FS
Forest Simmons
Mon, Oct 31, 2016 11:12 PM
Good suggestion. Here's my first crack at it:
If you write 100 next to a candidate's name, your ballot will contribute
max support (one point) to that candidate's approval, unconditionally.
If you write zero next to the name of candidate X, your ballot will not
contribute anything to X's approval.
If you write 70 next to a candidate, your ballot will contribute one point
to that candidate's approval unless more than 70 percent of the voters
write a larger number next to her name.
[Your ballot does not contribute to over zealous support.]
If you write 50 next to the name of candidate X, your ballot will
contribute one point to X's approval, unless moe than 50% of the ballots
write a number larger than 50 next to her name.
If you write 20 next to the name of candidate X, your ballot will
contribute one point to X's approval, unless moe than 20% of the ballots
write a number larger than 50 next to her name.
Etc.
The candidate with the greatest total approval is elected.
On Sat, Oct 29, 2016 at 5:09 PM, Michael Ossipoff email9648742@gmail.com
wrote:
It seems to me that if there's a way to introduce & explain XA to people,
it starts out like this:
"With XA, when you assign a number, it isn't just a merit-rating.
"If you write ".9" next to a candidate's name, you're saying that you want
for .......to........"
.That's as far as I got.
The left-out parts should refer to something directly affecting the matter
of who wins. It should be brief & simple, for a clear, easy, natural &
intuitive introduction & explanation.
Michael Ossipoff
On Thu, Oct 27, 2016 at 5:56 PM, Forest Simmons fsimmons@pcc.edu wrote:
It turns out that Chiastic Approval is a good method in the context of
the Chicken Dilemma, much better than ordinary Approval, Majority Judgment,
or plain Range.
Ballots are score/range style ratings. Let x be the greatest number for
which there is some candidate that is given a rating of at least x percent
on at least x percent of the ballots. Elect the candidate X that is
given a rating of at least x percent on the greatest number of ballots.
The Greek letter Chi corresponds to the Roman letter X,, hence the name
Chiastic Approval or XA for short. Furthermore, when the method is
described graphically, the value of x is found by intersecting two graphs
whose union looks like the letter Chi.
Andy Jennings came up with XA while thinking about how to improve
Majority Judgement. Since we were both familiar with ancient literary
structures called Chiasms (identified in the Book of Mormon about 15
decades after its first publication) the name came naturally.
Skip the following technical paragraph unless you are very curious about
the graphical description.
[Let f be the function given by f(x) = the percentage of ballots on which
X is given a rating of at least x percent. Then f is a decreasing
function whose graph looks like the downward stroke of the letter Chi. The
graph of y = x looks like the stroke with positive slope. These two
graphs cross at the point (x, x) which yields the Chiastic Approval cutoff
x.]
Now consider the following ballot profile …
41 C
31 A>B(33%)
28 B>A(50%)
Note that A is the only candidate with a rating of at least 50% on at
least 50% of the ballots, so A is the XA winner.
We could lower the 50% to 42%, and raise the 33% to 40%, and A would
still be the XA winner, as the only candidate with a rating of at least 42%
on at least 42% of the ballots.
In fact we could go further than that by splitting up the the 28 B>A
faction with some die hard defectors:
41 C
31 A>B(40%)
11 B>A(42%)
17 B
Candidate A is still the only candidate given a rating of at least 42% on
at least 42 percent of the ballots.
But if two more B faction voters defect, then C is elected as the only
candidate given a rating of at least 41 percent on at least 41 percent of
the ballots:
41 C
31 A>B(40%)
8 B>A(42%)
20 B
In the general CD set up we have three factions with sincere preference
profiles
P: C
Q: A>B
R: B>A
Where P > Q > R>0, and P+Q+R=100
Under Chiastic Approval there is a Nash equilibrium that protects the
sincere CW candidate A :
P: C
Q: A>B(33%)
R: B>A(50%)
Candidate A is the only candidate rated at a level of at least 50% on at
least 50% of the ballots.
As in the first example, the equilibrium is preserved if the 33% is
raised to any value less than P%, and/or the 50% is lowered to any value
greater than P percent.
P: C
Q: A>B(P%-epsilon)
R: B>A(P%+epsilon)
Furthermore part of the B>A faction can defect without destroying this
equilibrium:
P: C
Q: A>B(P%-epsilon)
R1: B>A(P%+epsilon)
R2: B
For R=R1+R2 as long as R1 > P – Q .
So we see that XA has a rather robust Nash equilibrium that protects the
CWs in the context of a Chicken Dilemma threat. The threatened faction
down-rates the candidate of the potential defectors to any value less than
P%. Since (in this context) P is always greater than 33 (otherwise it
could not be the largest of the three factions), the 33 percent rating can
always be safely used to deter the defection. Mainly psychological
reasons would make it more satisfactory to raise that 33% closer to P%.
