[EM] Comparing ranked versus unranked methods

[Adam Tarr]

Forest Wrote:

> n A and C
> n B and C
> 51-n A only
> 49-n B only
>
> As long as 2*n is greater than 51, the winning
> combination is {A,C} in sequential PAV.
>
> But when 2*n is not too much greater than 51 [up until 65], 
> the winning combination is still {A,B} in regular PAV.

So I was wrong... good thing I admitted that possibility!

This is a really interesting result, since this preference structure 
could be easily produced by the classic "left, middle, right" three 
candidate race.  C, the middle candidate, approval winner, and likely 
Condorcet winner, is NOT elected in PAV.  This actually makes sense, 
as it strikes a better balance between extremes to elect left and 
right.  Only when middle has much stronger support than the weaker 
extreme (65 vs. 49) does the popularity of middle override the desire 
for a balanced council.

As you mentioned, d'Hondt was the rule used for discounting ballots 
that had already elected one candidate.  I have observed that using 
Webster's rule can create some strategic problems.  Specifically, in 
a perfect-information election, a faction can do better for itself by 
arranging to split its votes up among candidates, rather than having 
the entire faction vote the entire slate of candidats from that 
faction.  For example, say there are 4 seats to be filled.  One 
faction has 1203 voters, and the other faction has 1200 voters.  
Faction 1 votes for all four of its candidates with every vote.  
Faction B devises a scheme where the voters split into three groups.  
So the votes go,

1200 A,B,C,D
401 E,F
401 G,H
401 I,J

The first candidate elected is one of A,B,C, and D.  But the next 
three candidates elected are one out of each of the sub-factions of 
the second group.  The same outcome occurs if you apply PAV with 
Webster's rule in the non-sequential fashion.  (Side note: I'm almost 
sure sequential and non-sequential PAV are equivalent if there is no 
overlap in the votes between various voting factions.)  Using 
d'Hondt's rule, this vote-splitting technique nearly backfires, but 
produces the proper 2-2 council split.

Even if we try to combat this by using modified Webster's 
(1.5,3,5,... in stead of 1,3,5,...), these manipulations are still 
possible, although more difficult.  Here's an example, with a 6-seat 
race, where one faction has 1202 votes and the other has 1000.

1000 A,B,C,D,E,F
601 G,H,I
601 J,K,L

In this scenario, two candidates are elected out of each sub-faction, 
which gives four seats to the strategically split faction where three 
seats for each faction is more fair (it is what would be given by 
either Webster's OR d'Hondt if the voting were simply 1202 to 1000).  
With d'Hondt's rule, the election has the proper 3-3 split regardless 
of whether the larger faction splits up or not.

Using d'Hondt's rule, this sort of offensive strategic manipulation 
by clever vote-splitting appears to be impossible... it seems obvious 
from playing with examples, although I'm having trouble coming up 
with a clean way to explain it.  So, it looks like d'Hondt might be 
the better choice for PAV for strategic reasons.  I consider this to 
be a shame, since Webster's is the most proportional method we could 
choose.  But when it comes to campaigning for the methods, d'Hondt 
has two advantages: it seems more intuitive, and it can be called 
Jefferson's method.  And no good American can disagree with our 
illustrious third president.

-Adam