[EM] Free riding

[Jonathan Lundell]

On Sep 3, 2008, at 12:28 AM, Juho wrote:

> I hope this speculation provided something useful. And I hope I got  
> the Meek's method dynamics right.

Meek completely fixes Woodall free riding. That strategy takes  
advantage of the fact that most STV methods (to the extent we're in a  
STV/Meek/etc context) are sensitive to elimination order in how they  
distribute surpluses. In most other STV methods, if I vote for my  
first and second preferences AB first, and A has a surplus, then only  
a fraction of my vote (or a probabilistic whole) transfers to B. But  
if I rank hopeless candidate Z first: ZAB, then (hopefully) A gets  
elected before Z is eliminated, and my whole vote goes to B. If Z gets  
eliminated first, no harm done, I'm left with AB. The hazard, of  
course, is that so many voters do this that Z gets elected and/or AB  
eliminated.

Meek cures this entirely via its principle that when Z is eliminated,  
the ballots are counted *as if Z had never run*. There's no advantage  
to me in ranking Z first.


Hylland is another kettle of fish. Here, I vote BA instead of my  
sincere AB, because I "know" that A will be elected without my help,  
and I can afford to spend my entire vote on B.

This is only useful, of course, if I'm competing with other A  
supporters who have some second choice, say AC voters. They will have  
only a fraction of their votes transfer to C, while I will have my  
entire vote counted for B because I didn't bother to rank A first,  
even though A is my first choice (I'd better be very confident).


There's a risk to the Hylland strategy, of course, if I make a mistake  
in judging that A will be elected without my help. Other than that,  
though, I don't offhand see a way of defending against Hylland free  
riding.