You guys have forgotten a real classic; the ol' forum destroyer. I suppose it doesn't quite fit the riddle-like format that has been popular so far, but I still think it's worth a mention.
Does 0.999... = 1?
And the answer:
Yes.
YOU AND YOUR BOUGEOIS "ROUNDING"
Actual answer: the question is retarded and has no meaning as phrased. Precisely phrasing the question requires some understanding of limits and sums.
No, it doesn't. I have a perfectly good understanding of limits and sums. 0.999... (said "point nine nine repeating", or even "point nine repeating") is in fact exactly equal to one. No limits or sums required (although proving the equality does require those).
I refuse to accept repeating decimals that are not the result of integer divisions.
Here's a fun one that usually causes a big argument.
An alien walks up to you one day and offers you two suitcases. You choose one and open it, and it contains $50. He then tells you that one of the suitcases has twice the money as the other, and that you can switch to the other one if you want. What's the best course of action?
If $50 is the larger amount, you end up with $25. -$25
If $50 is the smaller amount, you end up with $100. +$50
The problem is clear cut. The only questions is HOLY SHIT AN ALIEN!
Capture it. Set for life.
Ok. Assume that you get given the case and are asked if you want to switch before opening it. You realise you have half a chance of doubling your money and half a chance of halving it, leading to an expected profit. So you switch. Then the alien asks if you'd like to switch back. You have half a chance of doubling your money and half a chance of halving it by doing this, leading to an expected profit. So you should switch back.
No matter which suitcase you have, according to your reasoning, it is right to switch to the other.
Solution/Response:
The reason the second swap doesn't work that way is because it's not the same situation. In the first we have a deterministic amount of money in the envelope we are holding, and the amount in the other envelope is a random variable with an expected value of $62.50. For the second trade, the amount in the envelope you are holding is a random variable (with expectation $62.50) and the amount in the other envelope is deterministic (and equal to $50). The key point here is that in the first case you can consider it two random variables -- the first being the deterministic amount in the envelope and the second being the "double or half" variable. Then simply use independence to find expectation of the product of the two. In the second case you can't use independence, because it isn't independent -- you only double if the envelope you're holding has $25, and you only half if the envelope you're holding has $100.
Also, a new puzzle:
a and b are two numbers (integers) in the interval [2,99]. Mr. P and Mr. S are told a*b and a+b respectively. They have the following conversation:
Mr. P: I don't know the two numbers.
Mr. S: I already knew you didn't.
Mr. P: Ah, now I know them!
Mr. S: And now, so do I.
Assuming they aren't lying or incorrect, what are a and b?
Note: Both Mr. P and Mr. S know that the numbers are in the interval [2,99]. That's important, if I recall correctly. Also, this can be done on pencil and paper but some very basic programming helps. If you think you have a method of getting the solution and don't want to program it or do it by hand, just post a description of the method and I'll tell you if it's correct or not.
For 99 nights, nobody leaves. On night 100, all the blues leave en masse.
Each day everyone watches one specific blue-eye to see if they leave. When they don't, okay, the guru's not talking about that guy, so I'll watch this other one over here.
On Day 100, if you're a blue-eye, you've now looked at all the other blue-eyes and nobody's left, so the guru must have been referring to you. You go bye-bye.
If you're a brown-eye or the Guru, on Day 100 you still have one more blue-eye to look at, so when they all go on Day 100, you know you're not it.
Yep.
Correct. Logical Induction.
Edit: That's the right answer but wrong reason.
You aren't watching each blue-eyed person in turn. You're figuring out how many blue-eyed people there are on the island. If there were only one, he would leave on the first night. If there were only two, then each one figures out that there must be two on the second day because otherwise the other guy would have left the previous night, and so on until the hundredth day.
But everyone know's there's at least 99, and they also know that everyone else know's there's at least 98, so everyone knows that no-one has any reason to leave on any of those days, and everyone knows that everyone knows that. And why is there even a Guru?
This took me a minute to resolve also. To make is easier, just say that everyone starts with the assumption that they have brown eyes, then think the case where only three people have blue eyes. On the first night, no one leaves and no one is surprised. But on the second night, still no one leaves because each of the three assumes the other two are the only two with blue eyes. But when morning of the third day roles around, each realizes that there can't just be two - otherwise both of the other two would have left the previous night. Therefore, since everyone else is accounted for, the extra blue eyes must be you, and each blue-eyed person will leave on that third night.
Now you can just add a fourth person. Now every blue-eyes on the third day will still assume that the other three are the only blue-eyes on the island, so no one leaves that night. The next day they will all realize that there must be four, and the fourth is them. And so, up to however many blue-eyed people you want.
This works for three people, but I don't see how it can work for four or more. It work for three people because each one knows that if he has brown eyes, then each blue guy will only see one other blue guy on the island and that he'll be able to tell what colour eyes the other guy has based on that. But if there are four blue guys, then each blue guy knows that the other person can see at least two blue guys and will be able to deduce whether the other guy is in a three-guy situation or a four-guy situation based on his behaviour okay now i understand it.
There's still something I don't understand about this: the piece of information given as a catalyst for this whole thing ("I see someone with blue eyes!") was available all along. Everyone knows there's at least one other person with blue eyes, so shouldn't this whole exodus have happened already?
There's still something I don't understand about this: the piece of information given as a catalyst for this whole thing ("I see someone with blue eyes!") was available all along. Everyone knows there's at least one other person with blue eyes, so shouldn't this whole exodus have happened already?
Think back to the base case: If I'm the only person with blue eyes, why is the statement necessary? If there is one other person with blue eyes, why is it still necessary?
You guys have forgotten a real classic; the ol' forum destroyer. I suppose it doesn't quite fit the riddle-like format that has been popular so far, but I still think it's worth a mention.
