And for today, a new original hat guessing puzzle!

CHALLENGE: Guess the hats of the 4 prisoners.

SPECIAL RULE: One of the prisoners is lying

picture of the 4 prisoners

DESCRIPTION: Each one can see the hats that are in front of them, and not their own.

They give some tips about what they see (remember one of them is lying):

1- A sees B, C and D, but he says he only sees 1 green hat.

2- B sees C and D, and he assures he sees a green hat on both of them.

3- D, who is looking in the opposite direction, can see A, B and C, and claims to see 2 green hats together.

There are 2 green hats and 2 black hats, so everybody must wear a hat.

Good luck!


A and B have green hats, C and D have black hats.


If A is lying, then B and D must be telling the truth. If B is telling the truth, then C and D must be the two green hats, but then D would be lying because he can't see his own hat. Therefore, A must be telling the truth, and since he and B are looking the same direction, B must be lying. That means D must be a truth teller. So A can only see one green, therefore his own hat must be green (since he can see everybody but his own). Since D is telling the truth and the two greens are next to each other, B must have the second green. That leaves C and D as black.

  • $\begingroup$ well done! nice reasoning $\endgroup$ – Guess Hat Oct 31 '20 at 16:51

Here is a simpler way to identify the liar. The end is like Anthony's answer.

- A and B contradict each other. If B sees 2 green hats, A should too.
- B and D contradict each other. If B sees 2 green hats, D should see only one.
- So B is the liar. The others tell the truth.
- A sees 1 green hat, so he must wear the other one.
- D sees 2 green hats together, that must be A's and B's. C and D wear black hats.


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