Interesting Logic Problem

[smitchell333]

I have a couple more thoughts on this:

- The doctor's comment and the timing of it is irrelevant unless for some reason everyone was waiting for him to speak some day in hopes of revealing critical information. They should have commenced the process on day 1 and not wait for the doctor to speak

- Stampede Theorem - I still feel that logically one can deduce that in fact everyone or at least some will also consider and try my proposed solution; and if that is the case then the whole system/solution described by TacLuv and Nick the Greek will break down and therefore the only logical solution is to implement my scheme. If even one person implements my scheme and get off the island "early" then the whole other scheme breaks down and doesn't work any longer.

[Franz Klammer]

Ah, all you guys should do some game theory and learn about common knowledge. A similar puzzle already shows up in one of Aumann's papers (Economics Nobel Prize Winner in 2005).

Also, as soon as you google blue eyes island you find this:

It is common to introduce the idea of common knowledge by some variant of the following logic puzzle: On an island, there are k people who have blue eyes, and the rest of the people have green eyes. There is at least one blue-eyed person on the island (k >= 1). If a person ever knows herself to have blue eyes, she must leave the island at dawn the next day. Each person can see every other persons' eye color, there are no mirrors, and there is no discussion of eye color. At some point, an outsider comes to the island and makes the following public announcement, heard and understood by all people on the island: "at least one of you has blue eyes". The problem: Assuming all persons on the island are truthful and completely logical, what is the eventual outcome?

The answer is that, on the kth dawn, all the blue-eyed people will leave the island.

This can be easily seen with an inductive argument. If k = 1, the person will recognize that he or she has blue eyes (by seeing only green eyes in the others) and leave at the first dawn. If k = 2, no one will leave at the first dawn. The blue-eyed people, seeing only one person with blue eyes, and that no one left on the 1st dawn, will leave. So on, it can be reasoned that no one will leave at the first k-1 dawns if and only if there are at least k blue-eyed people. Those with blue eyes, seeing k-1 blue-eyed people among the others and knowing there must be at least k, will reason that they have blue eyes and leave.

What's most interesting about this scenario is that, for k > 1, the outsider is only telling the island citizens what they already know: that there are blue-eyed people among them. However, before this fact is announced, the fact is not common knowledge; it is merely "first-order" knowledge. The notion of common knowledge therefore has a palpable effect. Knowing that everyone knows does make a difference. When the outsider's public announcement (a fact already known to all) becomes common knowledge, the blue-eyed people on this island eventually deduce their status, and leave.

[72Twenty]

If you are stranded on an island and only eat coconuts and drink coconut milk and nothing else, after a few days you will get severe diarrhea and eventually die. I learned that watching Survivor man.

[iceman]

Well, since that's settled, it's time to figure this fucking thing out again:

http://www.andkon.com/arcade/puzzle/rivergame/

[72Twenty]

Well, since that's settled, it's time to figure this fucking thing out again:

http://www.andkon.com/arcade/puzzle/rivergame/

I'm.....so............confused.

[iceman]

I'm.....so............confused.

You gotta get everyone across the river. Dad can't be left alone with the girls, Mom can't be left alone with the boys and the prisoner can't be without the guard on the near side, that's about all I remember right now. Kids can't drive the boat.

Dammit I've solved this a bunch of times, my mind is such a piece of shit today...

[pappag]

- Stampede Theorem - I still feel that logically one can deduce that in fact everyone or at least some will also consider and try my proposed solution; and if that is the case then the whole system/solution described by TacLuv and Nick the Greek will break down and therefore the only logical solution is to implement my scheme. If even one person implements my scheme and get off the island "early" then the whole other scheme breaks down and doesn't work any longer.

Please explain. Again, simplify by using less people to start and work your way up.. It works.

