
between two of {X,A,B}?
See above
This follows the logic of Forest Simmons' Score Chain Climbing to
resolve
cycles.
The clone problem is that A and B could be clones. Removing A's ballot
contributors finds the non-clone while avoiding pushover incentive. If X
defeats both A and B, it's likely the CW. Otherwise, whichever of A or
B is
defeated by X is "weaker" (low probability, but possible in cycles).
Using a lower-scoring candidate as an eliminator reduces burial
incentive.
Not sure I follow. The claim is counter-intuitive because usually a
candidate to
be given artificial preferences is a weaker one (lower-scoring) because
the
insincere voters believe this candidate can't win, and don't wish for him
to win.
For example in your case if either A or B (doesn't matter) believes they
will
beat X pairwise then they are able to try to use X to knock the other of
A/B
out. Just rate X a 1/10, and the other 0/10. No?
As I say, I think it's not chain climbing generally, but TACC(implicit)
specifically, that has anti-burial value.
I'd love to see an example of what you're suggesting. Can you think of
one? I don't necessarily need my proposal to win, I'm just looking for the
best way to cloneproof STAR while still selecting the best candidates.
Why do I propose finding X that way instead of by a Hare or Droop quota?
Well, for one thing, it's a summable process. Next, if A has >50%
approval,
A and B are probably the two candidates to choose from anyway. If A has
<50% approval, X is being found with something like a Hare quota, moving
more toward Droop as A's approval decreases.
Your thoughts?
I think the goal of decloning the election is a bit different from the
goal of
selecting the "best" two finalists for a runoff (if this method actually
has a
runoff, wasn't sure there). So I'm not so worried if you don't use some
more
"proper" decloning method like a quota.
Kevin
Hi Ted,
Le mardi 15 février 2022, 14:52:03 UTC−6, Ted Stern dodecatheon@gmail.com a écrit :
The clone problem is that A and B could be clones. Removing A's ballot
contributors finds the non-clone while avoiding pushover incentive. If X
defeats both A and B, it's likely the CW. Otherwise, whichever of A or B is
defeated by X is "weaker" (low probability, but possible in cycles).
Using a lower-scoring candidate as an eliminator reduces burial incentive.
Not sure I follow. The claim is counter-intuitive because usually a candidate to
be given artificial preferences is a weaker one (lower-scoring) because the
insincere voters believe this candidate can't win, and don't wish for him to win.
For example in your case if either A or B (doesn't matter) believes they will
beat X pairwise then they are able to try to use X to knock the other of A/B
out. Just rate X a 1/10, and the other 0/10. No?
As I say, I think it's not chain climbing generally, but TACC(implicit)
specifically, that has anti-burial value.
I'd love to see an example of what you're suggesting. Can you think of one? I
don't necessarily need my proposal to win, I'm just looking for the best way to
cloneproof STAR while still selecting the best candidates.
It might not be easy to make a natural-looking scenario. I think this one is at
least correct:
0.235: A 10 -------> A 10, X 1
0.260: A 10, B 10
0.237: B 10, A 1
0.265: X 10
I believe the vote change of the first bloc moves the win from B to A. If this
is wrong, I'll take another stab at it.
I'm not saying the method is flawed, I just wanted to scrutinize the motivation.
The burial analysis is a little complicated since B can be X, and it's not a
Condorcet method.
Kevin
Thanks, Kevin, that's an interesting scenario. Back to the drawing board ...
On Wed, Feb 16, 2022 at 12:58 AM Kevin Venzke stepjak@yahoo.fr wrote:
Hi Ted,
Le mardi 15 février 2022, 14:52:03 UTC−6, Ted Stern dodecatheon@gmail.com
a écrit :
The clone problem is that A and B could be clones. Removing A's ballot
contributors finds the non-clone while avoiding pushover incentive. If
X
defeats both A and B, it's likely the CW. Otherwise, whichever of A or
B is
defeated by X is "weaker" (low probability, but possible in cycles).
Using a lower-scoring candidate as an eliminator reduces burial
incentive.
