[EM] A more efficient strategy-free ratings-based method than Hay voting

[Jobst Heitzig]

Dear Forest,

you wrote:

> I have one question, though. If best strategy is to report true
> utilities, then what do you mean by encouraging compromise?
>
> You must mean compromise in the outcome as opposed to compromise in the
> ballots.
That's true. By "compromise" I mean the transfer of two voters' share of 
the winning probability from their favourite options to a common 
"compromise" option in case that transfer increases both voters' 
expected utility according to the ratings they provided.

> In other words, you want the lottery to favor centrist candidates as
> much as possible without giving incentive for the voters to compromise
> through distorted ratings.
>
> Is that right?
I'm not sure. My intention was just to construct a democratic (in the 
sense of equal power for all voters) method under which it was optimal 
to reveal true utilities and which was hopefully more efficient than 
both Hay voting and Random Ballot. Having studied D2MAC before, the idea 
was natural to use two randomly drawn ballots for this. The third drawn 
ballot is only used for providing the potential compromise option in a 
clone-proof way, so that the whole method becomes clone-proof (another 
advantage over Hay voting).

Of course, it would be even more efficient to not draw the potential 
compromise option at random (by means of a third randomly drawn ballot) 
but try to find a "best" compromise given the first two randomly drawn 
ballots, and also to try to find "optimal" instead of random numbers x,y 
for the probability transfers. That would however destroy the incentive 
to vote sincerely and introduce strategy.

Jobst

>> From: Jobst Heitzig
>
>> Dear friends,
>>
>> Hay voting was supposedly the first known method under which it is
>> always optimal (as judged from expected utility) to vote sincere
> ratings
>> (i.e. ratings proportional to true utility). However, it seems that it
>> is a rather inefficient method (as judged from total expected utility),
>> even less efficient than Random Ballot.
>>
>> Here's a different, more efficient method under which it is also always
>> optimal (as judged from expected utility) to vote sincere ratings. It
> is
>> also based on Random Ballot, but in a very different way. It is
>> essentially a Random Ballot method with an added mechanism of automatic
>> cooperation for compromise. The basic idea is that when there is a pair
>> of ballots showing preferences A>...>C>...>B and B>...>C>...>A, those
>> two voters can profit from cooperating and transferring part of "their"
>> share of the winning probability from A and B to the compromise option
> C.
>> Here's the method, I call it...
>>
>>
>> RANDOM BALLOT WITH AUTOMATIC COOPERATION, Version 1 (RBAC1):
>> ------------------------------------------------------------
>> Voters rate each option.
>> Three ballots i,j,k and two numbers x,y between 0 and 1/2 are drawn at
>> random.
>> Assume that the top-ranked options of i,j,k are A,B,C, and that i and j
>> have assigned to A,B,C the ratings ri(A),ri(B),ri(C) and
>> rj(A),rj(B),rj(C), respectively.
>> Now check whether the inequalities
>> y * (ri(C) - ri(B)) > x * (ri(A) - ri(C))
>> and
>> x * (rj(C) - rj(A)) > y * (rj(B) - rj(C))
>> both hold.
>> If so, elect A, B, or C with probabilities 1/2 - x, 1/2 - y, x + y,
>> respectively.
>> Otherwise, elect A or B each with probability 1/2.
>>
>>
>> Why should it be optimal to vote sincere ratings under this method?
>>
>> Consider an arbitrary voter i with favourite option A, and some
>> arbitrary options B,C and numbers x,y between 0 and 1/2.
>> Let us designate the A,B,C-lottery with probabilities 1/2 - x, 1/2 -
>> y, x + y by L, and the A,B-lottery with probabilities 1/2 and 1/2 by
> M.
>> The only thing i can do about the election outcome is by influencing
>> whether or not "her" inequality
>> y * (ri(C) - ri(B)) > x * (ri(A) - ri(C))
>> holds, and the only situations in which this matters at all are those
> in
>> which i is among the first two drawn ballots, the other of the two has
> B
>> top-ranked, and the third has C top-ranked.
>> As it is equally likely for i's ballot to be drawn as the first or the
>> second ballot, and as i cannot influence whether or not the other
> inequality
>> x * (rj(C) - rj(A)) > y * (rj(B) - rj(C))
>> holds, i would therefore want "her" inequality
>> y * (ri(C) - ri(B)) > x * (ri(A) - ri(C))
>> to hold if and only if she prefers lottery L to lottery M.
>> But the latter is the case if and only if
>> y * (ui(C) - ui(B)) > x * (ui(A) - ui(C))
>> where ui(A),ui(B),ui(C) are i's evaluations of the true utility of the
>> options A,B,C.
>> Now x and y were arbitrary numbers, so the only way to get this
>> equivalence is to put ri(A),ri(B),ri(C) proportional to
>> ui(A),ui(B),ui(C), and perhaps adding some irrelevant constant. Q.E.D.
>>
>>
>> Note that it doesn't matter from which precise distribution x and y are
>> drawn as long as all values from 0 to 1/2 are possible. For the sake of
>> efficiency, one should therefore use a distribution that strongly
>> favours values near 1/2, so that cooperation will be more likely. Also,
>> the winning probabilities can safely be changed to
>> 1/2 - x/z, 1/2 - y/z, (x+y)/z,
>> where z := 2 * max(x,y). This will increase the probability of good
>> compromises further.
>>
>> Finally, note the following important fact about the method: It is
>> perfectly democratic since it distributes power equally in the
> following
>> sense: Any faction of m voters can give "their" share m/n of the
> winning
>> probability to any option they like by simply "bullet-rating" that
>> option at one and all others at zero.
>>
>> Please send comments!
>> Jobst
>>
>
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