oxbridge interview's logic questions Watch

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hitchhiker_13
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#21
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#21
(Original post by llama boy)
I'm having a go...dunno how i could do that in an interview situation though..

What can the pirate do? If they knew they weren't going to be next (i.e. C and D) they would vote against any proposal made by A. But A and B will vote in favour, because B will know he's screwed if A gets chucked off. As it's a tie, the vote will pass. So really A could say anything...not sure about this last point. Because whatever A says, B will agree, because if A is chucked off, the B will be left 1 against 2, so he'll be chucked off. So A can just say I'll have it all.

If there were 5, then it would be 2 against 3 so it wouldn't work, but after that return to the same, so B can do what they want.

If it was random, then no-one will know who will have to make the next proposal, so hopefully that would allow an equal sharing of the money.

This could be all wrong, it's quite late and I'm tired...
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hitchhiker_13
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#22
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#22
(Original post by ZJuwelH)
Try this one, I don't think it's an Oxbridge thing, I read it in a book on Fermat's Last Theorem (don't ask)...

Mr Red, Mr Green and Mr Blue are in a truel (a duel with three people). It involves guns, and a shot that hits its target kills the target.

Mr Red will hit his target a third of the time. Mr Blue will hit his target half the time. Mr Green is the deadliest, he hits the target every time.
Since Mr Red is the least inaccurate he is allowed first shot, followed by Blue (provided Blue isn't dead), then Green and round again, and the winner is the last man alive.

Question: Where does Mr Red aim his first shot?

Answer to come later.


Tendency would be to say Mr Green, to take him out. But I then thought about it for an extra 2 seconds or so, and thought if I was red I would deliberately misfire, as if he killed Mr Blue, then Mr Green would shoot him, but if he leaves Mr BLue intact, Mr Blue has a better chance of killing Mr Green.

Probably haven't thought about this enough.
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Tifa
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#23
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#23
(Original post by powerball)
no ones tried the pirate question?
EDIT: this is SO wrong as it answers a different question, but how about if you try reconsidering the effect of making it have to be a majority to passs the vote... ummm.. solution below - lol

Okay, the pirate question is actually really difficult. imho. Umm, okay, I'm working on the basis that even thought pirates value their life more than gold, they really really value gold and wouldn't want to lose it, and also that these pirates are all really really brainy and could think ahead. Working backwards.

If only pirate D were left he would get everything.

If only pirate C and D were left then pirate C is dead, regardless of what he suggests pirate D will vote to throw him overboard to get all the gold. So it doesn't matter what pirate C suggests.

If only pirates B,C, and D were left then pirate B could count on pirate C's vote as long as pirate C got at least 1 coin. This is because if pirate B goes overboard pirate C knows he will get nothing. So pirate B would say 0,1,999 to which he and pirate C would vote 'yes.'

If only pirates A,B,C, and D were left then pirate A could count on pirate B voting 'no' unless pirate B got more than 999 coins, which would leave everybody else with 0, which wouldn't be passed. So pirate A wouldn't count on pirate B's vote so he won't give any gold to him. If pirate A goes overboard he knows the gold will be split 0,1,999. So if he offers 2 gold to pirate C and 1 gold to pirate D then he will get their vote since they will come out better. Therefore pirate A would divide the gold 1,2,0,997 to which pirates A, C and D would vote 'yes.'

Then if there were 5 pirates it would be, 2,0,1,0,997 and so on…

I *think* that works. There is no snowball in hell's chance I would have been able to work that out convincingly in the pressure of an interview I think… did you get that in the interview???????
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Juwel
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#24
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#24
(Original post by hitchhiker_13)
Tendency would be to say Mr Green, to take him out. But I then thought about it for an extra 2 seconds or so, and thought if I was red I would deliberately misfire, as if he killed Mr Blue, then Mr Green would shoot him, but if he leaves Mr BLue intact, Mr Blue has a better chance of killing Mr Green.

