My brother just created this puzzle for me for Halloween, at first I thought it looked simple, but then it turned out to be very challenging (at least for me!) I'll add it to my collection/app soon. I hope you like it:

CHALLENGE: Guess the hats of the prisoners.

SPECIAL RULE: One of them is short-sighted, which means he can only see one hat in front of him. He is aware he is short-sighted, but nobody else knows who this person is. Everyone does know there is one short-sighted amongst them.

enter image description here

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

  • A sees B,C and D; B sees C and D; C sees A and B; D sees A, B and C
  • This last sentence is not true for the short-sighted one, which can only see the first one in front of him

They give tips one by one, pay attention to the order in which they speak:

1- A is the first one to talk. He sees more green hats than black ones. (This would mean he sees 2 green hats and 1 black hat, or only one green hat if A were the short-sighted one, which you will have to figure out)

2- Then D talks: he sees at least 1 black hat.

3- C speaks third and says that, from what he can see and what he has heard, B doesn't know the hat he is wearing

4- Lastly, B says that only after listening to C (and not before), he can then figure out which hat he is wearing.

There are 2 green hats and 2 black hats, and all of them know this.

They know who talks, where they are sitting, and what direction they're facing.

Good luck!

  • 1
    $\begingroup$ Upvote for the cool graphic. $\endgroup$
    – PDT
    Nov 1, 2020 at 1:23
  • 1
    $\begingroup$ Sorry I presume they know how many hats there are of each colour right? $\endgroup$
    – PDT
    Nov 1, 2020 at 1:35
  • $\begingroup$ Yes, I'll make sure to mention that, thanks! $\endgroup$
    – Guess Hat
    Nov 1, 2020 at 2:18
  • 1
    $\begingroup$ Also do they know where people are sitting in relation to one another? $\endgroup$
    – PDT
    Nov 1, 2020 at 2:32
  • 1
    $\begingroup$ @msh210 There's no way he can see 2 greens only. He can either see 3 people (if he's not short sighted) or 1 person (if he's short sighted).There is no possible situation where he can only see two people $\endgroup$ Nov 1, 2020 at 13:52

3 Answers 3


New answer:

The arrangement is BBGG, b is short sighted


If B is short sighted he knows that the rest are not. So from A's statement he would know that A is black because there are only 3 possibilities BBGG, BGGB or BGBG, all of which makes A Black.

All he sees is Green in front of him because if he saw black he would know that he was Green, seeing only Green his colour is still equivocal. So C must be Green.

D's statement does not make it any clearer for B because it does not disclose D's colour it fits both BBGG and BGGB.

If the arrangement was BGGB then C would not know if it was BGBG or BGGB or if A or B or D was shortsighted. If A was short sighted then B might know his hat once A said his statement because if A was such and the arrangement happened to be BGBG then b would have known he was green from A's statement alone for if seeing BG in front of him means that regardless of whether A is short sighted or not he must be green.

If C sees BBGG he knows that A cannot be short sighted and nor could D because D sees green in front of him and he would have realised that B is short sighted, saw him as Green and so for sure did not know the arrangements of hats.

So after C says his statement b knows that BBGG is the only possibility left and so he knows that he must be Black.

  • 2
    $\begingroup$ Is there any way you can make the explanation better formatted? As of right now I see a giant block of text and my eyes glaze over. $\endgroup$
    – bobble
    Nov 1, 2020 at 4:08
  • $\begingroup$ Sure I will try to break it up! $\endgroup$
    – PDT
    Nov 1, 2020 at 6:36
  • $\begingroup$ @Prince Deepthinker .well done again Prince, I challenge you to the next one.. work in progress $\endgroup$
    – Guess Hat
    Nov 3, 2020 at 0:45

I confused myself a bit on this one, but I think I have it.

The solution is:

Black, Black, Green, Green, and B is short sighted


Starting with A: If A was shortsighted, then B must be green as A would see 1 green and nothing else. If A is not short sighted, then A must have a black hat, since he would be able to see 2 green hats and 1 black, so the remaining hat (his) would have to be black.

D goes next. If D was short sighted, then the black hat he sees would be C's. Also, that would tell us A wasn't short sighted, so A would be Black, B and D would be green. If he isn't short sighted, this doesn't tell us anything yet.

C is next, and based on what he sees and what A and D said, he can claim that B doesn't know his hat. Here's where things got a little confusing for me. First I determined B could not have green, because if his hat was green and he was not short sighted, he'd either see 1 green and 1 black, or 2 blacks. If he saw 2 blacks, he would know his hat was green. If he saw 1 and 1, then he'd know A would have had to have seen his green hat to see more greens then blacks. If B was short sighted, then he'd either see a green or a black hat on C. If he saw black, he'd know he was green (because of A), if he saw green he wouldn't know, so if he was green, and was short sighted, then C would have to be green. But C doesn't know if B is short sighted or not, so he couldn't be confident that this one situation would be true and B wouldn't know his hat.

So we know B must have a black hat, and A saw more green hats than black hats, so A must not be short sighted, and must have seen C and D with green hats. C can see A and B with black hats, so he knows that neither A or D can be short sighted, since A could see greens and D could see blacks. Since he knows he isn't short sighted, he can conclude B must be short sighted, and can only see a single green hat.

  • 2
    $\begingroup$ Lol I beat you to it $\endgroup$
    – PDT
    Nov 1, 2020 at 4:02

B is shortsighted because otherwise B knows one of the following:
- B sees GG and wears black
- B sees BB and wears green
- B sees GB or BG and wears green because A sees more green than black

C must know at least this to make his statement
He can only know A is not shortsighted if B wears black
He can only know D is not shortsighted if C wears green, to be sure of that D must also wear green (which C only knows if A wears black).

So the solution is BBGG, and Cs statement is correct since C knows B is shortsighted and cannot exclude BGGB (before C speaks).
Note: after C speaks BGGB can be excluded because C cannot differentiate with BGBG, so Ds statement is true too.


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