Unutterably Silly Quiz

[HomeInMyShoes]

At least there's good company.

[orlok]

....

[pdurrant]

Originally Posted by norway1456

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?

This is a different statement of the prisoners with coloured dots on their foreheads problem.

The interesting thing is that it seems that none of the prisoners have been given new information, as they all already know that there are red hats in the room.

What is more, since they can each see four red hats, they all know that everyone else in the room can see at least three red hats.

What they do not know is that everyone else can also see four red hats, they gain this knowledge with the passage of time, by reasoning this way:

If I were seeing only blue hats, I would be able to leave at 12:00, as I'd know that my hat was red. But I can see some red hats, so I can't leave at 12:00.

If I were seeing three blue hats and one red hat, and my hat was also blue, then the person with the only red hat would leave at 12:00, as they could not see any other red hats, and I could then leave at 13:00. But I can see more than one red hat, so I can't leave at 13:00.

If I were seeing two blue hats and two red hats, and my hat was also blue, then the two people with red hats would leave at 13:00, as each of them would only be able to see one red hat, and I could leave at 14:00. But I can see more than two red hats, so I can't leave at 14:00.

If I were seeing one blue hat and three red hats, and my hat was also blue, then the three people with red hats would would leave at 14:00, as each of them would see two blue hats and two red hats, and I could leave at 15:00. But I can see more than three red hats, so I can't leave at 15:00.

I am seeing four red hats. If my hat was blue, everyone else would see one blue hat and three red hats, and would all leave at 15:00. But they haven't left. So my hat must be red, and we can all leave at 16:00.

[orlok]

Thank you for the clearly articulated explanation, Paul. I am still struggling, however...

"If I were seeing two blue hats and two red hats, and my hat was also blue, then the two people with red hats would leave at 13:00, as each of them would only be able to see one red hat, and I could leave at 14:00. But I can see more than two red hats, so I can't leave at 14:00."

Why could the two leave at 13:00? Even if they could only see one red hat, how would they know theirs was also red? All the person who came in said was that they could see a red hat. This could belong to the person they can see with the red hat, and they still wouldn't know what colour their own was.

[issybird]

Forget about what you see. The problem is solved by eliminating the possibilities, one each hour.

At 12:00, you can eliminate four blue, one red, since the person with the red hat would have left the room.

At 13:00, you eliminate three blue, two red, because two people with red hats would have left the room: each knows he has a red hat, since the other person with a red hat didn't leave an hour ago.

And so forth. By the last hour, the only possibility left is all red hats, and every leaves at once.

[orlok]

Originally Posted by issybird

Forget about what you see. The problem is solved by eliminating the possibilities, one each hour.

At 12:00, you can eliminate four blue, one red, since the person with the red hat would have left the room.

At 13:00, you eliminate three blue, two red, because two people with red hats would have left the room: each knows he has a red hat, since the other person with a red hat didn't leave an hour ago.

And so forth. By the last hour, the only possibility left is all red hats, and every leaves at once.

I had an "Aha!" moment there, thank you.

And then I began to think again, which in my case is a dangerous thing to do...

The elimination theory is all very well, but as they can actually see that everyone else has a red hat right from the start, then to me the logic falls down. The reasoning doesn't seem to work as there is no process of elimination to go through. The first two people would never have left, as there were no blue hats on show for them to start going through the reasoning process.

[Stitchawl]

... only in math problems can there be five people in a room, all wearing hats, and none of them knowing the color of their own hat! Isn't that a little strange? All my hats are the same color. It makes life so much easier, especially if I get stuck in a room with four other people and told I can't leave until I know my own hat color.

Besides knowing where your towel is, it's a good idea to know the color of your hat.


Stitchawl

[Bookpossum]

... and don't forget the packet of salted peanuts either.

[Iznogood]

As Issybird says, this is a process where the prisoners get information over time by the actions of the others in the room; in this case the action is that none of them is able to leave the room. And something has to start that process, and that is the man that comes in at 11.55. This is because the prisoners get information as time passes by, so they need a common reference for time passed. Without this man, none of them could ever leave. I'll explain that part in a while.

