[EM] Some myths about voting methods

[Jobst Heitzig]

Hello Warren,

you're right in pointing out that the notion of a mere Nash equilibrium is not particularly fruitful in the analysis of voting systems. This does not mean, however, that game theory is fruitless, too. Game theory provides many valuable concepts, in particular those specializations of Nash equilibria which involve coalitions have been applied quite successfully on this list, often termed "group strategy equilibrium" or the like, although often not explicitly defined thoroughly in every detail. 

A recent example of a more thorough game theoretic analysis is Forest's and my paper titled (http://www.fair-chair.de/some_chance_for_consensus.pdf) in which we use the concept of "correlated equilibrium" and "very strong correlated equilibrium" (which is quite similar to Moreno's "coalition-proof correlated equilibrium"). Anyone interested in these concepts may study section 2.2 of that paper for starters. As the paper is still under review, we also would be glad to receive any further comments or suggestions, by the way.

Yours, Jobst


> -----Ursprüngliche Nachricht-----
> Von: "Warren Smith" <warren.wds at gmail.com>
> Gesendet: 06.06.09 00:54:17
> An: Raph Frank <raphfrk at gmail.com>
> CC: election-methods at lists.electorama.com
> Betreff: Re: [EM] Some myths about voting methods


> >> --"Nash equilibria" are an attempt to salvage game theory in N-player
> >> games with N>2.
> >> But it works badly for voting purposes.
> >> My usual example is, suppose everybody realizes Adolf Hitler is the
> >> worst candidate but still (idiotically) everybody votes for Hitler,
> >> who wins.
> >> OK, this election is a Nash equilibrium representing, in the sense of
> >> Nash equilibria,
> >> "best voting strategy" for all.
> >
> > Huh?
> >
> > If everyone else follows the "vote for Hitler" strategy, your optimal
> > strategy isn't vote for Hitler.  (Well assuming that the secret ballot
> > hasn't been compromised).
> >
> > You might as well vote for someone else.
> 
> --A game player situation is a "Nash equilibrium" if no player can get
> more reward by changing their strategy.  In the all-vote-for-Hitler
> situation, no player can alter the
> election result by changing their vote.  Therefore this is a Nash equilibrium.
> 
> Is this "best strategy"?  Well, only in the sense that Nash-advocates
> proclaim that
> in a Nash equilibrium situation, each player's strategy is "optimal"!
> 
> The point is, as you yourself have just observed, this whole
> conception often is silly and useless when applied to voting.
> 
> > In most real elections, you don't know the exact way the others are
> > going to vote.  If all the other voters were using "vote for Hitler
> > with a 99% probability and a random candidate otherwise", then your
> > optimal vote is to vote for your favourite.
> 
> --true.   This actually sounds like a good way to perturb the Nash equilibrium
> notion/definition to make it become more sensible than the official
> definition.
> 
> So the new improved Raphfrk+Nash notion would be, assume each player will
> play whatever strategy they select, or with probability epsilon they play
> a random strategy.   Now we only have equilibrium if no player can
> improve their expected reward, and this includes improvements by very
> tiny amounts
> proportional to epsilon^4 or whatever.
> 
> This would get rid of stupid equilibria like "all vote for worst."
> 
> Here's another nasty Nash equilibrium which still applies for the
> Raphfrk-Nash version:
> In a plurality election consider say, "all vote for Gore or Bush about 50-50"
> but a higher reward would come if Nader won.   In this scenario  I guess
> you cannot improve expected reward, and in fact will worsen it, by
> switching your vote to Nader.
> 
> 
> -- 
> Warren D. Smith
> http://RangeVoting.org  <-- add your endorsement (by clicking
> "endorse" as 1st step)
> and
> math.temple.edu/~wds/homepage/works.html
> ----
> Election-Methods mailing list - see http://electorama.com/em for list info
>