Does 0.999... = 1?
And the answer:
Yes.
YOU AND YOUR BOUGEOIS "ROUNDING"
Actual answer: the question is retarded and has no meaning as phrased. Precisely phrasing the question requires some understanding of limits and sums.
No, it doesn't. I have a perfectly good understanding of limits and sums. 0.999... (said "point nine nine repeating", or even "point nine repeating") is in fact exactly equal to one. No limits or sums required (although proving the equality does require those).
I refuse to accept repeating decimals that are not the result of integer divisions.
But 0.999... is the result of integer division. It is the result of 1/1!
There is an island that is considered to be paradise. All the inhibitants of the island are Perfect Logicians, and every knows of every that they are Perfect Logicans.
Exactly 100 of these persons have blue eyes, 100 have brown eyes, and 1 has green eyes. The inhibitants do not know what his/her color eyes is. Everyone is constantly aware of everyone elses eye color but no person knows that there are 100 blue eyed, 100 brown eyed, and 1 green eyed person on the island.
If a person finds out his/her own eye color she/he must leave the island at midnight of the day she/he finds out! There are no mirrors or reflections of any kind on the island. Also, nobody on the island ever speaks except the Guru, who is the person with the green eyes, (she does not know her eye color and if she found out she would have to leave the island at midnight). The Guru says one sentence every fifty years. One day the Guru arrives and tells everyone on the island the following: “I SEE SOMEONE WITH BLUE EYES.”
Who (if anyone) leaves the island and when?
I'll be back to clean up when you're all done killing each other in 5 pages when the arguments heat up.
There's still something I don't understand about this: the piece of information given as a catalyst for this whole thing ("I see someone with blue eyes!") was available all along. Everyone knows there's at least one other person with blue eyes, so shouldn't this whole exodus have happened already?
Think back to the base case: If I'm the only person with blue eyes, why is the statement necessary? If there is one other person with blue eyes, why is it still necessary?
It isn't. That's my point. If you see one other person with blue eyes who does not leave on the second day, then you know you have brown eyes and must leave yourself...
Wait a minute. I don't see how any number of people with any combination of brown and blue eye colors is stable in this situation.
There's still something I don't understand about this: the piece of information given as a catalyst for this whole thing ("I see someone with blue eyes!") was available all along. Everyone knows there's at least one other person with blue eyes, so shouldn't this whole exodus have happened already?
Yeah, the Guru thing seems unnecessary and potentially misleading.
There's still something I don't understand about this: the piece of information given as a catalyst for this whole thing ("I see someone with blue eyes!") was available all along. Everyone knows there's at least one other person with blue eyes, so shouldn't this whole exodus have happened already?
Yeah, the Guru thing seems unnecessary and potentially misleading.
No, you need something to start it off. If there was only one person with blue eyes and no one said anything about blue eyes he'd never know and nothing would happen.
There's still something I don't understand about this: the piece of information given as a catalyst for this whole thing ("I see someone with blue eyes!") was available all along. Everyone knows there's at least one other person with blue eyes, so shouldn't this whole exodus have happened already?
Think back to the base case: If I'm the only person with blue eyes, why is the statement necessary? If there is one other person with blue eyes, why is it still necessary?
It isn't. That's my point. If you see one other person with blue eyes who does not leave on the second day, then you know you have brown eyes and must leave yourself...
Wait a minute. I don't see how any number of people with any combination of brown and blue eye colors is stable in this situation.
If you see one person with blue eyes, you know that there is either one or two people with blue eyes. You believe that the blue monk sees nobody with blue eyes, so he'll never leave.
When the person comes along and says "someone has blue eyes," you 'know' that the monk with blue eyes will see nobody with blue eyes, realize that it must be him, and leave. When he does not leave, you realize that he must see someone with blue eyes, and since you know it is not any of the other monks, it must be you. Iterate.
There's still something I don't understand about this: the piece of information given as a catalyst for this whole thing ("I see someone with blue eyes!") was available all along. Everyone knows there's at least one other person with blue eyes, so shouldn't this whole exodus have happened already?
Think back to the base case: If I'm the only person with blue eyes, why is the statement necessary? If there is one other person with blue eyes, why is it still necessary?
It isn't. That's my point. If you see one other person with blue eyes who does not leave on the second day, then you know you have brown eyes and must leave yourself...
Wait a minute. I don't see how any number of people with any combination of brown and blue eye colors is stable in this situation.
If you see one person with blue eyes, you know that there is either one or two people with blue eyes. You believe that the blue monk sees nobody with blue eyes, so he'll never leave.
When the person comes along and says "someone has blue eyes," you 'know' that the monk with blue eyes will see nobody with blue eyes, realize that it must be him, and leave. When he does not leave, you realize that he must see someone with blue eyes, and since you know it is not any of the other monks, it must be you. Iterate.
Still not getting it.
Every piece of information contained in the statement "I see someone with blue eyes" was already known to the whole population before. How then can this statement, which contains no information, set off the exodus 100 days later? It is equivalent to the guru having said nothing at all.
CycloneRanger on
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GoslingLooking Up Soccer In Mongolia Right Now, ProbablyWatertown, WIRegistered Userregular
There is an island that is considered to be paradise. All the inhibitants of the island are Perfect Logicians, and every knows of every that they are Perfect Logicans.
Exactly 100 of these persons have blue eyes, 100 have brown eyes, and 1 has green eyes. The inhibitants do not know what his/her color eyes is. Everyone is constantly aware of everyone elses eye color but no person knows that there are 100 blue eyed, 100 brown eyed, and 1 green eyed person on the island.
