Unutterably Silly Quiz

[orlok]

Originally Posted by Stitchawl

It's the 'Baranoski Sequence!!!'



Stitchawl

A seismic guess, but no.

Clue time: try sequencing them correctly and think about time and how it is measured in different parts of the world.

[Iznogood]

If we write the times in the 24 hour format, we get

01.14 -> 13.14
02.15 -> 14.15
03.16 -> 15.16
04.17 -> 16.17
05.18 -> 17.18
06.19 -> 18.19
07.20 -> 19.20
08.21 -> 20.21
09.22 -> 21.22
10.23 -> 22.23
11.24 -> 23.24
12.25 -> 00.25

Every minute is the hour plus one, and the next hour is the same as the last minute (except from the last one, but let's ignore it 24.25 is not the correct way of writing 25 minutes past midnight, but it is possible to understand what is means)

I admit it is a bit farfetched, but I'll take my chances

[Stitchawl]

.....

[orlok]

Originally Posted by norway1456

If we write the times in the 24 hour format, we get

01.14 -> 13.14
02.15 -> 14.15
03.16 -> 15.16
04.17 -> 16.17
05.18 -> 17.18
06.19 -> 18.19
07.20 -> 19.20
08.21 -> 20.21
09.22 -> 21.22
10.23 -> 22.23
11.24 -> 23.24
12.25 -> 00.25

Every minute is the hour plus one, and the next hour is the same as the last minute (except from the last one, but let's ignore it 24.25 is not the correct way of writing 25 minutes past midnight, but it is possible to understand what is means)

I admit it is a bit farfetched, but I'll take my chances

That's pretty much it. It's a numerical progression using the 24 hour clock (I wasn't sure about the 24 thing). Looking at it written out like you have, it would have been better to leave every alternating time out of the sequence, so it flows better. Well I did say it came to me in a dream...

Over to you, norway1456.

[Stitchawl]

Originally Posted by Stitchawl
It's the 'Baranoski Sequence!!!'

A seismic guess, but no.

It was worth a try.

If we write the times in the 24 hour format, we get

01.14 -> 13.14
..............
11.24 -> 23.24
12.25 -> 00.25

I admit it is a bit farfetched, but I'll take my chances

Hmmmm.... this guy is good! ....


Stitchawl

[Iznogood]

Originally Posted by orlok

Well I did say it came to me in a dream...

It doesn't sound like a good dream

I'm not a mathematician, but I have a long education containing some mathematics, and when I dream about numbers it's usually nightmares...

However, since we're on a subject that in the wide sense concerns mathematics, lets continue with that - my question is not about numbers or sequences, but, if there are any mathematicians out there, this task can be solved by mathematical induction. For all non-mathematicians out there, this task can be solved with logical reasoning. This is not a trick. The task has a solution, and it can be found by using logic. I admit that the solution may be a wee bit farfetched, but here goes:

Imagine a room with five people in it, all of them are wearing a red hat. None of the persons know the color of their own hat, but they can see the color of all the other hats. There are no mirrors or reflecting surfaces in the room they can use to look at their hats. They are strictly forbidden to communicate about the subject of hats and color, so asking each other (in any way) about the color of the hats they are wearing is not an option. The persons can only leave the room if the know for certain the color of their own hat. A person can only leave the room at a whole hour (i.e. 12.00, 13.00, 14.00 etc.).

Imagine furter: The time is now 11.55. A person comes in the door, says: "I can see a red hat" and leaves the room. No other person enters the room and the five persons does not receive any more information about their hats. We have to assume that all the persons in the room reasons in the same way, that they observe the same things and draws the same conclusions.

Question:
When can the last of the persons leave the room?

A short summary of the problem:

• None of the five persons know the color of their own hat, but they can see the color of all other hats except their own.
• They know they can only leave the room when they know for certain the color of their own hat. Guessing is not allowed.
• They cannot ask each other in any way or communicate with the outside world to discover the color of their hat. There are no mirrors or other tools available. The only way to find out is by deduction.
• They can only leave at a whole hour (12.00, 13.00, 14.00 etc.)
• At 11.55 there is at least one red hat in the room
• All persons observe the same things, draw the same conclusion and reasons the same way. I.e., if person A thinks of something, he knows that all the other persons knows/thinks exactly the same.

Only we, who are to solve the problem, knows that all the five hats in the room are red. The persons in the room must deduce, from the information given above, the color of their own hat. The problem has a logical solution. When can the last of the persons leave the room?

[HomeInMyShoes]

At 11:55, provided everyone heard what was said, they would all know that at least one hat is red and they would all see four red hats and not know what there’s was. If someone saw all blue hats, they would know their hat is red and leave the room at 12:00. But at 12:00, no one leaves the room. That means they know they all saw at least one red hat. But if we look at two people we see that red=4 for both of them, but neither leaves the room. Which means that I know if no one left the room that that person saw a red hat on mine as well and I can leave the room at 13:00. Of course, everyone else should be able to follow that logic and everyone would leave the room at 13:00.

