[Election-Methods] [english 98%] Re: [english 89%] Re: [english 95%] Re: [english 95%] Re: [english 94%]Re: method designchallenge+new method AMP

[Jobst Heitzig]

Dear Raphfrk,

if I understand you right, your idea is to start with a Random Ballot 
lottery and then, as long as some option is preferred to the current 
lottery by at least 90% of the voters (as judged from their ratings), 
slightly advance that option which is preferred to the current lottery 
by the largest number of voters.

That would be clone-proof, I guess.

In the situation at hand, if all voters were sincere, that would indeed 
advance C until C is the sure winner. If the A voters were to decrease 
their rating of C to less than 51, they would instead end up with the 
Random Ballot lottery which they don't prefer to C.

So that seems to be OK so far unless some other strategy makes sincerity 
a non-equilibrium.


In other situations with even much better compromise options, however, 
your method would perform poorly: If the ratings were
   45: A 100 > C 90 > B,D 0
   44: B 100 > C 90 > A,D 0
   11: D 100 > A,B,C 0
for example, the result would be Random Ballot since only 89% prefer C 
to it.

The problem is that you cannot really reduce the threshold too much (say 
to 50%) without making the method more and more majoritarian. Perhaps a 
compromise threshold could be 67%. But that would still mean that 70% of 
the voters could force their will upon the other 30% without any need to 
cooperate.

The method AMP would do the following with the above ballots: it would 
match the 44 B voters with 44 of the A voters and transfer "their" share 
of the winning probability from A and B to C, resulting in a lottery in 
which C/D/A win with 88/11/1% probability, and there would be no obvious 
strategy for anyone to improve upon this.

For the sake of completeness, let's see also what D2MAC would do with 
the above situation: With D2MAC, the equilibrium ballots would look like 
this:
   45: A favourite, C also approved
   44: B favourite, C also approved
   11: D favourite.
The result would be a lottery in which C wins with probability 89%*89% 
(which is approx. 80%) while D/A/B would win with approximately 
11/4.5/4.5% probability. Considering that D2MAC is monotonic and so much 
simpler than the other methods, this is still quite good performance I 
guess.

There is another method I have in mind, but that will be a new posting...

Yours, Jobst

> I was thinking of a possible modification to your system that may take 
> the chances of C to 100%.
> 
> 1) Each voter submits a ballot rating each candidate.
> 
> 2) The initial probability of each candidate is set at the proportion of 
> highest ratings they get (equal top ratings split equally)
> 
> 3) For each candidate:
> 
> Add a small increase in their probability
> 
> Renormalise probabilities (so bring total back to 1)
> 
> Determine number of voters who think this probability set is an improvement
> 
> 4) If the change that has the greatest support has the support of more 
> than (high)% of the voters, update the probabilities and goto 3
> 
> 5) Randomly draw a candidate based on the final probabilities
> 
> The threshold for step 4 might be set at 90%.
> 
> 
> So under your system, the voters might submit:
> 
> 51: A(100) B(52) C(0)
> 49: A(0) B(52) C(0)
> 
> The initial probabilities are
> 
> A: 51%
> B: 0%
> C: 49%
> 
> Adding 1% to A would have the support of the A voters
> Adding 1% to B would have the support of the B voters
> 
> Adding 1% to C would give
> 
> A: 50.495%
> B: 1%
> C: 49.505%
> 
> 100% voters consider that an improvement.
> 
> This should be true the whole way until C gets to 100%
> A: 0.51%
> B: 0.49%
> C: 99%
> 
> A voter's utility => 0.51*100 + 99*52 = 5199
> B voter's utility => 0.49*100 + 99*52 = 5197
> 
> After
> 
> A: 0%
> B: 0%
> C: 100%
> 
> A voter's utility => 100*52 = 5200
> B voter's utility => 100*52 = 5200
> 
> 100% of voters think it is an improvement.
> 
> If A voters voted
> 
> A(100) B(0) C(0)
> 
> the the results would be
> 
> Initial
> 
> A: 51%
> B: 49%
> C: 0%
> 
> Reducing the probability of A winning would represent a decrease in the 
> utility from the perspective of the A voters (according to their votes).
> 
> This means that no change would be approved by 90%+ of the voters and 
> thus the initial probabilities would be the final ones.
> 
> This is not as good from the A supporters, so they would be advised to 
> vote honestly.
> 
> The 90% rule prevents hold-outs and increases the chances that a single 
> winner occurs with 100% probability.
> 
> I am not sure if there a way for the A supporters to shift the result to 
> a non-certain winner that they like better.
> 
> 
> 
> Raphfrk
> --------------------
> Interesting site
> "what if anyone could modify the laws"
> 
> www.wikocracy.com
> 
> 
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