If a person finds out his/her own eye color she/he must leave the island at midnight of the day she/he finds out! There are no mirrors or reflections of any kind on the island. Also, nobody on the island ever speaks except the Guru, who is the person with the green eyes, (she does not know her eye color and if she found out she would have to leave the island at midnight). The Guru says one sentence every fifty years. One day the Guru arrives and tells everyone on the island the following: “I SEE SOMEONE WITH BLUE EYES.”
Who (if anyone) leaves the island and when?
I'll be back to clean up when you're all done killing each other in 5 pages when the arguments heat up.
Clean up then, because I won that one.
Gosling on
I have a new soccer blog The Minnow Tank. Reading it psychically kicks Sepp Blatter in the bean bag.
There's still something I don't understand about this: the piece of information given as a catalyst for this whole thing ("I see someone with blue eyes!") was available all along. Everyone knows there's at least one other person with blue eyes, so shouldn't this whole exodus have happened already?
Think back to the base case: If I'm the only person with blue eyes, why is the statement necessary? If there is one other person with blue eyes, why is it still necessary?
It isn't. That's my point. If you see one other person with blue eyes who does not leave on the second day, then you know you have brown eyes and must leave yourself...
Wait a minute. I don't see how any number of people with any combination of brown and blue eye colors is stable in this situation.
I think you misread the bolded. If I'm the only one with blue eyes, how do I know my own eye color without a guru? If I see someone with blue eyes, how can I expect them to figure theirs out?
There is an island that is considered to be paradise. All the inhibitants of the island are Perfect Logicians, and every knows of every that they are Perfect Logicans.
Exactly 100 of these persons have blue eyes, 100 have brown eyes, and 1 has green eyes. The inhibitants do not know what his/her color eyes is. Everyone is constantly aware of everyone elses eye color but no person knows that there are 100 blue eyed, 100 brown eyed, and 1 green eyed person on the island.
If a person finds out his/her own eye color she/he must leave the island at midnight of the day she/he finds out! There are no mirrors or reflections of any kind on the island. Also, nobody on the island ever speaks except the Guru, who is the person with the green eyes, (she does not know her eye color and if she found out she would have to leave the island at midnight). The Guru says one sentence every fifty years. One day the Guru arrives and tells everyone on the island the following: “I SEE SOMEONE WITH BLUE EYES.”
Who (if anyone) leaves the island and when?
I'll be back to clean up when you're all done killing each other in 5 pages when the arguments heat up.
Clean up then, because I won that one.
See, but I knew the fighting wouldn't start until the solution was declared and arguments popped up over the logic of it.
For 99 nights, nobody leaves. On night 100, all the blues leave en masse.
Each day everyone watches one specific blue-eye to see if they leave. When they don't, okay, the guru's not talking about that guy, so I'll watch this other one over here.
On Day 100, if you're a blue-eye, you've now looked at all the other blue-eyes and nobody's left, so the guru must have been referring to you. You go bye-bye.
If you're a brown-eye or the Guru, on Day 100 you still have one more blue-eye to look at, so when they all go on Day 100, you know you're not it.
Yep.
Correct. Logical Induction.
Edit: That's the right answer but wrong reason.
You aren't watching each blue-eyed person in turn. You're figuring out how many blue-eyed people there are on the island. If there were only one, he would leave on the first night. If there were only two, then each one figures out that there must be two on the second day because otherwise the other guy would have left the previous night, and so on until the hundredth day.
But everyone know's there's at least 99, and they also know that everyone else know's there's at least 98, so everyone knows that no-one has any reason to leave on any of those days, and everyone knows that everyone knows that. And why is there even a Guru?
This took me a minute to resolve also. To make is easier, just say that everyone starts with the assumption that they have brown eyes, then think the case where only three people have blue eyes. On the first night, no one leaves and no one is surprised. But on the second night, still no one leaves because each of the three assumes the other two are the only two with blue eyes. But when morning of the third day roles around, each realizes that there can't just be two - otherwise both of the other two would have left the previous night. Therefore, since everyone else is accounted for, the extra blue eyes must be you, and each blue-eyed person will leave on that third night.
Now you can just add a fourth person. Now every blue-eyes on the third day will still assume that the other three are the only blue-eyes on the island, so no one leaves that night. The next day they will all realize that there must be four, and the fourth is them. And so, up to however many blue-eyed people you want.
This works for three people, but I don't see how it can work for four or more. It work for three people because each one knows that if he has brown eyes, then each blue guy will only see one other blue guy on the island and that he'll be able to tell what colour eyes the other guy has based on that. But if there are four blue guys, then each blue guy knows that the other person can see at least two blue guys and will be able to deduce whether the other guy is in a three-guy situation or a four-guy situation based on his behaviour okay now i understand it.
There's still something I don't understand about this: the piece of information given as a catalyst for this whole thing ("I see someone with blue eyes!") was available all along. Everyone knows there's at least one other person with blue eyes, so shouldn't this whole exodus have happened already?
Think of the case where there is one person with blue eyes. He would look around, see everyone else has non-blue eyes,... but that tells him nothing of himself. Now add a guru, and blue eyes leaves (of course). Now consider the case of two people with blue eyes and no guru. They each look around, see one person with blue eyes and a bunch of people with non-blue eyes, but that tells them nothing of themselves so no one leaves. Now add a guru. On the first day no one leaves, because Blue1 thinks Blue2 could be the only person with blue eyes, and Blue2 thinks Blue1 could be the only person with blue eyes. On the second day, Blue1 realizes that since Blue2 didn't leave and everyone else besides Blue1 and Blue2 have non-blue eyes that Blue1 must have blue eyes, and vice versa. Thus they leave.