Originally Posted by orlok

Thank you for the clearly articulated explanation, Paul. I am still struggling, however...

"If I were seeing two blue hats and two red hats, and my hat was also blue, then the two people with red hats would leave at 13:00, as each of them would only be able to see one red hat, and I could leave at 14:00. But I can see more than two red hats, so I can't leave at 14:00."

Why could the two leave at 13:00? Even if they could only see one red hat, how would they know theirs was also red? All the person who came in said was that they could see a red hat. This could belong to the person they can see with the red hat, and they still wouldn't know what colour their own was.

See point 2) below. I have tried to explain it, but I'm not so sure I have made a good explanation. This problem is as difficult to explain as it is to solve since it involves "person A thinks that B thinks that C thinks etc.", and is further complicated by "since A didn't leave the room at 13.00, that means that A also thinks that..."


1)
If there were only one red hat, the bearer of that hat would only see non-red hats and leave the room at 12.00.

2)
Since nobody leaves at 12.00, all of them knows that there are more than one red hat in the room.
If there were two red hats (on person A and B) and three non-red (on person C, D and E), we would have the following situation:
A can see one red and three non-red
B can see one red and three non-red
C can see two red and two non-red
D can see two red and two non-red
E can see two red and two non-red
Person A knows that there are more than one red hat, but (s)he can only see one. Which means that the hat not visible to person A must be his/her own. B reasons in the same way, and both can leave at 14.00. When A and B leaves at 13.00, this tells the other that they have non-red hats.

3)
Since nobody leaves at 13.00, all of them knows that there are more than two red hat in the room.
If there were three red hats (on person A, B and C) and two non-red (on person D and E), we would have the following situation:
A can see two red and two non-red
B can see two red and two non-red
C can see two red and two non-red
D can see three red and one non-red
E can see three red and one non-red
Person A knows that there are at least three red hats, but (s)he can only see two. Which means that the red hat that A cannot account for must sit on his/her own head. B and C reasons in the same way, and all three can leave at 14.00.

4)
Since nobody leaves at 14.00, all of them knows that there are more than three red hats in the room.
If there were four red hats (on person A, B, C and D) and one non-red (on person E), we would have the following situation:
A can see three red and one non-red
B can see three red and one non-red
C can see three red and one non-red
D can see three red and one non-red
E can see four red and none non-red
Person A knows that there are at least four red hats, but (s)he can only see three. Which means that the red hat that A cannot see sit on his/her own head. B, C and D reasons in the same way, and all four can leave at 15.00.

5)
Since nobody leaves at 15.00, all of them knows that there are more than four red hat in the room. Since they are only five, all knows that they all have red hats and can leave at 16.00.

Originally Posted by orlok

I had an "Aha!" moment there, thank you.

And then I began to think again, which in my case is a dangerous thing to do...

The elimination theory is all very well, but as they can actually see that everyone else has a red hat right from the start, then to me the logic falls down. The reasoning doesn't seem to work as there is no process of elimination to go through. The first two people would never have left, as there were no blue hats on show for them to start going through the reasoning process.

None of them can "see that everyone else has a red hat right from the start". At 11.55, all can see four red hats, and they know that everybody else can see at least three red hats.

The easiest way to solve this problem is not by finding out what your own hat color are, but finding out how many red hats there are in total in the room. When you know that there are five red hats and that there are five people wondering about their hat color, all can leave. The only way to get the information about the total number of hats is by the sequence of deductions I have outlined.

When you have eliminated the possibility of one red hat in the room, and the possibility of two red hats in the room, and three and four red hats, you only have the possibility of five red hats left.

Originally Posted by pdurrant

This is a different statement of the prisoners with coloured dots on their foreheads problem.

The interesting thing is that it seems that none of the prisoners have been given new information, as they all already know that there are red hats in the room.

What is more, since they can each see four red hats, they all know that everyone else in the room can see at least three red hats.

What they do not know is that everyone else can also see four red hats, they gain this knowledge with the passage of time, by reasoning this way:

(...)