If 1 out of 200 has blue eyes (call him Bob), Bob will leave on the first night because he sees no blue eyes and there is at least one person with blue eyes. If 2 out of 200, Bob and Bill (another blue-eyed person) will leave on the 2nd night because neither of them can be 100% positive until the 2nd night. It is after the 1st night that Bob and Bill see each other and both think simultaneously that if the other guy didn't leave that 1st night, then I MUST HAVE blue eyes.

If 3 out of 200 have blue eyes, then Bob, Bill, and now Brad won't know that they have blue eyes until after the 2nd night. All three people will think, "Hey, if Bob and Bill (or Bob and Brad or Bill and Brad) are still here after the 2nd night, that means they each must see 2 other people with blue eyes. If they only saw 1 other person with blue eyes, they would have both left on the 2nd night. But because it is now the 3rd day, and both are still here with me, that MUST mean that I, too, have blue eyes."

---

Taking this further, let's play the part of a non-blue eyed person and go through the above scenarios once more. If 1 of 200 have blue eyes, every non-blue-eyed person will know with 100% certainty that they do not have blue eyes because they all see the only person that has blue eyes. If 2 of 200 have blue eyes, and using the logic described above, both blue-eyed people will remain until the 2nd night. Of course, a non-blue-eyed person should be unsure of his eye color, but should the 2 people with blue eyes leave on the 2nd night, that can only mean that they each see 1 and ONLY ONE other blue-eyed person. If there were more blue-eyed people than just those 2, then each blue-eyed person would have to wait an additional day for each additional blue-eyed person that he saw. Repeat until you get to 100 of 200 people have blue eyes.

[BeanDip4All]

Please explain. Again, simplify by using less people to start and work your way up.. It works.

Seconded.

Also, I was correct in my prediction that this thread would go to 5 pages.

[uncle crud]

If that's an interesting logic problem, I take it you and your class and its professor don't care much about syntax or meaning in the English language, and have your own secret code to divine the meaning of a sentence that lacks a critical word, as well as a secret meaning for the nonsense word "derivine."

Nice try. Next time, how about something that (1) makes sense, and (2) provides challenge for someone beyond 3d graders.

Wogga wogga.

[berko]

If that's an interesting logic problem, I take it you and your class and its professor don't care much about syntax or meaning in the English language, and have your own secret code to divine the meaning of a sentence that lacks a critical word, as well as a secret meaning for the nonsense word "derivine."

Nice try. Next time, how about something that (1) makes sense, and (2) provides challenge for someone beyond 3d graders.

Wogga wogga.

I take it you couldnt solve it?

[uncle crud]

I take it you couldnt solve it?

HAH! I didn't even try because I got to the confusing sentences and the non-word and wondered how many other errors there are. In any case I don't see what's so hard about it. It's just a math problem. 100 each of each of the 2 eye colors. Round up everyone and count off the first 100 or 99 of a given color. If it's 99 then you know what color your eyes are -- you're the 100th of that color. Same if it's 100, you're the other color.

But maybe I'm missing something.

[pechelman]

has everyone just ignored the other logic probleM?

i gave up on trying to figure out a formula

the hallway lights people thing that is

[AbsolutStoli]

But the class is "Applied Econometrics and Business Forecasting."

if this is the class, i have a feeling that your professor had other motives in mind for giving you this problem. in business, if you are waiting for 99 idiots to make a decision before doing something, nothing ever gets accomplished. just a thought...

[Sirshredalot]

Applied Econometrics and Business Forecasting?

Are you studying with Silberberg et al. at UW, TacomaLuv?

[BeanDip4All]

Applied Econometrics and Business Forecasting?

Are you studying with Silberberg et al. at UW, TacomaLuv?

Any syntax errors were on my behalf as I typed the problem up (he gave it to us on a handout).

Uncle Crud, since this was so elementary for you, perhaps I could PM you to look over some of my econometrics papers? We're heading into midterms and I could probably use the help

[iceman]

I knew you bitches would ignore my river-ferry hijack, it's way too hard for you foos.

I've solved that thing at least ten times!

unfortunately I can't remember how I did it right now

[iceman]

Uncle Crud, since this was so elementary for you...

