MAM vs Schulze
Do I remember correctly that MAM is just Ranked Pairs with a better tie
breaker?
It seems like MAM and Ranked Pairs put in as many defeats as possible
without creating cycles, but like Juho says do we really need all of those
defeats to decide the winner?
River recognizes that we just need one defeat for each non-winner. It seems
to me that the one defeat for each non-winner should be as strong as
possible, but (having already been eliminated) their other defeats don't
matter.
I want to propose a new Condorcet Method below, but first a simple three
slot method inspired by Jameson's MAS, but one that truly satisfies the
Chicken Defense Criterion:
Ballots are scores or ratings on a scale of zero to three.
The Smith candidate with the highest average score wins.
In other words, the method is Smith//Score (or is it Smith\Score ?).with
three slot ballots.
Example:
49 C
27 A>B
24 B (sincere is B>A)
With sincere votes A is elected as the only member of Smith.
Under the B faction defection C wins as the Smith candidate with the
highest Score.
Now, how do we adapt this to general rankings? We assume that equal top
rankings and equal bottom or multiple truncations are allowed.
For each ballot on which a candidate is ranked above bottom but below top
that candidte receives one point. For each ballot on which the candidate
is ranked top or equal top that candidate receives two points.
The Smith candidate with the greatest number of points wins.
[End of definition]
Note that the method does satisfy CD unlike Smith//ImplicitApproval.
Jameson's idea of three slot scores makes it work.
How does it do on burial?
Forest
It's a slight annoyance of mine that the name MAM is used. Like someone can tidy up a method slightly and just reclaim it for their own. I don't know why people call it MAM to be honest.
From: Forest Simmons <fsimmons@pcc.edu>
To: EM election-methods@lists.electorama.com
Sent: Sunday, 9 October 2016, 22:13
Subject: [EM] MAM vs Schulze
Do I remember correctly that MAM is just Ranked Pairs with a better tie breaker?
On 10/09/2016 11:13 PM, Forest Simmons wrote:
Do I remember correctly that MAM is just Ranked Pairs with a better tie
breaker?
Let "descending pairwise" (DP) be the spanning tree-like algorithm at
the basis of Ranked Pairs.
Then the original interpretation of the names, as far as I know, is:
Ranked Pairs is DP (margins).
MAM is DP (wv) with a random voter hierarchy tiebreak.
Sometimes, "Ranked Pairs" is used for DP in general. (I've done that,
myself, at times.)
On 10/10/2016 7:43 AM, Forest Simmons wrote:
I want to propose a new Condorcet Method below, but first a simple
three slot method inspired by Jameson's MAS, but one that truly
satisfies the Chicken Defense Criterion:
Ballots are scores or ratings on a scale of zero to three.
The Smith candidate with the highest average score wins.
Forest,
What is the "Chicken Defense Criterion"?
Your suggested 3-slot Smith//Score doesn't meet the Chicken Dilemma
criterion.
http://wiki.electorama.com/wiki/Chicken_Dilemma_Criterion
33: A>B
32: B
34: C
A>B>C>A , all candidates in the Smith set. 0-1-2 Scores: B 97 > C 68
A 66.
B easily wins, but the Chicken Dilemma criterion specifies that B must
not win.
Chris Benham
Do I remember correctly that MAM is just Ranked Pairs with a better
tie breaker?
C: MAM ("Maximum Affirmed Majorities") is Ranked Pairs (Winning
Votes) with a specific random-ballot based tie-breaker.
http://wiki.electorama.com/wiki/Maximize_Affirmed_Majorities
It seems like MAM and Ranked Pairs put in as many defeats as possible
without creating cycles, but like Juho says do we really need all of
those defeats to decide the winner?
River recognizes that we just need one defeat for each non-winner. It
seems to me that the one defeat for each non-winner should be as
strong as possible, but (having already been eliminated) their other
defeats don't matter.
