Here's a nice one for you:
A deck of 52 cards is randomly shuffled, and 10 of the cards a randomly face up somewhere in the deck. Deal the deck into 2 piles such that each pile has the same number of face up cards. You are blind.

I've never been blind, so forgive me for asking, but is there a way you can feel the difference between the two sides or not at all?

If not, are you allowed to talk to someone who's not blind, or are you alone in a room or something?

Well that leaves me with one trick: after dealing into two piles, hold the piles on their side, so no card can be considered face up (or face down). I hope that's not the answer though.

Hm, how about this: first you divide the deck into two piles of 26 cards. Then you move the top card of pile A to the top of pile B, after which you move the bottom card of pile B to the bottom of pile A. Keep repeating that process. There will eventually be a time when pile A and pile B both have five face-up cards in them, even though you yourself will not know when that is.

Nah. Avistew's is incorrect. Harald's is... well he probably knows it's cheating. It does technically answer the question, but I'm not gonna let him have it. You have to declare when the tasks been achieved, let's say.

I got it! First put 10 cards in pile A and the rest in pile B. Let n be the number of face up cards in pile A. Now flip pile A. The number of face up cards in it becomes 10-n, the same as in pile B.

Right now I'm thinking something about turning some cards, and if they were face down it meant the face up one was in the other deck to it evens it, and if they were face up then the face up one isn't in the other deck and it evens it.
I'm not sure if I'm on to something or if it's just stupid...

Maybe take ten out of the pile and turn them all? Would that work? That's your pile one. the rest is pile two.
Normally since there are 10 cards, there are 10-x in that pile, where x is how many are in the other pile.
When turning all of them you should get the same amount in both.

All the prisoners in a jail are supremely logical (even though they've wound up in jail). The warden speaks to them and says "One of you lot has blue eyes. If, on any night, you know the colour of your own eyes, you can leave. No talking to each other though. That's naughty."
23 nights go by with nobody leaving.
On the 24th night, half the prison population leaves.
How many prisoners are left?

All the prisoners in a jail are supremely logical (even though they've wound up in jail). The warden speaks to them and says "One of you lot has blue eyes. If, on any night, you know the colour of your own eyes, you can leave. No talking to each other though. That's naughty."
23 nights go by with nobody leaving.
On the 24th night, half the prison population leaves.
How many prisoners are left?

Hum. Well I'm assuming you're have blind people in the lot. Otherwise everyone would know the colour of their own eyes, wouldn't they?

The "half the population leaves" makes me lean towards "one person left, one person is left", but on the other hand, that wouldn't have taken 24 nights. Plus why do half of them still not know? So I guess I'm lost on this one.

In assuming they can see, say there are four people. As 1 person has blue eyes, they could see that no one else has blue eyes, and would leave immediately. Therefore you must assume that they are blind, or at least the one with blue eyes is blind.

So, perhaps it has something to do with them all leaving on the 24th night?

If there were only one person with blue eyes, then he would know and leave on the first night.

Since nobody leaves on the first night, there must be at least two people with blue eyes. After the first night, this becomes common knowledge.

Suppose there only two people with blue eyes. Then on the second night, each would see only one other person with blue eyes. Since they know there must be two, they would both leave.

Etcetera. Therefore since they all leave on the 24th night, there are exactly 24 people with blue eyes. (You cannot know your own eye color otherwise; too many options.)

Since these 24 are half the population, there are now 24 prisoners left.

Harald B has solved it Your proof isn't very good, but you have the right answer, so I'm guessing the logic in your head is probably right. Very impressed. I took a long time on that puzzle, thinking it was impossible.

Yeah, I thought about something like "one more person every day" but it doesn't make sense if you assume people are blind. They can't see other people's eyes, therefore they can't base their leaving on that. People would need to see 23 people with blue eyes, notice that people aren't leaving on the 23rd night, and therefore leave on the 24th.
Also, why would he tell them "one of you has blue eyes" if they can already see that?

And finally, since it's been determined that they're not blind, how on Earth did you gather 48 people who don't know what colour their own eyes are? Were they born in prison and are mirrors prohibited?

Harald B has solved it Your proof isn't very good, but you have the right answer, so I'm guessing the logic in your head is probably right. Very impressed. I took a long time on that puzzle, thinking it was impossible.

Damn right the logic is right! I guess I glossed over a few steps, so feel free to ask about any steps you have difficulty following. And I should mention that I have some formal training and have have seen riddles like this before.

but why can they only figure out one more person with blue eyes per day?

Excellent question. That's indeed a big problem with a lot of these type of riddles. There's this unstated assumption that people have one shot at guessing their own color and leaving per night, and have to do it at the same time or something.
It would be better to phrase things in terms of opportunities or question rounds. With this poorly phrasing they might as well all figure it out after 24 hours or 24 minutes.

Yeah, I thought about something like "one more person every day" but it doesn't make sense if you assume people are blind.

Yeah, you kind of have to assume people aren't blind. Note to Telltale: be very careful about the way you phrase things. Run it by a critic or internet panel.

I've remembered a new riddle for you people, presented in a topical version in my next post.

There are a group of Hidden People with different colored hats: all have either a red or a white hat. Assume that everyone can see everyone else's hat but nobody knows the color of his own. Without (directly) informing anyone of their hat color, how can the Hidden People arrange to stand in a line with all the red hatted ones on one end and all the white hatted ones on the other?
I won't take an answer that comes down to playing the prisoner game above. Assume they're not that logically perfect. I also won't accept "they should punch all the white hatted ones" and the like.