He doesn't need me to stick up for him, but he said he didn't try to solve it because it was ambiguous and poorly worded and I agree.

Another reason I didn't try to solve it is that I am not so logical.

But the first two reasons still get me and Crud off the hook.

[doublediamond223]

Well, if we want to play with LOGIC problems, how about some symbolic logic?

(> means horseshoe, can't find the font)

1. A≡W
2. ~A v ~W
3. R > A / ~(W v R)


Prove the conclusion using the 18 rules of inference.

[Sir Jongalot]

My answer was, on the 100th day, all 100 blue eyed people will leave the island.

Think about if there was just one person with blue eyes on the island- he would know that they would leave the first night, becuase he knows he is the only person the Dr. was talking about. He would look around, see no one else, and know he should leave. So (THEOREM 1) if there is one blue eyed person, they would leave the first night.

Now consider there are two blue eyed people. On day 1, X thinks to himself, "Suppose that I don't have blue eyes. Since I can see one person (Y) with blue eyes, this must be the one-person case. Thus, Y, seeing nobody else with blue eyes, will leave tonight." (THEOREM 1) But Y doesn't leave that night, so on day 2, X thinks to himself, "Well, Y didn't leave, so Y must seen somebody else with blue eyes. That is, because they are all perfect logicians, Y must not have gone through the reasoning I went through. Thus, I must also have blue eyes." X does the same thing with Y, and they both leave the second night.

Thus, (THEOREM 2): If there are two blue-eyed people on the island, they will each leave the 2nd night.

This continues with the group leaving en mass on the corresponding night of the # of people in their group. When you reach the 99th night and no one has left, all 100 blue eyed people will leave together on the 100th night. Think of it as all the blue eyed people on the island playing a big game of chicken with each other, right up until the 100th night. Good thing they are all "perfect logicians!"

Am I the only Jong who's loins long to inseminate a smart woman like this ASAP?

[iceman]

Lay off the sauce & the river fairy problem gets easier.

Wait, one of them turned gay?

[smitchell333]

Please explain. Again, simplify by using less people to start and work your way up.. It works.

I understand the 100 days answer and its explanation. It makes some sense in a purely mathematical way. I have a couple problems with it:

I. Lets take the logic/answer given and start to play it out from 1 person's perspective (lets assume they are blue eyed). The single person with green eye color or in fact the fact that any one person must consider the possibility that they have a unique eye color in addition to more than one other option messes it up:

On day one (I think of being on the island, although you assert its after the professor makes his irrelevant comment, no matter) this person sees 1 green eye, 99 blue eyes, 100 browns. According to the schema/answer since they see the green eye person there on day 2 and don't see any others it must mean they have green eyes!!!! In fact EVERYONE else thinks they have green eyes if they use your logic - they "know" they have green eyes (assuming that the DR sees their green eyes and "knows" not to leave on the 1st night), according to your logic. Whereas the truth is the Dr has no partner to signal him by staying as well.

In fact each person must also consider the possibility that they, just like the DR is the only person with eyes of their color, and therefore lacking any true signalling partner(s). That is they may not belong to a group that can signal its eye color to each other mathematically.

II. In addition even if you have just two eye colors someone mentioned game theory and in that context I just simply think its not in any way realistic other than as a purely theoretical construct. If these are people and not simply mathematical variables (N+1) simple understanding of human nature will tell you that they will not act this way. If the question were worded as computers that were not aware of their power light color but aware of the other computer's power light color you could count on this behavior. With humans they'll act both in ways that will defeat the mathematical prediction based on how "perfect logician" is interpreted to mean perfect=identical.

