Nov 17, 2009

5 Pirates Puzzle

Perhaps the most common of all math/logic puzzles being discussed in forums on the internet, yet an interesting one. Here it goes...



There are five rational pirates, A, B, C, D and E. They find 100 gold coins. They must decide how to distribute them.
The Pirates have a strict order of seniority: A is superior to B, who is superior to C, who is superior to D, who is superior to E.
The Pirate world's rules of distribution are thus: that the most senior pirate should propose a distribution of coins. The pirates, including the proposer, then vote on whether to accept this distribution. If the proposed allocation is approved by a majority or a tie vote, it happens. If not, the proposer is thrown overboard from the pirate ship and dies, and the next most senior pirate makes a new proposal to begin the system again.
Pirates base their decisions on three factors. First of all, each pirate wants to survive. Secondly, each pirate wants to maximize the number of gold coins he receives. Thirdly, each pirate would prefer to throw another overboard, if all other results would otherwise be equal




Source:  Stewart, Ian (1999-05), "A Puzzle for Pirates", Scientific American: 98–99

10 comments:

  1. This puzzle is too complicated for a written explanation: for each position, the game tree must be built for all possible positions that can result from it - so a recursive function is indicated. Here's a simplified version of the outcome, though (which undoubtedly contains tactical errors): E always votes no, D cannot allow himself to become leader, since he won't command a majority, so C knows that D will have to accept whatever he offers - so C is likely to vote no to A and B's offers. A's best chance would therefore seem to be to make a reasonable offer to B and D. From: Moby Dick

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  2. ans 1s 97 gold coins taken by 1st pirate who proposes the solution. distribution is 1 gold coin to 3rd pirate and 2 to any of the 4th or 5th.

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  3. @piyush...the ans should be A-98,C-0,C-1,D-0,E-1
    u cant give same no to 4 and 5,else they'll vote against the first one

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  4. Why would E settle for one coin? B will be obliged to make E an offer (because C will automatically vote against him) - and that offer must be at least one. - Moby Dick

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  5. @ Moby Dick

    Read Carefully, in case of a tie, the distribution is passed...

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  6. C could have the max no of gold coins if A & D are thrown overboard & if C offers E 1 gold coin ... hence the possible distribution cld be A,B,D=0 & C=99, E=1

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  7. Let me try to put this in a table. Had E been the only pirate left, he would obviously take all the 100 coins. If D & E would have been the ones remaining, E would have taken 100 and given nothing to E (tie; so distribution accepted). Extending the logic to C, B, and A gives the final result (98, 0, 1, 0, 1).


    A B C D E
    _________________________________
    100
    ---------------------------------
    100 0
    ---------------------------------
    99 0 1
    ---------------------------------
    99 0 1 0
    ---------------------------------
    98 0 1 0 1
    ---------------------------------

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  8. case 1: D&E only remains. Whatever D proposes will be accepted(if E opposes it is still a tie and passes). So D says D-100 & E-0.
    case 2:C,D,E remains.C foresees D's move and gives E coins > 0 (ie 1) so that he votes for him because he gets atleast one coin compared to case 1.
    case 3: B,C,D,E remains. If B give coins > 1 to E ,he will vote for B and vote passes. So B-98,C-0,D-0,E-2.
    case 4: A,B,C,D,E remains. Since in the above case C doesnt get any coin A will give one to C and 2 to E. For both of them this is the best proposal. So they will vote for A.
    Thus the ans is A-97, B-0, C-1, D-0, E-2.
    Since the same case of D is applicable to D, the last 2 coins could be given to D also ie, it goes to either E or D.

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    Replies
    1. Case 3:

      B,C,D,E - 99,0,1,0........

      B need not give E 2 coins. Just one coin to D would do for him. Just think about it... Because, in this case, C votes against B proposal, E is indifferent as either B or C gives him just 1 coin. Now, had D voted against B proposal, the distribution responsibility goes to C in which case D gets 0 coins. So, he foresees all this and simply accepts B proposal where he gets 1 coin.. So, B wins both his & D vote...

      Extending this logic

      A - 98, B - 0, C - 1, D -0, E - 1......... This will do.... think about it... Or am I wrong somewhere??

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  9. the ans should be 97 0 1 0 2 as every1 wants to survive

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