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  1. #41
    Superwoman Red Herring's Avatar
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    I really have to get back to work, but want a shot at this one later on.

    A short question though beforehand: In the theoretical case that the highest rankimg pirate does get killed, how do the remaining 4 (or any even number of) pirates determine a majority?
    The good life is one inspired by love and guided by knowledge. Neither love without knowledge, nor knowledge without love can produce a good life. - Bertrand Russell
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  2. #42
    null Jonny's Avatar
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    Even - n/2 + 1
    Odd - n/2 + .5
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  3. #43
    Superwoman Red Herring's Avatar
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    ...and where do I get half a pirate?
    The good life is one inspired by love and guided by knowledge. Neither love without knowledge, nor knowledge without love can produce a good life. - Bertrand Russell
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  4. #44
    null Jonny's Avatar
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    lol, you don't. I just mean that when there are an odd number of pirates, say 5, the majority is n/2 + .5 → 5/2 + .5 → 2.5 + .5 → 3; and when there are an even number of pirates, say 4, the majority is n/2 + 1 → 4/2 + 1 → 2 + 1 → 3.
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  5. #45
    Superwoman Red Herring's Avatar
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    So the proposal is considered rejected in case of a 2:2 stalemate?
    The good life is one inspired by love and guided by knowledge. Neither love without knowledge, nor knowledge without love can produce a good life. - Bertrand Russell
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  6. #46
    Senior Member esidebill's Avatar
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    Quote Originally Posted by Red Herring View Post
    ...and where do I get half a pirate?
    Rofl. Most hilarious response.
    "Others should not judge what you truly are, instead you should find yourself. You may find yourself in a bowl of cereal or dreaming of the unknown, but make sure it is you who finds you." - Myself


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  7. #47
    null Jonny's Avatar
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    Quote Originally Posted by Red Herring View Post
    So the proposal is considered rejected in case of a 2:2 stalemate?
    Precisely. It is accepted if and only if a majority of pirates vote for it; rejected otherwise.
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  8. #48
    Senior Member matmos's Avatar
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    @Johnny

    My bad - I totally misread your OP. Apologies, good sir.

    If I understand your take, the results are:

    Pirate 5: 97
    Pirate 4: 0
    Pirate 3: 1
    Pirate 2: 0
    Pirate 1: 2

    However, there's a problem. We assume that all 5 agents are "perfectly rational". Therefore all five agents would be able to independently work out the mathematical equilibrium you posit. Agents 1-4 would be anxious to avoid this scenario as the possibility of more gold for each is available. This would be perfectly rational.

    All know this is the likely outcome - because all are brilliant game theorists. Therefore the perfectly rational approach would be for pirates 1 - 4 to feign irrationality and demand nothing less than a fair split (20 coins is significantly better than zero, one or two) at the cost of killing pirate 5 and obtaining a larger, equal, split between the remaining four pirates.

    Pirates 1-4, being perfectly rational, realise that Pirate 5 will realise this and, rather than risk rejection, accept an equal cut or a bargained offer, depending on how highly rank is enforced.

    So a "rational" agent at best can hope for zero, one or two coins, but an "irrational" agent can obtain 20 or even 25. - Some rationality!

    The rational approach would be to forget the rational approach.

  9. #49
    Dreaming the life onemoretime's Avatar
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    Are you taking the LSAT or something?

  10. #50
    Senior Member esidebill's Avatar
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    Quote Originally Posted by onemoretime View Post
    Are you taking the LSAT or something?
    I used to think I would :p but no.
    "Others should not judge what you truly are, instead you should find yourself. You may find yourself in a bowl of cereal or dreaming of the unknown, but make sure it is you who finds you." - Myself


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    Intuitive (N) 64.29% Sensing (S) 35.71%
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