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Email your solutions to solution@math.colostate.edu
One winner each week is eligible for a free ice cream and topping, courtesy of Cold Stone Creamery.
Update: Despite the heroic stand of Scott Lundberg,
the sole CSU student who submitted a solution to
Challenge 5, CSU students got shown up this week by Colorado School
of Mines students Jim Howard, Dmitriy Mestets, Scott Danford,
Vincent Gonzales, Greg Gerou, and Ted Archuletta, and Virginia Willett,
as well as Liz Boese (CSU faculty), Nicolae Popescu (Fort Collins),
Byungsoo Kim (South
Korea), and Bogdan Chlebus (C.U. Denver), who submitted clear and clever
strategies. Optimum strategies were given by Scott,
Jim, Dmitriy, Liz, Nicolae, and
Byungsoo.
The problem again:Five hundred gold coins come into the possession of five pirates. They agree on a strategy for dividing up the money.
They draw straws to assign numbers from 1 to 5 to the five pirates. Pirate 5 will propose a solution to how to divide up the coins. If he gets more than half of the pirates (counting himself) to agree, then that will be the final distribution. If he fails to get this majority, he must walk the plank, and it is Pirate 4's turn to try to forge a majority among the remaining pirates. If 4 fails, then he walks the plank and it's Pirate 3's turn, etc.
These pirates all think its great fun to throw somebody overboard. If they can do this for free, they won't hesitate. However, they are so greedy that, given a choice between throwing somebody overboard and getting some money, they will always opt for the money. Needless to say, none of them will never spend more money than he has to in order to forge a majority. They all know these things about each other.
One last thing: they are graduates of CSU. While they were here, they found out about mathematical induction, and they know this about each other.
The problem: Suppose you are Pirate 5. What is the greediest distribution that you can get away with?
Hints: Here are some observations to get you started. Suppose Pirates 5, 4, and 3 go overboard. What will happen next to Pirate 2? (It's worth mentioning that Pirate 1 is big.)
The pirates are smart enough to figure this out. They can also figure out the answer to the following question: If it gets to his turn, how can Pirate 3 exploit what Pirate 2 knows will happen if 3 goes overboard?