A better math problem.

what does this have to do with monty hall?

is this a real problem?

if it is ...interesting... I have no idea what the solution is

There are exactly 20 students in the classroom. 14 have mud on there face, 6 have clean faces. Every student can see every other student at all times. And yes, what does the ringing of the bell have to do with it? That is the question.

Just a few clarifications:

- Are we assuming that no students raised their hands after the previous 13 chimes?
- Also, if any students did raise their hands on the previous chimes, were they always correct?
- Finally, all of the dirty kids raised their hands at the end... did any of the clean kids?

Also, does the fact that it''s winter have something to do with the answer?

quote:

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Original post by GaulerTheGoat
There are exactly 20 students in the classroom. 14 have mud on there face, 6 have clean faces. Every student can see every other student at all times. And yes, what does the ringing of the bell have to do with it? That is the question.

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I would look at my reflection in the bell, but I'll bet that's not the sollution

EDIT: Hmm... the 14th chime...

EDIT2: Meh, sleep is more important thatn this problem...

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-~-The Cow of Darkness-~-

If you see the image I am online

[edited by - cowsarenotevil on May 18, 2003 12:39:05 AM]

[edited by - cowsarenotevil on May 18, 2003 12:45:44 AM]

I''ve heard something very similar to this before. There are two very important premises that you didn''t make entirely clear:

1) The children have perfect reasoning skills.
2) The children all know that the other children have perfect reasoning skills.

Te number of bell rings tells the children how many have dirty faces. If a child sees 13 dirty faces on the 14th bell ring, then the child knows he must have a dirty face. If the child sees 14 dirty faces on the 14th bell ring, then the child is still not sure because there could be more bell rings, and on the 14th bell ring the 14 dirty faced children all only see 13 so all 14 know they have dirty faces and raise their hands.

I think that is what went on, but I dunno, I hope it''s clear.

quote:

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Original post by GaulerTheGoat
Here is a very debatable NEW problem, and if there is a god, Monty Hall will die.

There are 20 kids in an class for gifted children. One winter morning, 14 come to school with mud on their faces, while the other 6 are clean. The teacher says, "We''re going to play a game. Look at each other''s faces. I''ll ring this bell every 10 seconds and if you know that your own face is dirty, raise your hand after the bell." So after awhile and nothing has happened, the teacher says, "I''ll give you a hint. At least one of you has a dirty face." As every child already sees at least 13 dirty mugs, this would seem redundant. However, when the game starts over, all the dirty-faced children raise their hands on the 14th chime.

What''s up?

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Hmm... it''s late so I''m not going to try to solve it, but first thoughts.

- 14 dirty faces, 14 rings of the bell, interesting...
- a dirty face sees 13 dirty faces + know at least one has a dirty face = something about 14, interesting...

I''d start by trying to find a way to eliminate one dirty face per ring of the bell. Induction and the pigeon-hole principle look promising. I''d be interested if the 6 clean faces are significant or not (i.e. 5 clean, 15 dirty, 15 hands belonging to dirty faced children go up on the 15th chime). Same with the 10 seconds (which may just mean they had time to think between each ring). Also, I''m not sure whether the fact that without the hint they couldn''t figure it out is important or just means the hint is important. Of course, it''s not stated whether or not any hands were raised before the 14th chime, but this line of thinking would lead to hands only being raised on the 14th chime. Anyway, I''ll look into it, but it''ll probably be solved when (if) I find the solution.

Yeah, that''s what I thought, but wouldn''t the teacher just keep on ringing it?

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-~-The Cow of Darkness-~-

If you see the image I am online

quote:

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Original post by EAX
Te number of bell rings tells the children how many have dirty faces. If a child sees 13 dirty faces on the 14th bell ring, then the child knows he must have a dirty face. If the child sees 14 dirty faces on the 14th bell ring, then the child is still not sure because there could be more bell rings, and on the 14th bell ring the 14 dirty faced children all only see 13 so all 14 know they have dirty faces and raise their hands.

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That''s the general idea, but a little more detail wouldn''t hurt:

If only one child had a dirty face, that child will obviously raise its hand on the first ring (since no one else has a dirty face, and from the premise that there was at least one dirty face).

If two children have dirty faces, they won''t raise their hands on the first turn -- each will watch the other on the first turn. On the second turn, they both raise their hands, having realised that their faces are dirty.

...and the pattern continues up to 14 children on the 14th turn.

To RabidChild
No student raised their hand on the first 13 chimes and the students how finally did raise their hands are exactly those students that had dirty faces.

To Beer Hunter (First post)
The children are all capable of logical thinking and have confidence that their peers are, too.