So we see that high resolution ratings are not needed. Four levels will
suffice nicely if they are 0, 33%, 50%, and 100%. Grade ballots like those
used for Majority Judgement could be adapted to XA.
As an approval variant like Bucklin, XA has no vulnerability to burial
tactics.
Unlike MMPO it also satisfies Plurality.
It is monotone and clone independent (in the sense that Approval and
Range are clone independent).
It is efficiently summable, but is it precinct consistent? i.e. does a
candidate that wins in every precinct win over-all?
Does it satisfy Participation?
We need to explore it, and learn how to explain it as simply as possible,
so we can persuade people to use it.
Forest
Good suggestion. Here's my first crack at it:
If you write 100 next to a candidate's name, your ballot will contribute
max support (one point) to that candidate's approval, unconditionally.
If you write zero next to the name of candidate X, your ballot will not
contribute anything to X's approval.
If you write 70 next to a candidate, your ballot will contribute one point
to that candidate's approval unless more than 70 percent of the voters
write a larger number next to her name.
[Your ballot does not contribute to over zealous support.]
If you write 50 next to the name of candidate X, your ballot will
contribute one point to X's approval, unless moe than 50% of the ballots
write a number larger than 50 next to her name.
If you write 20 next to the name of candidate X, your ballot will
contribute one point to X's approval, unless moe than 20% of the ballots
write a number larger than 50 next to her name.
Etc.
The candidate with the greatest total approval is elected.
On Sat, Oct 29, 2016 at 5:09 PM, Michael Ossipoff <email9648742@gmail.com>
wrote:
> It seems to me that if there's a way to introduce & explain XA to people,
> it starts out like this:
>
> "With XA, when you assign a number, it isn't just a merit-rating.
>
> "If you write ".9" next to a candidate's name, you're saying that you want
> for .......to........"
>
> .That's as far as I got.
>
> The left-out parts should refer to something directly affecting the matter
> of who wins. It should be brief & simple, for a clear, easy, natural &
> intuitive introduction & explanation.
>
> Michael Ossipoff
>
>
> On Thu, Oct 27, 2016 at 5:56 PM, Forest Simmons <fsimmons@pcc.edu> wrote:
>
>> It turns out that Chiastic Approval is a good method in the context of
>> the Chicken Dilemma, much better than ordinary Approval, Majority Judgment,
>> or plain Range.
>>
>>
>> Ballots are score/range style ratings. Let x be the greatest number for
>> which there is some candidate that is given a rating of at least x percent
>> on at least x percent of the ballots. Elect the candidate X that is
>> given a rating of at least x percent on the greatest number of ballots.
>>
>>
>> The Greek letter Chi corresponds to the Roman letter X,, hence the name
>> Chiastic Approval or XA for short. Furthermore, when the method is
>> described graphically, the value of x is found by intersecting two graphs
>> whose union looks like the letter Chi.
>>
>>
>> Andy Jennings came up with XA while thinking about how to improve
>> Majority Judgement. Since we were both familiar with ancient literary
>> structures called Chiasms (identified in the Book of Mormon about 15
>> decades after its first publication) the name came naturally.
>>
>>
>> Skip the following technical paragraph unless you are very curious about
>> the graphical description.
>>
>>
>> [Let f be the function given by f(x) = the percentage of ballots on which
>> X is given a rating of at least x percent. Then f is a decreasing
>> function whose graph looks like the downward stroke of the letter Chi. The
>> graph of y = x looks like the stroke with positive slope. These two
>> graphs cross at the point (x, x) which yields the Chiastic Approval cutoff
>> x.]
>>
>>
>> Now consider the following ballot profile …
>>
>> 41 C
>>
>> 31 A>B(33%)
>>
>> 28 B>A(50%)
>>
>>
>> Note that A is the only candidate with a rating of at least 50% on at
>> least 50% of the ballots, so A is the XA winner.
>>
>>
>> We could lower the 50% to 42%, and raise the 33% to 40%, and A would
>> still be the XA winner, as the only candidate with a rating of at least 42%
>> on at least 42% of the ballots.
>>
>>
>> In fact we could go further than that by splitting up the the 28 B>A
>> faction with some die hard defectors:
>>
>> 41 C
>>
>> 31 A>B(40%)
>>
>> 11 B>A(42%)
>>
>> 17 B
>>
>>
>> Candidate A is still the only candidate given a rating of at least 42% on
>> at least 42 percent of the ballots.
