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If you're bored see if you can answer this one...

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  • Still 50%.
  • Leuth said:

    Fiiish obviously googled the working-out before I did though

    Err what?
  • foxjam said:

    Let's see if I can get this one right...

    There are 7 pirates on a ship including the captain. They have an order of rank that they all know. So the captain is rank 1 and the least important is rank 7. At any time the pirates can take a vote to remove the Captain, under which circumstances the rank 2 pirate becomes captain. If the vote is tied then the current captain has the deciding vote.

    They come across 100 coins (of equal value), the captain must decide how to distribute these coins among the crew. What is the most number of coins he can take for himself without a mutiny?

    I should add that the pirates are all mathematical geniuses and completely ruthless with no loyalty whatsoever.

    Feels like there is a bit of this question missing.
    Do all the ranks have to have a descending number of coins. What relevance do the coins have?
  • Leuth said:

    Fiiish obviously googled the working-out before I did though

    Leuth I have been trying to find out the probability if the person took the test a second time and came out positive. Nobody on here has answered that. Can you enlighten us?
    Assuming each test is independent of the other (ie that the probability of the test being accurate is not affected by any personal characteristics) then I think the probability of being positive given two positive results is 75%. Not done the maths yet.
  • MrOneLung said:

    foxjam said:

    Let's see if I can get this one right...

    There are 7 pirates on a ship including the captain. They have an order of rank that they all know. So the captain is rank 1 and the least important is rank 7. At any time the pirates can take a vote to remove the Captain, under which circumstances the rank 2 pirate becomes captain. If the vote is tied then the current captain has the deciding vote.

    They come across 100 coins (of equal value), the captain must decide how to distribute these coins among the crew. What is the most number of coins he can take for himself without a mutiny?

    I should add that the pirates are all mathematical geniuses and completely ruthless with no loyalty whatsoever.

    Feels like there is a bit of this question missing.
    Do all the ranks have to have a descending number of coins. What relevance do the coins have?
    I don't think there's anything missing. Each pirate obviously wants as many coins as possible for themselves, the captain chooses how the coins are distributed. However, if enough of the pirates are unhappy, they will vote the captain out and the number 2 will then become the captain. So then this pirate will be in charge of how the coins are distributed.

    Apologies if this isn't clear, all the information is there though I believe.
  • But how do we know if they would be happy with ten each and captain taking forty or if happy is 12 each and captain having 28 ?
  • IA said:

    Leuth said:

    Fiiish obviously googled the working-out before I did though

    Leuth I have been trying to find out the probability if the person took the test a second time and came out positive. Nobody on here has answered that. Can you enlighten us?
    Assuming each test is independent of the other (ie that the probability of the test being accurate is not affected by any personal characteristics) then I think the probability of being positive given two positive results is 75%. Not done the maths yet.
    Sorry, done the maths and it's 99%

    The independence condition mentioned above would not be the case for a real-life medical test. The kind of situation that would fulfil the condition would be if the reason the result is 99% accurate is because the test is 100% accurate, but before giving the result the machine generates a random number from 1 to 100 and if that number is 57 it gives the wrong result

    In a real-world medical situation, performing the same test on the same person twice would often generate the same result, whether it's correct or incorrect.
  • foxjam said:

    MrOneLung said:

    foxjam said:

    Let's see if I can get this one right...

    There are 7 pirates on a ship including the captain. They have an order of rank that they all know. So the captain is rank 1 and the least important is rank 7. At any time the pirates can take a vote to remove the Captain, under which circumstances the rank 2 pirate becomes captain. If the vote is tied then the current captain has the deciding vote.

    They come across 100 coins (of equal value), the captain must decide how to distribute these coins among the crew. What is the most number of coins he can take for himself without a mutiny?

    I should add that the pirates are all mathematical geniuses and completely ruthless with no loyalty whatsoever.

    Feels like there is a bit of this question missing.
    Do all the ranks have to have a descending number of coins. What relevance do the coins have?
    I don't think there's anything missing. Each pirate obviously wants as many coins as possible for themselves, the captain chooses how the coins are distributed. However, if enough of the pirates are unhappy, they will vote the captain out and the number 2 will then become the captain. So then this pirate will be in charge of how the coins are distributed.

    Apologies if this isn't clear, all the information is there though I believe.
    What's the relevance of the rank to the vote or the coins? Do higher ranked pirates have more votes or deserve more coins?

    The pirates are "ruthless" but what is their commitment to democracy? Do they mutiny if they're on the losing side?

    What happens to the captain if he/she is replaced? Does he/she go to the end of the line or walk the plank?

    What are their names, how many eyes does each have and do they have a parrot? Should the parrot be included in the coin-sharing?

    So many questions...
  • Early but obviously wrong answer: the captain gives three of them nothing and 25 each to himself and the other three.
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  • Take 97 coins and give 1 each to pirates 3, 5 and 7
  • If the person took the test a second time you really would need to know a bit more about the test itself. Is it something about a particular individual that makes it give an incorrect result?

    For many tests (such as a temperature reading) it is likely that the result of a repeat will be exactly the same as the first test and tell you nothing!
  • What's the Chinese car answer and why
  • 87 - Look at the numbers upside down - i.e. the way someone pulling into the space would see them.
  • Take 97 coins and give 1 each to pirates 3, 5 and 7

    Absolutely right, care to explain your answer?
  • edited June 2015
    foxjam said:

    Take 97 coins and give 1 each to pirates 3, 5 and 7

    Absolutely right, care to explain your answer?
    Solve it by working backwards. (They're all mathematicians so they can all foresee these scenarios) also remembering that the rebellion needs a clear majority to oust the captain.
    If it got to pirates 6 and 7, captain 6 can't be outvoted so he'd take 100 coins, so number 7 doesn't want this, therefore he won't vote against captain 5 as long as he's given 1 coin. Therefore, 6 has no chance of out voting captain 5, because he knows pirate 7 will not rebel, as a result, 6 will vote for captain 4 as long as he offers 1 coin.
    This pattern repeats: 5 will support captain 3, 4 will support captain 2, 3 will support captain 1 and all evens/odds will follow evens/odds respectively. That is, pirates 3, 5 and 7 know it is in their interest to stick with the captain as long as they are offered at least 1 coin.
    So that's what Captain 1 does.

    I thank you.
    :smiley:

  • But if they are considering mutiny, why would they stick to their democratic rules at all?
  • IdleHans said:

    But if they are considering mutiny, why would they stick to their democratic rules at all?

    When did any brainteasing riddle ever have any relationship to the real world?
  • Probability can not be greater than 1 or less than 0, therefore I think .01
  • edited June 2015
    Rossman92 said:

    Probability can not be greater than 1 or less than 0, therefore I think .01

    Can be negative (inverse) up to -1.0 - so less than zero ;-)

    I'm being obtuse and attempting to be a smart ass - which always fails. It's possible in quasiprobability distributions but not real life probabilities.
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  • I love a bit of quasiprobability distribution on a Friday night
  • bobmunro said:

    Rossman92 said:

    Probability can not be greater than 1 or less than 0, therefore I think .01

    Can be negative (inverse) up to -1.0 - so less than zero ;-)

    I'm being obtuse and attempting to be a smart ass - which always fails. It's possible in quasiprobability distributions but not real life probabilities.
    Lol touché my friend. What exactly is Quasiprobability? We didn't learn that in Statistics
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