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[DStu]

The Axiom of Chioce is trivial for all countable sets, but I don't see how that matters?

I just meant you don't need to assume AC as the part of it that you (probably) need, you get for free...

[Grujah]

They all know in advance.  Or any other answer that prevents them passing information that way: for example, they could each write down their guesses on a piece of paper that nobody else can see.

my plan is that they somehow, make black hats "ones" and white hats "zeroes", and than all should after X seconds (X being the number that they binary represent) somehow led by the last person in line. But there is no way for him to lead it that way.

[shMerker]

Does everyone know how far they are from the back of the line?

[qmech]

Does everyone know how far they are from the back of the line?

There are four possible variations:

1) The line has a start point
2) The line is doubly-infinite

a) The positions are labelled, and each person knows the label of their position
b) The positions are unlabelled

Each statement of the problem has an answer: starting with 1a may well make you more comfortable (it did me).

[pacovf]

I guess you cannot give a clue to those of us that are clueless...?

I cannot fathom how:
-Infinite people discuss a strategy;
-Devil puts hats on people according to the strategy they have agreed upon;
-Without any means to communicate (since they have to shout simultaneously, which negates the possibility that waiting "t" seconds before shouting means anything) after the hats are on, they have to guess the colour of their hat;

can somehow lead to a finite amount of errors. Afterall, the impossibility to communicate once the hats are on means that they cannot react to the devil's strategy, while the devil can react to their strategy, which would mean that their best shot at surviving would be to guess randomly...

[Kuildeous]

The hats are put on them in the order Black/White/Black|White. Who is able to free them all (none of them, one, or multiple people; if more than 0, specify which)?

I'm a little confused by this. If the first person who says something is correct, then they all go free, but if he is incorrect, then they are all executed. So I'm not sure what it means that "multiple people" is an option here….unless it's for the sake of completeness.

I had misread this puzzle at first and thought that everyone had to get their hat color correct. I was able to figure out how #2 and #3 could get their colors, but I was at a loss for the two outside people. But if the outcome hinges on the first guy to say something, then I agree that person #2 can set them free.

In fact, I think it only takes one guy to free them all no matter how the hats are distributed. It just depends on how they're arranged to determine if the savior is person #1 or person #2.

[Ozle]

For the infinite people in a line question (using Red or Blue Hats)

For the infinite people in a line puzzle, you can save all but the first one can't you?

The first person shouts out BLUE if he can see an odd number of red hats, and RED if he can see an even number.

The second person can then count how many red hats he can see, if it is an odd number and the person has shouted BLUE, then he knows he is wearing a RED hat.

The next person can then count the Red hats, he knows there is an odd or even number depending on what the two people before him shouted, so therefore knows what he is wearing.

This then follows on down the line for ever?

I am unsure what the infinity adds to the puzzle, makes it needlessly complicated.

[Drab Emordnilap]

In the Devil's Infinity hats, don't they ALL have to guess at the same time?

[DStu]

@Ozle. They all shout out at the same time, so that doesn't work.

I am unsure what the infinity adds to the puzzle, makes it needlessly complicated.[/spoiler]

Even if they wouldn't shout out at the same time, they will in almost all cases (and certainly in the worst case for your strategy) see infinite Blues and Reds, which is neither odd nor even, so infinity adds that this strategy does not work.

[Grujah]

For the infinite people in a line question (using Red or Blue Hats)

For the infinite people in a line puzzle, you can save all but the first one can't you?

The first person shouts out BLUE if he can see an odd number of red hats, and RED if he can see an even number.

The second person can then count how many red hats he can see, if it is an odd number and the person has shouted BLUE, then he knows he is wearing a RED hat.

The next person can then count the Red hats, he knows there is an odd or even number depending on what the two people before him shouted, so therefore knows what he is wearing.

This then follows on down the line for ever?

I am unsure what the infinity adds to the puzzle, makes it needlessly complicated.

Yeah, this is the solution I posted for last similar riddle, but here they need to guess all at the same time.

[Ozle]

Oh, bugger.

Back to the drawing board then

[qmech]

Yes, simultaneous guessing, and everyone is looking in the direction that lets them see infinitely many others.

Here's another way to think about it for those worried about the mechanics of communication.  A strategy is just a huge lookup table for each person, mapping the infinite pattern of hats they see to a guess of red or blue.  And the strategy is assigned to each person by you, the God of the Matrix, who has absolute freedom, unconstrained by time or space.

[Grujah]

Well,

there is no true random so one could see a pattern over time?

[DStu]

Well,

there is no true random so one could see a pattern over time?

It's not even stated that it's random...

edit: Beside this, of course there exists true random.
edit2 scnr

[Grujah]

Well,

there is no true random so one could see a pattern over time?

It's not even stated that it's random...

If it is not random, than it has a pattern, than it is easy to solve. \o/.

[mith]

I desperately want to give a hint on that one in light of one of DStu's posts...

In the meantime, here's my favourite hat puzzle:

There are four people. In five minutes, each will be given a hat which is black or white, each with probability 1/2, with each chance independent. Each may submit a guess if he chooses. If at least one person guesses correctly, and nobody guesses incorrectly, they will be released. Otherwise, all will be killed.

What is their optimal strategy?

And my second favourite:

10 people are captured by some psychopath, who offers them a chance of survival based the color of their hat, as is usual with psychopaths. All 10 of them get to confer, with the bad guy listening. After they're done talking or five minutes elapse (this psychopath is not big on loopholes) he magically makes either a white hat or a black hat appear on the head of each of them, with all hats appearing at the same time. Then they each write "black" or "white" on a piece of paper, give him the pieces of paper (all this without them talking to each other or seeing what other people wrote), he kills all those whose hat color doesn't match their and lets the rest go.

