[EM] Richard, re: Nash equilibrium for voting systems

Richard Moore rmoore4 at cox.net
Fri Jul 19 19:11:25 PDT 2002

> Richard replied:
> I would only
>                object to the idea that this definition is useful in 
> real-life
>                public elections, where many if not most voters will 
> strategize
>                independently from the rest of their factions.
> I reply:
> The objection stated in that last sentence is the one that I'm
> talking about. Saying that many voters won't vote exactly the
> same as others in their faction misses the point here.
> My point
> is that if, with a certain votes configuration and outcome, some
> voters can improve the outcome for themselves by changing their
> vote, then that existing votes configuration and outcome
> is obviously unstable.

Voters can team with other voters in any number of ways, but they have
to be cognizant of the other voters and in communication with them.
If a voter John Smith in San Diego isn't aware of voter Jane Brown
in San Francisco, then they don't team together (unless they do so
through some organized group). Perhaps John and Jane have similar
views but belong to separate organizations that don't coordinate
political strategies, or perhaps they are loners who belong to no
organizations at all. That is the reality of public elections.

> Besides, even with Blake's most unlenient equilibrium definition,
> Condorcet(wv) and Approval, when there's a CW, always have
> equilibria in which the CW wins and no one order-reverses.
> So, whichever equilibrium definition you use, margins is the
> method that fails by sometimes having no equilibria in which
> defensive order-reversal isn't used.

Well, I wasn't commenting on how this applies to the margins debate;
I was focusing on Approval. I have no quarrel with your observation
about equilibria in margins.

> By the way, I disagree when Richard calls Blake's definition
> a definition of Nash equilibrium. Blake's definition may have
> uses, but it certainly isn't a definition of Nash equilibrium.
> It  violates the intent of Nash equilibrium. Nash's
> definition speaks of one person changing his vote.

Does it refer to a "person" or to a "player"? (I know it doesn't
refer to "votes" but I'm sure you didn't mean to say that it did;
of course you were thinking of Nash equilibria in the context of
election theory.)


The definitions I've seen all refer to a "player" changing his
or her strategy. So how do you define a "player"? I don't know
how Nash defines the term but I suspect a player is an atomic
entity that will choose one of a set of strategies available to
it. If a bloc of voters (with or without identical preferences,
but presumably with enough in common to *want* to work together)
agrees that it will choose a strategy, and that all members will
execute that strategy in solidarity (and this is agreed to in
advance of the strategy decision), then that bloc is acting as
a single player, for the intents of defining a Nash equilibrium.

 > If Nash had
 > meant any set of people with different sincere rankings and different
 > current strategies, then he would have said that. You could say
 > we're overextending Nash's definition by letting more than 1 voter
 > change their votes, but we change it beyond recognition if we
 > have different-voting voters with different sincere rankings all
 > change their vote.

The tricky part is determining in advance how the people will group
themselves. Sure, maybe only a very few voters will make a sincere
commitment to vote with the rest of their blocs regardless of the
strategy, but nevertheless if there are even tiny groups of two or
three voters who do, then you can have multi-voter players that
don't consist strictly of voters with identical preferences.

Of course, each time you regroup the voters to form different sets
of players, you've defined a whole new game. I'm not considering those
different groupings to be different analyses of the *same* game; they
are different games with different sets of equilibria.

  -- Richard

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