More formally, take P(n) to mean "If there are exactly n people with blue eyes on the island, they will leave on the nth night." P(1) is true, I hope you will agree. Now we want to show that, if P(n) is true then P(n+1) is true for any n.
To see this, consider that there are n+1 blue-eyed people on the island. One of them, let's say Bob, sees that there are n other blue-eyed people on the island, and the rest (besides himself, as he doesn't know his eye color) are not blue-eyed. Thus he knows there are either n or n+1 blue-eyed people on the island. He waits n days, and on the (n+1)th day the n blue-eyed people are still there. Since P(n) is true by assumption, those people would have left if Bob were not blue eyed (making the total n+1), so on the (n+1)th night, Bob and all of the other blue-eyes leave.
Then, since P(1) is true and P(n) implies P(n+1), P(n) is true for any n.
Without the guru you have P(n) implies P(n+1) -- but not P(1).
a and b are two numbers (integers) in the interval [2,99]. Mr. P and Mr. S are told a*b and a+b respectively. They have the following conversation:
Mr. P: I don't know the two numbers.
Mr. S: I already knew you didn't.
Mr. P: Ah, now I know them!
Mr. S: And now, so do I.
Assuming they aren't lying or incorrect, what are a and b?
Note: Both Mr. P and Mr. S know that the numbers are in the interval [2,99]. That's important, if I recall correctly. Also, this can be done on pencil and paper but some very basic programming helps. If you think you have a method of getting the solution and don't want to program it or do it by hand, just post a description of the method and I'll tell you if it's correct or not.
Can we assume a & b are different from one another?
I'm going to have to think about this one, but I suspect that we can narrow the options a little bit:
it cannot be the case that both a and b are primes, because otherwise Mr P would know the two numbers immediately
we can also throw out any pairs where both numbers are so high or so low as to have the only other possible choices that match the product fall outside of the given range
There's still something I don't understand about this: the piece of information given as a catalyst for this whole thing ("I see someone with blue eyes!") was available all along. Everyone knows there's at least one other person with blue eyes, so shouldn't this whole exodus have happened already?
Yeah, the Guru thing seems unnecessary and potentially misleading.
The guru is the weakness of the problem. In reality, the countdown would start from the first day everyone was on the island.
EDIT: no, wait, it wouldn't. It just seems very weird.
a and b are two numbers (integers) in the interval [2,99]. Mr. P and Mr. S are told a*b and a+b respectively. They have the following conversation:
Mr. P: I don't know the two numbers.
Mr. S: I already knew you didn't.
Mr. P: Ah, now I know them!
Mr. S: And now, so do I.
Assuming they aren't lying or incorrect, what are a and b?
Note: Both Mr. P and Mr. S know that the numbers are in the interval [2,99]. That's important, if I recall correctly. Also, this can be done on pencil and paper but some very basic programming helps. If you think you have a method of getting the solution and don't want to program it or do it by hand, just post a description of the method and I'll tell you if it's correct or not.
Can we assume a & b are different from one another?
I'm going to have to think about this one, but I suspect that we can narrow the options a little bit:
it cannot be the case that both a and b are primes, because otherwise Mr P would know the two numbers immediately
we can also throw out any pairs where both numbers are so high or so low as to have the only other possible choices that match the product fall outside of the given range
a and b could be the same.
Your reasoning in the spoiler sounds good so far, but you (hint)
missed one other similar case.
Real spoiler:
If a and b are p and p^2 for a prime p, then Mr. P would know the two numbers off the bat.
Also, I didn't mention this in my first post but it's noteworthy: If a and b are distinct, you don't have to know which is which (and, indeed, it's pretty easy to show it's impossible for you or Mr. P or Mr. S to know which is which).
There's still something I don't understand about this: the piece of information given as a catalyst for this whole thing ("I see someone with blue eyes!") was available all along. Everyone knows there's at least one other person with blue eyes, so shouldn't this whole exodus have happened already?
Yeah, the Guru thing seems unnecessary and potentially misleading.
The guru is the weakness of the problem. In reality, the countdown would start from the first day everyone was on the island.
EDIT: no, wait, it wouldn't. It just seems very weird.
Again, the inductive proof makes it obvious why the guru is necessary.
a and b are two numbers (integers) in the interval [2,99]. Mr. P and Mr. S are told a*b and a+b respectively. They have the following conversation:
Mr. P: I don't know the two numbers.
Mr. S: I already knew you didn't.
Mr. P: Ah, now I know them!
Mr. S: And now, so do I.
Assuming they aren't lying or incorrect, what are a and b?
Note: Both Mr. P and Mr. S know that the numbers are in the interval [2,99]. That's important, if I recall correctly. Also, this can be done on pencil and paper but some very basic programming helps. If you think you have a method of getting the solution and don't want to program it or do it by hand, just post a description of the method and I'll tell you if it's correct or not.
Can we assume a & b are different from one another?
I'm going to have to think about this one, but I suspect that we can narrow the options a little bit:
it cannot be the case that both a and b are primes, because otherwise Mr P would know the two numbers immediately
we can also throw out any pairs where both numbers are so high or so low as to have the only other possible choices that match the product fall outside of the given range
And by the same reasoning, any two numbers that have the same sum as the sum of two primes.
There's still something I don't understand about this: the piece of information given as a catalyst for this whole thing ("I see someone with blue eyes!") was available all along. Everyone knows there's at least one other person with blue eyes, so shouldn't this whole exodus have happened already?
Think back to the base case: If I'm the only person with blue eyes, why is the statement necessary? If there is one other person with blue eyes, why is it still necessary?
It isn't. That's my point. If you see one other person with blue eyes who does not leave on the second day, then you know you have brown eyes and must leave yourself...
Wait a minute. I don't see how any number of people with any combination of brown and blue eye colors is stable in this situation.