Your job is to find the strategy that ensures the most people survive in the worst case, and prove that this strategy is indeed the best (for the worst case).

[DStu]

If it is not random, than it has a pattern, than it is easy to solve. \o/.

That's not true, it could be the binary of an unknown irrational number for example...

[Grujah]

I desperately want to give a hint on that one in light of one of DStu's posts...

In the meantime, here's my favourite hat puzzle:

There are four people. In five minutes, each will be given a hat which is black or white, each with probability 1/2, with each chance independent. Each may submit a guess if he chooses. If at least one person guesses correctly, and nobody guesses incorrectly, they will be released. Otherwise, all will be killed.

Can they actually do better than 15/16 ?

They can all see eachother, right?

10 people are captured by some psychopath, who offers them a chance of survival based the color of their hat, as is usual with psychopaths. All 10 of them get to confer, with the bad guy listening. After they're done talking or five minutes elapse (this psychopath is not big on loopholes) he magically makes either a white hat or a black hat appear on the head of each of them, with all hats appearing at the same time. Then they each write "black" or "white" on a piece of paper, give him the pieces of paper (all this without them talking to each other or seeing what other people wrote), he kills all those whose hat color doesn't match their and lets the rest go.

Your job is to find the strategy that ensures the most people survive in the worst case, and prove that this strategy is indeed the best (for the worst case).

Switch papers?

Seriously, what are they able to do after hats are given? I guess they can't talk otherwise it is trivial. Can they move? Can they "form pairs"? Can they see each other?

[DStu]

I desperately want to give a hint on that one in light of one of DStu's posts...

ok, we have some nitpicking with axiom of choice, that can't be it. Don't see anyway what that can have to do with it.
The non-euklidian geometry was a joke but of course would work.

so my best guess is "infinity is neither odd nor even".  But even if we could get this principle to infinty, you can pass no info, that doesn't help.
What else did I say?  The post with random? You could argue that the sequence set up by the devil must have some finite complexity (but why should that be true), and that in this sense there is some finiteness in the series, but even if you would limit the devil to finiteness you yourself would need at least as much capacity as the devil to find the regularity, which I don't think is reasonable to assume...

[jonts26]

I desperately want to give a hint on that one in light of one of DStu's posts...

In the meantime, here's my favourite hat puzzle:

There are four people. In five minutes, each will be given a hat which is black or white, each with probability 1/2, with each chance independent. Each may submit a guess if he chooses. If at least one person guesses correctly, and nobody guesses incorrectly, they will be released. Otherwise, all will be killed.

What is their optimal strategy?

Can they see each other but not their own hats? Do they know if someone has already made a guess? I'm assuming no communication. I need constraints.

[Galzria]

What if they agree to all yell "Black" if anywhere in front of them they see even a single black hat, otherwise they yell "White"?

This would mean that any person wearing a Black Hat will guess Black if any person in front of him has a Black hat on. And if nobody does, then every person in front of him will guess correctly (even though he is wrong). This guarantees some number X of infinity will be correct.

[DStu]

But you nevertheless have infinitely many people who guess wrong.

[Tables]

What if they agree to all yell "Black" if anywhere in front of them they see even a single black hat, otherwise they yell "White"?

This would mean that any person wearing a Black Hat will guess Black if any person in front of him has a Black hat on. And if nobody does, then every person in front of him will guess correctly (even though he is wrong). This guarantees some number X of infinity will be correct.

The difficulty isn't getting an infinite number of correct solutions. It's getting a finite number of incorrect solutions.

[jonts26]

Here's my favourite hat problem.  It might not be quite right for this thread, but I'll throw it out there anyway.

A countable infinity of people are all standing in a line and wearing a black or white hat.  Everybody can see the colours of the hats of the (infinitely many) people standing in front of them.  They must all simultaneously shout out a guess as to the colour of their own hat.  Can they ensure that only finitely many people guess wrong?

Everyone is standing on some sort of non-euclidean plane such that each can see himself from behind and thus just says what color hat he's wearing. 100% success rate.

[mith]

I desperately want to give a hint on that one in light of one of DStu's posts...

In the meantime, here's my favourite hat puzzle:

There are four people. In five minutes, each will be given a hat which is black or white, each with probability 1/2, with each chance independent. Each may submit a guess if he chooses. If at least one person guesses correctly, and nobody guesses incorrectly, they will be released. Otherwise, all will be killed.

What is their optimal strategy?

Can they see each other but not their own hats? Do they know if someone has already made a guess? I'm assuming no communication. I need constraints.

10 people are captured by some psychopath, who offers them a chance of survival based the color of their hat, as is usual with psychopaths. All 10 of them get to confer, with the bad guy listening. After they're done talking or five minutes elapse (this psychopath is not big on loopholes) he magically makes either a white hat or a black hat appear on the head of each of them, with all hats appearing at the same time. Then they each write "black" or "white" on a piece of paper, give him the pieces of paper (all this without them talking to each other or seeing what other people wrote), he kills all those whose hat color doesn't match their and lets the rest go.

Your job is to find the strategy that ensures the most people survive in the worst case, and prove that this strategy is indeed the best (for the worst case).

Switch papers?

Seriously, what are they able to do after hats are given? I guess they can't talk otherwise it is trivial. Can they move? Can they "form pairs"? Can they see each other?

For both puzzles: They are given the opportunity to see each others hats (but not their own), and then are taken away to isolated cells. No communication once the hats are given out.