What I called Stampede Theorem is that in addition to considering the schema for "knowing" your eye color all the "perfect logicians" must consider others such as that which I constructed. As perfect logicians I think they too would come to the conclusion that there is a possibility they are the only one with their eye color and if they try the schema answer they could end up days or in fact months later without "knowing" their eye color . Everyone will stampede to the ferry so to speak, coming up with the schema that will tell them for certain in the most rapid time what their eye color is.Therefore they will decide the only way to confirm and to know their eye color is by trying to get off the island with my previous answer:

A) each person will count the # of others with blue eyes (ie for those with blue eyes = N). And the # of others with Green, and Brown eyes).

B) once you know these you will logically know that theoretically you have 4 options - you have:

1) Blue eyes, 2) Green eyes, 3) Brown eyes, 4) you have some other eye color and are the only one. There is no way of ever knowing for CERTAIN whether you have any eye color, other than guessing and presenting yourself as knowing you have an eye color and if you get off - you know.

C) Therefore you (assuming you want off the island) logically construct a plan to escape in the shortest possible time:

1) Night one - present yourself as either randomly blue or brown - you have 0 information to suggest either way but you know that statistically brown and blue are best represented and therefore your best guess - so by laws of chance approximately 50 blue and 50 brown will get off.

2) Night two - present yourself as the other option of the blue/brown from what you presented the day before -you have 0 information to suggest either way but you know that statistically brown and blue are best represented and therefore your best guess - so the remaining 50 blue and 50 brown will get off.

3) Night three - Doctor is the only one remaining - he will present himself as randomly chosen color other than blue, brown, green - "I have grey eyes" - the doctor could get off, but only by a lucky guess.

5) Day 4, etc - Dr presents himself as randomly chosen color other than blue, brown, green - "I have grey eyes" - the doctor could get off, but only by a lucky guess - he could be here forever but probably will be off by day 8 or 9 as there are really only a handful of common eye colors to guess.

[pappag]

I understand the 100 days answer and its explanation. It makes some sense in a purely mathematical way. I have a couple problems with it:

I. Lets take the logic/answer given and start to play it out from 1 person's perspective (lets assume they are blue eyed). The single person with green eye color or in fact the fact that any one person must consider the possibility that they have a unique eye color in addition to more than one other option messes it up:

On day one (I think of being on the island, although you assert its after the professor makes his irrelevant comment, no matter) this person sees 1 green eye, 99 blue eyes, 100 browns. According to the schema/answer since they see the green eye person there on day 2 and don't see any others it must mean they have green eyes!!!! In fact EVERYONE else thinks they have green eyes if they use your logic - they "know" they have green eyes (assuming that the DR sees their green eyes and "knows" not to leave on the 1st night), according to your logic. Whereas the truth is the Dr has no partner to signal him by staying as well.

I now understand the importance of the Doctor's statement and value it now and don't regard it as irrelevant and redundant.

Going more on what you say here, this person that you propose to think that he has green eyes on day two will know that he has blue eyes on the 100th day because if he DID have green eyes (or any non-blue-colored-eyes), then the 99 people he sees with blue eyes would have left on the 99th midnight because they each see only 98 others with blue eyes and logic dictates that they can leave with 100% certainty. But should the 99 others with blue eyes remain after the 99th midnight, then each person that sees 99 blue eyes can leave in full confidence. Follow?

II. In addition even if you have just two eye colors someone mentioned game theory and in that context I just simply think its not in any way realistic other than as a purely theoretical construct. If these are people and not simply mathematical variables (N+1) simple understanding of human nature will tell you that they will not act this way. If the question were worded as computers that were not aware of their power light color but aware of the other computer's power light color you could count on this behavior. With humans they'll act both in ways that will defeat the mathematical prediction based on how "perfect logician" is interpreted to mean perfect=identical.

That's more of a philosophical approach rather than a mathematical one and one where you are introducing "what ifs". If these were humans, then pick any scenario provided from the peanut gallery. I think you answered your qualm.