I think you misread the bolded. If I'm the only one with blue eyes, how do I know my own eye color without a guru? If I see someone with blue eyes, how can I expect them to figure theirs out?
You don't; I wasn't actually answering your question directly. Sorry; I just changed to another objection that popped up in my mind while thinking about what you said.
Consider a situation with 1 brown-eyed person and one blue eyed person (plus the guru). The guru says: I see someone with blue eyes! The blue-eyed fellow must leave, because he sees no one else with blue eyes.
If we start with 2 blues and 1 brown (plus guru), and the guru reads his lines, the situation is different. The blue-eyed fellow now sees someone else with blue eyes, and assumes the guru is talking about him. When he realizes that the other guy has the same idea (on day two), he leaves (along with the other guy).
If we start with 3 blues and 1 brown (plus guru), and the guru reads his lines, the same thing happens. The blue-eyed fellow sees two others with blue eyes, and assumes the guru was talking about each of them in turn and only realizes they are thinking the same thing when they both stay for two nights.
But now let's change it. Say we keep 3 blues and 1 brown, but the Guru forgets what he is about and says nothing. Each blue knows there are two other blues. Each thinks to himself "the guru sees someone with blue eyes", and each assumes he is referring to the other blues. Each guy thinks "those other two blue-eyed guys are the ones he's seeing. Blue #2 will see Blue #3 and assume that he's the blue guy in question, and vice versa. Tomorrow, they will realize that neither has left, and so each must be thinking the same thing, and they will leave that night." Of course, this doesn't happen, because they're all actually seeing two other blues—a fact which becomes apparent on the third day, when all blues leave. No guru intervention required.
But now let's change it. Say we keep 3 blues and 1 brown, but the Guru forgets what he is about and says nothing. Each blue knows there are two other blues. Each thinks to himself "the guru sees someone with blue eyes", and each assumes he is referring to the other blues. Each guy thinks "those other two blue-eyed guys are the ones he's seeing. Blue #2 will see Blue #3 and assume that he's the blue guy in question, and vice versa. Tomorrow, they will realize that neither has left, and so each must be thinking the same thing, and they will leave that night." Of course, this doesn't happen, because they're all actually seeing two other blues—a fact which becomes apparent on the third day, when all blues leave. No guru intervention required.
Where is the flaw in my reasoning, here?
In order for someone to leave, they must know their own eye color. Seeing someone else with blue eyes tells me nothing about my own eye color. The fact that they don't leave after n days tells me nothing, because they have no more information about their own eye color than I do about mine.
Again, step back to the base case and work the inductive proof. The guru's statement is not purely obvious information. Consider the fact that the guru tells the same thing to everyone, and everyone is aware that the guru has told everyone.
Imagine there are two people. I see someone with blue eyes. I know that there is someone with blue eyes, however I do not know that they know that there is someone with blue eyes, as my eyes (to the best of my knowledge) could be brown. Even if my eyes are blue, they can only think the same thing.
When the guru speaks, I know that they know that someone has blue eyes, and they vice versa.
Oh, I found the flaw in my logic. In the case without the guru, every person may be silently thinking "the guru sees someone with blue eyes", but they do not know that everyone else also thinks that. The "knowing that the other guy also knows" is the hidden piece of information in the Guru's speech.
Each blue knows that there must be either two or three blues. However, without the guru they never have any means of figuring out which it is. Think of the case of two blues and a bunch of browns. Each blue will know that the other blue doesn't realize his eye color, and therefore when that blue fails to leave at night it doesn't reveal anything.
Oh, I found the flaw in my logic. In the case without the guru, every person may be silently thinking "the guru sees someone with blue eyes", but they do not know that everyone else also thinks that. The "knowing that the other guy also knows" is the hidden piece of information in the Guru's speech.
But in the case with lots of blue-eyed people, everyone does know that everyone else knows that there are people with blue eyes. So there is literally no information added when the Guru says her piece.
Oh, I found the flaw in my logic. In the case without the guru, every person may be silently thinking "the guru sees someone with blue eyes", but they do not know that everyone else also thinks that. The "knowing that the other guy also knows" is the hidden piece of information in the Guru's speech.
But in the case with lots of blue-eyed people, everyone does know that everyone else knows that there are people with blue eyes. So there is literally no information added when the Guru says her piece.
So then the original system isn't stable at all, is it? The guru is unnecessary.
Oh, I found the flaw in my logic. In the case without the guru, every person may be silently thinking "the guru sees someone with blue eyes", but they do not know that everyone else also thinks that. The "knowing that the other guy also knows" is the hidden piece of information in the Guru's speech.
But in the case with lots of blue-eyed people, everyone does know that everyone else knows that there are people with blue eyes. So there is literally no information added when the Guru says her piece.
It doesn't stop the first time. Imagine three blues: I see two blues. I have to assume each of them sees one blue, and therefore I have to assume they have to assume that person doesn't know that there are any blue eyes. The guru's statement tells me that they know that that person would know that they have blue eyes, and therefore when the first night passes I know that they would know that they have blue eyes unless I have blue eyes, so when the second night passes I know that I have blue eyes and I (and they) leave on the third night.
Oh, I found the flaw in my logic. In the case without the guru, every person may be silently thinking "the guru sees someone with blue eyes", but they do not know that everyone else also thinks that. The "knowing that the other guy also knows" is the hidden piece of information in the Guru's speech.
But in the case with lots of blue-eyed people, everyone does know that everyone else knows that there are people with blue eyes. So there is literally no information added when the Guru says her piece.
