Re: [Election-Methods] [english 94%] Re: method design challenge +new method AMP
On May 3, 2008, at 19:08 , [EMAIL PROTECTED] wrote: One possible solution to the clone issue is to scale the number of candidates to first choice vote. One option for that would be to allow people cast a nominating vote as well as the ranking. The total number of nominee votes would become the quota for each candidate. If a reasonable number of people (5%) recognised C as a compromise, then he would win. Yes, use of the first positions and nominations are good approaches to eliminate the clone problems in the STV based method (to make it more applicable to typical real life elections (not necessarily for the challenge of Jobst)). One approach would be to count all the candidates that are ranked above the planned winner and share the support of the voter between all of them. Juho ___ All new Yahoo! Mail The new Interface is stunning in its simplicity and ease of use. - PC Magazine http://uk.docs.yahoo.com/nowyoucan.html Election-Methods mailing list - see http://electorama.com/em for list info
Re: [Election-Methods] [english 94%] Re: method design challenge +new method AMP
Juho wrote: Here's an example on how the proposed method might work. I'll use your set of votes but only the rankings. 51: ACB 49: BCA Let's then reverse the votes to see who the voters don't like. 51: BCA 49: ACB Then we'll use STV (or some other proportional method) to select 2 ? (=3-1) candidates. STV would elect B and A. B and A are thus the ? worst candidates (proportionally determined) that will be eliminated. ? Only C remains and is the winner. This is not? clone independent. 52: ACB 48: BCA B+A 'elected', so C wins However, if it is changed to 26: A1A2CB 26: A2A1CB 48: BCA1A2 Since 3 are now elected, it requires 25% of the vote per candidate elected. the 52 block can 'elect' B and C and the 48 block elects A2. This means that A1 wins as he isn't picked. One possible solution to the clone issue is to scale the number of candidates to first choice vote. In effect, in the STV stage, the quota for each candidate would be equal to the number of first choice votes the candidate received. To be 'elected', the candidate would have to exceed the quota. The first candidate to be eliminated becomes the winner. Reversed votes: 26: BCA1A2 26: BCA2A1 48: A2A1CB Quotas (number of first choices in original ballots): A2: 26 A1: 26 C: 0 B: 48 Round 1: A1: 0 A2: 48 B: 52 C: 0 B exceeds quota by 4 and A2 exceeds quota by 22 Round 2: A1: 22 (-26) A2*: 26 (+26) B*: 48 (-4) C: 4 (+4) C is elected though, so A1 still wins. Note this is clone independent though: Quotas A: 52 B: 48 C: 0 Round 1: A: 48 B: 52 C: 0 B exceeds quota Round 2: A: 48 B: 48 C: 4 C exceeds quota A wins. However, if 5 voters voted C first choice, then C would be eliminated as being on the lowest total. One option for that would be to allow people cast a nominating vote as well as the ranking.? The total number of nominee votes would become the quota for each candidate.? If a reasonable number of people (5%) recognised C as a compromise, then he would win. I am not sure of the tactical issues associated with the 2 votes though. Also, it is majority compliant.? If a majority support a candidate first choice (i.e. first choice and nominate him), then he cannot lose. Another issue is how to actually layout the ballot.? It might be worth having voters enter the reversed ballot order.? In most practical cases, voters would need to enter their lowest ranked candidates, unlike in normal STV where it would be their most ranked. The ballot instructions could be something like: Place an X beside the candidate you wish to nominate in the nominate column In the rank column, rank the candidates in order of your preference giving a rank of 1 to your least favourite, 2 to your next least favourite and so on You do not have to rank all the candidates and any you do not rank will be considered preferred to any ranked candidate Raphfrk Interesting site what if anyone could modify the laws www.wikocracy.com AOL's new homepage has launched. Take a tour at http://info.aol.co.uk/homepage/ now. Election-Methods mailing list - see http://electorama.com/em for list info
Re: [Election-Methods] [english 94%] Re: method design challenge + new method AMP
Dear Juho,
I'm not sure what you mean by
How about using STV or some other proportional method to select the
n-1 worst candidates and then elect the remaining one?
