Manipulability stats for more poll methods (fixed footnotes)
Oops, I numbered my footnotes incorrectly. Let's do that again. (Ignore
my last post.)
Here are voter manipulability stats for most of the cardinal and
cardinal hybrid methods. The exceptions are:
- MJ, because I'm not confident enough about how I implemented it (in
particular, what tiebreaker it uses),
- Margins-Sorted Approval, because I'm not sure how it works, and
- Approval with manual runoff, because it's difficult to model the
effects of further discussion between the rounds.
Same testing parameters as the other stats: spatial (Gaussian) model, 4
dimensions, 4 candidates, 99 voters, 500k elections tested, and 32k
strategy attempts per election.
As in the last post, I've marked the non-poll methods with an asterisk.
The manipulability values are:
0.937 *Range(0-5, absolute scale)
0.928 Approval (absolute scale)[1]
0.710 *Range (0-10, normalized)
0.708 *Range(0-5, normalized)
0.705 Smith//Range(0-5, absolute scale)
0.666 Approval (mean utility cutoff)
0.655 Smith//Range(0-10, absolute scale)
0.645 STAR
0.564 Smith//Range (0-5, normalized)
0.557 Smith//Range (0-10, normalized)
0.514 Smith//Approval (explicit, mean utility cutoff)
0.490 Smith//Approval (implicit, mean utility cutoff)
0.443 Smith//DAC (mean utility truncation)[2]
Some values from my last post for reference:
0.480 Copeland//Borda (Ranked Robin)
0.417 Plurality
0.333 Schulze, minmax
0.074 Condorcet-IRV
and for verification, James Green-Armytage's results are:[3]
0.710 Range (normalized)
0.668 Approval (mean utility cutoff)
Range is not part of the poll, but it serves to show the differences
between absolute and relative (normalized) scales, and to show that my
results are similar to JGA's.
"Absolute scale" gives the voters a common scale to rate on, to model
the Range component passing IIA. The voters' utilities in this model are
maximum if the candidate is at the same point in opinion space as they
are, and minimum at a utility that the spatial model (with random
candidates and voters) would exceed 90% of the time. Since this 90%
quantile doesn't depend on the candidates who were selected, it's an
absolute scale, and 10% of the voter-candidate judgements would be
clamped to zero, on average.
On the other hand, "normalized" has the voters rate their least favorite
zero and their favorite to maximum.
"Mean utility cutoff" is the (relative scale) Approval guideline where
the voter approves every candidate above mean utility and disapproves
everybody else. Though a relative scale, it's not quite the same thing
as "normalized".
STAR uses a scale of 0-5 inclusive. Since the official STAR ballot text
tells the voters to normalize,[4] I've only included the normalized
manipulability value.
For most of the other cardinal methods and their hybrids, I've given
both 0-5 and 0-10 ballot formats. The 11-slot ballot makes it easier to
show a difference of preference, which helps identify the honest Smith
set in Smith//Range. However, there's not otherwise much of a difference.
-km
[1] Absolute scale approval has a high tie rate of 5%, so it's possible
that it should "really" be worse than Range. My simulator deliberately
only checks elections with unique honest winners.
[2] The detailed stats suggest that pushover is a problem with
Smith//DAC. However, getting a per-strategy breakdown for cardinal
methods is hard due to limitations of my simulator, so it would still
have to be verified by other means. The "mean utility truncation" is
what makes it cardinal in my simulator's eyes.
[3] Green-Armytage, James (2011). "Four Condorcet-Hare hybrid methods
for single-winner elections". Voting matters (29): p. 7;
https://www.votingmatters.org.uk/ISSUE29/I29P1.pdf
[4] https://www.starvoting.org/paper_ballots Step 3.
Kristofer,
Thanks for this, but a few things leave me a bit confused and/or disturbed.
"Mean utility cutoff" is the (relative scale) Approval guideline where
the voter approves every candidate above mean utility and disapproves
everybody else. Though a relative scale, it's not quite the same thing
as "normalized".
How is it different? I assume you never have a voter approving all or
none of the candidates, right?
4 "dimensions" sounds like a lot. What are the "strategy attempts" ?
How much and what information do the strategists have? Are the
strategists confined to just trying to get their favourites elected, or
any candidate they prefer to the initial winner?
[2] The detailed stats suggest that pushover is a problem with Smith//DAC
You don't have enough candidates for a sub-cycle, and so the method
can't fail mono-raise. How can it have a Pushover problem?
- Margins-Sorted Approval, because I'm not sure how it works
(I struggle to take this at face value. Probably my promotion of MSA
has convinced you that it is the best method and you were concerned that
your simulation wouldn't do it justice. But our expert doing the
simulation claiming he can't understand the method isn't a good look for
its proposability.)
Why didn't you simply ask me to explain it to you?
What happened to separate entries for BTR, Woodall and Benham?
Chris B.
On 4/05/2024 5:48 am, Kristofer Munsterhjelm wrote:
Oops, I numbered my footnotes incorrectly. Let's do that again.
(Ignore my last post.)
Here are voter manipulability stats for most of the cardinal and
cardinal hybrid methods. The exceptions are:
- MJ, because I'm not confident enough about how I implemented it
(in particular, what tiebreaker it uses),
- Margins-Sorted Approval, because I'm not sure how it works, and
- Approval with manual runoff, because it's difficult to model the
effects of further discussion between the rounds.
Same testing parameters as the other stats: spatial (Gaussian) model,
4 dimensions, 4 candidates, 99 voters, 500k elections tested, and 32k
strategy attempts per election.
As in the last post, I've marked the non-poll methods with an asterisk.
The manipulability values are:
0.937 *Range(0-5, absolute scale)
0.928 Approval (absolute scale)[1]
0.710 *Range (0-10, normalized)
0.708 *Range(0-5, normalized)
0.705 Smith//Range(0-5, absolute scale)
0.666 Approval (mean utility cutoff)
0.655 Smith//Range(0-10, absolute scale)
0.645 STAR
0.564 Smith//Range (0-5, normalized)
0.557 Smith//Range (0-10, normalized)
0.514 Smith//Approval (explicit, mean utility cutoff)
0.490 Smith//Approval (implicit, mean utility cutoff)
0.443 Smith//DAC (mean utility truncation)[2]
Some values from my last post for reference:
0.480 Copeland//Borda (Ranked Robin)
0.417 Plurality
0.333 Schulze, minmax
0.074 Condorcet-IRV
and for verification, James Green-Armytage's results are:[3]
0.710 Range (normalized)
0.668 Approval (mean utility cutoff)
Range is not part of the poll, but it serves to show the differences
between absolute and relative (normalized) scales, and to show that my
results are similar to JGA's.