>>
>>
>> But if two more B faction voters defect, then C is elected as the only
>> candidate given a rating of at least 41 percent on at least 41 percent of
>> the ballots:
>>
>> 41 C
>>
>> 31 A>B(40%)
>>
>> 8 B>A(42%)
>>
>> 20 B
>>
>>
>> In the general CD set up we have three factions with sincere preference
>> profiles
>>
>> P: C
>>
>> Q: A>B
>>
>> R: B>A
>>
>>
>> Where P > Q > R>0, and P+Q+R=100
>>
>>
>> Under Chiastic Approval there is a Nash equilibrium that protects the
>> sincere CW candidate A :
>>
>> P: C
>>
>> Q: A>B(33%)
>>
>> R: B>A(50%)
>>
>>
>> Candidate A is the only candidate rated at a level of at least 50% on at
>> least 50% of the ballots.
>>
>>
>> As in the first example, the equilibrium is preserved if the 33% is
>> raised to any value less than P%, and/or the 50% is lowered to any value
>> greater than P percent.
>>
>> P: C
>>
>> Q: A>B(P%-epsilon)
>>
>> R: B>A(P%+epsilon)
>>
>>
>> Furthermore part of the B>A faction can defect without destroying this
>> equilibrium:
>>
>>
>>
>> P: C
>>
>> Q: A>B(P%-epsilon)
>>
>> R1: B>A(P%+epsilon)
>>
>> R2: B
>>
>> For R=R1+R2 as long as R1 > P – Q .
>>
>>
>> So we see that XA has a rather robust Nash equilibrium that protects the
>> CWs in the context of a Chicken Dilemma threat. The threatened faction
>> down-rates the candidate of the potential defectors to any value less than
>> P%. Since (in this context) P is always greater than 33 (otherwise it
>> could not be the largest of the three factions), the 33 percent rating can
>> always be safely used to deter the defection. Mainly psychological
>> reasons would make it more satisfactory to raise that 33% closer to P%.
>>
>>
>> So we see that high resolution ratings are not needed. Four levels will
>> suffice nicely if they are 0, 33%, 50%, and 100%. Grade ballots like those
>> used for Majority Judgement could be adapted to XA.
>>
>>
>> As an approval variant like Bucklin, XA has no vulnerability to burial
>> tactics.
>>
>>
>> Unlike MMPO it also satisfies Plurality.
>>
>>
>> It is monotone and clone independent (in the sense that Approval and
>> Range are clone independent).
>>
>>
>> It is efficiently summable, but is it precinct consistent? i.e. does a
>> candidate that wins in every precinct win over-all?
>>
>>
>> Does it satisfy Participation?
>>
>>
>> We need to explore it, and learn how to explain it as simply as possible,
>> so we can persuade people to use it.
>>
>>
>> Forest
>>
>>
>>
>>
>>
>>
>>
>>
>
FS
Forest Simmons
Mon, Oct 31, 2016 11:27 PM
And any method that fails Participation also fails Consistency.
However, this example is not as bad as it looks:
(1) It shows that Majority Judgment and other forms of Bucklin fail
Participation in exactly the same way.
(2) Obviously, the voters were not aware of minimum strategy, i.e. to give
max support to Favorite, and no support to Worst; No two-candidate
election can fail Participation if the ballots are normalized.
(3) If the purpose of the ballots is to estimate the "worth" of the
respective candidates on a scale of zero to 100, then we don't talk about
"winner" or "loser."
Before the additional participants joined in, A's estimated worth (50%) was
within one standard deviation (about 11 points) of B's estimated worth of
40%.
After the additional ballots are added, the estimation intervals overlap
even more.
So the results are not inconsistent with the expected errors of estimation.
On Fri, Oct 28, 2016 at 8:56 AM, Toby Pereira tdp201b@yahoo.co.uk wrote:
I think Chiastic Approval would fail participation, assuming I've done
this right. Take the following ballots with scores out of 100:
2 voters: A=50, B=40
1 voter: A=50, B=60
A would have a score of 50. B would have a score of 40. Everyone gives B a
score of at least 40, and only a third give B a score higher. Now imagine
there are two extra voters and we have these ballots:
2 voters: A=50, B=40
1 voter: A=50, B=60
2 voters: A=100, B=60
A still has a score of 50, but B now has a score of 60. So these two
ballots cause B to overtake A despite them both preferring A to B.
From: Forest Simmons fsimmons@pcc.edu
Does it satisfy Participation?
And any method that fails Participation also fails Consistency.
However, this example is not as bad as it looks:
(1) It shows that Majority Judgment and other forms of Bucklin fail
Participation in exactly the same way.
(2) Obviously, the voters were not aware of minimum strategy, i.e. to give
max support to Favorite, and no support to Worst; No two-candidate
election can fail Participation if the ballots are normalized.