Probably haven't thought about this enough.
Well you've got it. If he aims at either and scores, he faces a great chance of being shot by the remaining person.
Whereas if he misfires, Blue will aim at Green (the more dangerous). If he scores, it's Red's turn again. If Blue misses, Green will shoot the more dangerous Blue dead. Either way Red ends up with the first shot in a duel and one clear target.
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llama boy
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#25
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#25
(Original post by hitchhiker_13)
What can the pirate do? If they knew they weren't going to be next (i.e. C and D) they would vote against any proposal made by A. But A and B will vote in favour, because B will know he's screwed if A gets chucked off. As it's a tie, the vote will pass. So really A could say anything...not sure about this last point. Because whatever A says, B will agree, because if A is chucked off, the B will be left 1 against 2, so he'll be chucked off. So A can just say I'll have it all.

If there were 5, then it would be 2 against 3 so it wouldn't work, but after that return to the same, so B can do what they want.

If it was random, then no-one will know who will have to make the next proposal, so hopefully that would allow an equal sharing of the money.

This could be all wrong, it's quite late and I'm tired...
You're assuming that C and D will always team up, which is erronous, I think.

Hmm, what about if you look at this backwards.

CD - C can make a proposal in which C gets 100% of the treasure and D gets nothing, because the vote will still be tied and the deal will go through. Also, neither C or D have any reason to fear for their life.

BCD - C will know that he can get the scenario above by rejecting any deal offered by B. So he will, short of being offered 100%. But, it seems that B could get away with offering 1% to D (who stands to be screwed in the CD situation) and 99% to himself, thereby carrying the proposal 2 to 1. Hmm, but then there is nothing stopping C turning round and offering a premlininary deal of 2% to D, then more etc to save himself from getting nothing. This is the bit I'm not sure about. :confused:

ABCD - OK, so C has no incentive to accept anything below 100%. B, likewise, can get 99% for himself by rejecting it and doing the deal outlined in the paragraph above. But D is still in a ****ty position, likely to get either 1% or 0% from either of the later scenarios.

So, in theory, A can propose to split the cash between himself and D, presumably with terms favourable to himself.

Then A and D will vote for it, B and C won't, it'll be 2/2, a tie, and therefore it'll be carried.

Right? :confused:

(OK, its late, I'm tired, I know there are errors in this, but everyone else is putting put their theories, so i'll put my half finished labours up and skulk off to bed...)
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hitchhiker_13
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#26
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#26
(Original post by ZJuwelH)
Well you've got it. If he aims at either and scores, he faces a great chance of being shot by the remaining person.
Whereas if he misfires, Blue will aim at Green (the more dangerous). If he scores, it's Red's turn again. If Blue misses, Green will shoot the more dangerous Blue dead. Either way Red ends up with the first shot in a duel and one clear target.

Ooh, I'm quite proud of myself! Not sure how I'd cope in a pressurised interview situation though.
Was that the Fermat's Last Theorem by Simon Singh or a different one? I keep meaning to read it.
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llama boy
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#27
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#27
(Original post by Tifa)
If only pirate D were left he would get everything.

If only pirate C and D were left then pirate C is dead, regardless of what he suggests pirate D will vote to throw him overboard to get all the gold. So it doesn't matter what pirate C suggests.
No...

Remember, if the vote is a tie, then the proposal goes through. In a CD situation, you can assume that C will vote for his own proposal, therefore it will always go through.
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Juwel
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#28
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#28
(Original post by hitchhiker_13)
Ooh, I'm quite proud of myself! Not sure how I'd cope in a pressurised interview situation though.
Was that the Fermat's Last Theorem by Simon Singh or a different one? I keep meaning to read it.
It was Singh's book, a good read I thought. It doesn't show you the solution of the theorem but the whole 360-year-old story of people trying to solve it.
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Tifa
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#29
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#29
(Original post by llama boy)
No...