"Absolute scale" gives the voters a common scale to rate on, to model
the Range component passing IIA. The voters' utilities in this model
are maximum if the candidate is at the same point in opinion space as
they are, and minimum at a utility that the spatial model (with random
candidates and voters) would exceed 90% of the time. Since this 90%
quantile doesn't depend on the candidates who were selected, it's an
absolute scale, and 10% of the voter-candidate judgements would be
clamped to zero, on average.
On the other hand, "normalized" has the voters rate their least
favorite zero and their favorite to maximum.
"Mean utility cutoff" is the (relative scale) Approval guideline where
the voter approves every candidate above mean utility and disapproves
everybody else. Though a relative scale, it's not quite the same thing
as "normalized".
STAR uses a scale of 0-5 inclusive. Since the official STAR ballot
text tells the voters to normalize,[4] I've only included the
normalized manipulability value.
For most of the other cardinal methods and their hybrids, I've given
both 0-5 and 0-10 ballot formats. The 11-slot ballot makes it easier
to show a difference of preference, which helps identify the honest
Smith set in Smith//Range. However, there's not otherwise much of a
difference.
-km
[1] Absolute scale approval has a high tie rate of 5%, so it's
possible that it should "really" be worse than Range. My simulator
deliberately only checks elections with unique honest winners.
[2] The detailed stats suggest that pushover is a problem with
Smith//DAC. However, getting a per-strategy breakdown for cardinal
methods is hard due to limitations of my simulator, so it would still
have to be verified by other means. The "mean utility truncation" is
what makes it cardinal in my simulator's eyes.
[3] Green-Armytage, James (2011). "Four Condorcet-Hare hybrid methods
for single-winner elections". Voting matters (29): p. 7;
https://www.votingmatters.org.uk/ISSUE29/I29P1.pdf
[4] https://www.starvoting.org/paper_ballots Step 3.
Election-Methods mailing list - see https://electorama.com/em for list
info
On 2024-05-04 11:22, Chris Benham wrote:
Kristofer,
Thanks for this, but a few things leave me a bit confused and/or disturbed.
"Mean utility cutoff" is the (relative scale) Approval guideline where
the voter approves every candidate above mean utility and disapproves
everybody else. Though a relative scale, it's not quite the same thing
as "normalized".
How is it different? I assume you never have a voter approving all or
none of the candidates, right?
Yes, that's right. But consider a voter with the following utilities:
A: 0.57
B: 0.32
C: 0.23
D: 0.08
Normalization to two steps fixes the highest value (0.57) to 1 and the
lowest value (0.08) to 0 and rounds off the intermediate values after
linearly scaling them. This in essence says that a value is rounded off
to 1 if it's greater than or equal to 0.325 (the midpoint between 0.08
and 0.57), so the 0-1 normalized ballot is
A: 1, B: 0, C: 0, D: 0
On the other hand, the mean utility is 0.3. So the mean utility approval
ballot is
A: 1, B: 1, C: 0, D: 0.
4 "dimensions" sounds like a lot. What are the "strategy attempts" ?
How much and what information do the strategists have? Are the
strategists confined to just trying to get their favourites elected, or
any candidate they prefer to the initial winner?
The method works pretty much like this, for generating and testing a
single election. (I've simplified the exact order that strategies are
called upon, but this is in effect what happens.)
==== (Algorithm start) =====
Draw candidate positions for each candidate (in this case, each is a
point on a 4D normal distribution with mean 0 and variance 1).
Draw voter positions for each voter, and create their honest ballots
based on the distances between the voter and candidates.
Pass the resulting ballots through the method to establish the honest
outcome.
If there's a tie, skip (because deciding what a strict improvement is
when there's a honest tie is ambiguous). Otherwise let the winner be W.
For each candidate X who is not the winner W:
For i = 1 to number of strategy attempts / number of candidates
Set the strategic ballots to the honest ballots.
For every voter who prefers X to W:
Replace that voter's strategic ballot with a
ballot according to a strategy that depends on
i.
Pass the modified strategic ballots through the method.
If X is now a winner, the method is manipulable in
this election. Return success.
If we reach this point without any success, return failure; the method
is (probably) not manipulable in this election.
==== (Algorithm end) =====
The indexed strategies are
i=0: Compromising (raise X to unique top)
i=1: Burial (lower W to unique bottom)
i=2: Two-sided (do both at once)
i>2: Coalitional strategy
The compromising, burial, and two-sided strategies modify the voters'
otherwise honest ballots - for instance, compromising changes a
strategist's ballot so that X is at unique top and the rest of the
ballot is unchanged.
The first time the coalitional strategy is called for a particular
election, candidate to strategize for, and value of i, it chooses a
random number of strategic ballots (between 1 and 3 inclusive). Each
strategic voter then picks one of these ballots at random. This
simulates strategies where every strategist ballot is equal, as well as
ones where there are a few groups each with their own ballot type, thus
covering more than JGA's simulations without becoming too
computationally expensive.
So with the setup for the stats that I gave, the full setup for a single
method is like this:
for j = 1 to 500k
Run the algorithm detailed above.
It returns one of three states: honest tie, success, or failure.
Increment the corresponding counter, call it TIES, SUCCESSES or FAILURES.
manipulability = SUCCESSES/(500k - TIES)
So to answer your questions:
The strategists don't adapt their strategy to the information available
to them, even though they strictly speaking have full information.
However, they get to try over and over again until they win. If there is
a full information strategy with not too many distinct ballots, then
this random sampling will eventually find it, given a high enough
strategy attempts value.
For each non-winner X, everybody who prefers X to the current winner
gets to have a go. So not just their favorites: anybody they all prefer
to the current winner.
[2] The detailed stats suggest that pushover is a problem with Smith//DAC
You don't have enough candidates for a sub-cycle, and so the method
can't fail mono-raise. How can it have a Pushover problem?
I did a bit more checking, and the full preference version doesn't have
this high an "other strategy" count. Since I think it's unlikely that
the version with truncation would have more pushover than the fully
ranked one, I'm going to retract this; most likely it's just an artifact
of the simulator's ballot reduction process that falsely attributes the
strategy to the "other" category for cardinal methods.
- Margins-Sorted Approval, because I'm not sure how it works
(I struggle to take this at face value. Probably my promotion of MSA
has convinced you that it is the best method and you were concerned that
your simulation wouldn't do it justice.