To EAX
Then why did they need the hint?

To Beer Hunter (Second Post)
Congrats you were the first person to get it.

The Solution:
Suppose you are some kid in this class and you have a dirty face. Now suppose there are only 2 dirty-faced kids instead of 14 and call the other dirty kid A. Then you would only see 1 other dirty face.

If you DON''T have the hint, then you KNOW that kid A will not raise his hand on the first turn because even if you are clean he doesn''t know that there must be 1 dirty kid. This follows for all other rounds. If you do have the hint, however, then you can reason that kid A will raise his hand on the first ring only if your face is clean. Therefore, if he doesn''t, then you know your face is dirty. He uses the same logic and you both raise your hand on the 2nd ring.

In the case there are 3 dirty kids: you; kid A; and kid B. You see 2 dirty faces. On the first ring, you know that kid A will see only 1 dirty face if your face is clean. This reduces the problem to the previous case so that both kid A and kid B will aise their hands on turn 2. Therefore, if they don''t, then you know your face is dirty and raise your hand on turn 3. Kid A and B have the same reasoning and both raise their hands on turn 3.

Hopefully, you can see how this continues to the case where 14 of the kids have dirty faces.

I know a similar problem which goes like that:

Three explorers are captured in the jungle and each is attached against a tree. They can see each other. They are told the following:

We put each of you a hat onto the head. If you can guess what color your hat has, you are free to go. Otherwise you''ll have to die. But we give you a hint: There are 3 white hats and 2 black hats.

After 5 minutes the 3 explorers get free.

What hat color did they wear?

(Hope I told the story in the right way...)

quote:

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Original post by MrFreeze
What hat color did they wear?

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Three whites.
It''s too early to spoil it completely, so I''ll leave it there.

That problem with the dirty faces was a really tough one and I don''t udnerstand it. They know at least one have a dirty face, but what information does the bell add?

Ok, thanks AP, I think I''m starting to understand this one now. I sort of got stuck by assuming the bell added something to the problem. As I said above, tough one, but clever too.

Wouldn''t these so-called "gifted" kids know if they fell in the mud on the way to school? Why couldn''t all 14 just raise their hands with no bells? .

Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.

This reminds me (sorta) of a game we played in the first meeting of our Coding Theory class in the spring: we were broken up into groups of 4 with 3 people designated as those playing the game and one as the one assigning colored hats, either red or blue, to each player. Each player can look at the others'' hats but not at their own, and then each person can either guess the color of their hat or pass. If at least one person gets it right and nobody guesses wrong, the team wins; if anyone gets it wrong or nobody guesses correctly (ie they all passed) the team loses.

The strategy then is to look at the other two players'' hats. If theirs are the same color you guess opposite, if they are opposite colors then you pass. Assuming everyone follows that strategy you can only lose 25% of the time, if everybody''s hats are the same color.


Heh...forgive me if this has already been covered in another thread or something, I just saw this in the recent posts on the front page.

quote:

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Original post by MrFreeze
I know a similar problem which goes like that:

Three explorers are captured in the jungle and each is attached against a tree. They can see each other. They are told the following:

We put each of you a hat onto the head. If you can guess what color your hat has, you are free to go. Otherwise you''ll have to die. But we give you a hint: There are 3 white hats and 2 black hats.

After 5 minutes the 3 explorers get free.

What hat color did they wear?

(Hope I told the story in the right way...)

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I think I''ve got it figured out, but I''m not completely sure my reasoning is right.

Right off the bat, you can eliminate the possibility that there are two black hats being worn, because if that were the case then the one white hat would see both black hats, and know his color. Because nobody immediately guessed white, that is not an option.

Having established that there is at most one black hat, you can immediately eliminate the option that there is only one black hat. If there were one black hat, then the other two explorers would see one white hat and one black hat, and know that they are wearing white hats. Again, because nobody guesses, you can eliminate one black hat. This only leaves that all three explorers are wearing white hats.

Of course, if there were two black hats, then one of the explorers would guess white immediately, leaving the two black hats to figure it out. But because the one explorer knew he had on a white hat, the other two must be wearing black hats.

If there is only one black hat, then nothing would happen at first for the same reason given initially. But then the two white hats could guess correctly and be let free. Because the two explorers knew they wore white hats, there must have been one black hat visible. And since there is only one explorer left, he will know he is wearing it.

So to answer the question, you need to know how many guessed first. 1 initial guess indicates two black hats and one white hat. 2 initial guesses indicates one black hat and two white hats. 3 initial guesses indicates three white hats. 3-n black hats, and n white hats.

CM