All of them knows, as you say, that everybody in that room knows there are at least three red hats. It is tempting to go directly to step 3. However, going directly to step three is impossible. It is the passage of time that makes it possible to solve this problem, and that means they need a common reference, a common time to start their observations and deductions. They also need to go through the entire chain of deduction since it all is based upon the fact that at 12.00, the person wearing the only red hat would have been able to leave the room. Without this fact, step two wouldn't have made any sense, and even less step three.

Or, put in another way, it is only possible to go from step three to step four when you know that there are exactlythree red hat and two non-red. And you don't get this information until 13.00, and you need the information from the man at 11.55 to get there.

[Iznogood]

Originally Posted by Stitchawl

... only in math problems can there be five people in a room, all wearing hats, and none of them knowing the color of their own hat!

I agree. When I took math at high school (or rather, the Norwegian equivalent of high school) we had a problem involving dressing in the morning when the lights are out. The person dressing drew clothes blindly out of a drawer with n socks, m underpants, etc. How many combinations were there? What was the possibility of drawing socks that matched, and what was the possibility of drawing a jacket that matched the trousers.

I suggested that we rather calculated the possibility of a power breach, AND not having any candles in the house AND it was dark outside (this must have been in the two months of winter) GIVEN that it was not a week-end. What would the possibility of these conditions have been...

Originally Posted by Stitchawl

Isn't that a little strange? All my hats are the same color. It makes life so much easier, especially if I get stuck in a room with four other people and told I can't leave until I know my own hat color.

Besides knowing where your towel is, it's a good idea to know the color of your hat.


Stitchawl

Who said it was their own hat? Who knows, they could have been to a party and wearing party hats. The light could have gone out when they put on their party-hats so they couldn't see it. And a lunatic mathematician with guns could have taken all of them hostage and given them the ultimate challenge: leaving the room without knowing the color of your hat is punishable by death... Anything is possible in this world, even not knowing the color of ones hat.

The possibility of that is maybe equal (if not bigger) than having to dress blindly in the dark without having any light sources in the house available to you. Mathematicians are a strange species...

Maybe the next generation teachers will be more realistic? Say, what is the possibility of that

[Stitchawl]

Originally Posted by norway1456

Maybe the next generation teachers will be more realistic? Say, what is the possibility of that

BobbyGee runs six girls outa his stable. Each carries a Nine with a 13 round clip and one up the spout. Now, if two of the girls find tricks that carry Nines and four of the girls turn tricks that carry the bling new Glock 40mm's, what's the odds that in a drive-by three of the....



Stitchawl

[SneakySnake]

I turn my back for a day and miss all the fun. I am stuck on why knowing me knowing you aha... was not ABBA?

And @stichy is BobbyGee in the car?

[dreams]

Originally Posted by Stitchawl

BobbyGee runs six girls outa his stable. Each carries a Nine with a 13 round clip and one up the spout. Now, if two of the girls find tricks that carry Nines and four of the girls turn tricks that carry the bling new Glock 40mm's, what's the odds that in a drive-by three of the....



Stitchawl

Hey, don't stop! I almost had that one figured out.

Let's see... "what's the odds that in a drive-by three of the...."

girls have just jacked AR-M4SF's from the Bulgarian 7-11 owner and lay a line in front of the car before...

Well at this point, I'd have to guess 2:3, but if I take into account the speed of the car and number of speed bumps it would have to be 5:6.

[Stitchawl]

Originally Posted by SneakySnake

And @stichy is BobbyGee in the car?

He is, but he has an AK-47 with the folding stock.

Originally Posted by dreams

Hey, don't stop! I almost had that one figured out.

Let's see... "what's the odds that in a drive-by three of the...."
girls have just jacked AR-M4SF's from the Bulgarian 7-11 owner and lay a line in front of the car before...
Well at this point, I'd have to guess 2:3, but if I take into account the speed of the car and number of speed bumps it would have to be 5:6.

Damn! It was my teacher's question and I still don't understand. What about the elevator operator?!?



Stitchawl

[Bookpossum]

I always thought the final question was about the colour of the fireman's hat - which might get us back into that room again!