It doesn't stop the first time. Imagine three blues: I see two blues. I have to assume each of them sees one blue, and therefore I have to assume they have to assume that person doesn't know that there are any blue eyes. The guru's statement tells me that they know that that person would know that they have blue eyes, and therefore when the first night passes I know that they would know that they have blue eyes unless I have blue eyes, so when the second night passes I know that I have blue eyes and I (and they) leave on the third night.
However, this doesn't seem to be true for four blues.
Oh, I found the flaw in my logic. In the case without the guru, every person may be silently thinking "the guru sees someone with blue eyes", but they do not know that everyone else also thinks that. The "knowing that the other guy also knows" is the hidden piece of information in the Guru's speech.
But in the case with lots of blue-eyed people, everyone does know that everyone else knows that there are people with blue eyes. So there is literally no information added when the Guru says her piece.
It doesn't stop the first time. Imagine three blues: I see two blues. I have to assume each of them sees one blue, and therefore I have to assume they have to assume that person doesn't know that there are any blue eyes. The guru's statement tells me that they know that that person would know that they have blue eyes, and therefore when the first night passes I know that they would know that they have blue eyes unless I have blue eyes, so when the second night passes I know that I have blue eyes and I (and they) leave on the third night.
Ah, I see! You have to consider what each person will think each other person will think each other person will think... iterated (number of blues - 1) times. Without the guru, your conclusion is that someone will think that someone will think that... there is no one with blue eyes.
So I guess the proof posted before does correspond directly to reality. The guru is very literally giving us P(1), from which we derive P(n) as we all know from pure logic that P(n+1) is true. It's weird to see the mathematics line up like that.
Oh, I found the flaw in my logic. In the case without the guru, every person may be silently thinking "the guru sees someone with blue eyes", but they do not know that everyone else also thinks that. The "knowing that the other guy also knows" is the hidden piece of information in the Guru's speech.
But in the case with lots of blue-eyed people, everyone does know that everyone else knows that there are people with blue eyes. So there is literally no information added when the Guru says her piece.
It doesn't stop the first time. Imagine three blues: I see two blues. I have to assume each of them sees one blue, and therefore I have to assume they have to assume that person doesn't know that there are any blue eyes. The guru's statement tells me that they know that that person would know that they have blue eyes, and therefore when the first night passes I know that they would know that they have blue eyes unless I have blue eyes, so when the second night passes I know that I have blue eyes and I (and they) leave on the third night.
However, this doesn't seem to be true for four blues.
But it is, because if there were only three blues, everyone would have left on the third night. So on the fourth day, each blue knows there must be four and also knows he must be one of them.
Oh, I found the flaw in my logic. In the case without the guru, every person may be silently thinking "the guru sees someone with blue eyes", but they do not know that everyone else also thinks that. The "knowing that the other guy also knows" is the hidden piece of information in the Guru's speech.
But in the case with lots of blue-eyed people, everyone does know that everyone else knows that there are people with blue eyes. So there is literally no information added when the Guru says her piece.
It doesn't stop the first time. Imagine three blues: I see two blues. I have to assume each of them sees one blue, and therefore I have to assume they have to assume that person doesn't know that there are any blue eyes. The guru's statement tells me that they know that that person would know that they have blue eyes, and therefore when the first night passes I know that they would know that they have blue eyes unless I have blue eyes, so when the second night passes I know that I have blue eyes and I (and they) leave on the third night.
However, this doesn't seem to be true for four blues.
Oh, I found the flaw in my logic. In the case without the guru, every person may be silently thinking "the guru sees someone with blue eyes", but they do not know that everyone else also thinks that. The "knowing that the other guy also knows" is the hidden piece of information in the Guru's speech.
But in the case with lots of blue-eyed people, everyone does know that everyone else knows that there are people with blue eyes. So there is literally no information added when the Guru says her piece.
It doesn't stop the first time. Imagine three blues: I see two blues. I have to assume each of them sees one blue, and therefore I have to assume they have to assume that person doesn't know that there are any blue eyes. The guru's statement tells me that they know that that person would know that they have blue eyes, and therefore when the first night passes I know that they would know that they have blue eyes unless I have blue eyes, so when the second night passes I know that I have blue eyes and I (and they) leave on the third night.
However, this doesn't seem to be true for four blues.
No, it's true. The Blue 1 sees three blues, and assumes each of them sees two blues. Blue 1 must therefore assume that Blue 2 shall in turn assume that Blue 3 will see one blue and will therefore assume that Blue 4 sees no blues.
It's an iterated function of assumptions, sort of. Each scenario (3 blues, 2 blues, etc.) exists as an assumption within the scenario with more blues.
Oh, I found the flaw in my logic. In the case without the guru, every person may be silently thinking "the guru sees someone with blue eyes", but they do not know that everyone else also thinks that. The "knowing that the other guy also knows" is the hidden piece of information in the Guru's speech.
But in the case with lots of blue-eyed people, everyone does know that everyone else knows that there are people with blue eyes. So there is literally no information added when the Guru says her piece.
It doesn't stop the first time. Imagine three blues: I see two blues. I have to assume each of them sees one blue, and therefore I have to assume they have to assume that person doesn't know that there are any blue eyes. The guru's statement tells me that they know that that person would know that they have blue eyes, and therefore when the first night passes I know that they would know that they have blue eyes unless I have blue eyes, so when the second night passes I know that I have blue eyes and I (and they) leave on the third night.
However, this doesn't seem to be true for four blues.
A sees 3 people with blue eyes, and is therefore thinking, "If I have brown eyes, B sees 2 people with blue eyes, and is therefore thinking, "If I have brown eyes, C sees 1 person with blue eyes, and is therefore thinking, "If I have brown eyes, D sees nobody with blue eyes."""
Without the base case in there, no conclusion may be drawn.