(3) If the purpose of the ballots is to estimate the "worth" of the
respective candidates on a scale of zero to 100, then we don't talk about
"winner" or "loser."
Before the additional participants joined in, A's estimated worth (50%) was
within one standard deviation (about 11 points) of B's estimated worth of
40%.
After the additional ballots are added, the estimation intervals overlap
even more.
So the results are not inconsistent with the expected errors of estimation.
On Fri, Oct 28, 2016 at 8:56 AM, Toby Pereira <tdp201b@yahoo.co.uk> wrote:
> I think Chiastic Approval would fail participation, assuming I've done
> this right. Take the following ballots with scores out of 100:
>
> 2 voters: A=50, B=40
> 1 voter: A=50, B=60
>
> A would have a score of 50. B would have a score of 40. Everyone gives B a
> score of at least 40, and only a third give B a score higher. Now imagine
> there are two extra voters and we have these ballots:
>
> 2 voters: A=50, B=40
> 1 voter: A=50, B=60
> 2 voters: A=100, B=60
>
> A still has a score of 50, but B now has a score of 60. So these two
> ballots cause B to overtake A despite them both preferring A to B.
>
> ------------------------------
> *From:* Forest Simmons <fsimmons@pcc.edu>
>
>
> Does it satisfy Participation?
>
>
>
FS
Forest Simmons
Mon, Oct 31, 2016 11:40 PM
One nice advantage of XA over Bucklin is that the XA score of a candidate X
varies continuously with the ballot ratings, while the median score does
not:
Profile at time t for 0<t<1:
49 A(100), B(0)
49 B(100), A(0)
2 A(100 - 100t), B(100t)
The respective medians of the candidates jump discontinuously from 100 to
zero and from 0 to 100, respectively, near t = 1/2.
The XA scores vary continuously in the same time interval.
On Mon, Oct 31, 2016 at 4:27 PM, Forest Simmons fsimmons@pcc.edu wrote:
And any method that fails Participation also fails Consistency.
However, this example is not as bad as it looks:
(1) It shows that Majority Judgment and other forms of Bucklin fail
Participation in exactly the same way.
(2) Obviously, the voters were not aware of minimum strategy, i.e. to give
max support to Favorite, and no support to Worst; No two-candidate
election can fail Participation if the ballots are normalized.
(3) If the purpose of the ballots is to estimate the "worth" of the
respective candidates on a scale of zero to 100, then we don't talk about
"winner" or "loser."
Before the additional participants joined in, A's estimated worth (50%)
was within one standard deviation (about 11 points) of B's estimated worth
of 40%.
After the additional ballots are added, the estimation intervals overlap
even more.
So the results are not inconsistent with the expected errors of estimation.
On Fri, Oct 28, 2016 at 8:56 AM, Toby Pereira tdp201b@yahoo.co.uk wrote:
I think Chiastic Approval would fail participation, assuming I've done
this right. Take the following ballots with scores out of 100:
2 voters: A=50, B=40
1 voter: A=50, B=60
A would have a score of 50. B would have a score of 40. Everyone gives B
a score of at least 40, and only a third give B a score higher. Now imagine
there are two extra voters and we have these ballots:
2 voters: A=50, B=40
1 voter: A=50, B=60
2 voters: A=100, B=60
A still has a score of 50, but B now has a score of 60. So these two
ballots cause B to overtake A despite them both preferring A to B.
From: Forest Simmons fsimmons@pcc.edu
Does it satisfy Participation?
One nice advantage of XA over Bucklin is that the XA score of a candidate X
varies continuously with the ballot ratings, while the median score does
not:
Profile at time t for 0<t<1:
49 A(100), B(0)
49 B(100), A(0)
2 A(100 - 100t), B(100t)
The respective medians of the candidates jump discontinuously from 100 to
zero and from 0 to 100, respectively, near t = 1/2.
The XA scores vary continuously in the same time interval.
On Mon, Oct 31, 2016 at 4:27 PM, Forest Simmons <fsimmons@pcc.edu> wrote:
>
> And any method that fails Participation also fails Consistency.
>
> However, this example is not as bad as it looks:
>
> (1) It shows that Majority Judgment and other forms of Bucklin fail
> Participation in exactly the same way.
>
> (2) Obviously, the voters were not aware of minimum strategy, i.e. to give
> max support to Favorite, and no support to Worst; No two-candidate
> election can fail Participation if the ballots are normalized.
>
> (3) If the purpose of the ballots is to estimate the "worth" of the
> respective candidates on a scale of zero to 100, then we don't talk about
> "winner" or "loser."
>
> Before the additional participants joined in, A's estimated worth (50%)
> was within one standard deviation (about 11 points) of B's estimated worth
> of 40%.