This is of course all subject to the five persons in the room having any brain. It is probably just a likely to have a room of five dolts that couldn't figure this out and they would all stave to death in the room, or that the world is eaten by a giant space goat prior to them figuring it out.

[Iznogood]

The principle you are using is correct, but none can leave the room at 13.00.

Originally Posted by HomeInMyShoes

But at 12:00, no one leaves the room. That means they know they all saw at least one red hat.

I agree. Good thinking.

Originally Posted by HomeInMyShoes

But if we look at two people we see that red=4 for both of them, but neither leaves the room.

If I understand you correctly, you imply that both these persons know what the other see, i.e. that person A knows that person B sees exactly 4 red hats, and not two, three or four red hats?

None of the persons knows what the others are seeing, they only knows that all the other persons are observing the same tings (the color of the other hats) and draws the same conclusions based on what they see. Person A sees four red hats, but he doesn't know that person B also sees four red hats. For all A knows, person B may see three red hats, and that the hat of person A is in fact blue

Originally Posted by HomeInMyShoes

This is of course all subject to the five persons in the room having any brain. It is probably just a likely to have a room of five dolts that couldn't figure this out and they would all stave to death in the room, or that the world is eaten by a giant space goat prior to them figuring it out.

A very good point. I only said that they all thinks the same, but you are totally correct: they must also be intelligent enough to use what they are thinking. If I was in the same situation, I would not have been able to solve the problem

[issybird]

Is it something along these lines:

If there were four blue hats, then the person with the red hat would be able to leave the room at 12:00.

If there were three blue hats, when no one left the room at 12:00, the two people wearing red hats could leave at 13:00 (because the other person wearing a red hat would have figured out he wasn't in blue when I didn't leave an hour earlier).

If there were only two blue hats, then the three people wearing red hats could leave at 14:00.

If there was only one blue hat (mine), then everyone else could leave the room at 15:00.

So everyone knows s/he's wearing a red hat and they all leave at 16:00.

[Iznogood]

Originally Posted by issybird

So everyone knows s/he's wearing a red hat and they all leave at 16:00.

Indeed they leave at 16.00. That was quickly solved.

At 11.55 everybody knows there is at least one red hat in the room. If it were the only red hat, the person wearing it would leave at 12.00.

Since nobody leaves at 12.00, there must be at least two red hats in the room. If it were only two red hats, the two persons wearing them would leave at 13.00

etc.

Over to you, Issybird

[HomeInMyShoes]

Originally Posted by norway1456

None of the persons knows what the others are seeing, they only knows that all the other persons are observing the same tings (the color of the other hats) and draws the same conclusions based on what they see. Person A sees four red hats, but he doesn't know that person B also sees four red hats. For all A knows, person B may see three red hats, and that the hat of person A is in fact blue

I messed that up. Today I am too stupid to leave the room because I don't believe I can determine my hat colour, no matter how long I wait.

[issybird]

Anyone should feel free to post a quiz. Dang work.

[Iznogood]

Originally Posted by HomeInMyShoes

I messed that up. Today I am too stupid to leave the room because I don't believe I can determine my hat colour, no matter how long I wait.

Are you alone in your room? In that case I can pop in and say that "I see a red hat", if that is of any help.

And there's no such thing as "too stupid". I can't stress that enough with my students, and I can't stress that enough here either. This is a problem I use sometimes teaching students at university level

[HomeInMyShoes]

Originally Posted by issybird

Is it something along these lines:
...
If there were three blue hats, when no one left the room at 12:00, the two people wearing red hats could leave at 13:00 (because the other person wearing a red hat would have figured out he wasn't in blue when I didn't leave an hour earlier).

But what I don't get is that I don't see any red hats, I don't know what the other people see, only that they would see at least three red hats and at most two blue hats since of the people I look at the others have three people in common. If I'm person 1, I see R{2,3,4,5}, person 2 would have to see R{3,4,5} and I can deduce that they would think B?R?{1,2} since I don't actually know what they saw. That makes the 13:00 departure for anyone impossible for my mind to grasp, but I'm a bit dim sometimes. That's why I think I'm trapped in the room.

[orlok]

Originally Posted by HomeInMyShoes

But what I don't get is that I don't see any red hats, I don't know what the other people see, only that they would see at least three red hats and at most two blue hats since of the people I look at the others have three people in common. If I'm person 1, I see R{2,3,4,5}, person 2 would have to see R{3,4,5} and I can deduce that they would think B?R?{1,2} since I don't actually know what they saw. That makes the 13:00 departure for anyone impossible for my mind to grasp, but I'm a bit dim sometimes. That's why I think I'm trapped in the room.

Don't feel bad. I'm one of the other numptys stuck in the room with you.