Not sure I follow. The claim is counter-intuitive because usually a
candidate to
be given artificial preferences is a weaker one (lower-scoring) because
the
insincere voters believe this candidate can't win, and don't wish for
him to win.
For example in your case if either A or B (doesn't matter) believes
they will
beat X pairwise then they are able to try to use X to knock the other
of A/B
out. Just rate X a 1/10, and the other 0/10. No?
As I say, I think it's not chain climbing generally, but TACC(implicit)
specifically, that has anti-burial value.
I'd love to see an example of what you're suggesting. Can you think of
one? I
don't necessarily need my proposal to win, I'm just looking for the best
way to
cloneproof STAR while still selecting the best candidates.
It might not be easy to make a natural-looking scenario. I think this one
is at
least correct:
0.235: A 10 -------> A 10, X 1
0.260: A 10, B 10
0.237: B 10, A 1
0.265: X 10
I believe the vote change of the first bloc moves the win from B to A. If
this
is wrong, I'll take another stab at it.
I'm not saying the method is flawed, I just wanted to scrutinize the
motivation.
The burial analysis is a little complicated since B can be X, and it's not
a
Condorcet method.
Kevin
Ted,
Your brainstorming, if may call it that, is infectious.
As you know Score Chain Climbing produces a maximal totally ordered subset
of candidates ordered by pairwise defeat ... every candidate in the chain
pairwise defeats all of the other chain members below it, and the chain
cannot be extended upward while retaining this total order.
In fact, the chain consists of the successive values of the candidate
variable X in the following formulation of Score Chain Climbing:
SCC:
Initialize a variable X as the (name of) the lowest score candidate. Then
...
While more than one candidate remains, eliminate all of the candidates
pairwise defeated by X, before storing a new name into X, the name of the
lowest score remaining candidate.
EndWhile
The last value of X (the SCC winner Xf) is one of the finalists.
The other finalist is the second to the last value of X, which we designate
Xf'.
But doesn't the last X defeat all of the previous X's?
Yes, according to the ballots. But there is a good chance that the only
reason Xf defeats Xf' on the ballots is that Xf' was insincerely buried
under Xf.
So how do we vindicate (or expose as fraudulent) the finalist Xf?
We could take another trip to the polls for a runoff between between Xf and
Xf'.
Otherwise, we can require voters to submit two ballots ... one to determine
the two finalists, and the other to choose between them.
Sincere voters simply duplicate their first ballot to produce their second
one. The strategy burdened voters adjust their insincerities to produce
their second ballot.
It is crucial that the second ballot be used exclusively for choosing the
winner between the two finalists.
However, once the final winner has been certified , these ballots can be
used for forensics.
-Forest
El mar., 15 de feb. de 2022 9:48 a. m., Ted Stern dodecatheon@gmail.com
escribió:
Here's a proposal for a STAR variant that handles clones:
Top two score winners (A and B), plus the score winner when you exclude
the top score winner ballots (by score weight), that's the clone-proof
part. Call that the exclusive winner, X. In other words, if a ballot gives
a score of 5 out of 10 to the top score winner, remove half that ballot's
weight.
Eliminate any candidates defeated by X. If more than one remains, the
winner is the one who defeats the other.
This follows the logic of Forest Simmons' Score Chain Climbing to resolve
cycles.
The clone problem is that A and B could be clones. Removing A's ballot
contributors finds the non-clone while avoiding pushover incentive. If X
defeats both A and B, it's likely the CW. Otherwise, whichever of A or B is
defeated by X is "weaker" (low probability, but possible in cycles). Using
a lower-scoring candidate as an eliminator reduces burial incentive.
Why do I propose finding X that way instead of by a Hare or Droop quota?
Well, for one thing, it's a summable process. Next, if A has >50% approval,
A and B are probably the two candidates to choose from anyway. If A has
<50% approval, X is being found with something like a Hare quota, moving
more toward Droop as A's approval decreases.
Your thoughts?
Hi Forest,
Le vendredi 11 mars 2022, 23:03:30 UTC−6, Forest Simmons forest.simmons21@gmail.com a écrit :
SCC:
Initialize a variable X as the (name of) the lowest score candidate. Then ...
