Re: [Election-Methods] A Better Version of IRV?
If QLTD isn't cloneproof (and it isn't), then the result won't be either, hence we could just as well go with first preference Copeland (unless that has a flaw I'm not seeing). What is supposed to be the attraction of first preference Copeland? And how do you define it exactly? The attraction of FPC (which is what I call Simmons' supposed Cloneproof extension of Copeland, but that wasn't cloneproof after all) is that it's extremely hard to do burial with it. The definition of first preference Copeland is: The candidate for which those who beat him pairwise gather fewest first-place votes, in sum, is the winner. Simmons invented the method, I just use that name instead of Simmons cloneproof method, as it isn't cloneproof. It's an extension of Copeland since the only information it takes from the pairwise matrix is binary; in this case, whether some candidate Y beats X, and in Copeland's case whether some candidate Y is beaten by X; thus first preference Copeland. I could also just call it Simmons, I suppose, since the other Simmons methods I know of have defined names of their own and so wouldn't be confused with it. First preference Copeland would be vulnerable to the situation where multiple candidates have equal first place rival scores. One way to solve this would be to use Schwartz, or Schwartz//. Another would be to use a positional system that counts second, third, etc, place votes also but only very weakly, like Nauru-Borda (or something going 1/10^p, p = 0..n); yet another would be to use Bucklin (if there are any ties, count first and second place votes of rivals, etc), and even another would be to have an approval cutoff and use Approval instead of first preferences (unless everybody bullet votes). I haven't tested the positional or Bucklin variants here so I don't know if those solutions would be any good. I'm not sure if it's possible to make a situation where two candidates are in the Schwartz set yet none of their rivals rank first or the rivals' ranked-first sum is equal for all. Perhaps that's possible if you make dummy candidates that collectively hog all the first place votes, but who each are ranked below the various other candidates enough times that they don't beat any of them? Something like Q1 A B C Q2 Q3 Q2 B C A Q3 Q1 Q3 B A C Q1 Q2 .. In the general case, such scaffolding will work (block winners) with Schwartz,. It won't work with Schwartz// unless you can somehow get all the Qs inside the Schwartz set yet still have them cover each candidate, first-preference wise, equally. But, (reading the cloneproof Copeland thread even as I'm writing this,) Schwartz//FPC would not be summable. Election-Methods mailing list - see http://electorama.com/em for list info
Re: [Election-Methods] delegate cascade
On Jul 23, 2008, at 10:59 , Michael Allan wrote: (ii) Otherwise, A is a mosquito voting for an elephant! You seem to assume that there is a hierarchy of voters that is used for communication in the political process, and that this hierarchy is determined (maybe even formally) by the voting behaviour, and that direct links between mosquitos and elephants are not the best working solution. Should I read this so that if a person has voted for a candidate that has then (surprisingly) become popular, and this voter doesn't have many indirect votes to carry from the other voters, then it would be better for this voter to change his vote and vote for some less popular (mouse size) intermediate candidate whose votes will cascade to the original most preferred candidate? If this is true then the voting process is quite strongly a communication hierarchy building process. I.e. voters do not vote their favourites but candidates that they think would be good enough, and right size contact points for them, and whose votes would cascade to the right candidate. I understood that the votes are public, so the candidates would know who their voters are. I understood it would be acceptable for a mosquito to vote for an elephant, but the mosquito could then assume that the elephant would not have much time to discuss with him (worth one vote only). I expect the cycles in opinions to potentially cause repeated changes in the cast votes (but since I don't know yet exactly how the voter will be cascaded I will not attempt to describe the details yet). http://zelea.com/project/votorola/d/theory.xht#cascade-cyclic I doubt Figure 9 will ever occur in a real election - it's very much an edge case - but if it does, it shouldn't cause any instability. Unless I've overlooked something... Let's say that in Figure 9 there are three candidates that are interested in getting lots of votes. They could be the very top candidate (T), the bottom left candidate (L) and the bottom right candidate (R). Candidate T prefers R to L. Candidate R prefers L to T. Candidate L prefers T to R. Voting will start by all voting for their favourite candidate. The result is as in Figure 9. Then candidate T abstains. As a result he will get lots of votes. Candidate L reacts to this by abstaining. As a result of this candidate L will get the highest number of votes. Now candidate T realizes that he needs to vote again (as in Figure 9) in order to avoid electing L. Candidate L still has most votes. But now candidate R can (and will) abstain, and will get more votes than L. Now candidate L is in the same position as candidate T was few moments ago. Candidate L votes again and thereby opens up the option for candidate T to abstain again and become the leader. Next it is candidate R's turn to do the same tricks and allow candidate L to become the leader. These cyclic changes could in principle continue forever. The point here is that group opinions may contain cycles (this is not dependent on what election method is used). Methods that allow votes to be changed continuously may end up in loops like this. If cycles are expected to cause problems (when they exist) one could develop some tricks that could slow down the cyclic changes or even ban them somehow. Juho ___ Now you can scan emails quickly with a reading pane. Get the new Yahoo! Mail. http://uk.docs.yahoo.com/nowyoucan.html Election-Methods mailing list - see http://electorama.com/em for list info
