[EM] FBC survey & Simmons latest lottery method
wds at math.temple.edu
Sat Nov 19 09:23:56 PST 2005
I have added some stuff to incorporate Venzke's comments.
I was going to add a bit on Simmons' latest FBC-obeying lottery method.
But I think he is wrong so I deleted that bit.
In Simmons's method, you vote with a rank-ordering & approval cutoff.
Simmons did not say whether rank-equalities were permitted. For simplicity
I assume not.
The winner is either the approval-winner A, or the candidate B who is
most-approved among those not majority-pairwise losing to A.
(If any such exist. Otherwise A just wins.)
You decide which by tossing a coin.
Now Simmons claims this method obeys FBC.
Now, it seems to me simmons's claims ought to be equally valid
no matter what the bias-value is on that coin.
In particular, the 100% and 0% biased coins (i.e. deterministic) are
also FBC-obeying methods. Eh? So this is a whole other kind
of FBC-obeying voting method, hello.
Why does it obey FBC? Well, betraying your favorite F obviously
has no effect on everybody's approval counts so A is unaffected.
Also, it obviously has no effect on the pairwise battles between A and X,
so B is unaffected. (Unless A or B is F itself, of course.)
So not only does this obey FBC, but it does so in
the "stronger sense" in which Range and Approval obey FBC.
Sounds good... EXCEPT...
There is one caveat we have to worry about: what if A is your favorite F,
and then A loses to B, whereas if you betray F then A becomes somebody else
so you get a winner you like better?
E.g. say there is a Condorcet cycle B>A>F>B with F the most-approved in the cycle.
So A wins assuming 100% coin-bias. But some voters betray F. Now B is the most-approved
member of the cycle and then F wins the election! (You can also consider 99% coin-bias
to get nearly the same effect... even 50% coins still are a counterexample
witht he right candidate utilities...) So betraying F is the way to make F win!
So... thanks to this counterexample, I disagree Simmons' method obeys FBC.
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