>
> After the additional ballots are added, the estimation intervals overlap
> even more.
>
> So the results are not inconsistent with the expected errors of estimation.
>
> On Fri, Oct 28, 2016 at 8:56 AM, Toby Pereira <tdp201b@yahoo.co.uk> wrote:
>
>> I think Chiastic Approval would fail participation, assuming I've done
>> this right. Take the following ballots with scores out of 100:
>>
>> 2 voters: A=50, B=40
>> 1 voter: A=50, B=60
>>
>> A would have a score of 50. B would have a score of 40. Everyone gives B
>> a score of at least 40, and only a third give B a score higher. Now imagine
>> there are two extra voters and we have these ballots:
>>
>> 2 voters: A=50, B=40
>> 1 voter: A=50, B=60
>> 2 voters: A=100, B=60
>>
>> A still has a score of 50, but B now has a score of 60. So these two
>> ballots cause B to overtake A despite them both preferring A to B.
>>
>> ------------------------------
>> *From:* Forest Simmons <fsimmons@pcc.edu>
>>
>>
>> Does it satisfy Participation?
>>
>>
>>
>
RL
Rob LeGrand
Mon, Oct 31, 2016 11:51 PM
"With XA, when you assign a number, it isn't just a merit-rating.
"If you write ".9" next to a candidate's name, you're saying that you want
for .......to........"
Here's my attempt:
You have 1 point that will be given to or withheld from each candidate,
fully or partially. If you assign a candidate a rating of x, your 1 point
is given to or withheld from the candidate in whichever way makes their
final average closest to x.
--
Rob LeGrand
rob@approvalvoting.org
Michael Ossipoff wrote:
> "With XA, when you assign a number, it isn't just a merit-rating.
> "If you write ".9" next to a candidate's name, you're saying that you want
> for .......to........"
Here's my attempt:
You have 1 point that will be given to or withheld from each candidate,
fully or partially. If you assign a candidate a rating of x, your 1 point
is given to or withheld from the candidate in whichever way makes their
final average closest to x.
--
Rob LeGrand
rob@approvalvoting.org
MO
Michael Ossipoff
Tue, Nov 1, 2016 1:56 AM
Thanks to Andy, Forest & Rob, for answering my questions, and making XA
more understandable & straightforward.
Michael Ossipodf
On Mon, Oct 31, 2016 at 7:51 PM, Rob LeGrand honky98@gmail.com wrote:
"With XA, when you assign a number, it isn't just a merit-rating.
"If you write ".9" next to a candidate's name, you're saying that you
Here's my attempt:
You have 1 point that will be given to or withheld from each candidate,
fully or partially. If you assign a candidate a rating of x, your 1 point
is given to or withheld from the candidate in whichever way makes their
final average closest to x.
Election-Methods mailing list - see http://electorama.com/em for list info
Thanks to Andy, Forest & Rob, for answering my questions, and making XA
more understandable & straightforward.
Michael Ossipodf
On Mon, Oct 31, 2016 at 7:51 PM, Rob LeGrand <honky98@gmail.com> wrote:
> Michael Ossipoff wrote:
> > "With XA, when you assign a number, it isn't just a merit-rating.
> > "If you write ".9" next to a candidate's name, you're saying that you
> want
> > for .......to........"
>
> Here's my attempt:
>
> You have 1 point that will be given to or withheld from each candidate,
> fully or partially. If you assign a candidate a rating of x, your 1 point
> is given to or withheld from the candidate in whichever way makes their
> final average closest to x.
>
> --
> Rob LeGrand
> rob@approvalvoting.org
> ----
> Election-Methods mailing list - see http://electorama.com/em for list info
>
MO
Michael Ossipoff
Tue, Nov 1, 2016 2:13 AM
So it's like Approval, except that you can vote for each candidate to have
some particular final score.
But pulling the candidate toward a particular final score reminds me of MJ.
What makes XA do that more effectively than MJ? What's the main advantage
that distinguishes how XA does that from how MJ does it, or the results,
from the voters' strategic standpoint?
Say someone gives Jill 100, and Hillary 70. Is that voter reliably voting
for Jill to beat Hillary? I guess if Jill only ends up with 60, and you've
drawn Hillary up to 70, then you've helped Hillary beat Jill.
If you want Jill to beat Hillary, then shouldn't you just top-rate Jill &
bottom-rate Hillary, just as in Approval or Score?
Michael Ossipoff
On Mon, Oct 31, 2016 at 7:51 PM, Rob LeGrand honky98@gmail.com wrote:
"With XA, when you assign a number, it isn't just a merit-rating.