I want to propose a new Condorcet Method below, but first a simple
three slot method inspired by Jameson's MAS, but one that truly
satisfies the Chicken Defense Criterion:
Ballots are scores or ratings on a scale of zero to three.
The Smith candidate with the highest average score wins.
In other words, the method is Smith//Score (or is it Smith\Score
?).with three slot ballots.
Example:
49 C
27 A>B
24 B (sincere is B>A)
With sincere votes A is elected as the only member of Smith.
Under the B faction defection C wins as the Smith candidate with the
highest Score.
Now, how do we adapt this to general rankings? We assume that equal
top rankings and equal bottom or multiple truncations are allowed.
For each ballot on which a candidate is ranked above bottom but below
top that candidte receives one point. For each ballot on which the
candidate is ranked top or equal top that candidate receives two points.
The Smith candidate with the greatest number of points wins.
[End of definition]
Note that the method does satisfy CD unlike Smith//ImplicitApproval.
Jameson's idea of three slot scores makes it work.
How does it do on burial?
Forest
Election-Methods mailing list - see http://electorama.com/em for list info
Chris,
I agree, just give points to the equal top candidates.
Or if more extensive rankings are wanted for the purpose of finding Smith,
use standard ordinal ballots with an extra check box beside each candidate
name. Check the box iff you want to contribute a point to that candidate
in the point count should she make it into Smith.
On Sun, Oct 9, 2016 at 8:18 PM, C.Benham cbenham@adam.com.au wrote:
On 10/10/2016 7:43 AM, Forest Simmons wrote:
I want to propose a new Condorcet Method below, but first a simple three
slot method inspired by Jameson's MAS, but one that truly satisfies the
Chicken Defense Criterion:
Ballots are scores or ratings on a scale of zero to three.
The Smith candidate with the highest average score wins.
Forest,
What is the "Chicken Defense Criterion"?
Your suggested 3-slot Smith//Score doesn't meet the Chicken Dilemma
criterion.
http://wiki.electorama.com/wiki/Chicken_Dilemma_Criterion
33: A>B
32: B
34: C
A>B>C>A , all candidates in the Smith set. 0-1-2 Scores: B 97 > C 68
A 66.
B easily wins, but the Chicken Dilemma criterion specifies that B must not
win.
Chris Benham
Do I remember correctly that MAM is just Ranked Pairs with a better tie
breaker?
C: MAM ("Maximum Affirmed Majorities") is Ranked Pairs (Winning Votes)
with a specific random-ballot based tie-breaker.
http://wiki.electorama.com/wiki/Maximize_Affirmed_Majorities
It seems like MAM and Ranked Pairs put in as many defeats as possible
without creating cycles, but like Juho says do we really need all of those
defeats to decide the winner?
River recognizes that we just need one defeat for each non-winner. It
seems to me that the one defeat for each non-winner should be as strong as
possible, but (having already been eliminated) their other defeats don't
matter.
I want to propose a new Condorcet Method below, but first a simple three
slot method inspired by Jameson's MAS, but one that truly satisfies the
Chicken Defense Criterion:
Ballots are scores or ratings on a scale of zero to three.
The Smith candidate with the highest average score wins.
In other words, the method is Smith//Score (or is it Smith\Score ?).with
three slot ballots.
Example:
49 C
27 A>B
24 B (sincere is B>A)
With sincere votes A is elected as the only member of Smith.
Under the B faction defection C wins as the Smith candidate with the
highest Score.
Now, how do we adapt this to general rankings? We assume that equal top
rankings and equal bottom or multiple truncations are allowed.
For each ballot on which a candidate is ranked above bottom but below top
that candidte receives one point. For each ballot on which the candidate
is ranked top or equal top that candidate receives two points.
The Smith candidate with the greatest number of points wins.
[End of definition]
Note that the method does satisfy CD unlike Smith//ImplicitApproval.
Jameson's idea of three slot scores makes it work.
How does it do on burial?
Forest
Election-Methods mailing list - see http://electorama.com/em for list info