A Pirate's Booty

Keith McGuigan posted a great puzzle on his blog. I liked it so much had to add it to my collection. If you enjoy logic puzzles, be sure to visit Keith's blog, as he will be posting new challenges now and then. Like the "3 bulb" puzzle I posted earlier, there is a "key" approach that unlocks this problem that you'll find useful in attacking other puzzles. Good luck!

Five greedy pirates follow a treasure map to a deserted island. They dig at the "X", and uncover 100 gold coins! The pirates are unionized and therefore seniority rules. However, they are also democratic, and the majority has veto and execute (as in: "off with his head") power... so the leader (the most senior pirate) must be careful.

The leader gets to decide how to divide up the booty amongst himself and the rest of the pirates. However, after the plan is presented, all the pirates (including the leader) votes on the plan. If less than 50% approve the plan, the leader is fed to the fish and the process repeats itself with the next most senior pirate.

Now, the pirates are all pretty smart and don't make rash or emotional decisions. All of the pirates use the following priorities (in the following order) to drive their voting:

  1. They don't want to get killed
  2. They want to get the most money possible
  3. They want to kill other pirates

So, how does the leader divide up the treasure such that he keeps as much as he possibly can for himself, and still survivie?

If you give up, here is the solution:
<check back soon>

Comments:

The senior Pirate would give 25 gold coins to each of the remaining 4.....then he wins by surviving due to the others killing of their couterparts

Posted by DJC on November 24, 2005 at 09:24 PM EST #

The senior pirate would take 96 for himself and 1 coins to each of the others. The 3 most senior pirates would vote in favor, and the rest will vote against.

Posted by Ahmad on July 03, 2006 at 04:35 PM EDT #

Give the pirates numbers from 1-5, 1 being the most senior, five being the least. Start at the scenario of there being only one pirate left alive (the least senior one). The following are the equalibrium points for each scenario. One Pirate 5. 100 Two Pirates 5. 0 (votes kill) 4. 100 Three Pirates 5. 1 4. 0 (votes kill) 3. 99 Four Pirates 5. 0 (votes kill) 4. 0 (votes kill) 3. 100 2. 0 Five Pirates 5. 0 (votes kill) 4. 1 3. 0 (votes kill) 2. 1 1. 98 although as long as he distributes one coin to any two pirates out of the set 5, 4, and 2, he should live.

Posted by Chris on October 02, 2006 at 06:13 PM EDT #

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