"If you write ".9" next to a candidate's name, you're saying that you
Here's my attempt:
You have 1 point that will be given to or withheld from each candidate,
fully or partially. If you assign a candidate a rating of x, your 1 point
is given to or withheld from the candidate in whichever way makes their
final average closest to x.
Election-Methods mailing list - see http://electorama.com/em for list info
So it's like Approval, except that you can vote for each candidate to have
some particular final score.
But pulling the candidate toward a particular final score reminds me of MJ.
What makes XA do that more effectively than MJ? What's the main advantage
that distinguishes how XA does that from how MJ does it, or the results,
from the voters' strategic standpoint?
Say someone gives Jill 100, and Hillary 70. Is that voter reliably voting
for Jill to beat Hillary? I guess if Jill only ends up with 60, and you've
drawn Hillary up to 70, then you've helped Hillary beat Jill.
If you want Jill to beat Hillary, then shouldn't you just top-rate Jill &
bottom-rate Hillary, just as in Approval or Score?
Michael Ossipoff
On Mon, Oct 31, 2016 at 7:51 PM, Rob LeGrand <honky98@gmail.com> wrote:
> Michael Ossipoff wrote:
> > "With XA, when you assign a number, it isn't just a merit-rating.
> > "If you write ".9" next to a candidate's name, you're saying that you
> want
> > for .......to........"
>
> Here's my attempt:
>
> You have 1 point that will be given to or withheld from each candidate,
> fully or partially. If you assign a candidate a rating of x, your 1 point
> is given to or withheld from the candidate in whichever way makes their
> final average closest to x.
>
> --
> Rob LeGrand
> rob@approvalvoting.org
> ----
> Election-Methods mailing list - see http://electorama.com/em for list info
>
RL
Rob LeGrand
Tue, Nov 1, 2016 2:45 AM
But pulling the candidate toward a particular final score reminds me of
MJ. What makes XA do that more effectively than MJ? What's the main
advantage that distinguishes how XA does that from how MJ does it, or the
results, from the voters' strategic standpoint?
My dissertation is available at
http://www.cs.angelo.edu/~rlegrand/research/dissertation.pdf
One chapter describes a parameterized class of nonmanipulable ratings-
summarizing systems that includes plain median on one hand and chiastic
median (which we called Average-Approval-Ratings DSV) on the other. We used
movie-ratings data from metacritic.com to find the parameters whose
corresponding system gave results nearest to averaging the inputs
(essentially minimizing the Bayesian regret). So you can use real data to
help you decide on median, XM or something in between.
Here's one intuitive reason to favor XM over median: Say the ratings of one
candidate are all at the extremes: [0, 0, 0, 100, 100]. The median is 0,
making 3 of the voters maximally happy and completely alienating the others.
The XM outcome is 40, perhaps more fairly reflecting the voters' wishes. In
fact, when the voters' ratings are all at the extremes, XM always agrees
with the average rating.
Say someone gives Jill 100, and Hillary 70. Is that voter reliably voting
for Jill to beat Hillary? I guess if Jill only ends up with 60, and you've
drawn Hillary up to 70, then you've helped Hillary beat Jill.
That's correct. My dissertation also pointed out that basing a
multicandidate, single-winner voting system on a nonmanipulable
ratings-summarizing system such as median or XM does not give you a
nonmanipulable voting system. I gave a specific example of manipulating an
election that used AAR DSV for each candidate (equivalent to chiastic
approval) and recommended instead using plain approval or approval DSV
(which I analyzed in a separate chapter).
--
Rob LeGrand
rob@approvalvoting.org
Michael Ossipoff wrote:
> But pulling the candidate toward a particular final score reminds me of
> MJ. What makes XA do that more effectively than MJ? What's the main
> advantage that distinguishes how XA does that from how MJ does it, or the
> results, from the voters' strategic standpoint?
My dissertation is available at
http://www.cs.angelo.edu/~rlegrand/research/dissertation.pdf
One chapter describes a parameterized class of nonmanipulable ratings-
summarizing systems that includes plain median on one hand and chiastic
median (which we called Average-Approval-Ratings DSV) on the other. We used
movie-ratings data from metacritic.com to find the parameters whose
corresponding system gave results nearest to averaging the inputs
(essentially minimizing the Bayesian regret). So you can use real data to
help you decide on median, XM or something in between.
Here's one intuitive reason to favor XM over median: Say the ratings of one
candidate are all at the extremes: [0, 0, 0, 100, 100]. The median is 0,
making 3 of the voters maximally happy and completely alienating the others.
The XM outcome is 40, perhaps more fairly reflecting the voters' wishes. In
fact, when the voters' ratings are all at the extremes, XM always agrees
with the average rating.