Re: [Election-Methods] delegate cascade
Juho wrote: On Jul 22, 2008, at 14:26 , Michael Allan wrote: I'm grateful I was directed to this list. You're clearly experts. I wish I could reply more completely right away (I should know better than to start 2 separate threads). I'll just reply to Juho's questions today, and tomorrow I'll look at Abd's work. (You've been thinking about this longer than I have, Abd, and I need to catch up.) 1) All voters are candidates and it is possible that all voters consider themselves to be the best candidate. Therefore the method may start from all candidates having one vote each (their own vote). Maybe only after some candidates have numerous votes and the voter himself has only one vote still, then the voter gives up voting for himself and gives his vote to some of the frontrunners. How do you expect the method to behave from this point of view? The basic rule of vote flow is: a vote stops *before* it encounters a voter for a second time, and it remains held where it is. A vote is always considered to have encountered its original caster beforehand. So it is not possible to vote for oneself. It is permitted, but the vote stops before it is even cast - there is no effect. Ok, not allowing voters to vote for themselves may to some extent solve the problem. (Some voters may however decide to abstain for a while.) This is a bit offtopic (again), but another idea that might be less prone to strategy in the case of cyclical proxy candidacy occurred to me: use eigenvector or Markov-based methods to distribute the deferred power smoothly over the candidates in the cycle. At this point, the method looks similar to the original PageRank used to vote on web pages, where various web pages vote for the importance of each other - and such voting chains may be cyclical. Election-Methods mailing list - see http://electorama.com/em for list info
Re: [Election-Methods] a strategy-free range voting variant?
Range Voting selects the option with the highest average rating. Jobst has found a method that selects the option with the highest average rating by a random subset of the voters, while (totally?) discouraging the exageration of preferences that tends to happen in ordinary Range Voting. It seems to me that it should be even easier to find a similar strategy free method that selects the option with the highest median rating; when a vote is above or below the median it makes no difference on the value of the median how far above or below (at least in the case of an odd number of voters). The simplest idea is just to charge one voter grickle against the account of each voter that voted above the median of the winner, and redistribute these evenly to the accounts of the voters that voted below median. Of course, lots of technical details would have to be worked out, e.g. to take care of the case where several options have the same median, and the case where nobody voted above median. This version would end up being similar to some version of Bucklin with a tax for winning and a compensation for losing. More analogous to Jobst's idea would be a method where a random ballot benchmark lottery is used, but instead of using the expected ratings of that lottery on the various ballots, use the rating R for which it is equally likely that the lottery winner would be rated above or below R (on ballot i). If (on ballot i) the winner X is rated above R, then the probability P of the lottery winner being between R and X is the tax paid (by the compensating voters) on behalf of i into the accounts of the other voters. Instead of voters with higher accounts having greater range possibilities, they would have greater weight in determining medians. Also, the Random Ballot Lottery would take into account these weights. Essentially, if your virtual bank account is 30, it is like having thirty votes, whether in the Bucklin aspect, or in the RB Lottery aspect. I know that social scientists addicted to utility will prefer the mean approach over the median approach, but this makes more sense to me, because the money has a more direct relation to probability. What do you think? Can something along these lines be worked out? Forest Election-Methods mailing list - see http://electorama.com/em for list info
[Election-Methods] Representative Range Voting with Compensation - a new attempt
Dear folks,
I must admit the last versions of RRVC (Representative Range Voting with
Compensation) all had a flaw which I saw only yesterday night. Although
they did achieve efficiency and strategy-freeness, they did not achieve
my other goal: that voters who like the winner more than the random
ballot lottery compensate voters who liked the random ballot lottery
more than the winner. In short, the flaw was to use the three randomly
drawn voter groups for only one task each, either for the benchmark, or
the compensation, or the decision.
I spare you the details and just give a new version which I think may
finally achieve all three goals: efficiency, strategy-freeness, and
voter compensation.
