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    (Original post by Tifa)
    ok, *deep breath* I'm sorry about my editing, the previous post was a complete mess cos I got my letters mixed up... lets try again

    working from backwards

    1. Pirate D needs zero other people to vote for him. So he votes for himself and takes all the gold.
    2. Pirate C needs zero other people to vote for him. So he votes for himself and takes all the gold. Pirate D gets zero.
    3. Pirate B needs one other person to vote for him. He gives one coin to Pirate D for his vote - if we are reduced to two pirates, Pirate D gets zero so Pirate D knows one is better than none. Pirate B takes ninety-nine. Pirate C gets zero.
    4. Pirate A needs one other person to vote for him. He gives one coin to Pirate C - if we reduce to three pirates, Pirate C gets zero so Pirate C knows one is better than none. Pirate A takes ninety-nine. Pirate B gets zero. Pirate D gets 0.

    99,0,1,0

    ???

    I don't think working backwards works? Not sure.
    Really leaving this time!
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    (Original post by Tifa)
    ok, *deep breath* I'm sorry about my editing, the previous post was a complete mess cos I got my letters mixed up... lets try again

    working from backwards

    1. Pirate D needs zero other people to vote for him. So he votes for himself and takes all the gold.
    2. Pirate C needs zero other people to vote for him. So he votes for himself and takes all the gold. Pirate D gets zero.
    3. Pirate B needs one other person to vote for him. He gives one coin to Pirate D for his vote - if we are reduced to two pirates, Pirate D gets zero so Pirate D knows one is better than none. Pirate B takes ninety-nine. Pirate C gets zero.
    4. Pirate A needs one other person to vote for him. He gives one coin to Pirate C - if we reduce to three pirates, Pirate C gets zero so Pirate C knows one is better than none. Pirate A takes ninety-nine. Pirate B gets zero. Pirate D gets 0.

    99,0,1,0

    ???
    Yes, this seems very close.

    BUT...in point 3 (when B,C,D are left), surely C would surely attempt to buy off D by offering him 2 in a future C/D deal (as oppose to the 1 that B is offering him). Then a bidding war for D's vote might ensue?
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    LOL - me with my working backwards - ok, last word I promise... an elaboration of what I said before, with the implications of extra pirates

    With Pirate E (5 pirate scenario) he needs two other people to vote for him. And the two people he needs to convince are the two who get screwed in the four pirate scenario - Pirate B and Pirate D. So he can give them each one coin (which is better than zero - what they would get otherwise) and keep ninety-eight for himself…. And THEN so on and so forth

    Eg. (using numbers this time, cos letters are crazy - silly alphabet) with 15 pirates, pirate 15 needs seven other people to vote for him, so he bribes Pirates 13, 11, 9, 7, 5, 3, and 1 with one coin each and keeps ninety three coins for himself. Those pirates will all vote for him because they know that they get zero coins if he dies and Pirate 14 is in charge.

    God, I bet I'm *so* wrong... btw I think if a tie doesn't count as a win my very first solution might have worked... which is actually more complex than this problem in a fashion.

    eep - Powerball save us...
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    (Original post by llama boy)
    Yes, this seems very close.

    BUT...in point 3 (when B,C,D are left), surely C would surely attempt to buy off D by offering him 2 in a future C/D deal (as oppose to the 1 that B is offering him). Then a bidding war for D's vote might ensue?
    I would say no, because even if C says 'oh yes i promise I'll give you more gold' *flutters eyelashes* because they're bloodthirsty pirates, D can't trust him, because as soon as B 's been chucked, C will keep all the gold for himself... there is no reason for C to keep the deal and D knows that
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    (Original post by Tifa)
    I would say no, because even if C says 'oh yes i promise I'll give you more gold' *flutters eyelashes* because they're bloodthirsty pirates, D can't trust him, because as soon as B 's been chucked, C will keep all the gold for himself... there is no reason for C to keep the deal and D knows that
    yeah ok, that's probably true.

    so, er, i think we've solved it!
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    Yup tifa's got it 99 0 1 0... i was asked it on my Economics and management interview.... D wont agree on a future proposal with C as there is no way to trust C would honour the deal.

    They asked me an almost identical question to follow it up to test if I was guessing:

    There are 5 lions and 1 sheep. Again lions are lettered A B C D E. If a lion ate the sheep it turns into a sheep. A lion values it's own life more than eating the sheep. Will the sheep be eaten?

    Basically you use the same idea and get a similar conclusion...

    All 3 of my interviews were these logic questions. Although in some of them they made me draw diagrams.... glad its over
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    (Original post by powerball)
    Yup tifa's got it 99 0 1 0... i was asked it on my Economics and management interview.... D wont agree on a future proposal with C as there is no way to trust C would honour the deal.

    They asked me an almost identical question to follow it up to test if I was guessing:

    There are 5 lions and 1 sheep. Again lions are lettered A B C D E. If a lion ate the sheep it turns into a sheep. A lion values it's own life more than eating the sheep. Will the sheep be eaten?

    Basically you use the same idea and get a similar conclusion...

