- #1

- 22,089

- 3,296

- Thread starter micromass
- Start date

- #1

- 22,089

- 3,296

- #2

- 402

- 120

2 see's that 3 has a black hat

IF 2 was also wearing a black hat 1 could yell out he has a white hat since he see's both black hats and would know he must have a white hat.

Since 1 doesn't yell. He can assume that 1 see's both a white and black hat in front of him

Hope that's right :) makes sense though

- #3

Samy_A

Science Advisor

Homework Helper

- 1,241

- 510

If after a while man 1 remains silent, man 2 understands that man 1 is seeing hats of different colors on him and on man 3. As man 2 sees the color of man 3's hat, he knows that his hat has the other color, and can shout it.

In the setup of the question, it would thus be man 2 who shouts first.

- #4

Borek

Mentor

- 28,635

- 3,110

Are the hats placed as drawn, or is it just an example?

- #5

- 22,089

- 3,296

As drawn. But you can of course try to find the solution of every permutation of hats!Are the hats placed as drawn, or is it just an example?

- #6

- 117

- 43

2) Given the picture, man 2 will yell first

3) This will indicate that man 1 and 3 have the same color hat, prompting man 1 to yell

4) Man 3 will then yell

5) Man 4 is sh*t out of luck because even if he did yell he is still stuck behind that wall

- #7

Borek

Mentor

- 28,635

- 3,110

I think there are just two solutions.As drawn. But you can of course try to find the solution of every permutation of hats!

- #8

- 489

- 189

There's a minor mistake in there.

2) Given the picture, man 2 will yell first

3) This will indicate that man 1 and 3 have the same color hat, prompting man 1 to yell

4) Man 3 will then yell

5) Man 4 is sh*t out of luck because even if he did yell he is still stuck behind that wall

In fact the only one that can yell (with certainty) is number 3 knowing he has the opposite color of 2.

Let me introduce some shorthand notation for the setup, the n

The answer has been given above, number 1 and 4 have to gamble if they want to go free (I think?)

Case 2: "WWB | B" or "BBW | W"

See case 1 :P

Case 3: "WBB | W" or "BWW | B"

- Person 1 can yell out its color immediately since he sees both hats of the color opposite to his in front of him.

- Person 2 and 3 know their hat has the opposite color

- Person 4 knows he has the same color as person 2

That's about it. Although I recall a different puzzle, those that call out their color correctly go free, those that are wrong or silent get sentenced to death.

How many of you clicked to see the 3 cases?

- #9

- 117

- 43

I disagree with you.There's a minor mistake in there.

In fact the only one that can yell (with certainty) is number 3 knowing he has the opposite color of 2.

Let me introduce some shorthand notation for the setup, the n^{th}letter in the string "BWB | W" indicates the color of person n.

The answer has been given above, number 1 and 4 have to gamble if they want to go free (I think?)

Case 2: "WWB | B" or "BBW | W"

See case 1 :P

Case 3: "WBB | W" or "BWW | B"

- Person 1 can yell out its color immediately since he sees both hats of the color opposite to his in front of him.

- Person 2 and 3 know their hat has the opposite color

- Person 4 knows he has the same color as person 2

That's about it. Although I recall a different puzzle, those that call out their color correctly go free, those that are wrong or silent get sentenced to death.

How many of you clicked to see the 3 cases?

- #10

- 402

- 120

So at this point Guy 1 knows the colour of #3 and #2. He has no additional info than he did at the start....

- #11

- 117

- 43

Oops. I saw it in my head somehow though. Maybe I was thinking of the picture as I said it lol.

- #12

- 4,636

- 1,778

If prisoner number one sees that two and three have hats of different colours, the probability of a correct guess on his part is 0.5, so he will not shout.

Hearing no shout (after a long enough period of time), prisoner number two can conclude that his hat is a different colour than that on prisoner number three. So, once he has not heard anything from number one (assuming that he can count on number one to be timely with an accurate answer if one is available), he will be the first to shout out the colour of his hat - opposite to that of prisoner number three.

As drawn, prisoner number two would be the first to shout.

- #13

- 402

- 120

Number 3 can assume since 1 didn't yell he has a different colour hat than 2, and can yell opposite from whatever 2 had.

1 and 4 are up the creek with out a paddle

- #14

- 489

- 189

Or do they go free upon shouting the colour of their own hat?