[EM] SARA and a center squeeze scenario pair

Jameson Quinn jameson.quinn at gmail.com
Sun Oct 23 12:47:02 PDT 2016

The tld̦r (too long, didn't read; that is, summary) on my previous message

In SARA, in a center squeeze situation where the sympathies of the center
group aren't obvious, the strategic equilibrium is for the center group to
win through "accept" votes from the two wings. This works even if voters
only strategize against their third choice, not their second. This latter
characteristic does not hold for most other voting systems, even really
good ones like ICT or MAM.

2016-10-23 14:43 GMT-04:00 Jameson Quinn <jameson.quinn at gmail.com>:

> Let's say that the honest preferences are one of the following two
> scenarios, and the voters don't know which:
> 35 or 35: A>B>C
> 10 or 20: B>A>C
> 15 or 05: B>C>A
> 40 or 40: C>B>A
> Under SARA, the most naive/honest heuristic would probably be to rate the
> middle preference "abstain", yielding the following:
> 35 or 35: A>>B>C
> 10 or 20: B>>A>C
> 15 or 05: B>>C>A
> 40 or 40: C>>B>A
> Points: A70, B50, C80.
> This gives a win to C or A, depending on which scenario is true.
> Clearly, if all the B voters truncate, that's sufficient to elect B. But
> let's say that the B voters are not actually that highly motivated to
> strategize against their second choice, so they don't do that.
> In scenario 2, the C voters can ensure that A can't win if 25 of them
> switch to C>B>>A. This would give B 75 points, enough to beat A in scenario
> 2 but not enough to beat C in scenario 1.
> But if the A voters anticipate this, then they have no hope of winning,
> even in scenario 2; and thus, to avoid loss in scenario 1, all of them will
> switch to A>B>>C. In that case, even if none of the C voters actually
> strategize, B will still get at least 85; enough to win outright. So the
> following ballots will be stable:
> p1 or (1-p1)
> 35 or 35: A>B>>C at p2 or A>>B>C at (1-p2)
> 10 or 20: B>>A>C
> 15 or 05: B>>C>A
> 15 or 15: C>>B>A
> 25 or 25: C>B>>A at p3 or C>>B>A at (1-p3).
> Results: (p1, p2, p3: result)
> n,n,n: A
> y,n,n: C
> n,y,n: B
> y,y,n: B
> n,n,y: B
> y,n,y: C
> n,y,y: B
> y,y,y: B
> So if the payoff for B is 0, the A voters face a choice between a payoff
> of Aa(1-p1)(1-p3)-Ac(p1)(p3). Insofar as B is over 50% of the utility of A,
> so that Aa (the benefit they get from A winning) is less than Ac (the
> penalty for C winning), the C voters must leave p3 less than .5 if they are
> to tempt the A voters to not to give B points. Yet the payoff for the C
> voters is strictly better the higher p3 is; leaving p low enough to tempt
> the A voters not to give B points is dominated. Thus, (p2=p3=1 and so B
> wins with certainty) is the subgame-perfect equilibrium.
> If you're looking for a "trembling hand" equilibrium, B will only lose
> insofar as the hands are very trembly and/or B is the score loser. This
> seems to me to be a good outcome.
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