I'd like to believe both that I have enough scientific integrity not to
do that, and that people know I have, too :-)
Actually, I was planning on putting MSA at the same level as the other
"I don't know enough about these or their dynamics" methods (double
defeat Hare, MSMLV, and Max Strength Transitive Beatpath).
But our expert doing the
simulation claiming he can't understand the method isn't a good look for
its proposability.)
Why didn't you simply ask me to explain it to you?
I think it's the sorting phase that does it. My vague idea of how it
works is that you essentially run a sorting algorithm on intermediate
values, and that seems a little too complex to me. But I might just have
got it wrong and then the initial impression of it as an intimidating
method stuck.
Ted Stern pointed me at the Electowiki article for MSA, which in turn
led me to his Python implementation. I might port it if I have time, but
I feel a bit exhausted after gathering all these stats. We'll see :-)
What happened to separate entries for BTR, Woodall and Benham?
They're in the other post. I didn't want to add them all to the post
that was intended to focus on the new results. That's why I said "some
for comparison" - the others are here:
http://lists.electorama.com/pipermail/election-methods-electorama.com/2024-April/006029.html
I could post all the stats - ordinal and cardinal methods' - in a
summary post if you or other EM members would like.
-km
On Sat, May 4, 2024 at 14:45 Kristofer Munsterhjelm km_elmet@t-online.de
wrote:
Yes, that's right. But consider a voter with the following utilities:
A: 0.57
B: 0.32
C: 0.23
D: 0.08
Normalization to two steps fixes the highest value (0.57) to 1 and the
lowest value (0.08) to 0 and rounds off the intermediate values after
linearly scaling them.
Yes. So far, so good. But…
This in essence says that a value is rounded off
to 1 if it's greater than or equal to 0.325 (the midpoint between 0.08
and 0.57)
What? You didn’t average the normalized values. You averaged two of the
values before normalization. The midrange isn’t usually the same as the
mean. You used the midrange as the mean.
If you call the top value 1, & the bottom value 0,
then a rating’s new value is the number that’s the same % of the way from 0
to 1 as the old number’s % from.08 to .57
Average of those new values: .4475
You still approve the best two.
so the 0-1 normalized ballot is
A: 1, B: 0, C: 0, D: 0
On the other hand, the mean utility is 0.3. So the mean utility approval
ballot is
A: 1, B: 1, C: 0, D: 0.
4 "dimensions" sounds like a lot. What are the "strategy attempts" ?
How much and what information do the strategists have? Are the
strategists confined to just trying to get their favourites elected, or
any candidate they prefer to the initial winner?
The method works pretty much like this, for generating and testing a
single election. (I've simplified the exact order that strategies are
called upon, but this is in effect what happens.)
==== (Algorithm start) =====
Draw candidate positions for each candidate (in this case, each is a
point on a 4D normal distribution with mean 0 and variance 1).
Draw voter positions for each voter, and create their honest ballots
based on the distances between the voter and candidates.
Pass the resulting ballots through the method to establish the honest
outcome.
If there's a tie, skip (because deciding what a strict improvement is
when there's a honest tie is ambiguous). Otherwise let the winner be W.
For each candidate X who is not the winner W:
For i = 1 to number of strategy attempts / number of candidates
Set the strategic ballots to the honest ballots.
For every voter who prefers X to W:
Replace that voter's strategic ballot with a
ballot according to a strategy that depends on
i.
Pass the modified strategic ballots through the method.
If X is now a winner, the method is manipulable in
this election. Return success.
If we reach this point without any success, return failure; the method
is (probably) not manipulable in this election.
==== (Algorithm end) =====
The indexed strategies are
i=0: Compromising (raise X to unique top)
i=1: Burial (lower W to unique bottom)
i=2: Two-sided (do both at once)
i>2: Coalitional strategy
The compromising, burial, and two-sided strategies modify the voters'
otherwise honest ballots - for instance, compromising changes a
strategist's ballot so that X is at unique top and the rest of the
ballot is unchanged.
The first time the coalitional strategy is called for a particular
election, candidate to strategize for, and value of i, it chooses a
random number of strategic ballots (between 1 and 3 inclusive). Each
strategic voter then picks one of these ballots at random. This
simulates strategies where every strategist ballot is equal, as well as
ones where there are a few groups each with their own ballot type, thus
covering more than JGA's simulations without becoming too
computationally expensive.
So with the setup for the stats that I gave, the full setup for a single
method is like this:
for j = 1 to 500k
Run the algorithm detailed above.
It returns one of three states: honest tie, success, or failure.
Increment the corresponding counter, call it TIES, SUCCESSES or
FAILURES.
manipulability = SUCCESSES/(500k - TIES)
So to answer your questions:
The strategists don't adapt their strategy to the information available
to them, even though they strictly speaking have full information.
However, they get to try over and over again until they win. If there is
a full information strategy with not too many distinct ballots, then
this random sampling will eventually find it, given a high enough
strategy attempts value.
For each non-winner X, everybody who prefers X to the current winner
gets to have a go. So not just their favorites: anybody they all prefer
to the current winner.
[2] The detailed stats suggest that pushover is a problem with
Smith//DAC
You don't have enough candidates for a sub-cycle, and so the method
can't fail mono-raise. How can it have a Pushover problem?
I did a bit more checking, and the full preference version doesn't have
this high an "other strategy" count. Since I think it's unlikely that
the version with truncation would have more pushover than the fully
ranked one, I'm going to retract this; most likely it's just an artifact
of the simulator's ballot reduction process that falsely attributes the
strategy to the "other" category for cardinal methods.
- Margins-Sorted Approval, because I'm not sure how it works
(I struggle to take this at face value. Probably my promotion of MSA
has convinced you that it is the best method and you were concerned that
your simulation wouldn't do it justice.
I'd like to believe both that I have enough scientific integrity not to
do that, and that people know I have, too :-)
Actually, I was planning on putting MSA at the same level as the other
"I don't know enough about these or their dynamics" methods (double
defeat Hare, MSMLV, and Max Strength Transitive Beatpath).
But our expert doing the
simulation claiming he can't understand the method isn't a good look for
its proposability.)
Why didn't you simply ask me to explain it to you?
I think it's the sorting phase that does it. My vague idea of how it
works is that you essentially run a sorting algorithm on intermediate
values, and that seems a little too complex to me. But I might just have
got it wrong and then the initial impression of it as an intimidating
method stuck.