Could you give an example or show how this would work out in the
situation under consideration?
Yours, Jobst
Juho
On Apr 28, 2008, at 20:58 , Jobst Heitzig wrote:
Hello folks,
over the last months I have again and again tried to find a
solution to
a seemingly simple problem:
The Goal
-
Find a group decision method which will elect C with near certainty in
the following situation:
- There are three options A,B,C
- There are 51 voters who prefer A to B, and 49 who prefer B to A.
- All voters prefer C to a lottery in which their favourite has 51%
probability and the other faction's favourite has 49% probability.
- Both factions are strategic and may coordinate their voting
behaviour.
Those of you who like cardinal utilities may assume the following:
51: A 100 C 52 B 0
49: B 100 C 52 A 0
Note that Range Voting would meet the goal if the voters would be
assumed to vote honestly instead of strategically. With strategic
voters, however, Range Voting will elect A.
As of now, I know of only one method that will solve the problem (and
unfortunately that method is not monotonic): it is called AMP and is
defined below.
*** So, I ask everyone to design some ***
*** method that meets the above goal! ***
Have fun,
Jobst
Method AMP (approval-seeded maximal pairings)
-
Ballot:
a) Each voter marks one option as her favourite option and may name
any number of offers. An offer is an (ordered) pair of options
(y,z). by offering (y,z) the voter expresses that she is willing to
transfer her share of the winning probability from her favourite
x to
the compromise z if a second voter transfers his share of the winning
probability from his favourite y to this compromise z.
(Usually, a voter would agree to this if she prefers z to
tossing a
coin between her favourite and y).
b) Alternatively, a voter may specify cardinal ratings for all
options.
Then the highest-rated option x is considered the voter's favourite,
and each option-pair (y,z) for with z is higher rated that the mean
rating of x and y is considered an offer by this voter.
c) As another, simpler alternative, a voter may name only a
favourite
option x and any number of also approved options. Then each
option-pair (y,z) for which z but not y is also approved is
considered
an offer by this voter.
Tally:
1. For each option z, the approval score of z is the number of
voters
who offered (y,z) with any y.
2. Start with an empty urn and by considering all voters free for
cooperation.
3. For each option z, in order of descending approval score, do the
following:
3.1. Find the largest set of voters that can be divvied up into
disjoint
voter-pairs {v,w} such that v and w are still free for cooperation, v
offered (y,z), and w offered (x,z), where x is v's favourite and y is
w's favourite.
3.2. For each voter v in this largest set, put a ball labelled with
the
compromise option z in the urn and consider v no longer free for
cooperation.
4. For each voter who still remains free for cooperation after this
was
done for all options, put a ball labelled with the favourite option of
that voter in the urn.
5. Finally, the winning option is determined by drawing a ball from
the
urn.
(In rare cases, some tiebreaker may be needed in step 3 or 3.1.)
Why this meets the goal: In the described situation, the only
strategic
equilibrium is when all B-voters offer (A,C) and at least 49 of the
A-voters offer (B,C). As a result, AMP will elect C with 98%
probability, and A with 2% probability.
Election-Methods mailing list - see http://electorama.com/em for
list info
___
All new Yahoo! Mail The new Interface is stunning in its simplicity and ease
of use. - PC Magazine
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Election-Methods mailing list - see http://electorama.com/em for list info
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Re: [Election-Methods] [english 94%] Re: method design challenge + new method AMP
Here's an example on how the proposed method might work.
I'll use your set of votes but only the rankings.
51: ACB
49: BCA
Let's then reverse the votes to see who the voters don't like.
