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This time I added a special rule, please let me know what you think about it!

CHALLENGE: You have to guess which ones are wearing green hats and which ones are wearing black:

SPECIAL RULE: All the prisoners wearing black hats now have a special condition. They see black hats as green, and green ones as black. They are not aware of this condition.

This are the 6 prisoners

DESCRIPTION: Each one can see the hats that are in front of them, and not their own. F is hiding behind a wall, and no one can see them. Pay attention to the order, they give tips one by one:

1- A sees 4 hats and can figure out their hat

2- B sees 2 green hats (they talk after A, but the order is not important here)

3- C sees 2 black hats (again the order is not important here)

4- F, after having listened to their three companions, still can't figure out what hat they're wearing

You don’t know how many green and black hats there are of each color (6 in total, all of them must be wearing one), but they do know how many hats there are of each color.

We already know F is wearing a black hat (F has the special condition, but they can't see anything, so it doesn't matter). You have to guess the other 5 hats! Of course, there is only one possible solution.

Good luck!

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We need to reason about two worlds: reality, and what the prisoners believe is happening.

Based on A's statement,

The prisoners conclude that A and F have the same colored hat. We can also conclude that A's hat is black, because if A's hat was green, A wouldn't have believed they knew their hat color.

Therefore,

Either there are 2 black and 2 green hats among B,C,D,E, and A believes their hat is black, or there are 1 black and 3 green hats in that group, and A sees the opposite, and believes their hat is green. Any other possibility, and A would either realise something is wrong, or not believe they know their hat color.

From B's comment,

Either B has a green hat and there are 2 green hats among C,D,E, or B has a black hat and sees 2 black hats. Only the former is possible based on As comment, so B has a green hat, and A has a black hat.

From C's comment,

C has a black hat, and D and E both have green hats.

Now, something important happens:

B and C's comments are inconsistent with each other, to the prisoners, because the prisoners don't know about the color-changing. Thus, the prisoners realise they've been lied to, but they don't know how they've been lied to.

As a result,

F doesn't know what color their hat is, because they don't know what is truth and what is lies.

In sum, the color sequence is:

A: black, B: green, C: black, D: green, E: green, F: black

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Since they know how many of each color there are, and B nor C statement doesn't tell F what he is wearing, there must be three of each color.

Edited to include F.

Since A knows his hat color, or at least thinks he does, he sees either three black or three green. If he is wearing green, there aren't enough of either color for him to see three. Therefore, he is wearing black, there are three green in front of him, he sees them as black, and thinks he is wearing green.

C also sees two black hats, but there is only one unaccounted for, so C is wearing black, looking at two green, and seeing two black.

This leaves:
A, C, and F wear black while B, D, and E wear green

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  • $\begingroup$ "and neither B nor C tells us he know what he is wearing": Nothing can be concluded from this, because it is never stated the prisoners tell everything useful they know. So your reasoning is flawed here. $\endgroup$ – Retudin Sep 19 at 12:56
  • $\begingroup$ @Retudin OK, maybe we also need to involve F. rot13(Vs gurer jrer bayl gjb bs bar pbybe ung, gur fgngrzragf sebz O naq P gryy S jung pbybe ung ur unf.) $\endgroup$ – David G. Sep 19 at 13:10
  • $\begingroup$ Better, but there still is the "something important happens" part of the other answer you may want to take into consideration. $\endgroup$ – Retudin Sep 19 at 13:44
  • $\begingroup$ Except that it really is NOT relevant. The only people who can realize it are A and B, and they don't speak after C speaks. F doesn't necessarily know who can see what, just that someone knows his color, someone sees two black, someone sees two green, and how many of each color there are. F's knowledge gets my first conclusion. $\endgroup$ – David G. Sep 19 at 13:57
  • $\begingroup$ Agreed, I incorrectly assumed F knew the sitting order. $\endgroup$ – Retudin Sep 19 at 14:07

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