Ted Stern pointed me at the Electowiki article for MSA, which in turn
led me to his Python implementation. I might port it if I have time, but
I feel a bit exhausted after gathering all these stats. We'll see :-)
What happened to separate entries for BTR, Woodall and Benham?
They're in the other post. I didn't want to add them all to the post
that was intended to focus on the new results. That's why I said "some
for comparison" - the others are here:
http://lists.electorama.com/pipermail/election-methods-electorama.com/2024-April/006029.html
I could post all the stats - ordinal and cardinal methods' - in a
summary post if you or other EM members would like.
-km
Election-Methods mailing list - see https://electorama.com/em for list
info
There’s no reason for the renormalization. Among A, B, C & D (in that order
of magnitude) if B is at the mean, then, with the A=0 & D=1
renormalization, B’s renormalized value is the mean of all of the
renormalized values.
The position of the mean among the candidates doesn’t change with
renormalization.
On Sat, May 4, 2024 at 15:25 Michael Ossipoff email9648742@gmail.com
wrote:
On Sat, May 4, 2024 at 14:45 Kristofer Munsterhjelm km_elmet@t-online.de
wrote:
Yes, that's right. But consider a voter with the following utilities:
A: 0.57
B: 0.32
C: 0.23
D: 0.08
Normalization to two steps fixes the highest value (0.57) to 1 and the
lowest value (0.08) to 0 and rounds off the intermediate values after
linearly scaling them.
Yes. So far, so good. But…
This in essence says that a value is rounded off
to 1 if it's greater than or equal to 0.325 (the midpoint between 0.08
and 0.57)
What? You didn’t average the normalized values. You averaged two of the
values before normalization. The midrange isn’t usually the same as the
mean. You used the midrange as the mean.
If you call the top value 1, & the bottom value 0,
then a rating’s new value is the number that’s the same % of the way from
0 to 1 as the old number’s % from.08 to .57
Average of those new values: .4475
You still approve the best two.
so the 0-1 normalized ballot is
A: 1, B: 0, C: 0, D: 0
On the other hand, the mean utility is 0.3. So the mean utility approval
ballot is
A: 1, B: 1, C: 0, D: 0.
4 "dimensions" sounds like a lot. What are the "strategy attempts" ?
How much and what information do the strategists have? Are the
strategists confined to just trying to get their favourites elected, or
any candidate they prefer to the initial winner?
The method works pretty much like this, for generating and testing a
single election. (I've simplified the exact order that strategies are
called upon, but this is in effect what happens.)
==== (Algorithm start) =====
Draw candidate positions for each candidate (in this case, each is a
point on a 4D normal distribution with mean 0 and variance 1).
Draw voter positions for each voter, and create their honest ballots
based on the distances between the voter and candidates.
Pass the resulting ballots through the method to establish the honest
outcome.
If there's a tie, skip (because deciding what a strict improvement is
when there's a honest tie is ambiguous). Otherwise let the winner be W.
For each candidate X who is not the winner W:
For i = 1 to number of strategy attempts / number of candidates
Set the strategic ballots to the honest ballots.
For every voter who prefers X to W:
Replace that voter's strategic ballot with a
ballot according to a strategy that depends on
i.
Pass the modified strategic ballots through the method.
If X is now a winner, the method is manipulable in
this election. Return success.
If we reach this point without any success, return failure; the method
is (probably) not manipulable in this election.
==== (Algorithm end) =====
The indexed strategies are
i=0: Compromising (raise X to unique top)
i=1: Burial (lower W to unique bottom)
i=2: Two-sided (do both at once)
i>2: Coalitional strategy
The compromising, burial, and two-sided strategies modify the voters'
otherwise honest ballots - for instance, compromising changes a
strategist's ballot so that X is at unique top and the rest of the
ballot is unchanged.
The first time the coalitional strategy is called for a particular
election, candidate to strategize for, and value of i, it chooses a
random number of strategic ballots (between 1 and 3 inclusive). Each
strategic voter then picks one of these ballots at random. This
simulates strategies where every strategist ballot is equal, as well as
ones where there are a few groups each with their own ballot type, thus
covering more than JGA's simulations without becoming too
computationally expensive.
So with the setup for the stats that I gave, the full setup for a single
method is like this:
for j = 1 to 500k
Run the algorithm detailed above.
It returns one of three states: honest tie, success, or failure.
Increment the corresponding counter, call it TIES, SUCCESSES or
FAILURES.
manipulability = SUCCESSES/(500k - TIES)
So to answer your questions:
The strategists don't adapt their strategy to the information available
to them, even though they strictly speaking have full information.
However, they get to try over and over again until they win. If there is
a full information strategy with not too many distinct ballots, then
this random sampling will eventually find it, given a high enough
strategy attempts value.
For each non-winner X, everybody who prefers X to the current winner
gets to have a go. So not just their favorites: anybody they all prefer
to the current winner.
[2] The detailed stats suggest that pushover is a problem with
Smith//DAC
You don't have enough candidates for a sub-cycle, and so the method
can't fail mono-raise. How can it have a Pushover problem?
I did a bit more checking, and the full preference version doesn't have
this high an "other strategy" count. Since I think it's unlikely that
the version with truncation would have more pushover than the fully
ranked one, I'm going to retract this; most likely it's just an artifact
of the simulator's ballot reduction process that falsely attributes the
strategy to the "other" category for cardinal methods.
- Margins-Sorted Approval, because I'm not sure how it works
(I struggle to take this at face value. Probably my promotion of MSA
has convinced you that it is the best method and you were concerned
that
your simulation wouldn't do it justice.
I'd like to believe both that I have enough scientific integrity not to
do that, and that people know I have, too :-)
Actually, I was planning on putting MSA at the same level as the other
"I don't know enough about these or their dynamics" methods (double
defeat Hare, MSMLV, and Max Strength Transitive Beatpath).
But our expert doing the
simulation claiming he can't understand the method isn't a good look
for
its proposability.)
Why didn't you simply ask me to explain it to you?
I think it's the sorting phase that does it. My vague idea of how it
works is that you essentially run a sorting algorithm on intermediate
values, and that seems a little too complex to me. But I might just have
got it wrong and then the initial impression of it as an intimidating
method stuck.
Ted Stern pointed me at the Electowiki article for MSA, which in turn
led me to his Python implementation. I might port it if I have time, but
I feel a bit exhausted after gathering all these stats. We'll see :-)
What happened to separate entries for BTR, Woodall and Benham?