While more than one candidate remains, eliminate all of the candidates pairwise
defeated by X, before storing a new name into X, the name of the lowest score
remaining candidate.
EndWhile
The last value of X (the SCC winner Xf) is one of the finalists.
The other finalist is the second to the last value of X, which we designate Xf'.
For the case that the initial value of X is the CW, should an elimination order
be specified?
But doesn't the last X defeat all of the previous X's?
Yes, according to the ballots. But there is a good chance that the only reason
Xf defeats Xf' on the ballots is that Xf' was insincerely buried under Xf.
In my terminology, that would mean Xf' is the sincere CW and Xf is the "pawn."
The strategists' own candidate (the "rival") has been eliminated, so their
strategy failed (and would be a backfire, if the last X simply won).
This probably implies that the sincere CW was unexpectedly the Score loser.
So how do we vindicate (or expose as fraudulent) the finalist Xf?
We could take another trip to the polls for a runoff between between Xf and Xf'.
Otherwise, we can require voters to submit two ballots ... one to determine the
two finalists, and the other to choose between them.
Sincere voters simply duplicate their first ballot to produce their second one.
The strategy burdened voters adjust their insincerities to produce their second
ballot.
It is crucial that the second ballot be used exclusively for choosing the winner
between the two finalists.
However, once the final winner has been certified , these ballots can be used
for forensics.
All true. It seems like the effect of this is to make "backfired strategy"
outcomes impossible. Is that the goal? It seems like that might risk encouraging
voters to try burial strategies, unless it's sufficient to "name and shame"
strategists through the forensics performed afterwards.
It seems like this proposal could even prevent a backfire when both of two
major factions are ranking the same pawn insincerely high, so that the pawn
becomes the voted CW.
Kevin
Thanks, Kevin. It was a comment of yours that made me realize that burial
punishment (via chain climbing) was not enough ... but of course, looking
away and pretending the burier was probably sincere ... that is no good
either.
So a sincerity check is natural ... if the sincere ballots contradict the
strategic ballots, then in this case you have both detection and correction.
Perhaps we could forget Chain Climbing and just use the sincerity check on
the weakest defeat that was critical in determining the winner.
El vie., 11 de mar. de 2022 11:42 p. m., Kevin Venzke stepjak@yahoo.fr
escribió:
Hi Forest,
Le vendredi 11 mars 2022, 23:03:30 UTC−6, Forest Simmons <
forest.simmons21@gmail.com> a écrit :
SCC:
Initialize a variable X as the (name of) the lowest score candidate.
Then ...
While more than one candidate remains, eliminate all of the candidates
pairwise
defeated by X, before storing a new name into X, the name of the lowest
score
remaining candidate.
EndWhile
The last value of X (the SCC winner Xf) is one of the finalists.
The other finalist is the second to the last value of X, which we
designate Xf'.
For the case that the initial value of X is the CW, should an elimination
order
be specified?
But doesn't the last X defeat all of the previous X's?
Yes, according to the ballots. But there is a good chance that the only
reason
Xf defeats Xf' on the ballots is that Xf' was insincerely buried under
Xf.
In my terminology, that would mean Xf' is the sincere CW and Xf is the
"pawn."
The strategists' own candidate (the "rival") has been eliminated, so their
strategy failed (and would be a backfire, if the last X simply won).
This probably implies that the sincere CW was unexpectedly the Score loser.
So how do we vindicate (or expose as fraudulent) the finalist Xf?
We could take another trip to the polls for a runoff between between Xf
and Xf'.
Otherwise, we can require voters to submit two ballots ... one to
determine the
two finalists, and the other to choose between them.
Sincere voters simply duplicate their first ballot to produce their
second one.
The strategy burdened voters adjust their insincerities to produce their
second
ballot.
It is crucial that the second ballot be used exclusively for choosing
the winner
between the two finalists.
However, once the final winner has been certified , these ballots can be
used
for forensics.
All true. It seems like the effect of this is to make "backfired strategy"
outcomes impossible. Is that the goal? It seems like that might risk
encouraging
voters to try burial strategies, unless it's sufficient to "name and
shame"
strategists through the forensics performed afterwards.
