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RenaissanceV2 loves solving puzzles at PuzzleFry.com. I am proud PuzzleFry member and like my time invested in solving brain teasers.
  • 96 coins

    The problem has a recursive framework, such that the pirates could continue to rebel against the current captain. The ending condition is when only two pirates are left because one of their votes will meet the 50% required vote and no more mutiny can occur. Due to the leadership specification (where each successive group of pirates will lose their highest ranked pirate), these are the possible subsets of pirates throughout the recursive process: {1,2,3,4,5}, {2,3,4,5}, {3,4,5}, {4,5}

    As you can see, the ending condition is when only pirates 4 and 5 are left, so lets work backwards through the logic from this point!

    {4,5}
    Distribution: 4=100, 5=0
    Reasoning: If only pirates 4 and 5 are left, pirate 4 can give himself all the coins because his vote will meet the 50% requirement.

    {3,4,5}
    Distribution: 3=99, 5=1
    Reasoning: Pirate 4 will certainly vote against pirate 3 because he always stands to gain far more through mutiny, so pirate 3 needs pirate 5’s vote to avoid mutiny. All he has to do is offer a single coin because it’s better than the zero coins pirate 5 will earn if he rebelled.

    {2,3,4,5}
    Distribution: 2=98, 5=2
    Reasoning: Pirate 2 needs only a single vote to help him avoid mutiny, so he needs only offer pirate 5 a coin more than he’d earn in a mutinous situation.

    {1,2,3,4,5}
    Distribution: 1=96, 3=1, 5=3
    Reasoning: The captain, pirate 1, needs two additional votes to avoid mutiny. He can buy pirate 5’s vote by offering him a coin more than he’d earn in his next best situation. Since pirate 3 will earn zero coins in the next situation, he needs only offer a single coin to earn his vote.

    So the captain should give himself 96, the 3rd pirate 1, and the 5th pirate 3.

    I had initially guessed 34 based on my intuition, but you can see how wrong that really is!

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