    All 3 of my interviews were these logic questions. Although in some of them they made me draw diagrams.... glad its over
    im glad there's no logic in the med interviews....

    Sorry, but what incentive is there for a lion to eat the sheep?
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    (Original post by DanMushMan)
    im glad there's no logic in the med interviews....

    Sorry, but what incentive is there for a lion to eat the sheep?
    heehee, because it's hungry?

    I guess it goes like this? (haha backwards *again*)

    Just E, - E would eat the sheep
    Just D,E - D wouldn't eat the sheep (def get eaten)
    Just C,D,E - C would eat the sheep (because D wouldn't eat him - previous step)
    Just B,C,D,E - B wouldn't eat the sheep (because C would eat him - previous step)
    All 5 - A would eat the sheep (because C wouldn't eat him - previous step)

    Even number = happy, frolicking, alive sheep
    Odd number = mangled, sad, dead sheep
    All this assuming the lion's think these things through!!

    And if anyone's interested, consider the pirate question so that it had to be a majority for the vote to be passed. (I posted my solution in my first post on this thread, would like to know if anyone else agrees with it)

    Okay, here's a possible question… a man has two cubes on his desk. Every day he arranges the cubes so that the front faces show the current date (day of the month). What numbers are on the faces of the cubes to let this happen every day of the year?
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    Majority idea...

    If CD C will definitely lose so C will go overboard and wont let the situation go pass BCD. B will give himself a 100 as C will rather get 0 than be thrown overboard. Hence C and D will not want to get to that situation so will accept 1 from A so I reckon 98 0 1 1...
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    as for the cubes im guess 1,2,3,4,5,6 and 1,2,7,8,9,0
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    (Original post by powerball)
    4 pirates have a 100 gold and need to share it by a voting system. Pirates are A B C D. If A makes a proposal and wins.... thats how they would share the gold. If A loses the proposal he is thrown overboard and then B makes the proposal... if he loses then C e.t.c..... A himself has a vote. If the vote is a tie proposal goes through. Pirates value their lives more than gold. What proposal would A make?

    Extend the idea to 5 pirates.

    What would be the effect if the person chosen to do it after the first is overboard is chosen at random.

    Then some crap about risk takers.
    Am I stupid or wouldn't A just give all the coins (or whatever shape this gold takes) to either B or C or D. If he values his life more than gold, all he needs to succeed in his proposal is the vote of ONE other pirate. Therefore by giving all hundred to one pirate that guarantees a vote, and his vote makes two - therefore a tie and A gets to live and stay on board.

    If he only gives one coin to another they may think they can get a better deal and vote against him.

    If A is not willing to give up all the gold, he should offer one of the others at least half....an offer that is irresistable would do.

    Have i got it wrong?
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    (Original post by powerball)
    Majority idea...

    If CD C will definitely lose so C will go overboard and wont let the situation go pass BCD. B will give himself a 100 as C will rather get 0 than be thrown overboard. Hence C and D will not want to get to that situation so will accept 1 from A so I reckon 98 0 1 1...
    U assume the pirates are clever. But even if they are, then surely A will give himself all coins and C and D will vote for him anyway cos they dont want to be thrown over if they get a BCD situation??? This idea involves far more risk for A (but greater reward). I think almost certainly, if A wants to live more than he wants the gold, he will just give one pirate the majority to secure their vote.
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    thank god they asked me no logic. I have none.
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    (Original post by Mentally Ill)
    U assume the pirates are clever. But even if they are, then surely A will give himself all coins and C and D will vote for him anyway cos they dont want to be thrown over if they get a BCD situation??? This idea involves far more risk for A (but greater reward). I think almost certainly, if A wants to live more than he wants the gold, he will just give one pirate the majority to secure their vote.
    In the majority idea... A needs the support of 2 people.

    Because it was economics it was assumed that everyone was acting rationally.
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    at my interview at Magdalene i was asked..

    if u had a box on a table, what sort of things could u do to find out what was in it.

    ps: u cant open it/burn it or do anything to look inside

    not too difficult really
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    (Original post by llama boy)
    A man has three cards. One is red on both sides. One is white on both sides. One is white on one side and red on the other.

    He places a card on the table - you can see the top side is red.

    He tells you that as it obviously isn't the white/white card then it is either the red/red or the white/red therefore he'll give you odds of evens on which card it is.

    What are the true odds for this bet?
    No one going to have a go at this one?
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    (Original post by Snowqueen)
    at my interview at Magdalene i was asked..

    if u had a box on a table, what sort of things could u do to find out what was in it.

    ps: u cant open it/burn it or do anything to look inside

    not too difficult really
    do you mean i can't look inside at all? what kind of box is it? cos if it's a glass box then i don't have to do anything...

    anyways, what's the answer?
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    (Original post by llama boy)
    No one going to have a go at this one?
    2/3 RR, 1/3 RW
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    This one was not asked on an Oxbridge interview, it's just a brainteaser:

    There is one in a minute and two in a moment, but only one in a million years. What are we talking about?
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    the letter M?
 
 
 
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