Monday, May 11, 2009

Going to the Movies

Here's a situation we've all seen before; suppose Alice and Bob have to decide whether to go to the movies to see a chick flick, and that each has the liberty to decide whether to go themselves. If the personal preferences are based on Alice first wanting to be with Bob, then thinking it is a good film, and on Bob first wanting Alice to see it but then not wanting to go himself, then the personal preference orders might be:
  • Alice wants: both to go > neither to go > Alice to go > Bob to go
  • Bob wants: Alice to go > both to go > neither to go > Bob to go
What should they choose? One thing that they shouldn’t choose is Bob to go alone--this is everybody's least favored outcome. There are good arguments to make for both going or just Alice going, but they shouldn't choose neither going. Both prefer going together to not going. Any option where there are other possibilities that all parties prefer is ‘Pareto dominated’. It seems obvious that whatever system we want making our choices for us shouldn’t be choosing options that are Pareto dominated.

How will these preferences play out? Bob will not go on his own: he would not set off alone, but if for some reason he did, then Alice would follow because she prefers both to go > Bob to go. Alternatively, if Alice decided to go alone, Bob would not join because that is his most desired outcome because he prefers Alice to go > both to go. However if Bob chooses not to go, Alice will want to stay home too because she prefers neither to go > Alice to go.

Herein lies the rub: even though Alice and Bob prefer both to go > neither to go, if Alice and Bob choose individually, neither will end up going because neither prefers going alone to both staying home. Bob might try to convince Alice to go, since both scenarios he prefers over neither going have Alice go; but Alice can't convince Bob to go because as soon she's going Bob can achieve his optimal outcome by staying home.
  • Probable outcomes: neither go > Alice goes > both go > Bob goes.

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