> Say someone gives Jill 100, and Hillary 70. Is that voter reliably voting
> for Jill to beat Hillary? I guess if Jill only ends up with 60, and you've
> drawn Hillary up to 70, then you've helped Hillary beat Jill.
That's correct. My dissertation also pointed out that basing a
multicandidate, single-winner voting system on a nonmanipulable
ratings-summarizing system such as median or XM does not give you a
nonmanipulable voting system. I gave a specific example of manipulating an
election that used AAR DSV for each candidate (equivalent to chiastic
approval) and recommended instead using plain approval or approval DSV
(which I analyzed in a separate chapter).
--
Rob LeGrand
rob@approvalvoting.org
AJ
Andy Jennings
Tue, Nov 1, 2016 1:49 PM
What makes XA do that more effectively than MJ? What's the main advantage
that distinguishes how XA does that from how MJ does it, or the results,
from the voters' strategic standpoint?
Michael,
As Rob said, the median is not terribly robust if the distribution of votes
is two-peaked:
http://www.rangevoting.org/MedianVrange.html#twopeak
And I'm afraid many of our contentious political elections are two-peaked,
at least in the current environment.
With MJ, I like the fact that if the medians for all candidates will fall
between B and D, then I can use the range outside that for honest
expression. Yet in the back of my head, I know that if everyone tries to
"use the range outside that for honest expression", then the medians won't
be in that range anymore and it seems like a slippery slope to everyone
using only the two extreme grades.
XA solves this problem by making the more extreme grades more difficult to
achieve. As Rob said, in the case where everyone grades at the extremes,
the XA will match the mean.
On the other hand, I admit that:
- with the median, 50% would have to give the top grade for a candidate to
receive that grade. And 50% would have to give the bottom grade for a
candidate to receive that grade. I consider both of these very unlikely.
- MJ is not just "the median", it has a tie-breaking scheme which
mitigates this somewhat.
~ Andy
On Mon, Oct 31, 2016 at 7:13 PM, Michael Ossipoff <email9648742@gmail.com>
wrote:
> What makes XA do that more effectively than MJ? What's the main advantage
> that distinguishes how XA does that from how MJ does it, or the results,
> from the voters' strategic standpoint?
Michael,
As Rob said, the median is not terribly robust if the distribution of votes
is two-peaked:
http://www.rangevoting.org/MedianVrange.html#twopeak
And I'm afraid many of our contentious political elections are two-peaked,
at least in the current environment.
With MJ, I like the fact that if the medians for all candidates will fall
between B and D, then I can use the range outside that for honest
expression. Yet in the back of my head, I know that if everyone tries to
"use the range outside that for honest expression", then the medians won't
be in that range anymore and it seems like a slippery slope to everyone
using only the two extreme grades.
XA solves this problem by making the more extreme grades more difficult to
achieve. As Rob said, in the case where everyone grades at the extremes,
the XA will match the mean.
On the other hand, I admit that:
1) with the median, 50% would have to give the top grade for a candidate to
receive that grade. And 50% would have to give the bottom grade for a
candidate to receive that grade. I consider both of these very unlikely.
2) MJ is not just "the median", it has a tie-breaking scheme which
mitigates this somewhat.
~ Andy
MO
Michael Ossipoff
Tue, Nov 1, 2016 6:50 PM
XA beats MJ at its own game.
Your answers have clarified XA for me, & made it straightforward.
At first it looked like a difficult mathematical or logical puzzle.
But it needn't, & I thank you for clarifying that.
XA needs a name that explicitly tells what it's about.
How about:
"Approve To Desired Total" (ATDT)?
That tells it, right in the name.
I agree with Rob: ATDT wasn't designed & introduced for single-winner
elections, & isn't to be recommended for them.
Forest--
You spoke of ATDT for chicken dilemma, & mentioned a Nash equilibrium.
But Approval has that Nash equilibrium too. The numbers are different, but
if seems qualitatively the same.
In ATDT, when the A voters give B a 33, they ensure that B can't win. When
the B voters give A 50, they ensure that A outpolls C.
...as in Approval, when the A voters say, "You know we're bigger. We aren't
approving B, because you should approve us, the bigger faction.". ...& the
B voters approve A.
Your suggested chicken dilemma solution for Approval is much better than
what ATDT offers, when it might not be obvious which faction is bigger:
Sincere preferrences:
35: A>B>>C
25: B>A>>C
40: C>>A=B
The A voters say:
"The best available estimate is that C has 40%. Suppose we're nearly equal:
You have 31%, & we have 29%.
" If we give you 10%, you beat C & win.
That means, our 29% each give B (10/29) of a vote.
You should do the same, each of you giving A 10/29 of a vote.
