[EM] Comparing ranked versus unranked methods
Adam Tarr
atarr at ecn.purdue.edu
Wed Feb 6 13:38:32 PST 2002
I wrote and Forest responded:
>> It seems to me that rather than talking about "satisfaction points" it
>> makes more sense to decay the value of the ballots that voted for a
>> winner. The effect is the same, it just makes more sense by my way of
>> thinking. For example, if we wanted to use Webster's method, then divide
>> the value of ballots that voted for the most recent winner by (2n+1), where
>> n is the number of candidates that ballot has elected. Or multiply by
>> (2n-1)/(2n+1) if you prefer a recursive definition. For d'Hondt, use
>> 1/(n+1) at each stage, or n/(n+1) to work recursively.
>The council head H is taken to be the Approval winner from the Approval
>ballots. Then (using the same ballots) the PAV procedure is applied to
>all three member subsets of the form {H, X1, X2}. The subset with the
>highest PAV score determines who the two counselors C1 and C2 will be.
>
>Then (using the same ballots) the PAV procedure is applied to all subsets
>of the form {H, C1, C2, Y1, Y2, ... Yn} where n is the number of other
>members to be chosen for the council.
>
>In sequential PAV the successively conditioned subsets are of the
>respective forms {C1, X}, {C1, C2, X}, {C1,C2,C3,X}, etc. where one member
>is added at a time.
I _think_ I follow what you are saying. What I am arguing (perhaps
incorrectly!) is that my method of iteratively electing and removing candidates
is exactly equivalent to trying to maximize the PAV score of an entire set of
candidates. That is, repeat the following process k times for a k-seat council:
1) Tally the ballots.
2) Select the Approval winner. This candidate is elected.
3) For each ballot that contained the Approval winner, decay all of the
remaining votes on that ballot by n/(n+1) (d'Hondt) or (2n-1)/(2n+1) (Webster),
where n is the number of candidates that ballot has helped elect so far. (For a
manual count, simply mark the ballot each time it elects a candidate.)
4) Remove the elected candidate from the ballot.
Again, my contention is that this simpler process yields exactly the same
result. I may be wrong, but it seems to be the case. I think this can be
proven by induction (the methods elect the same candidate in the one-candidate
case; assume they elect the same n-candidate slate; show this implies they elect
the same n+1 candidate slate).
>This is appropriate if there is supposed to be a pecking order or
>seniority system in the elected body.
I agree, this is one of the nice advantages of this approach.
>Right. It turns out that since the sequence 1 + 1/2 + ... + 1/n is
>asymptotic to ln(n) that there is a simple way of getting a Borda version
>of PR:
...
>The three member subset with the highest product of ballot totals is the
>winning combination.
While I agree that this asymptotic approximation would work in with a very large
pool of candidates, I am not sure that it would be reliable under rounding
errors for a small number of candidates. Moreover, multi-winner Borda is likely
to be even more strategy-ridden than single-winner. Still, it's a slick,
elegant extension of Borda, and it's nice to see that all the methods have a
logical extension into multi-winner (a sort of "super symmetry" if you like).
Which brings us to...
>I posted a few attempts at Condorcet PR last spring or summer. The best I
>could come up with was a sequential method analogous to sequential PAV
>that went something like this:
>
>When comparing two candidates X and Y head-to-head to see which one should
>be added to the list of winners, discount the influence of ballot B by
>giving it a weight W to be determined as follows:
>
>Let K be the number of already determined winners above the
>preferred of X and Y. Let J be the number of already determined winners
>between X and Y in preference. Then take W to be 2/(2+J+2*K).
>
>In other words, the more satisfaction a voter already has relative to the
>two candidates in the current contest, the less the voter has to say about
>that current contest.
>
>I had mixed feelings about putting the J term into the formula. I would
>like to hear your ideas, perhaps stimulated by this partial progress.
I think you're definitely on the right track. I agree that the (J+2K) term is a
little incongruous, but my instinct is actually (2J+2K), not 2K. Or 4J+4K for
Webster's. I have a feeling that there's a right answer here, and looking at
enough concrete voting examples will reveal it. I will definitely take a look.
It will be very cool if we can make this work out.
-Adam
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