They're in the other post. I didn't want to add them all to the post
that was intended to focus on the new results. That's why I said "some
for comparison" - the others are here:
http://lists.electorama.com/pipermail/election-methods-electorama.com/2024-April/006029.html
I could post all the stats - ordinal and cardinal methods' - in a
summary post if you or other EM members would like.
-km
Election-Methods mailing list - see https://electorama.com/em for list
info
Of course, in Approval, if there aren’t other perceived reasons for
choosing whom to approve, one could approve above the mean…if you have a
feel for what’s average among the candidates. I guess that’s the usual
assumption for simulations.
If the candidate-lineup is so good that you do above-mean voting, then
you’re indeed fortunate.
If you’d do that, but you don’t have a feel about the average, & don’t
perceive cardinal-merits, then of course you could just approve the best
half of the candidates.
Maybe that was the assumed Approval strategy to which you were referring.
Approval is particularly perfectly matched for an election with
unacceptable candidates:
Just approve (only) all of the Acceptables.
On Sat, May 4, 2024 at 19:27 Michael Ossipoff email9648742@gmail.com
wrote:
There’s no reason for the renormalization. Among A, B, C & D (in that
order of magnitude) if B is at the mean, then, with the A=0 & D=1
renormalization, B’s renormalized value is the mean of all of the
renormalized values.
The position of the mean among the candidates doesn’t change with
renormalization.
On Sat, May 4, 2024 at 15:25 Michael Ossipoff email9648742@gmail.com
wrote:
On Sat, May 4, 2024 at 14:45 Kristofer Munsterhjelm km_elmet@t-online.de
wrote:
Yes, that's right. But consider a voter with the following utilities:
A: 0.57
B: 0.32
C: 0.23
D: 0.08
Normalization to two steps fixes the highest value (0.57) to 1 and the
lowest value (0.08) to 0 and rounds off the intermediate values after
linearly scaling them.
Yes. So far, so good. But…
This in essence says that a value is rounded off
to 1 if it's greater than or equal to 0.325 (the midpoint between 0.08
and 0.57)
What? You didn’t average the normalized values. You averaged two of the
values before normalization. The midrange isn’t usually the same as the
mean. You used the midrange as the mean.
If you call the top value 1, & the bottom value 0,
then a rating’s new value is the number that’s the same % of the way from
0 to 1 as the old number’s % from.08 to .57
Average of those new values: .4475
You still approve the best two.
so the 0-1 normalized ballot is
A: 1, B: 0, C: 0, D: 0
On the other hand, the mean utility is 0.3. So the mean utility approval
ballot is
A: 1, B: 1, C: 0, D: 0.
4 "dimensions" sounds like a lot. What are the "strategy attempts" ?
How much and what information do the strategists have? Are the
strategists confined to just trying to get their favourites elected,
or
any candidate they prefer to the initial winner?
The method works pretty much like this, for generating and testing a
single election. (I've simplified the exact order that strategies are
called upon, but this is in effect what happens.)
==== (Algorithm start) =====
Draw candidate positions for each candidate (in this case, each is a
point on a 4D normal distribution with mean 0 and variance 1).
Draw voter positions for each voter, and create their honest ballots
based on the distances between the voter and candidates.
Pass the resulting ballots through the method to establish the honest
outcome.
If there's a tie, skip (because deciding what a strict improvement is
when there's a honest tie is ambiguous). Otherwise let the winner be W.
For each candidate X who is not the winner W:
For i = 1 to number of strategy attempts / number of candidates
Set the strategic ballots to the honest ballots.
For every voter who prefers X to W:
Replace that voter's strategic ballot with a
ballot according to a strategy that depends on
i.
Pass the modified strategic ballots through the method.
If X is now a winner, the method is manipulable in
this election. Return success.
If we reach this point without any success, return failure; the method
is (probably) not manipulable in this election.
==== (Algorithm end) =====
The indexed strategies are
i=0: Compromising (raise X to unique top)
i=1: Burial (lower W to unique bottom)
i=2: Two-sided (do both at once)
i>2: Coalitional strategy
The compromising, burial, and two-sided strategies modify the voters'
otherwise honest ballots - for instance, compromising changes a
strategist's ballot so that X is at unique top and the rest of the
ballot is unchanged.
The first time the coalitional strategy is called for a particular
election, candidate to strategize for, and value of i, it chooses a
random number of strategic ballots (between 1 and 3 inclusive). Each
strategic voter then picks one of these ballots at random. This
simulates strategies where every strategist ballot is equal, as well as
ones where there are a few groups each with their own ballot type, thus
covering more than JGA's simulations without becoming too
computationally expensive.
So with the setup for the stats that I gave, the full setup for a single
method is like this:
for j = 1 to 500k
Run the algorithm detailed above.
It returns one of three states: honest tie, success, or failure.
Increment the corresponding counter, call it TIES, SUCCESSES or
FAILURES.
manipulability = SUCCESSES/(500k - TIES)
So to answer your questions:
The strategists don't adapt their strategy to the information available
to them, even though they strictly speaking have full information.
However, they get to try over and over again until they win. If there is
a full information strategy with not too many distinct ballots, then
this random sampling will eventually find it, given a high enough
strategy attempts value.
For each non-winner X, everybody who prefers X to the current winner
gets to have a go. So not just their favorites: anybody they all prefer
to the current winner.
[2] The detailed stats suggest that pushover is a problem with
Smith//DAC
You don't have enough candidates for a sub-cycle, and so the method
can't fail mono-raise. How can it have a Pushover problem?
I did a bit more checking, and the full preference version doesn't have
this high an "other strategy" count. Since I think it's unlikely that
the version with truncation would have more pushover than the fully
ranked one, I'm going to retract this; most likely it's just an artifact
of the simulator's ballot reduction process that falsely attributes the
strategy to the "other" category for cardinal methods.
- Margins-Sorted Approval, because I'm not sure how it works
(I struggle to take this at face value. Probably my promotion of MSA
has convinced you that it is the best method and you were concerned
that
your simulation wouldn't do it justice.
I'd like to believe both that I have enough scientific integrity not to
do that, and that people know I have, too :-)
Actually, I was planning on putting MSA at the same level as the other
"I don't know enough about these or their dynamics" methods (double
defeat Hare, MSMLV, and Max Strength Transitive Beatpath).
But our expert doing the
simulation claiming he can't understand the method isn't a good look
for
its proposability.)
Why didn't you simply ask me to explain it to you?
I think it's the sorting phase that does it. My vague idea of how it
works is that you essentially run a sorting algorithm on intermediate
values, and that seems a little too complex to me. But I might just have
got it wrong and then the initial impression of it as an intimidating
method stuck.
