For Shiggily and math lovers, Prob1: Beam them up , scotty!

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brown-eyed people don't go first? or blue-eyed people do not go at all?

Actually try the puzzle

What would happen if there were 2 blue eye people and 1 brown eye person?

the brown eyed person would go an the blue eyed people would stay

^ Wrong.

Hint: use the "if.." logic.

How did the brown eye person deduce what colour his/her eyes are? (i agree that your answer is wrong)

In this case I am not sure anyone would go.

If you want the answer I can PM it.

I reckon the best way to go about it is to work out what happens when there are small numbers, and then to generalise it to there being 100 blue eyed people.

Also, it might take them more than one day to deduce the colour of their eyes, depending if anyone gets beamed up the day before or not (that adds new information)

I think there is a wording on the post that is hanging me up.

Ok, so if there were 2 people with blue eyes and 1 with brown eyes... each person would see at least one person with blue eyes. so, the everyone would remain after the first night. So, the people with blue eyes would both still see each other the next day and realize that since they are both still there, they must both have blue eyes. Otherwise, say there was 2 with brown eyes, the one with blue, would see the two brown eyed people and realize he has blue eyes. So in the original example, it would take 99 days for the blue eyed people to realize they all have blue eyes, using that logic. So then its down to the 100 with brown eyes. so, from the point of view of one of them, youd see that everyone has brown eyes. So youd have the option of thinking you have brown eyes like everyone else, or some random other color. Going by the same logic after 99 more days, when all the brown eyed people are still there, each person would realize that means they themselves must have brown eyes as well, so it would take them 100 days. Please tell me im right cuz i totally twisted my brain in a knot...lol

EDIT: That is, 100 days after the blue eyed people. So heres my final verdict...

Day 99 - Blue eyed people vanish
Day 199 - Brown eyed people vanish

Try this one
Three guardians A, B, and C each guard a door, one of which leads to wealth beyond measure. The others lead to certain death. One of them always speaks the truth, one always speaks lies, and the other always gives an answer of random truth. You dont know which guardian is which. Each of the guardians knows which door leads to the treasure. By asking 5 yes or no questions, figure out which door leads to the treasure. Each question can be asked to exactly one guardian. They can understand any language, but will always respond in their own in which their words for yes and no are Cal and Dal. You do not know which word means what.

Here's another version of the same puzzle, if it helps:

There is an island upon which a tribe resides. The tribe consists of 1000 people, with various eye colours. Yet, their religion forbids them to know their own eye color, or even to discuss the topic; thus, each resident can (and does) see the eye colors of all other residents, but has no way of discovering his or her own (there are no reflective surfaces). If a tribesperson does discover his or her own eye color, then their religion compels them to commit ritual suicide at noon the following day in the village square for all to witness. All the tribespeople are highly logical and devout, and they all know that each other is also highly logical and devout (and they all know that they all know that each other is highly logical and devout, and so forth).

[Added, Feb 15: for the purposes of this logic puzzle, "highly logical" means that any conclusion that can logically deduced from the information and observations available to an islander, will automatically be known to that islander.]

Of the 1000 islanders, it turns out that 100 of them have blue eyes and 900 of them have brown eyes, although the islanders are not initially aware of these statistics (each of them can of course only see 999 of the 1000 tribespeople).

One day, a blue-eyed foreigner visits to the island and wins the complete trust of the tribe.

One evening, he addresses the entire tribe to thank them for their hospitality.

However, not knowing the customs, the foreigner makes the mistake of mentioning eye color in his address, remarking “how unusual it is to see another blue-eyed person like myself in this region of the world”.

What effect, if anything, does this faux pas have on the tribe?



kalantir - well done
Except the brown eyed people survive, for the same reason everyone was happily surviving before someone came and opened their mouth

Correct conclusion for the blue eyed people.

But wrong for the Brown people because I said:



So that means the brown eyed people will be always confused. By the construction of the problem all the brown-eyed people know is that their eyes are not blue... could be green or purple... so they are stranded forever. That might be true if the Space ship talks again and says :"at least one of you is brown eyed".

I made the same mistake when I fist solved the problem .

yeah it was the allowing other eye colors into the equation I got screwy on.

So let me explain the solution for anyone who still doesn't get it.
Imagine a situation where 2 people have blue eyes. They each know that at least one in the group has blue eyes. However, on the second day, since no one has disappeared, no one knows their eye color. This means that each blue eyed person sees at least one other blue eyed person (meaning there are 2). Since each blue eyed person only sees one, they know they are the second. Both disappear and the brown eyed people have no way of knowing there eye color. Now if there were 3 blue eyed people, on the third day, still no one is gone. By the logic of the last example with two blue eyed people, if there were only two blue eyed people, they would be gone already, making each blue eyed person realize he is the third and disappear. The same goes for 4 blue eyed people, 5, 6, etc. As far as I can tell, the number of brown eyes/other eye colors is irrelevant.(Someone please correct me if I'm wrong about that and explain.)

the thing I didn't understand was that they didn't know the eyes were blue and brown only. And in the assumption of only 2 choices and the statement that "at least one of you has blue eyes" I think would have caused the people to choose brown eyed over blue eyed, causing the brown eyed people to go first. Then once the brown eyed people went, the blue-eyed people would only see blue eyes and figure out that all the people left are blue eyed.

^^ Since they know that "at least one of you has blue eyes" and since they don't know their own eye color and since each person doesn't know the exact number of brown-eyed and blue-eyed then each person will think:

if there are only 2 blue-eyed person , each one will think:
-"If I am non-blue then that other blue person is the only blue person , then he'll be beamed tomorrow since he will realize that he's the only blue-eyed person."
But in the next day when they see that there's still a blue person then he'll think: "I must have blue eyes" and so both blue-eyed ppl will be beamed at the 2nd midnight.

if there are only 4 blue-eyed person , each one will think:

- "If I am non-blue then that other 3 blue persons are the only blue persons then one them be beamed tomorrow since they will think the same."
But after 4 days when they see that there's still a blue person then he'll think: "I must have blue eyes" and so the 4 blue-eyed ppl will be beamed at the 4th midnight.

-if there are N blue-eyed person: each one will think:

"If I am a non-blue then the other N-1 blue are the only blue person then they'll beamed on night N-1th but on the day N-1 at morning"
but since on day N-1 he'll see that there's still blue person then he'll think "I must have blue eyes too!" then all blue-eyed people are beamed on day Nth.

Answer: all 100 blue-eyed are beamed on 100th midnight.


in a better way:

4 blue eyed persons A B C and D
A will think:
"If i'm NE (not blue), then B will see 2 Es (blue): C and D and will think:
"If i'm NE, then C will see only 1 E and will think:
"If i'm NE, then D won't see any Es and he'll know he's an E, so he'll be beamed up tonight"""

day 2:
But D stays and A will think:
"B will think:
"C will think that "D didn't get beamed up, so i must be an E" and C should get beamed up the second night""

day 3:
C stays and A will think:
"B will think:
" C didn't get beamed up, so i must be an E" and B should get beamed up the third night"

day 4:
B stays and A will think:
"B didn't get beamed up, so i must be an E" and A will be beamed up on the forth night.

But all of them, being excellent thinkers, will go through the same thought process and will be beamed up on the forth night.