A sees 3 people with blue eyes, and is therefore thinking, "If I have brown eyes, B sees 2 people with blue eyes, and is therefore thinking, "If I have brown eyes, C sees 1 person with blue eyes, and is therefore thinking, "If I have brown eyes, D sees nobody with blue eyes."""
Without the base case in there, no conclusion may be drawn.
wouldn't you just count all the other fuckers on the island on day one, realise there's only 99 you can see with blue eyes, figure out that you must also have blue eyes, and then leave?
wouldn't you just count all the other fuckers on the island on day one, realise there's only 99 you can see with blue eyes, figure out that you must also have blue eyes, and then leave?
repeat for every individual.
i do not understaaaaaaand
There ARE 100 people with blue eyes, but nobody is aware of this fact. For all they know, it could be 99 blue and 101 brown.
wouldn't you just count all the other fuckers on the island on day one, realise there's only 99 you can see with blue eyes, figure out that you must also have blue eyes, and then leave?
A sees 3 people with blue eyes, and is therefore thinking, "If I have brown eyes, B sees 2 people with blue eyes, and is therefore thinking, "If I have brown eyes, C sees 1 person with blue eyes, and is therefore thinking, "If I have brown eyes, D sees nobody with blue eyes."""
Without the base case in there, no conclusion may be drawn.
Uh, no. Want to think that through again?
Uh, no. I could elaborate further, if it'd help:
Let's assume the Guru has said her line. Now, the reasoning goes as such:
Day 1:
A sees 3 people with blue eyes, and is therefore thinking, "If I have brown eyes, B sees 2 people with blue eyes, and is therefore thinking, "If I have brown eyes, C sees 1 person with blue eyes, and is therefore thinking, "If I have brown eyes, D sees nobody with blue eyes, and is therefore thinking, "I see nobody with blue eyes, therefore I must have blue eyes and will leave tonight.""""
Day 2:
A sees 3 people with blue eyes, and is therefore thinking, "If I have brown eyes, B sees 2 people with blue eyes, and is therefore thinking, "If I have brown eyes, C sees 1 person with blue eyes, and is therefore thinking, "If I had brown eyes, D would have left yesterday, therefore, I have blue eyes and will leave tonight."""
Day 3:
A sees 3 people with blue eyes, and is therefore thinking, "If I have brown eyes, B sees 2 people with blue eyes, and is therefore thinking, "If I had brown eyes, C would have left yesterday, therefore, I have blue eyes and will leave tonight.""
Day 4:
A sees 3 people with blue eyes, and is therefore thinking, "If I had brown eyes, B would have left yesterday, therefore, I have blue eyes and will leave tonight."
Day 5:
As every person can be arbitrarily chosen to be A, they all follow the exact same logic and therefore all have left.
Posts
I refuse to accept repeating decimals that are not the result of integer divisions.
Solution/Response:
Also, a new puzzle:
a and b are two numbers (integers) in the interval [2,99]. Mr. P and Mr. S are told a*b and a+b respectively. They have the following conversation:
Mr. P: I don't know the two numbers.
Mr. S: I already knew you didn't.
Mr. P: Ah, now I know them!
Mr. S: And now, so do I.
Assuming they aren't lying or incorrect, what are a and b?
Note: Both Mr. P and Mr. S know that the numbers are in the interval [2,99]. That's important, if I recall correctly. Also, this can be done on pencil and paper but some very basic programming helps. If you think you have a method of getting the solution and don't want to program it or do it by hand, just post a description of the method and I'll tell you if it's correct or not.
Think back to the base case: If I'm the only person with blue eyes, why is the statement necessary? If there is one other person with blue eyes, why is it still necessary?
...I should stop this.
Wait a minute. I don't see how any number of people with any combination of brown and blue eye colors is stable in this situation.
Yeah, the Guru thing seems unnecessary and potentially misleading.
No, you need something to start it off. If there was only one person with blue eyes and no one said anything about blue eyes he'd never know and nothing would happen.
If you see one person with blue eyes, you know that there is either one or two people with blue eyes. You believe that the blue monk sees nobody with blue eyes, so he'll never leave.
When the person comes along and says "someone has blue eyes," you 'know' that the monk with blue eyes will see nobody with blue eyes, realize that it must be him, and leave. When he does not leave, you realize that he must see someone with blue eyes, and since you know it is not any of the other monks, it must be you. Iterate.
Every piece of information contained in the statement "I see someone with blue eyes" was already known to the whole population before. How then can this statement, which contains no information, set off the exodus 100 days later? It is equivalent to the guru having said nothing at all.
Clean up then, because I won that one.
I think you misread the bolded. If I'm the only one with blue eyes, how do I know my own eye color without a guru? If I see someone with blue eyes, how can I expect them to figure theirs out?
See, but I knew the fighting wouldn't start until the solution was declared and arguments popped up over the logic of it.
But yes, you had it. They all leave on day 100.
More formally, take P(n) to mean "If there are exactly n people with blue eyes on the island, they will leave on the nth night." P(1) is true, I hope you will agree. Now we want to show that, if P(n) is true then P(n+1) is true for any n.
To see this, consider that there are n+1 blue-eyed people on the island. One of them, let's say Bob, sees that there are n other blue-eyed people on the island, and the rest (besides himself, as he doesn't know his eye color) are not blue-eyed. Thus he knows there are either n or n+1 blue-eyed people on the island. He waits n days, and on the (n+1)th day the n blue-eyed people are still there. Since P(n) is true by assumption, those people would have left if Bob were not blue eyed (making the total n+1), so on the (n+1)th night, Bob and all of the other blue-eyes leave.
Then, since P(1) is true and P(n) implies P(n+1), P(n) is true for any n.
Without the guru you have P(n) implies P(n+1) -- but not P(1).