The basic idea is still the same: Partition the voters randomly into
three groups, let one group decide via Range Voting, and use each group
to benchmark another group and to compensate still another group.
To make an analysis more easy, I write it down more formally this time
and assume the number of voters is a multiple of 3.
DEFINITION OF METHOD RRVC (Version 3)
=
Notation:
-
X,Y,Z are variables for options
i,j,k are variables for voters
f,g,h are variables for groups of voters
Input:
--
All voters give ratings and mark a favourit. Put...
R(X,i) := the rating voter i gave option X
F(i) := the option marked favourite on ballot of voter i
A(i) := balance on voter i's voting account before the decision
Tally:
--
Randomly partition the N voters into three groups of equal size.
The winner is the range voting winner of group 1.
The voting accounts are adjusted as follows. Put...
S := N/3
Q := (S-1)/S
G(i) := group in which voter i landed
T(X,f) := total rating group f gave option X
= sum { R(X,i) : i in group f }
W(g) := range voting winner of group g
= that W with T(W,g)T(X,g) for all X other than W
P(X,h) := proportion of group h favouring X
= probability of X in group h's random ballot lottery
= # { i in group h : F(i)=X } / S
D(f,g,i) := rating difference on voter i's ballot
between the range voting winner of group f
and the random ballot lottery of group g
= R(W(f),i) - sum { P(X,g)*R(X,i) : X }
E(f,g,h) := total rating difference in group h
between the range voting winner of group f
and the random ballot lottery of group g
= sum { D(f,g,i) : i in group g }
For each voter i, add the following amount to her voting account C(i):
If i is in group 1:
deltaC(i) := E(1,2,1)-D(1,2,i) - E(2,2,2) - Q*E(3,3,2) + E(3,3,3)
If i is in group 2:
deltaC(i) := E(3,3,2)-D(3,3,i) - E(3,3,3) - Q*E(1,1,3) + E(1,1,1)
If i is in group 3:
deltaC(i) := E(1,1,3)-D(1,1,i) - E(1,1,1) - Q*E(1,2,1) + E(2,2,2)
(Remark: E(1,2,1) and D(1,2,1) are not typos!)
(END OF METHOD RRVC)
Analysis:
-
1.
The sum of all C(i) remains constant, so voting money retains its
value. To see this, note that
sum { E(1,2,1)-D(1,2,i) - E(2,2,2) : i in group 1 }
= S*E(1,2,1) - E(1,2,1) - S*E(2,2,2) )
= S*( Q*E(1,2,1) - E(2,2,2) )
= sum { Q*E(1,2,1) - E(2,2,2) : i in group 3 }
and analogous for the other terms in the above sums.
2.
Note that the terms E(1,2,1)-D(1,2,i), E(3,3,2)-D(3,3,i), and
E(1,1,3)-D(1,1,i) in the above sums do not depend on voter i's ratings!
Hence the only way in which the ballot of voter i can affect her own
voting account is trough the dependency of W(1) on her ratings, and
this is only the case for voters in group 1, the deciding group.
So, as only voters in group 1 can influence their outcome, an analysis
of individual voting strategy is only required these voters. For such a
voter i the net outcome, up to some constant which is independent of
i's behaviour, is this:
O(i) := sum { R(W(1),j) : j other than i } + U(W(1),i)
where
U(X,i) := true value of X for i.
If voter i is honest and puts R(X,i)=U(X,i), this simply adds up to
O(i) = T(W(1),1) (if i is honest).
Now assume this honest voter i thinks about changing the winner from
W(1) to some other option Y by voting dishonestly. The net outcome for
i after this manipulation would be
O'(i) = sum { R(Y,j) : j other than i } + U(Y,i)
= T(Y,1)-R(Y,i) + U(Y,i)
= T(Y,1)
T(W(1),1) = O(i).
So after all, i has no incentive to manipulate the outcome because she
would have to pay more than she gains from this.
3.
Now consider a large electorate of honest voters, and think about what a
voter can expect, before the random process of drawing the three groups
is applied, of how much her voting account will be adjusted. If I got
it right this time, this expected value of deltaC(i) should be, up to
some constant term which is equal for all voters, just
the rating difference on voter i's ballot
between the random ballot lottery
and the winner of the decision,
Re: [Election-Methods] Representative Range Voting with Compensation -a new attempt
A first typo: It must read C(i) instead of A(i) under Input...
