# oxbridge interview's logic questionsWatch

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16 years ago
#81
(Original post by KaiserSoze)
Yeah well done! Took a very long time for me/other people to figure that out....
heehee, I think it's because I'm so cynical, usually when people ask me puzzle questions the first thing I think is... where's the trick? And this one was just screaming *LIES * at me... and Maths on the other hand never lies... dammit

island question

Ok, this one took me absolutely ages - its a bit long, but I want to make it clear... and I promise its a good one *grin*

There is an island with 10 people. One day a monster comes along and says 'I intend to eat every one of you but will give you a chance to survive in this way'
In the morning, the monster will line up all the people in single file so that the last person sees the remaining 9, and the first person sees no one in front. The monster will put black or red hats on their heads at random (they can be all red, all black or any combination).
The monster will ask the last person (who sees everyone else's hats) to guess the colour of their own hat. For their answer they can only say "red" or "black". The monster will eat him immediately if he is wrong, and will let him live if he gets it right. All the remaining people will hear the guess and whether he gets eaten. The monster will then go on to the next person (who only sees 8 people), and so on until the he gets to the front. The monster gives them the night to chew it over.

What is the best strategy that these natives could use to maximise their survival rate?

Remember that the only allowed response is a short, unemotional "red" or "black", if they say it in a particular way or add bits on they will be gobbled up. However they can put any meaning into these words if they so wish. E.g. "red" can mean "I have a really sore toe" or "the person in front has a black hat on"…
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16 years ago
#82
What is the quantity of zero to the zeroth power?
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16 years ago
#83
(Original post by Tifa)
There is an island with 10 people. One day a monster comes along and says 'I intend to eat every one of you but will give you a chance to survive in this way'
In the morning, the monster will line up all the people in single file so that the last person sees the remaining 9, and the first person sees no one in front. The monster will put black or red hats on their heads at random (they can be all red, all black or any combination).
The monster will ask the last person (who sees everyone else's hats) to guess the colour of their own hat. For their answer they can only say "red" or "black". The monster will eat him immediately if he is wrong, and will let him live if he gets it right. All the remaining people will hear the guess and whether he gets eaten. The monster will then go on to the next person (who only sees 8 people), and so on until the he gets to the front. The monster gives them the night to chew it over.

What is the best strategy that these natives could use to maximise their survival rate?

Remember that the only allowed response is a short, unemotional "red" or "black", if they say it in a particular way or add bits on they will be gobbled up. However they can put any meaning into these words if they so wish. E.g. "red" can mean "I have a really sore toe" or "the person in front has a black hat on"…
hmm.

maybe the first five people should pair up with the last five people, with the first five people saying the colour of their partner's hat. that way five are definitely saved, and the other 5 have a 50/50 chance?

first guess anyway...there must be a more efficient way of doing this...
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16 years ago
#84
(Original post by llama boy)
hmm.

maybe the first five people should pair up with the last five people, with the first five people saying the colour of their partner's hat. that way five are definitely saved, and the other 5 have a 50/50 chance?

first guess anyway...there must be a more efficient way of doing this...
ok, that's a definite strategy, and the idea of using what the first person sees is important, but obviously it only guarentees a 50% survival.

The best strategy will guarantee a minimum of 90% survival...

and yawn - I know the answer is infinity, but I have no idea how to prove it mathematically, something to do with it being an indeterminate form I seem to remember...
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16 years ago
#85
hmm, interesting question - the first one could take one for the team and give the next nine really fast then run for it - only he will be eaten! Doubt thats the answer though....

Will keep thinking...
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16 years ago
#86
Ok.

I think I've found a solution.

If you consider that the person at the back of the queue sees 9 people in front of them, then there must be an odd number of people wearing one colour of hat (be it red or black) and an even number of people wearing the other colour of hat.

It should be decided that the person at the back says the colour that the odd number's are wearing (for example, if there are three people with black hats and six with red hats then they would shout red). The next person will then see all of the eight people in front of them and will be able to deduce the colour of their hat because they know that there are an odd number or people in black hats, so if they see and even number of people in black hats ahead of them, they can infer that they must be wearing a black hat to make the total odd. However, if the see their is an odd number of people in black hats, their hat must be red. The next person in line knows that there were an even number of red hats (out of the nine people in front of the last person). So, they will have heard what the person behind them said their hat colour was. If they said it was red (they will have been correct either way by deduction as I will have already explained), then this means that there should be an odd number of people in red hats in front of and including themselves, so they will know they are wearing a red hat if there are an even number of red hats ahead, but a black hat if there are an odd number in red hats.

The rest of the people in the line can apply the same logic, and voila, you will have saved at least nine of them from death.
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16 years ago
#87
(Original post by Ralfskini)
Ok.

I think I've found a solution.

If you consider that the person at the back of the queue sees 9 people in front of them, then there must be an odd number of people wearing one colour of hat (be it red or black) and an even number of people wearing the other colour of hat....
*cheer* bravo *round of applause* tens upon tens of flowers shower the stage as Ralfskini humbly bows and acknowledges his fans with a wave...

very well done, that is the correct answer - you are indeed very very logical
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16 years ago
#88
(Original post by powerball)
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?
Did you apply to Hertford? As I got the same question (except there were 10 lions, numbered 1-10...)
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16 years ago
#89
(Original post by llama boy)
It's very simple. You can see one side of a card which is red. In total there are three red sides and three white sides. As you only need to worry about the three red sides that means there is a 1 in 3 chance that the side you see will be one of them;

Red side 1: 1 in 3
Red side 2: 1 in 3
Red side 3: 1 in 3

As two of the sides ( for our purposes red side 1 and red side 2 ) are on the same card the chances of it being that card are the chances that it is one of those sides added together ( P(R1+R2)=2/3 ) meaning there is a 2 in 3 chance of it being a red/red card.
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16 years ago
#90
This was one of the logic questions from the Sidney Sussex interviews this year;

There are three people at a table; you, Dr Ahmed and Dr Pearson. You originally had a full deck of cards but now you have discarded all cards from it except the four twos and the four eights. The cards are shuffled and without looking each of you put two cards in turn on your heads so the other two players can see but you cannot. Play then passes round the circle in turn until one of the players can deduce what cards are on their head. The two remaining cards ( ( 4 + 4 ) - ( 2 + 2 + 2 ) = 2 ) are in the centre of the table face down so no one can see.

Game One: You can see that Dr Ahmed has two twos on his head and Dr Pearson has two twos on his head. What do you have on your head.

Game Two: You can see that Dr Ahmed has two twos on his head and Dr Peason has two eights on his head. Dr Ahmed says that he doesn't know what he has, Dr Pearson says that he doesn't know what he has. What do you have on your head?

Game three: You can see that Dr Ahmed has two twos on his head and Dr Peason has a two and an eight on his head. You go first and say that you do not know what you have on your head. Dr Ahmed says he doesn't know what he has. Dr Pearson says he doesn't know what he has. What do you have on your head?
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16 years ago
#91
More please guys...my bro and I are really enjoying these!
Oops, how sad are we lol!
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16 years ago
#92
More please guys...my bro and I are really enjoying these!
Oops, how sad are we lol!
http://rec-puzzles.org/logic.html
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16 years ago
#93
(Original post by llama boy)
http://rec-puzzles.org/logic.html
These look fun! Thank you!
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