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Pirates’ Games

This is not a puzzle related post. 🙂 It’s a brain teaser (I used to post brain teasers years ago). I’ve been studying game theory lately and wanted to share one “classic” problem of this discipline with you. See if you can solve it by yourself. Post your solution as comments below this post. So… here we go…

Pirate Games

Five pirates have obtained 100 gold coins and have to divide up the loot. The pirates are all extremely intelligent, treacherous and selfish (especially the captain) each wanting to maximize the number of coins that he gets.
It is always the highest ranked pirate who proposes a distribution of the loot. All pirates vote on the proposal, and if half the crew or more go “Aye”, the loot is divided as proposed.

If the captain fails to obtain support of at least half his crew (which includes himself), all pirates turn against him and make him walk the plank. The pirates then start over again with the next most senior pirate as captain (the pirates have a strict order of seniority denoted by A, B, C, D and E).
Pirates’ preferences are ordered in the following way. First of all, each pirate wants to survive. Second, given survival, each pirate wants to maximize the number of gold coins he receives. Finally, each pirate would prefer to throw another overboard in the case of indifference.

What is the maximum number of coins that the original captain should propose for himself? And for others?

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  1. Deniz
    Posted January 29, 2013 at 7:30 am | Permalink


    • Deniz
      Posted January 29, 2013 at 7:32 am | Permalink

      may i change it to 58 (lets see how many time i will change? 😉 )

      • Posted January 29, 2013 at 9:56 am | Permalink

        Sure you can change, Deniz, but it’s still wrong! 🙂

  2. Aaron B.
    Posted January 29, 2013 at 11:53 am | Permalink

    Here’s my logic, working backwards:

    If it comes down to DE, D can split it 100-0, vote Aye, and take it all. So E should vote Aye if he’s offered at least 1 coin earlier.

    Therefore at CDE, C can split it 99-0-1, get E’s vote, and win. So D should vote Aye if he’s offered at least 1 coin earlier.

    Therefore at BCDE, B can split it 98-0-1-1, get D’s and E’s votes, and keep 98. So C should do like D and E and take 1 coin if it’s offered. But he won’t get a chance because:

    A can split it 98-0-0-1-1, get D’s and E’s votes, and take 98 for himself.

    One problem: maybe B has to offer E two coins, to make sure E won’t decide to wait until CDE when he’s still guaranteed to get at least one. Likewise, A might have to offer D 2 and E 3 to keep them from waiting until B’s turn.

    • Posted January 30, 2013 at 6:00 am | Permalink

      Aaron, your solution is very very close, but it is not exactly right! 🙂

      • Posted January 31, 2013 at 5:01 pm | Permalink

        Darn. I’ll look forward to the answer, then.

        • Posted February 1, 2013 at 1:47 am | Permalink

          Would you like a hint instead? 🙂 Check the rules again. The last one basically says this: when indifferent about the amount of gold he receives, each pirate prefers others dead!

  3. Posted February 1, 2013 at 3:11 pm | Permalink

    Ah, I’d missed the implication of that last rule; that eliminates the ties I was thinking of.

    Ok, so it starts out the same: At DE, D will take it all, so E should take 1 at CDE.

    At CDE, C can offer 99-0-1, so D should take 1 and E should take 2 at BCDE.

    At BCDE, B can offer 97-0-1-2, so C should take 1, D should take 2, and E should take 3 at ABCDE.

    So at ABCDE, A can offer 97-0-1-2-0. Better?

    • Posted February 2, 2013 at 3:15 am | Permalink

      Don’t rush it… 🙂 Think again whether B must offer 2 to E or does he have a “cheaper” alternative?

      • Posted February 2, 2013 at 9:46 am | Permalink

        Ah, of course: B can offer 98-1-1-0. So A can offer 98-1-0-0-1? (That was me posting last time, wrong account.)

        • Posted February 2, 2013 at 9:55 am | Permalink

          Oops, I forgot that B only needs 2 votes. So B can offer 99-0-1-0, and A can offer 98-0-1-0-1. Maybe that’s it, finally.

          • Posted February 3, 2013 at 2:08 am | Permalink

            Well done Aaron! 🙂

  4. Erica Kirby
    Posted February 3, 2013 at 12:51 am | Permalink

    I have worked out A’s proposal needs to be A=98,B=0,C=1,D=0,E=1
    However this ignores the “treachery factor”. It leaves A sitting pretty but with 4 disgruntled pirates on board who, even after agreement is reached, would go back on their word and bump A off in the blink of an eye if they thought they could do better. If A really wants to survive ( and that is his primary concern) he would do better to split it evenly five ways. 20 each would buy him a bit of loyalty and he might live to enjoy his (reduced) share!.

    • Posted February 3, 2013 at 2:04 am | Permalink

      Well, Erica, you are absolutely right about the solution and the epilogue thereof. 🙂 But remember, it’s a made up game with certain rules which should be followed. I don’t think that real pirates would reason this way at all. But then again, I might be wrong. 😀

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