Ted Stern pointed me at the Electowiki article for MSA, which in turn
led me to his Python implementation. I might port it if I have time, but
I feel a bit exhausted after gathering all these stats. We'll see :-)
What happened to separate entries for BTR, Woodall and Benham?
They're in the other post. I didn't want to add them all to the post
that was intended to focus on the new results. That's why I said "some
for comparison" - the others are here:
http://lists.electorama.com/pipermail/election-methods-electorama.com/2024-April/006029.html
I could post all the stats - ordinal and cardinal methods' - in a
summary post if you or other EM members would like.
-km
Election-Methods mailing list - see https://electorama.com/em for list
info
Or approve everyone on the better side of the widest gap among the
successive candidate-merits.
Even if you don’t have a feel for cardinal-merits, you likely know where
the biggest gap is.
On Sat, May 4, 2024 at 21:10 Michael Ossipoff email9648742@gmail.com
wrote:
Of course, in Approval, if there aren’t other perceived reasons for
choosing whom to approve, one could approve above the mean…if you have a
feel for what’s average among the candidates. I guess that’s the usual
assumption for simulations.
If the candidate-lineup is so good that you do above-mean voting, then
you’re indeed fortunate.
If you’d do that, but you don’t have a feel about the average, & don’t
perceive cardinal-merits, then of course you could just approve the best
half of the candidates.
Maybe that was the assumed Approval strategy to which you were referring.
Approval is particularly perfectly matched for an election with
unacceptable candidates:
Just approve (only) all of the Acceptables.
On Sat, May 4, 2024 at 19:27 Michael Ossipoff email9648742@gmail.com
wrote:
There’s no reason for the renormalization. Among A, B, C & D (in that
order of magnitude) if B is at the mean, then, with the A=0 & D=1
renormalization, B’s renormalized value is the mean of all of the
renormalized values.
The position of the mean among the candidates doesn’t change with
renormalization.
On Sat, May 4, 2024 at 15:25 Michael Ossipoff email9648742@gmail.com
wrote:
On Sat, May 4, 2024 at 14:45 Kristofer Munsterhjelm <
km_elmet@t-online.de> wrote:
Yes, that's right. But consider a voter with the following utilities:
A: 0.57
B: 0.32
C: 0.23
D: 0.08
Normalization to two steps fixes the highest value (0.57) to 1 and the
lowest value (0.08) to 0 and rounds off the intermediate values after
linearly scaling them.
Yes. So far, so good. But…
This in essence says that a value is rounded off
to 1 if it's greater than or equal to 0.325 (the midpoint between 0.08
and 0.57)
What? You didn’t average the normalized values. You averaged two of the
values before normalization. The midrange isn’t usually the same as the
mean. You used the midrange as the mean.
If you call the top value 1, & the bottom value 0,
then a rating’s new value is the number that’s the same % of the way
from 0 to 1 as the old number’s % from.08 to .57
Average of those new values: .4475
You still approve the best two.
so the 0-1 normalized ballot is
A: 1, B: 0, C: 0, D: 0
On the other hand, the mean utility is 0.3. So the mean utility
approval
ballot is
A: 1, B: 1, C: 0, D: 0.
4 "dimensions" sounds like a lot. What are the "strategy attempts" ?
How much and what information do the strategists have? Are the
strategists confined to just trying to get their favourites elected,
or
any candidate they prefer to the initial winner?
The method works pretty much like this, for generating and testing a
single election. (I've simplified the exact order that strategies are
called upon, but this is in effect what happens.)
==== (Algorithm start) =====
Draw candidate positions for each candidate (in this case, each is a
point on a 4D normal distribution with mean 0 and variance 1).
Draw voter positions for each voter, and create their honest ballots
based on the distances between the voter and candidates.
Pass the resulting ballots through the method to establish the honest
outcome.
If there's a tie, skip (because deciding what a strict improvement is
when there's a honest tie is ambiguous). Otherwise let the winner be W.
For each candidate X who is not the winner W:
For i = 1 to number of strategy attempts / number of candidates
Set the strategic ballots to the honest ballots.
For every voter who prefers X to W:
Replace that voter's strategic ballot with a
ballot according to a strategy that depends on
i.
Pass the modified strategic ballots through the method.
If X is now a winner, the method is manipulable in
this election. Return success.
If we reach this point without any success, return failure; the method
is (probably) not manipulable in this election.
==== (Algorithm end) =====
The indexed strategies are
i=0: Compromising (raise X to unique top)
i=1: Burial (lower W to unique bottom)
i=2: Two-sided (do both at once)
i>2: Coalitional strategy
The compromising, burial, and two-sided strategies modify the voters'
otherwise honest ballots - for instance, compromising changes a
strategist's ballot so that X is at unique top and the rest of the
ballot is unchanged.
The first time the coalitional strategy is called for a particular
election, candidate to strategize for, and value of i, it chooses a
random number of strategic ballots (between 1 and 3 inclusive). Each
strategic voter then picks one of these ballots at random. This
simulates strategies where every strategist ballot is equal, as well as
ones where there are a few groups each with their own ballot type, thus
covering more than JGA's simulations without becoming too
computationally expensive.
So with the setup for the stats that I gave, the full setup for a
single
method is like this:
for j = 1 to 500k
Run the algorithm detailed above.
It returns one of three states: honest tie, success, or failure.
Increment the corresponding counter, call it TIES, SUCCESSES or
FAILURES.
manipulability = SUCCESSES/(500k - TIES)
So to answer your questions:
The strategists don't adapt their strategy to the information available
to them, even though they strictly speaking have full information.
However, they get to try over and over again until they win. If there
is
a full information strategy with not too many distinct ballots, then
this random sampling will eventually find it, given a high enough
strategy attempts value.
For each non-winner X, everybody who prefers X to the current winner
gets to have a go. So not just their favorites: anybody they all prefer
to the current winner.
[2] The detailed stats suggest that pushover is a problem with
Smith//DAC
You don't have enough candidates for a sub-cycle, and so the method
can't fail mono-raise. How can it have a Pushover problem?
I did a bit more checking, and the full preference version doesn't have
this high an "other strategy" count. Since I think it's unlikely that
the version with truncation would have more pushover than the fully
ranked one, I'm going to retract this; most likely it's just an
artifact
of the simulator's ballot reduction process that falsely attributes the
strategy to the "other" category for cardinal methods.