Then you give A (10/29)X31 = 10.69% in total.
10.69 + 29 = 39.69, less than your 41%. You win.
We all should share that 10/29 of a vote, in case our own candidate isn't
the larger one, & can't win with the 10% help.
Michael Ossipoff
On Nov 1, 2016 6:50 AM, "Andy Jennings" elections@jenningsstory.com wrote:
What makes XA do that more effectively than MJ? What's the main advantage
that distinguishes how XA does that from how MJ does it, or the results,
from the voters' strategic standpoint?
Michael,
As Rob said, the median is not terribly robust if the distribution of
votes is two-peaked:
http://www.rangevoting.org/MedianVrange.html#twopeak
And I'm afraid many of our contentious political elections are two-peaked,
at least in the current environment.
With MJ, I like the fact that if the medians for all candidates will fall
between B and D, then I can use the range outside that for honest
expression. Yet in the back of my head, I know that if everyone tries to
"use the range outside that for honest expression", then the medians won't
be in that range anymore and it seems like a slippery slope to everyone
using only the two extreme grades.
XA solves this problem by making the more extreme grades more difficult to
achieve. As Rob said, in the case where everyone grades at the extremes,
the XA will match the mean.
On the other hand, I admit that:
- with the median, 50% would have to give the top grade for a candidate
to receive that grade. And 50% would have to give the bottom grade for a
candidate to receive that grade. I consider both of these very unlikely.
- MJ is not just "the median", it has a tie-breaking scheme which
mitigates this somewhat.
~ Andy
XA beats MJ at its own game.
Your answers have clarified XA for me, & made it straightforward.
At first it looked like a difficult mathematical or logical puzzle.
But it needn't, & I thank you for clarifying that.
XA needs a name that explicitly tells what it's about.
How about:
"Approve To Desired Total" (ATDT)?
That tells it, right in the name.
I agree with Rob: ATDT wasn't designed & introduced for single-winner
elections, & isn't to be recommended for them.
Forest--
You spoke of ATDT for chicken dilemma, & mentioned a Nash equilibrium.
But Approval has that Nash equilibrium too. The numbers are different, but
if seems qualitatively the same.
In ATDT, when the A voters give B a 33, they ensure that B can't win. When
the B voters give A 50, they ensure that A outpolls C.
...as in Approval, when the A voters say, "You know we're bigger. We aren't
approving B, because you should approve us, the bigger faction.". ...& the
B voters approve A.
Your suggested chicken dilemma solution for Approval is much better than
what ATDT offers, when it might not be obvious which faction is bigger:
Sincere preferrences:
35: A>B>>C
25: B>A>>C
40: C>>A=B
The A voters say:
"The best available estimate is that C has 40%. Suppose we're nearly equal:
You have 31%, & we have 29%.
" If we give you 10%, you beat C & win.
That means, our 29% each give B (10/29) of a vote.
You should do the same, each of you giving A 10/29 of a vote.
Then you give A (10/29)X31 = 10.69% in total.
10.69 + 29 = 39.69, less than your 41%. You win.
We all should share that 10/29 of a vote, in case our own candidate isn't
the larger one, & can't win with the 10% help.
Michael Ossipoff
On Nov 1, 2016 6:50 AM, "Andy Jennings" <elections@jenningsstory.com> wrote:
>
> On Mon, Oct 31, 2016 at 7:13 PM, Michael Ossipoff <email9648742@gmail.com>
> wrote:
>
>> What makes XA do that more effectively than MJ? What's the main advantage
>> that distinguishes how XA does that from how MJ does it, or the results,
>> from the voters' strategic standpoint?
>
>
> Michael,
>
> As Rob said, the median is not terribly robust if the distribution of
> votes is two-peaked:
> http://www.rangevoting.org/MedianVrange.html#twopeak
> And I'm afraid many of our contentious political elections are two-peaked,
> at least in the current environment.
>
> With MJ, I like the fact that if the medians for all candidates will fall
> between B and D, then I can use the range outside that for honest
> expression. Yet in the back of my head, I know that if everyone tries to
> "use the range outside that for honest expression", then the medians won't
> be in that range anymore and it seems like a slippery slope to everyone
> using only the two extreme grades.
>
> XA solves this problem by making the more extreme grades more difficult to
> achieve. As Rob said, in the case where everyone grades at the extremes,
> the XA will match the mean.
>
> On the other hand, I admit that:
> 1) with the median, 50% would have to give the top grade for a candidate
> to receive that grade. And 50% would have to give the bottom grade for a
> candidate to receive that grade. I consider both of these very unlikely.
> 2) MJ is not just "the median", it has a tie-breaking scheme which
> mitigates this somewhat.
>
> ~ Andy
>
>