It seems like this proposal could even prevent a backfire when both of
two
major factions are ranking the same pawn insincerely high, so that the pawn
becomes the voted CW.
Kevin
Reviving an old topic.
I had another thought for clone proofing STAR or Top Two Approval runoff,
based on SPAV or RRV. It is not summable.
Round one:
Approval or Score ballots.
Advance the top two approved or top two total score candidates.
Advance a third candidate using SPAV with Approval ballots or RRV if score
ballots, as if the two already advanced candidates were chosen by that
method. That is, if using approval, a ballot's weight for the third count
has weight 1 if it approved no previous winners, 1/2 if it approved one of
the first two winners, or 1/3 if it approved both the previous winners.
Similarly for score ballots using RRV.
If round one uses Approval ballots, a score ballot runoff will be held. If
using Score, the round one ballots can be recounted to find pairwise
preferences, using ratings to infer rankings. Or a separate score runoff
could also be held with the three winners, which I prefer.
In round two, use a cloneproof and burial resistant Condorcet method. Let's
say, score sorted margins or score chain climbing.
The difference from my first proposal is that there are always 3
candidates, representing either 2 factions, if top two are clones, or 3
factions, if not cloned.
My overall preference is for an approval first round, to eliminate most
candidates, then the cloneproof third candidate will generally represent a
different perspective to be debated before the runoff.
On Sat, Mar 12, 2022, 00:17 Forest Simmons forest.simmons21@gmail.com
wrote:
Thanks, Kevin. It was a comment of yours that made me realize that burial
punishment (via chain climbing) was not enough ... but of course, looking
away and pretending the burier was probably sincere ... that is no good
either.
So a sincerity check is natural ... if the sincere ballots contradict the
strategic ballots, then in this case you have both detection and correction.
Perhaps we could forget Chain Climbing and just use the sincerity check on
the weakest defeat that was critical in determining the winner.
El vie., 11 de mar. de 2022 11:42 p. m., Kevin Venzke stepjak@yahoo.fr
escribió:
Hi Forest,
Le vendredi 11 mars 2022, 23:03:30 UTC−6, Forest Simmons <
forest.simmons21@gmail.com> a écrit :
SCC:
Initialize a variable X as the (name of) the lowest score candidate.
Then ...
While more than one candidate remains, eliminate all of the candidates
pairwise
defeated by X, before storing a new name into X, the name of the lowest
score
remaining candidate.
EndWhile
The last value of X (the SCC winner Xf) is one of the finalists.
The other finalist is the second to the last value of X, which we
designate Xf'.
For the case that the initial value of X is the CW, should an elimination
order
be specified?
But doesn't the last X defeat all of the previous X's?
Yes, according to the ballots. But there is a good chance that the only
reason
Xf defeats Xf' on the ballots is that Xf' was insincerely buried under
Xf.
In my terminology, that would mean Xf' is the sincere CW and Xf is the
"pawn."
The strategists' own candidate (the "rival") has been eliminated, so their
strategy failed (and would be a backfire, if the last X simply won).
This probably implies that the sincere CW was unexpectedly the Score
loser.
So how do we vindicate (or expose as fraudulent) the finalist Xf?
We could take another trip to the polls for a runoff between between Xf
and Xf'.
Otherwise, we can require voters to submit two ballots ... one to
determine the
two finalists, and the other to choose between them.
Sincere voters simply duplicate their first ballot to produce their
second one.
The strategy burdened voters adjust their insincerities to produce
their second
ballot.
It is crucial that the second ballot be used exclusively for choosing
the winner
between the two finalists.
However, once the final winner has been certified , these ballots can
be used
for forensics.
All true. It seems like the effect of this is to make "backfired strategy"
outcomes impossible. Is that the goal? It seems like that might risk
encouraging
voters to try burial strategies, unless it's sufficient to "name and
shame"
strategists through the forensics performed afterwards.
It seems like this proposal could even prevent a backfire when both of
two
major factions are ranking the same pawn insincerely high, so that the
pawn
becomes the voted CW.
Kevin