Thursday, December 15, 2005
Arrrrr!!!! Is it more tea ye be needing? (More pirate economics)
(Metablogging 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 illformed because of the wildly differing equilibria that result from changes in assumptions.
Suppose we make the following assumptions
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 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.

I had a little time this morning to think over the pirate problem and I've decided the problem is illformed because of the wildly differing equilibria that result from changes in assumptions.
Suppose we make the following assumptions
 The pirates can't offer fractional gold pieces
 The pirates exhibit perfect Nash rationality
 Pirates weakly time discount, i.e., given two equivalent solutions, they prefer to take their first chance to take the equivalent solution
 Pirates don't die, they live on to fight ninjas
 Promises aren't binding
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
 If we remove weak time discounting, but keep the other assumptions, and assume that when pirates are indifferent they flip a coin (some mechanism must be in place for resolving indifference), then we get Thason's solution (by risking the indifference of pirate 5 with an offer of (100,0,0), pirate 3 reduces his expected payoff to 50 from 99).
 If we remove weak time discounting, but kill pirates when offers fail and assume that no pirate wishes to kill another pirate when indifferent to allocations (it would presumably reduce future booty), then we get the (100,0,0,...,0) solution again (thus ties are broken in the favor of the offerer).
 If we remove weak time discounting and kill pirates but don't care about other pirates dying, then Thason's solution is the right solution (pirate 3 can't risk the indifference of pirate 5).
 If we remove weak time discounting, and permit fractional gold, then a solution different from those presented so far emerges. (Would depend on the smallest fractional size allowable).
 If promises are binding, then we get more solutions than I care to think about as the game transforms into a quasicoalitional game which generally have a shload of equilibria (including Roland's) (mmmm.... Cores and Shapley values....)
 If the pirates aren't perfectly rational, then we're going to have to define their thought processes before I analyze the situation.
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.