What I called Stampede Theorem is that in addition to considering the schema for "knowing" your eye color all the "perfect logicians" must consider others such as that which I constructed. As perfect logicians I think they too would come to the conclusion that there is a possibility they are the only one with their eye color and if they try the schema answer they could end up days or in fact months later without "knowing" their eye color . Everyone will stampede to the ferry so to speak, coming up with the schema that will tell them for certain in the most rapid time what their eye color is.Therefore they will decide the only way to confirm and to know their eye color is by trying to get off the island with my previous answer:

So...are you saying that everyone goes to the ferry on the 1st night without ever knowing their eye color and just hope that they have blue eyes so that they can leave?

[FlatlandFlyfisher]

Doctor's statement is meaningless. I think it's put there to throw people off and get you to think about something you needn't spend time thinking about.

<snip>

Sigh. This question is so much intellectual masturbation. Better to chase the Mntlion's NSFW links....

[smitchell333]

Going more on what you say here, this person that you propose to think that he has green eyes on day two will know that he has blue eyes on the 100th day because if he DID have green eyes (or any non-blue-colored-eyes), then the 99 people he sees with blue eyes would have left on the 99th midnight because they each see only 98 others with blue eyes and logic dictates that they can leave with 100&#37; certainty. But should the 99 others with blue eyes remain after the 99th midnight, then each person that sees 99 blue eyes can leave in full confidence. Follow?

Well according to the logic you put forth - he (and everyone who can see a green eyed person) MUST "know" he has green eyes on day/night 2. And this does not end up being true. When that happens, then when all 200 non-green eyes show up saying they have green eyes. If this is allowed (that is false "knowing" events), well then my system would have a much higher certainty of getting them off ASAP. I suppose your answer is that they could deduce after day 2 that the doctor was the only one with green eyes and thus was reason they couldn't leave night 2.

But the core of my objection with the whole thing is that we are talking about humans here and the reality is that you cannot count on cipher-like thinking from 99 or 100 people. Its not philosophical - its REALISTIC - this is more the way that PEOPLE act. Philosophical objection would be something more like - there's no real free will here and therefore human spirit will rebel against the ferryman keeping them down and they take over the ferry boat and all get off day 1. I think its realistic to assume that there is a good chance that at least 1 person gets doubts about it and doesn't want to revist the failure/embarrassment that DID happen on night 2 and doesnt show at the ferry on night 100 or finds a way to beat the system - then total collapse.

I agree if you make them computers with no egos, doubts, etc instead of people - simple - yes your logic works as long as the computer doesn't blue screen on the green eyed false "knowing" event.

So...are you saying that everyone goes to the ferry on the 1st night without ever knowing their eye color and just hope that they have blue eyes so that they can leave?

I'm saying that they would deduce that they highest probability of getting off the island is to play my game - and that rather than hoping they are playing the odds - its approximately 100/201 that they have brown eyes, 99/201 they have blue eyes, and 1/201 they have green eyes.

Now if you re-word the question to limit somehow the people showing up without having a mathematical argument for how they "know" they are X eye color OK that would prohibit potentially them from coming. Although I still think I would show up on night one and say "to my calculations I am most likely to have have brown eyes since that appears to be the best odds as there is a 100/201 chance I do."

[uncle crud]

Uncle Crud, since this was so elementary for you, perhaps I could PM you to look over some of my econometrics papers? We're heading into midterms and I could probably use the help

Well, not to thrash an obviously mangled and rotting corpse, but if you're studying "econometrics," I feel sorry for you. First, because you're paying to study bullshit. Second, because based on your question you're obviously taking it rather seriously even though it's patent bullshit. And third, because "econometrics" is a pretentious fucking name.

But if you found the original question not only logically and coherently worded, but also difficult, that would maybe explain your sincere and serious study of "econometrics." Maybe.

Or maybe it's a sad requisite on the way to receiving your degree in some other bogus field like "business" or "economics" or "statistics"? Sorry, but to me the dearth of rigorous intellectual dissection of academic subjects is a laughingstock and a rather embarrassing feature of post-secondary education.

Sorry, but that's how I see it. Maybe you can get out of jail free on the excuse that you're a poor, dumb and rather gullible college student?