- Margins-Sorted Approval, because I'm not sure how it works
(I struggle to take this at face value. Probably my promotion of MSA
has convinced you that it is the best method and you were concerned
that
your simulation wouldn't do it justice.
I'd like to believe both that I have enough scientific integrity not to
do that, and that people know I have, too :-)
Actually, I was planning on putting MSA at the same level as the other
"I don't know enough about these or their dynamics" methods (double
defeat Hare, MSMLV, and Max Strength Transitive Beatpath).
But our expert doing the
simulation claiming he can't understand the method isn't a good look
for
its proposability.)
Why didn't you simply ask me to explain it to you?
I think it's the sorting phase that does it. My vague idea of how it
works is that you essentially run a sorting algorithm on intermediate
values, and that seems a little too complex to me. But I might just
have
got it wrong and then the initial impression of it as an intimidating
method stuck.
Ted Stern pointed me at the Electowiki article for MSA, which in turn
led me to his Python implementation. I might port it if I have time,
but
I feel a bit exhausted after gathering all these stats. We'll see :-)
What happened to separate entries for BTR, Woodall and Benham?
They're in the other post. I didn't want to add them all to the post
that was intended to focus on the new results. That's why I said "some
for comparison" - the others are here:
http://lists.electorama.com/pipermail/election-methods-electorama.com/2024-April/006029.html
I could post all the stats - ordinal and cardinal methods' - in a
summary post if you or other EM members would like.
-km
Election-Methods mailing list - see https://electorama.com/em for list
info
Oh, I see what you meant. You didn’t use the Above-Mean strategy. I just
assumed that that was your intention.
Instead you renormalized the ballots to 0-1, & rounded off a candidate’s
renormalized rating to the nearest integer.
Forgive me for jumping to conclusions. It was just that Above-Mean is the
assumed Approval-strategy that I’ve heard of in simulations. That of course
doesn’t make it better.
I hadn’t heard of it being done as you did it.
Both Above-Mean & Above-Midrange require cardinal estimates. One for the
overall (mean) merit of the candid-lineup; & the other for the halfway
point.
For one thing, Best-Half (Above-Median) is easier, not requiring any
estimate…nothing but your preference-ordering.
Of course the reason why Above-Mean is so much more popular for simulations
is because it’s a better strategy than Above-Midpoint,.
…when you don’t have information for:
Best-Frontrunner
Above-Expectation (based on which candidates you’d rather appoint than hold
the election, or on which candidates’ victories wouldn’t disappoint
you…etc.)
Where the largest merit-gap in the candidate-lineup is (“Largest-Gap”)
Which are acceptable & which are unacceptable
Which you like.
What makes Above-Mean better than Above-Midpoint is that, with no
information about the other voters, our expectation, or significant
differences in acceptability or likability, Above-Mean is what maxiimizes
your expectation.
In addition to Best-Half being the one that doesn’t require anything more
than your preference-ordering, it also has the advantage of maximizing the
number of pairwise-preferences that you vote.
If one doesn’t have the information, feel, or estimates that the other
methods need (including the perception of a particularly big merit-gap),
then Best-Half (Above-Median) would be a good choice.
Speaking for myself, there are unacceptable candidates, & it’s simply a
matter of Approving (only) the Acceptables.
On Sat, May 4, 2024 at 21:14 Michael Ossipoff email9648742@gmail.com
wrote:
Or approve everyone on the better side of the widest gap among the
successive candidate-merits.
Even if you don’t have a feel for cardinal-merits, you likely know where
the biggest gap is.
On Sat, May 4, 2024 at 21:10 Michael Ossipoff email9648742@gmail.com
wrote:
Of course, in Approval, if there aren’t other perceived reasons for
choosing whom to approve, one could approve above the mean…if you have a
feel for what’s average among the candidates. I guess that’s the usual
assumption for simulations.
If the candidate-lineup is so good that you do above-mean voting, then
you’re indeed fortunate.
If you’d do that, but you don’t have a feel about the average, & don’t
perceive cardinal-merits, then of course you could just approve the best
half of the candidates.
Maybe that was the assumed Approval strategy to which you were referring.
Approval is particularly perfectly matched for an election with
unacceptable candidates:
Just approve (only) all of the Acceptables.
On Sat, May 4, 2024 at 19:27 Michael Ossipoff email9648742@gmail.com
wrote:
There’s no reason for the renormalization. Among A, B, C & D (in that
order of magnitude) if B is at the mean, then, with the A=0 & D=1
renormalization, B’s renormalized value is the mean of all of the
renormalized values.
The position of the mean among the candidates doesn’t change with
renormalization.
On Sat, May 4, 2024 at 15:25 Michael Ossipoff email9648742@gmail.com
wrote:
On Sat, May 4, 2024 at 14:45 Kristofer Munsterhjelm <
km_elmet@t-online.de> wrote:
Yes, that's right. But consider a voter with the following utilities:
A: 0.57
B: 0.32
C: 0.23
D: 0.08
Normalization to two steps fixes the highest value (0.57) to 1 and the
lowest value (0.08) to 0 and rounds off the intermediate values after
linearly scaling them.
Yes. So far, so good. But…
This in essence says that a value is rounded off
to 1 if it's greater than or equal to 0.325 (the midpoint between 0.08
and 0.57)
What? You didn’t average the normalized values. You averaged two of the
values before normalization. The midrange isn’t usually the same as the
mean. You used the midrange as the mean.
If you call the top value 1, & the bottom value 0,
then a rating’s new value is the number that’s the same % of the way
from 0 to 1 as the old number’s % from.08 to .57
Average of those new values: .4475
You still approve the best two.
so the 0-1 normalized ballot is
A: 1, B: 0, C: 0, D: 0
On the other hand, the mean utility is 0.3. So the mean utility
approval
ballot is
A: 1, B: 1, C: 0, D: 0.
4 "dimensions" sounds like a lot. What are the "strategy attempts"
?
How much and what information do the strategists have? Are the
strategists confined to just trying to get their favourites elected,
or
any candidate they prefer to the initial winner?
The method works pretty much like this, for generating and testing a
single election. (I've simplified the exact order that strategies are
called upon, but this is in effect what happens.)
==== (Algorithm start) =====
Draw candidate positions for each candidate (in this case, each is a
point on a 4D normal distribution with mean 0 and variance 1).
Draw voter positions for each voter, and create their honest ballots
based on the distances between the voter and candidates.
Pass the resulting ballots through the method to establish the honest
outcome.
If there's a tie, skip (because deciding what a strict improvement is
when there's a honest tie is ambiguous). Otherwise let the winner be W.
