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

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A good problem to think about using Mathematical-like techniques .

A starship beams a group of settlers down to an uninhabited planet. 100 of them have exactly two brown eyes, and 100 of them have exactly two blue eyes.

Strangely, though, none knows his own eye color. Right after the starship has beamed the settlers down, it sends them all this message: "At least one of you has blue eyes." The settlers are all excellent thinkers -- if a conclusion can be deduced logically from the available information, each one will do it almost instantly.

Every night at midnight, the star ship passes overhead, quickly performs a remote mind-scan on the settlers, and instantly beams up anyone who has managed to logically deduce their own eye color. The remaining settlers all attend a community breakfast each morning, but they cannot communicate in any way. However, since they can see each other at breakfast, each knows how many of each eye color is present (excluding themeselves, of course). All the settlers know all the facts in this paragraph.

Because the settlers don't know their own eye color, they do not know the number of people having each eye color, nor do they know for certain all the colors that are represented. Any given blue-eyed person will see 100 people with brown eyes and 99 people with blue eyes, but that does not tell him his own eye color. As far as he knows the color totals could be 101 brown and 99 blue... or 100 brown, 99 blue, and he himself could have green eyes!

Question: Who is beamed up, and on what night(s)?

NB: There are no mirrors or reflecting surfaces of any kind. This is not a "trick" question; the answer is based entirely on logical reasoning. It doesn't depend on subtle wording or anyone lying or guessing, and it doesn't involve people doing something silly like creating a sign language or doing genetics.

And, lastly, the answer is not "no one leaves."

(NOTE: Anyone can answer.)


Note: this is not pure math but it requires mathematical logic to solve this. Good luck.

The first night, noone leaves because all the blue eyed people see 99 other blue eyed people, but they don't know for certain that they have blue eyes, so they don't leave. The next day they (the blue eyed ones) all realise that as noone left, they must all have blue eyes, so they all leave. The poor brown eyed people stay behind as there's no way for them to tell they have brown eyes and not purple.

...or I could be totally wrong.

Can you elaborate on this?

Insufficient information. Probably a transcription error occured, and the phrase "At least one of you has blue eyes" is incorrect. It doesn't add any information, since at the first breakfast, everyone will see at least 99 blue-eyed people, at which point they will all know that there is at least one blue eyed person.

Uh... no, I'm wrong, I think. I was thinking that each of them sees 99 people with blue eyes and 100 with brown, but none of them have enough information to conclude they're blue eyed, but the next day they could work that out because they all realised they were in the same position... but that's wrong. Hmm.

(edit) If it is a transcription error, it makes it more fun, in a twisted kind of way.

it would be my assumption that all the brown-eyed people would go and leave all the blue-eyed people(first).

then on the next day all the blue-eyed people would go.

But I may have read it wrong.

this is an actual logic problem.


There are 10 sets of 10 coins. You know how much the coins should weigh. You know all the coins in one set of ten are exactly a hundredth of an ounce off, making the entire set of ten coins a tenth of an ounce off. You also know that all the other coins weight the correct amount. You are allowed to use an extremely accurate digital weighing machine only once.

How do you determine which set of 10 coins is faulty?

How much can I weigh at once with the weighing device? One coin or as many coins at once as I'd like?

I like it better when they're on the island

I'm scared of my posts giving it away

But it won't be on the next day all the blue eyed people go. They don't know the fact that 100 people have blue eyes, 100 people have brown eyes. There could be 100 brown 99 blue and they have red eyes as far as a blue-eyed individual knows.

Label the coins from 0-9. Put that number of coins on the machine. (0 from the 0th stack, 1 from the 1st stack, ... 9 from the 9th stack)

From the amount of the difference, you can tell which stack has the light coins.

Same idea as me. (Maybe I should have posted it instead of asking about the weighing device!)

no set limit.

that is correct.

now... did you look it up or do it on your own?

On my own. But I'd heard problems kind of like that before, so I knew there was going to be a trick to it. I couldn't figure it out at first, but I checked back later, and Hector's question put me in the right frame of mind to figure it out.

I figured it out on my own as well, took a few minutes of brainstorming. My first instinct was to weigh all the coins, and I immediately realised that would give no useful information, but then it occurred to me that since there are ten sets of ten coins I could divide it in a way so that there would be useful information.

Mine is also a logic problem ...

C'mon! think! think!