--
Dear folks,
I must admit the last versions of RRVC (Representative Range Voting with
Compensation) all had a flaw which I saw only yesterday night. Although
they did achieve efficiency and strategy-freeness, they did not achieve
my other goal: that voters who like the winner more than the random
ballot lottery compensate voters who liked the random ballot lottery
more than the winner. In short, the flaw was to use the three randomly
drawn voter groups for only one task each, either for the benchmark, or
the compensation, or the decision.
I spare you the details and just give a new version which I think may
finally achieve all three goals: efficiency, strategy-freeness, and
voter compensation.
The basic idea is still the same: Partition the voters randomly into
three groups, let one group decide via Range Voting, and use each group
to benchmark another group and to compensate still another group.
To make an analysis more easy, I write it down more formally this time
and assume the number of voters is a multiple of 3.
DEFINITION OF METHOD RRVC (Version 3)
=
Notation:
-
X,Y,Z are variables for options
i,j,k are variables for voters
f,g,h are variables for groups of voters
Input:
--
All voters give ratings and mark a favourit. Put...
R(X,i) := the rating voter i gave option X
F(i) := the option marked favourite on ballot of voter i
A(i) := balance on voter i's voting account before the decision
Tally:
--
Randomly partition the N voters into three groups of equal size.
The winner is the range voting winner of group 1.
The voting accounts are adjusted as follows. Put...
S := N/3
Q := (S-1)/S
G(i) := group in which voter i landed
T(X,f) := total rating group f gave option X
= sum { R(X,i) : i in group f }
W(g) := range voting winner of group g
= that W with T(W,g)T(X,g) for all X other than W
P(X,h) := proportion of group h favouring X
= probability of X in group h's random ballot lottery
= # { i in group h : F(i)=X } / S
D(f,g,i) := rating difference on voter i's ballot
between the range voting winner of group f
and the random ballot lottery of group g
= R(W(f),i) - sum { P(X,g)*R(X,i) : X }
E(f,g,h) := total rating difference in group h
between the range voting winner of group f
and the random ballot lottery of group g
= sum { D(f,g,i) : i in group g }
For each voter i, add the following amount to her voting account C(i):
If i is in group 1:
deltaC(i) := E(1,2,1)-D(1,2,i) - E(2,2,2) - Q*E(3,3,2) + E(3,3,3)
If i is in group 2:
deltaC(i) := E(3,3,2)-D(3,3,i) - E(3,3,3) - Q*E(1,1,3) + E(1,1,1)
If i is in group 3:
deltaC(i) := E(1,1,3)-D(1,1,i) - E(1,1,1) - Q*E(1,2,1) + E(2,2,2)
(Remark: E(1,2,1) and D(1,2,1) are not typos!)
(END OF METHOD RRVC)
Analysis:
-
1.
The sum of all C(i) remains constant, so voting money retains its
value. To see this, note that
sum { E(1,2,1)-D(1,2,i) - E(2,2,2) : i in group 1 }
= S*E(1,2,1) - E(1,2,1) - S*E(2,2,2) )
= S*( Q*E(1,2,1) - E(2,2,2) )
= sum { Q*E(1,2,1) - E(2,2,2) : i in group 3 }
and analogous for the other terms in the above sums.
2.
Note that the terms E(1,2,1)-D(1,2,i), E(3,3,2)-D(3,3,i), and
E(1,1,3)-D(1,1,i) in the above sums do not depend on voter i's ratings!
Hence the only way in which the ballot of voter i can affect her own
voting account is trough the dependency of W(1) on her ratings, and
this is only the case for voters in group 1, the deciding group.
So, as only voters in group 1 can influence their outcome, an analysis
of individual voting strategy is only required these voters. For such a
voter i the net outcome, up to some constant which is independent of
i's behaviour, is this:
O(i) := sum { R(W(1),j) : j other than i } + U(W(1),i)
where
U(X,i) := true value of X for i.
If voter i is honest and puts R(X,i)=U(X,i), this simply adds up to
O(i) = T(W(1),1) (if i is honest).
Now assume this honest voter i thinks about changing the winner from
W(1) to some other option Y by voting dishonestly. The net outcome for
i after this manipulation would be
O'(i) = sum { R(Y,j) : j other than i } + U(Y,i)
= T(Y,1)-R(Y,i) + U(Y,i)
= T(Y,1)
T(W(1),1) = O(i).
So after all, i has no incentive to manipulate the outcome because she
would have to pay more than she gains from this.
3.
Now consider a large electorate of honest voters, and think about what a
voter can expect, before the random process of drawing the three groups
is applied, of how much her voting account will be adjusted. If I got
it right this time, this expected value of deltaC(i) should be, up to
some constant term which is equal for all voters, just
the rating difference on