There are a group of Hidden People with different colored hats...

Let's try this:

The Hidden People stand in a line (0...n)
Each hidden person starting with 0 walks down the line and cuts in when everyone else in front of him has hats of the same color.

It seems to work, but I don't have a formal proof right now...

Let's say they can all agree that white hats stand on the left and red hats stand on the right. (I'm assuming they can plan things in a group even if they're not allowed to discuss each other's hat color.)

Hidden Person 1 stands first.
If HP2 sees HP1 has a white hat, HP2 stands to the right of HP1, else to the left.
If HP1 sees HP2 has a white hat, HP1 moves to the right of HP2, else HP1 stays put. HP1 and HP2 are now in order.

HP3 comes along. If he sees HP1 and HP2 both have white hats, he stands to the right of both. If HP1 and HP2 both have red hats, he stands to the left of both. If there's a white hat on the left and a red hat on the right, he stands in between. If he stands on the far left or far right, HP1 and HP2 move again depending on what hat color they see HP3 wearing. If he stands in between, the HP on the left switches places with him if he sees that HP3 has a white hat, or the HP on the right switches places if he sees that HP3 has a red hat.

For HPn, he again sees all n-1 HPs standing, and if all have the same color hat, he stands to the left/right of all of them like HP3 did, and they all switch places if necessary. If HPn stands in between HPi and HPi+1, only HPi or HPi+1 need switch places based on the color of HPn's hat.

Then Nelson Tethers barges in and shoots them all, proving they should have stayed hidden instead of playing silly hat color games.

## Comments

I've never been blind, so forgive me for asking, but is there a way you can feel the difference between the two sides or not at all?

If not, are you allowed to talk to someone who's not blind, or are you alone in a room or something?

If the answer is no, I put the cards on their sides or something, so both piles have 0 facing up cards in them.

I'm not sure if I'm on to something or if it's just stupid...

Maybe take ten out of the pile and turn them all? Would that work? That's your pile one. the rest is pile two.

Normally since there are 10 cards, there are 10-x in that pile, where x is how many are in the other pile.

When turning all of them you should get the same amount in both.

Is that how it works?

Edit: Think avistew has it too. The idea comes from turning over 10 or 42 of the cards anyway.

All the prisoners in a jail are supremely logical (even though they've wound up in jail). The warden speaks to them and says "One of you lot has blue eyes. If, on any night, you know the colour of your own eyes, you can leave. No talking to each other though. That's naughty."

23 nights go by with nobody leaving.

On the 24th night, half the prison population leaves.

How many prisoners are left?

Hum. Well I'm assuming you're have blind people in the lot. Otherwise everyone would know the colour of their own eyes, wouldn't they?

The "half the population leaves" makes me lean towards "one person left, one person is left", but on the other hand, that wouldn't have taken 24 nights. Plus why do half of them still not know? So I guess I'm lost on this one.

So, perhaps it has something to do with them all leaving on the 24th night?

Also, why would he tell them "one of you has blue eyes" if they can already see that?

And finally, since it's been determined that they're not blind, how on Earth did you gather 48 people who don't know what colour their own eyes are? Were they born in prison and are mirrors prohibited?

Excellent question. That's indeed a big problem with a lot of these type of riddles. There's this unstated assumption that people have one shot at guessing their own color and leaving per night, and have to do it at the same time or something.

It would be better to phrase things in terms of opportunities or question rounds. With this poorly phrasing they might as well all figure it out after 24 hours or 24 minutes.

Yeah, you kind of have to assume people aren't blind. Note to Telltale: be very careful about the way you phrase things. Run it by a critic or internet panel.

I've remembered a new riddle for you people, presented in a topical version in my next post.

I won't take an answer that comes down to playing the prisoner game above. Assume they're not that logically perfect. I also won't accept "they should punch all the white hatted ones" and the like.

RRRRRRRRRRRRRRRRRRRRRRRRWWWWWWWWWWWWWWWW

or like this:

WWWWWWWWWWWWWWWWRRRRRRRRRRRRRRRRRRRRRRRR

(obviously the actual amount of Whites and Reds could be different). Does that help?

Let's try this:

The Hidden People stand in a line (0...n)

Each hidden person starting with 0 walks down the line and cuts in when everyone else in front of him has hats of the same color.

It seems to work, but I don't have a formal proof right now...

Hidden Person 1 stands first.

If HP2 sees HP1 has a white hat, HP2 stands to the right of HP1, else to the left.

If HP1 sees HP2 has a white hat, HP1 moves to the right of HP2, else HP1 stays put. HP1 and HP2 are now in order.

HP3 comes along. If he sees HP1 and HP2 both have white hats, he stands to the right of both. If HP1 and HP2 both have red hats, he stands to the left of both. If there's a white hat on the left and a red hat on the right, he stands in between. If he stands on the far left or far right, HP1 and HP2 move again depending on what hat color they see HP3 wearing. If he stands in between, the HP on the left switches places with him if he sees that HP3 has a white hat, or the HP on the right switches places if he sees that HP3 has a red hat.

For HPn, he again sees all n-1 HPs standing, and if all have the same color hat, he stands to the left/right of all of them like HP3 did, and they all switch places if necessary. If HPn stands in between HPi and HPi+1, only HPi or HPi+1 need switch places based on the color of HPn's hat.

Then Nelson Tethers barges in and shoots them all, proving they should have stayed hidden instead of playing silly hat color games.