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.
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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.
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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.
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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.
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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.
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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:
(...)
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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.