Remember, if the vote is a tie, then the proposal goes through. In a CD situation, you can assume that C will vote for his own proposal, therefore it will always go through.
ooooh yeah - I misread that a tie gets passed - hmm, duh
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hitchhiker_13
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#30
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#30
(Original post by llama boy)
You're assuming that C and D will always team up, which is erronous, I think.

You could be right, I have no experience with these things and am just throwing out some suggestions, but my reasoning was that by teaming up C and D try to get rid of A and B, therefore only have to share the money between 2, so why wouldn't they? I have gone on to say that this won't work, as B will vote with A, knowing he can't survive otherwise, unless offering C or D all the money basically.

Besides even if C and D didn't team up my original conclusion would still stand.

I think... :confused:
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Juwel
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#31
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#31
Why don't the pirates just split it equally? That's the simplest way, no one dies and no one gets a raw deal.
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hitchhiker_13
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#32
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#32
(Original post by ZJuwelH)
Why don't the pirates just split it equally? That's the simplest way, no one dies and no one gets a raw deal.

Yes that would be fair, but in my (possibly, ok probably, horribly incorrect) hypothesis, A can get away with the whole lot.
If it is made random, then it will be split equally, provided the pirates have the same (shaky?) logic as me.
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llama boy
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#33
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#33
(Original post by hitchhiker_13)
You could be right, I have no experience with these things and am just throwing out some suggestions, but my reasoning was that by teaming up C and D try to get rid of A and B, therefore only have to share the money between 2, so why wouldn't they?
Well, C might not want to, as he knows that once it gets down to just him and D, he can propose, and (through a tie) pass a proposal that gives him everything.

D might not support it because he knows that once it gets down to just C and D, C has no incentive to give him anything.

This is f***ing confusing, though!
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Juwel
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#34
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#34
(Original post by llama boy)
Well, C might not want to, as he knows that once it gets down to just him and D, he can propose, and (through a tie) pass a proposal that gives him everything.

D might not support it because he knows that once it gets down to just C and D, C has no incentive to give him anything.

This is f***ing confusing, though!
On that logic you can say C will go against both A and B unless either offers him 100. He would attempt to create a C vs D scenario.
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hitchhiker_13
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#35
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#35
(Original post by llama boy)
Well, C might not want to, as he knows that once it gets down to just him and D, he can propose, and (through a tie) pass a proposal that gives him everything.

D might not support it because he knows that once it gets down to just C and D, C has no incentive to give him anything.

This is f***ing confusing, though!

But it really makes no difference whether they both disagree with A or one agrees and one disagrees (following my logic), as B will vote with A, and therefore it will be at least a tie so will pass?

Actually, if B voted against A, and so did C and D, then he could just give one to D and keep the rest for himself. But C wouldn't do this so would vote for A....

Then there is still nothing to stop A getting it all?
I should definitely go to bed.
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Tifa
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#36
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#36
GODDAM MY READING SKILLS

i think this makes it a bit easier??
EDIT - thats wrong, me that is
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hitchhiker_13
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#37
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#37
(Original post by Tifa)
GODDAM MY READING SKILLS

i think this makes it a bit easier??

how about this

Pirate A needs one other person to vote for him. He gives one coin to Pirate B because if we go to three pirates, Pirate B gets zero so Pirate B knows one is better than none. Pirate D takes 99. Pirate B gets 1. Pirate A gets 0.

Do you mean pirate A gets 99? I'm confused?
But I'm saying B will vote for A anyway, beacause if he doesn't he's going over board, so A doesn't need to give him even one coin?
So A gets 100.
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llama boy
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#38
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#38
(Original post by ZJuwelH)
On that logic you can say C will go against both A and B unless either offers him 100. He would attempt to create a C vs D scenario.
If you read my main post, you'll see that's what I did say..
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hitchhiker_13
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#39
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#39
Apparently sleep is used as a time to solve problems, and the best thing to do is mull a problem over just before you drift off. So I'm off to bed to think about this...

Oiche Mhaith
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Tifa
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#40
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#40
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

???
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