Five pirates find 100 gold coins in their recent loot. In order to distribute the loot among them, they use the following approach:

The most senior pirate will propose a distribution. If at least half of them agree with the proposed distribution, the loot will be divided as proposed. Otherwise the most senior pirate (who made the proposal) will be killed. Then they start over with the next most senior pirate, and so on.

The question is: what distribution should the most senior pirate propose so that he won’t be killed? Assume that the pirates are intelligent, greedy and prefer not to die.

Update: The solution is posted in the comments section.


11 responses to “Pirates

  1. Anonymous

    33,3 gold for 2 of them and himself

    • It’s not that simple! πŸ™‚

      Are you suggesting 34 for pirate 1 (the most senior), 33 for 2, and 33 for 3? Or, between pirates 1, 4 and 5? Remember that the other two pirates may not vote for him if they think they can get a bigger share by refusing the proposed distribution.

  2. Floris

    Give the next two seniors 50 each?

    • But then he (the most senior pirate) doesn’t get anything. He (obviously) doesn’t want to get killed, but if there’s an alternative that won’t get him killed and get a share of the loot, he would opt for that.

      And there’s an alternative that can keep him alive, with some gold in his pocket!

  3. Jeroen Stet

    Hello Vishal,

    Just found your weblog after googling on Fermat’s Room and I really like it. I

    Now on to the puzzle. Please keep in mind that my native language is Dutch so keep that in mind if my english seems a little “broken”.

    Let’s call the oldest pirate 1 and the youngest 5.

    If Pirate 1 would propose, 100 – 0 – 0 – 0 – 0 he would get 4 nay’s which mean he would be dead. So my first response was: 34 – 33 – 33 – 0 – 0. An equal split between the three oldest pirates. But then I thought it through and if pirate number 2 – 3 would say nay against the proposal. Pirate 1 would be executed and pirate 2 – 3 could split the loot 50 / 50 (when there are 4 pirates left 2 votes is enough).

    So pirate 4 and 5 wouldn’t get anything if pirate 1 was to be executed. So if you offer them a token bribe they would have to accept it because it’s better then 0. So my answer is:

    98 – 0 – 0 – 1 – 1

    • You’re on the right track. But think about the situation when there are 4 pirates left. Is 50/50 split between the two senior pirates (and nothing for the two juniors) the most obvious outcome? Is there a way the senior pirate can manage to get more than 50 coins for him? (There is!)

      Glad you found my blog interesting! πŸ™‚

  4. Erik Beentjes

    Here’s my attempt:

    Scenario 1: Imagine there are 2 pirates left:
    Then a 100-0 split would pass (1 out of 2 agreeing)

    Scenario 2: Imagine there are 3 pirates left:
    Then a 99-0-1 split would pass. (2 out of 3 agreeing)

    Scenario 3: Imagine there are 4 pirates left:
    Then a 98-0-0-2 split would pass (2 out of 4 agreeing)

    Scenario 4: So with 5 pirates left:
    Then a 96-0-0-1-3 spilt would pass (3 out of 5 agreeing)

    Since 3 coins for pirate 5 is the max he could ever get, he will agree to this.
    Since 1 coin for pirate 4 is the max he could ever get (since with 4 left, he will get nothing and that will be accepted), he will agree.

  5. Where is the solution? I’ve been waiting for three months now! πŸ™‚

    • Hey Raja, I am traveling right now; will post the solution in few days.

      • Solution:

        Let’s see what happens when there are only two pirates left: D and E

        {100-0} would be the obvious outcome, because D needs only one vote. He will take take all coins.

        When there are three pirates left: C, D and E

        {99-0-1} because C would need two votes. He will have to spend only 1 coin (minimal amount) to buy E‘s vote.

        If E refuses to vote for this distribution, then guess what happens: C gets killed, and there will be only two pirates left. And we know what happens when there are only two pirates left – D takes everything, leaving nothing for E.

        When there are four pirates left: B, C, D and E

        {99-0-1-0} B needs only two votes. He can give a minimum amount to buy D‘s vote. C and E gets nothing.

        D will have to agree with this solution because if he doesn’t, then B gets killed, and as we saw above C will take 99, E will get 1 coin, and D will get nothing. For D, it’s better to accept B‘s offer and take 1 coin.

        When there are five pirates (initial condition): A, B, C, D and E

        {98-0-1-0-1} is the most logical outcome, because if C and E don’t agree with this then they will get zilch (because if A gets killed then the next most logical outcome leaves nothing for C and E).

        Erik was so close to the correct answer in his comment above.

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