Can we assume a & b are different from one another?
I'm going to have to think about this one, but I suspect that we can narrow the options a little bit:
we can also throw out any pairs where both numbers are so high or so low as to have the only other possible choices that match the product fall outside of the given range
The guru is the weakness of the problem. In reality, the countdown would start from the first day everyone was on the island.
EDIT: no, wait, it wouldn't. It just seems very weird.
a and b could be the same.
Your reasoning in the spoiler sounds good so far, but you (hint)
Real spoiler:
Also, I didn't mention this in my first post but it's noteworthy: If a and b are distinct, you don't have to know which is which (and, indeed, it's pretty easy to show it's impossible for you or Mr. P or Mr. S to know which is which).
Again, the inductive proof makes it obvious why the guru is necessary.
Consider a situation with 1 brown-eyed person and one blue eyed person (plus the guru). The guru says: I see someone with blue eyes! The blue-eyed fellow must leave, because he sees no one else with blue eyes.
If we start with 2 blues and 1 brown (plus guru), and the guru reads his lines, the situation is different. The blue-eyed fellow now sees someone else with blue eyes, and assumes the guru is talking about him. When he realizes that the other guy has the same idea (on day two), he leaves (along with the other guy).
If we start with 3 blues and 1 brown (plus guru), and the guru reads his lines, the same thing happens. The blue-eyed fellow sees two others with blue eyes, and assumes the guru was talking about each of them in turn and only realizes they are thinking the same thing when they both stay for two nights.
But now let's change it. Say we keep 3 blues and 1 brown, but the Guru forgets what he is about and says nothing. Each blue knows there are two other blues. Each thinks to himself "the guru sees someone with blue eyes", and each assumes he is referring to the other blues. Each guy thinks "those other two blue-eyed guys are the ones he's seeing. Blue #2 will see Blue #3 and assume that he's the blue guy in question, and vice versa. Tomorrow, they will realize that neither has left, and so each must be thinking the same thing, and they will leave that night." Of course, this doesn't happen, because they're all actually seeing two other blues—a fact which becomes apparent on the third day, when all blues leave. No guru intervention required.
Where is the flaw in my reasoning, here?
In order for someone to leave, they must know their own eye color. Seeing someone else with blue eyes tells me nothing about my own eye color. The fact that they don't leave after n days tells me nothing, because they have no more information about their own eye color than I do about mine.
Again, step back to the base case and work the inductive proof. The guru's statement is not purely obvious information. Consider the fact that the guru tells the same thing to everyone, and everyone is aware that the guru has told everyone.
When the guru speaks, I know that they know that someone has blue eyes, and they vice versa.
But in the case with lots of blue-eyed people, everyone does know that everyone else knows that there are people with blue eyes. So there is literally no information added when the Guru says her piece.
I'm having trouble with this one still, it seems.
It doesn't stop the first time. Imagine three blues: I see two blues. I have to assume each of them sees one blue, and therefore I have to assume they have to assume that person doesn't know that there are any blue eyes. The guru's statement tells me that they know that that person would know that they have blue eyes, and therefore when the first night passes I know that they would know that they have blue eyes unless I have blue eyes, so when the second night passes I know that I have blue eyes and I (and they) leave on the third night.
However, this doesn't seem to be true for four blues.
So I guess the proof posted before does correspond directly to reality. The guru is very literally giving us P(1), from which we derive P(n) as we all know from pure logic that P(n+1) is true. It's weird to see the mathematics line up like that.
But it is, because if there were only three blues, everyone would have left on the third night. So on the fourth day, each blue knows there must be four and also knows he must be one of them.
I will cut you.
It's an iterated function of assumptions, sort of. Each scenario (3 blues, 2 blues, etc.) exists as an assumption within the scenario with more blues.
A sees 3 people with blue eyes, and is therefore thinking, "If I have brown eyes, B sees 2 people with blue eyes, and is therefore thinking, "If I have brown eyes, C sees 1 person with blue eyes, and is therefore thinking, "If I have brown eyes, D sees nobody with blue eyes."""
Without the base case in there, no conclusion may be drawn.
Uh, no. Want to think that through again?
repeat for every individual.
i do not understaaaaaaand
There ARE 100 people with blue eyes, but nobody is aware of this fact. For all they know, it could be 99 blue and 101 brown.
never mind.
Uh, no. I could elaborate further, if it'd help:
Let's assume the Guru has said her line. Now, the reasoning goes as such:
Day 1:
A sees 3 people with blue eyes, and is therefore thinking, "If I have brown eyes, B sees 2 people with blue eyes, and is therefore thinking, "If I have brown eyes, C sees 1 person with blue eyes, and is therefore thinking, "If I have brown eyes, D sees nobody with blue eyes, and is therefore thinking, "I see nobody with blue eyes, therefore I must have blue eyes and will leave tonight.""""
Day 2:
A sees 3 people with blue eyes, and is therefore thinking, "If I have brown eyes, B sees 2 people with blue eyes, and is therefore thinking, "If I have brown eyes, C sees 1 person with blue eyes, and is therefore thinking, "If I had brown eyes, D would have left yesterday, therefore, I have blue eyes and will leave tonight."""
Day 3:
A sees 3 people with blue eyes, and is therefore thinking, "If I have brown eyes, B sees 2 people with blue eyes, and is therefore thinking, "If I had brown eyes, C would have left yesterday, therefore, I have blue eyes and will leave tonight.""
Day 4:
A sees 3 people with blue eyes, and is therefore thinking, "If I had brown eyes, B would have left yesterday, therefore, I have blue eyes and will leave tonight."
Day 5:
As every person can be arbitrarily chosen to be A, they all follow the exact same logic and therefore all have left.