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PolySciFi Blog

Thursday, December 15, 2005

 

Arrrrr!!!! Is it more tea ye be needing? (More pirate economics)

(Meta-blogging note - I'm having camera difficulties, or more accurately problems with msi - which says to me that it's time to reinstall the OS - so I still haven't gotten the pictures off my camera.)

I had a little time this morning to think over the pirate problem and I've decided the problem is ill-formed because of the wildly differing equilibria that result from changes in assumptions.

Suppose we make the following assumptions
Note that if I change some of those assumptions, I get different solutions, which I'll illustrate in a bit.

Under those assumptions, the following backwards induction makes use of Thason's nomenclature which would appear a bit odd after 5 pirates, but fortunately we don't have to get there. Also for brevity, I'm listing payoffs as vectors cause I'm a dork like that (and I think it's clearer).

3 Pirate Scenario (3,4,5)
Technically, in this scenario, there are some 1+ 100x101 possible allocations to consider, which is quite a few. But how many are Nash equilibria? Thason notes that (99,0,1) is an equilibrium under some assumptions, but (99,0,1) is not an equilibrium for weak time discounting. Why?

Because pirate 5 is obviously indifferent to (100,0,0) (the result if 5 votes yes) and (0,100,0) (the result if 5 votes no). If we assume weak time discounting (which I think is a fine assumption), by accepting (100, 0, 0) 5 is slightly better off by voting yes and and getting the process over with a stage sooner (unless, of course, there's a pressing ninja problem). Thus (99,0,1) is not an equilibrium as 3 can improve his payoff by offering (100,0,0) which will be accepted because of weak time discounting.

4 Pirate Scenario (2,3,4,5)
Reasoning backwards to the 4 pirate scenario and applying weak time indifference, pirate 4 can offer (100,0,0,0) and be ok as 4 and 5 are indifferent to that outcome to an outcome of (100,0,0) and would prefer the whole thing just end by voting yes.

With weak time discounting, the problem scales up rather quickly to the most senior pirate getting everything, i.e., the allocation vector (100,0,0,...,0).

Considering some changes in assumptions, we get the following
So if we tweak the assumptions around just a little, we get a bunch of different results. We've also implicitly assumed that there are no observable precommitments being employed (e.g., the burning dollar problem.)

So the takeaway from this little analysis is the following: because how to handle indifference is not clearly defined in the problem statement, changing reasonable assumptions results in radically different solutions.

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