[EM] Re: Approval cutoff AKA "None of the Below"

[Jobst Heitzig]

Dear Ted!

You wrote:
> I explored TACC with some enthusiasm over the last couple of days.  

It's astonishing that a method that simple could give such nice results,
isn't it? That was a fabulous idea by Forest to construct chains in
order of approval! I still want to explore his Needle method in more
detail to see whether it may be an interesting method if one does not
require Condorcet-efficiency.

You continued:
> First of all, I realized that TACC is equivalent to the following:
>       1) List the candidates from lowest to highest approval
>       2) Starting with the lowest approved candidate, eliminate anybody she
>          defeats.
>       Any remaining candidates must defeat her.
>       Pick the remaining candidate with next higher approval, and repeat.

Right. That's a bit like Pythagoras' sieve, isn't it?

> By the way, Jobst, what do you do about defeat ties or approval ties?

When I design a method I usually don't mind at first. However, ties can
always be resolved using a TBRO as in MAM without violating
monotonicity. Of course, ties will have to get more attention when we
design social choice methods for smaller groups, but here we deal with
general elections, right?

> Since members of the Smith set defeats everybody outside the Smith set, TACC
> would gets the same result if you start in the Smith set and ignore the
> others.  So consider the following classic case:
> 
>       40: A>B>>C
>       35: B>C>>A
>       25: C>A>>B
> 
> So you start with C.  C (approval 60)  defeats A, so A (approval 65) is
> eliminated.  B (approval 75) defeats C and wins via TACC.
> 
> This differs from RP and Beatpath.  

That's true. It resolves three-cycles in a fundamentally different way
than most Condorcet methods studied in the history of this list. It
doesn't care about defeat strengths but about approval scores instead.

> I don't think it encourages altruistic approval cutoff 

I missed the definition of that, so I will have to look that up in the
archives...

> -- the Nash equilibrium will occur with
> 
>       40: A>>B>C
>       35: B>>C>A
>       25: C>>A>B
> 
> which again has B winning.

Why is that? Can you argue for this equilibrium in more detail? It was
always my impression that in most situations there is either no
equilibrium of an interesting kind, or many of them, so why is there
exactly one in this situation? And what definition of "Nash equilibrium"
do you use, are the players (i) the individual voters, (ii) groups of
voters with identical sincere preferences [and approvals], or (iii)
arbitrary groups of voters?

Yours, Jobst