51: BCA
49: ACB
Then we'll use STV (or some other proportional method) to select 2
(=3-1) candidates. STV would elect B and A. B and A are thus the
worst candidates (proportionally determined) that will be eliminated.
Only C remains and is the winner.
- I used only rankings = also worse than 52 point compromise
candidates would be elected
- I didn't use any lotteries = C will be elected with certainty
Juho
On May 2, 2008, at 22:29 , Jobst Heitzig wrote:
Dear Juho,
I'm not sure what you mean by
How about using STV or some other proportional method to select
the n-1 worst candidates and then elect the remaining one?
Could you give an example or show how this would work out in the
situation under consideration?
Yours, Jobst
Juho
On Apr 28, 2008, at 20:58 , Jobst Heitzig wrote:
Hello folks,
over the last months I have again and again tried to find a
solution to
a seemingly simple problem:
The Goal
-
Find a group decision method which will elect C with near
certainty in
the following situation:
- There are three options A,B,C
- There are 51 voters who prefer A to B, and 49 who prefer B to A.
- All voters prefer C to a lottery in which their favourite has 51%
probability and the other faction's favourite has 49% probability.
- Both factions are strategic and may coordinate their voting
behaviour.
Those of you who like cardinal utilities may assume the following:
51: A 100 C 52 B 0
49: B 100 C 52 A 0
Note that Range Voting would meet the goal if the voters would be
assumed to vote honestly instead of strategically. With strategic
voters, however, Range Voting will elect A.
As of now, I know of only one method that will solve the problem
(and
unfortunately that method is not monotonic): it is called AMP and is
defined below.
*** So, I ask everyone to design some ***
*** method that meets the above goal! ***
Have fun,
Jobst
Method AMP (approval-seeded maximal pairings)
-
Ballot:
a) Each voter marks one option as her favourite option and may
name
any number of offers. An offer is an (ordered) pair of options
(y,z). by offering (y,z) the voter expresses that she is
willing to
transfer her share of the winning probability from her
favourite x to
the compromise z if a second voter transfers his share of the
winning
probability from his favourite y to this compromise z.
(Usually, a voter would agree to this if she prefers z to
tossing a
coin between her favourite and y).
b) Alternatively, a voter may specify cardinal ratings for all
options.
Then the highest-rated option x is considered the voter's
favourite,
and each option-pair (y,z) for with z is higher rated that the mean
rating of x and y is considered an offer by this voter.
c) As another, simpler alternative, a voter may name only a
favourite
option x and any number of also approved options. Then each
option-pair (y,z) for which z but not y is also approved is
considered
an offer by this voter.
Tally:
1. For each option z, the approval score of z is the number of
voters
who offered (y,z) with any y.
2. Start with an empty urn and by considering all voters free for
cooperation.
3. For each option z, in order of descending approval score, do the
following:
3.1. Find the largest set of voters that can be divvied up into
disjoint
voter-pairs {v,w} such that v and w are still free for
cooperation, v
offered (y,z), and w offered (x,z), where x is v's favourite and
y is
w's favourite.
3.2. For each voter v in this largest set, put a ball labelled
with the
compromise option z in the urn and consider v no longer free for
cooperation.
4. For each voter who still remains free for cooperation after
this was
done for all options, put a ball labelled with the favourite
option of
that voter in the urn.
5. Finally, the winning option is determined by drawing a ball
from the
urn.
(In rare cases, some tiebreaker may be needed in step 3 or 3.1.)
Why this meets the goal: In the described situation, the only
strategic
equilibrium is when all B-voters offer (A,C) and at least 49 of the
A-voters offer (B,C). As a result, AMP will elect C with 98%
probability, and A with 2% probability.
Election-Methods mailing list - see http://electorama.com/em for
list info
___ All
new Yahoo! Mail The new Interface is stunning in its simplicity
and ease of use. - PC Magazine http://uk.docs.yahoo.com/
nowyoucan.html
Election-Methods mailing list - see http://electorama.com/em for
list info
