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

[Forest W Simmons]

Jobst,

I'm still digesting your new method.  It is starting to make sense to 
me.  It is extremely creative and ingenious, in my opinion. I just hope 
that we haven't overlooked any subtle logical error.

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.

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?

Forest


>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
>