For each candidate X who is not the winner W:
For i = 1 to number of strategy attempts / number of candidates
Set the strategic ballots to the honest ballots.
For every voter who prefers X to W:
Replace that voter's strategic ballot with a
ballot according to a strategy that depends on
i.
Pass the modified strategic ballots through the method.
If X is now a winner, the method is manipulable in
this election. Return success.
If we reach this point without any success, return failure; the method
is (probably) not manipulable in this election.
==== (Algorithm end) =====
The indexed strategies are
i=0: Compromising (raise X to unique top)
i=1: Burial (lower W to unique bottom)
i=2: Two-sided (do both at once)
i>2: Coalitional strategy
The compromising, burial, and two-sided strategies modify the voters'
otherwise honest ballots - for instance, compromising changes a
strategist's ballot so that X is at unique top and the rest of the
ballot is unchanged.
The first time the coalitional strategy is called for a particular
election, candidate to strategize for, and value of i, it chooses a
random number of strategic ballots (between 1 and 3 inclusive). Each
strategic voter then picks one of these ballots at random. This
simulates strategies where every strategist ballot is equal, as well
as
ones where there are a few groups each with their own ballot type,
thus
covering more than JGA's simulations without becoming too
computationally expensive.
So with the setup for the stats that I gave, the full setup for a
single
method is like this:
for j = 1 to 500k
Run the algorithm detailed above.
It returns one of three states: honest tie, success, or
failure.
Increment the corresponding counter, call it TIES, SUCCESSES
or FAILURES.
manipulability = SUCCESSES/(500k - TIES)
So to answer your questions:
The strategists don't adapt their strategy to the information
available
to them, even though they strictly speaking have full information.
However, they get to try over and over again until they win. If there
is
a full information strategy with not too many distinct ballots, then
this random sampling will eventually find it, given a high enough
strategy attempts value.
For each non-winner X, everybody who prefers X to the current winner
gets to have a go. So not just their favorites: anybody they all
prefer
to the current winner.
[2] The detailed stats suggest that pushover is a problem with
Smith//DAC
You don't have enough candidates for a sub-cycle, and so the method
can't fail mono-raise. How can it have a Pushover problem?
I did a bit more checking, and the full preference version doesn't
have
this high an "other strategy" count. Since I think it's unlikely that
the version with truncation would have more pushover than the fully
ranked one, I'm going to retract this; most likely it's just an
artifact
of the simulator's ballot reduction process that falsely attributes
the
strategy to the "other" category for cardinal methods.
- Margins-Sorted Approval, because I'm not sure how it works
(I struggle to take this at face value. Probably my promotion of
MSA
has convinced you that it is the best method and you were concerned
that
your simulation wouldn't do it justice.
I'd like to believe both that I have enough scientific integrity not
to
do that, and that people know I have, too :-)
Actually, I was planning on putting MSA at the same level as the other
"I don't know enough about these or their dynamics" methods (double
defeat Hare, MSMLV, and Max Strength Transitive Beatpath).
But our expert doing the
simulation claiming he can't understand the method isn't a good look
for
its proposability.)
Why didn't you simply ask me to explain it to you?
I think it's the sorting phase that does it. My vague idea of how it
works is that you essentially run a sorting algorithm on intermediate
values, and that seems a little too complex to me. But I might just
have
got it wrong and then the initial impression of it as an intimidating
method stuck.
Ted Stern pointed me at the Electowiki article for MSA, which in turn
led me to his Python implementation. I might port it if I have time,
but
I feel a bit exhausted after gathering all these stats. We'll see :-)
What happened to separate entries for BTR, Woodall and Benham?
They're in the other post. I didn't want to add them all to the post
that was intended to focus on the new results. That's why I said "some
for comparison" - the others are here:
http://lists.electorama.com/pipermail/election-methods-electorama.com/2024-April/006029.html
I could post all the stats - ordinal and cardinal methods' - in a
summary post if you or other EM members would like.
-km
Election-Methods mailing list - see https://electorama.com/em for
list info
On 2024-05-05 00:25, Michael Ossipoff wrote:
On Sat, May 4, 2024 at 14:45 Kristofer Munsterhjelm
<km_elmet@t-online.de mailto:km_elmet@t-online.de> wrote:
Yes, that's right. But consider a voter with the following utilities:
A: 0.57
B: 0.32
C: 0.23
D: 0.08
Normalization to two steps fixes the highest value (0.57) to 1 and the
lowest value (0.08) to 0 and rounds off the intermediate values after
linearly scaling them.
Yes. So far, so good. But…
This in essence says that a value is rounded off
to 1 if it's greater than or equal to 0.325 (the midpoint between 0.08
and 0.57)
What? You didn’t average the normalized values. You averaged two of the
values before normalization. The midrange isn’t usually the same as the
mean. You used the midrange as the mean.
Yes, that's my point. What I call "normalize" uses the midrange, because
it's a generalization of the following approach for mapping utilities to
a cardinal scale:
Let min be the minimum utility (the utility of your worst candidate),
and max be the utility of your favorite. Then the normalized ratings
from 0 to k inclusive are given by
r_i = round( (x_i - min)/(max-min) * k), where x_i is the utility of the
ith candidate, and r_i is the rating of that candidate.
I thought this was the natural way to create Range ballots from
utilities. To my knowledge, there's no obvious way to generalize
above-mean utility thresholding to say, a 0-10 scale.
I'm aware that "normalize" as I've defined it above is different from
above-mean thresholding, and that the latter is more common in
treatments of approval voting in particular. For that reason, I use mean
utility cutoffs for the approval hybrids, and range-based normalization
for the Range hybrids. My point was just that one shouldn't confuse the
two and think I used the same approach for approval and Range.
-km
On 2024-05-05 12:50, Michael Ossipoff wrote:
Oh, I see what you meant. You didn’t use the Above-Mean strategy. I just
assumed that that was your intention.
Instead you renormalized the ballots to 0-1, & rounded off a candidate’s
renormalized rating to the nearest integer.
Forgive me for jumping to conclusions. It was just that Above-Mean is
the assumed Approval-strategy that I’ve heard of in simulations. That of
course doesn’t make it better.
I hadn’t heard of it being done as you did it.
I guess I was being imprecise. To restate what I said in the other post,
but in fewer words:
- For Range (Smith//Range, STAR, etc): I use a generalization of the
midpoint approach. - For Approval (Smith//Approval explicit implicit etc): I use mean
utility cutoffs.
These seem to be the obvious ways to do it for the respective domains.
-km