[EM] Equilibrium in Approval Voting

[Adam Tarr]

>I'm just an amateur, but isn't an Approval election just a game in which
>every player has a finite set of strategies, and payoffs for each
>strategy?  If so, then doesn't it have to have a Nash equilibrium?

I believe that is correct, although there can be many such equilibria 
depending on the situation.  For example, I cited

49 Reagan
33 Carter, Anderson
18 Anderson

As an equilibrium for the election.  I chose it because it was the 
"minimal" equilibrium; that is, the equilibrium where the least number of 
people expressed multiple preferences.  On the other hand, the

49:Reagan,Anderson
51 Carter,Anderson

Example I cited is also an equilibrium.  Note that both give the election 
to Anderson, the sincere Condorcet winner.  If you look at my first example:

49:Reagan
33:Carter
18:Anderson

The Anderson and Carter voters will both regret not casting votes for one 
another.  But say the Anderson voters decided, since they were in last 
place in the polls, to vote for Carter:

49:Reagan
33 Carter
18 Carter,Anderson

...then the Reagan voters will regret not casting a second vote for 
Anderson, and you would get

49 Reagan, Anderson
33 Carter
18 Carter, Anderson

Which is, again, a stable equilibrium.  It looks totally different from the 
other stable equilibria I cited, but Anderson is still the winner.  It 
seems clear that only the Condorcet winner (if one exists) will produce a 
Nash equilibrium.

>A separate question is equilibrium under repeated polling.  Brams and
>Fishburn show with examples that if after each poll people adjust their
>strategies to vote for their favorite of the top 2 (as well as any whom
>they prefer to him, and any candidate they rank between the top 2 whom they
>consider acceptable) cycles can occur.  e.g. First A and B are contenders,
>after adjustments A and C are, after more adjustments B and C are, and
>finally it's back to A vs. B.  They didn't lay out any general conditions
>for cycles, however.

Although certain dogmatic approaches in approval voting adjustment can 
produce a cycle, this doesn't mean a Nash equilibrium does not exist.  Take 
the following circular tie example, where everyone starts out by bullet voting:

10 A>BC
8  B>CA
6  C>AB

The BCA voters are the only ones who have reason to regret their vote, so 
the next round goes

10 A>BC
8  BC>A
6  C>AB

Now it is the ABC voters who have reason to regret their vote, so the next 
round goes

10 AB>C
8  BC>A
6  C>AB

And presto, we have a Nash Equilibrium.  B wins 10-18-14, and no voter has 
any reason to regret their vote; they cannot change their vote to get a 
better result.  The approach you mention from Brams and Fishburn indicates 
that C should now cast a vote for A, but that doesn't change the outcome.

I'm pretty sure that B is the only winner produced by a Nash equilibrium 
(Nash winner?) for this set of preferences and this voting method.  Note 
that in general, Nash equilibria are not guaranteed to be unique, and 
indeed this one is not, there are several slight variations here that still 
produce a Nash equilibrium with B winning.  It would be interesting to see 
if somebody can produce an example where there are two different possible 
Nash winners for the same set of preferences.  I'd be surprised if it's 
possible.

Also note that the ranked pairs and beatpath winner for a sincere Condorcet 
vote is A, not B.  Go figure.

-Adam