# A hat puzzle

#7

Hmm. I may be incorrect, but it seems to me…

[spoiler]… that there is only one circumstance in which the third logician may answer with certainty, and that is if both of the hats of the first two logicians are blue. Both blue hats must be removed from the pool for her to be sure that hers is red.

I make the assumption that the logicians have drawn their hats randomly from the pool. Therefore, the probability that the first logician will select a blue hat is 2/5 (two blue hats out of five chances). That leaves one blue hat out of four (1/4). Thus, 2/5 x 1/4 = 2/20, which reduces to 1/10. So I arrive at my conclusion that the condition for a correct answer (in which both blue hats have been selected first) is 1/10[/spoiler].

#8

What color the first two hats are also matter as otherwise either the third logician couldn’t have figured out her hat color, or one of the first two could have figured out theirs.

#9

Yes, I agree.

#10

Disagree.

There are four scenarios that all fit the answers described in the question above:

1:R 2:R 3:R
1: I see two red hats, which leaves one red and two blue that mine can be. "I don’t know."
2: I see two red hats, which leaves one red and two blue that mine can be. "I don’t know."
3. If my hat were blue, 2 would know her own hat colour. “Red.”

1:B 2:R 3:R
1: I see two red hats, which leaves one red and two blue that mine can be. "I don’t know."
2: I see one red and one blue hat, which leaves two red and one blue that mine can be. "I don’t know."
3. If my hat were blue, 2 would know her own hat colour. “Red.”

1:R 2:B 3:R
1: I see one red and one blue hat, which leaves two red and one blue that mine can be. "I don’t know."
2: I see two red hats, which leaves one red and two blue that mine can be. "I don’t know."
3. If my hat were blue, 2 would know her own hat colour. So would 1. “Red.”

1:B 2:B 3:R
1: I see one red and one blue hat, which leaves two red and one blue that mine can be. "I don’t know."
2: I see one red and one blue hat, which leaves two red and one blue that mine can be. "I don’t know."
3. If my hat were blue, 2 would know her own hat colour. Plus, I see two blue hats, and there are only two blue hats. “Red.”

Every scenario with the third logician wearing a red hat ends with the same deduction by person #3, and there’s a 60% chance of that happening.

#11

Oh I know this one!

The bear was white - they’re at the South Pole!

#12

So close! (North Pole)

#13

Spoilers!

#14

Brilliant proofs.

I can’t make the first scenario work, though. Why would Logician 2 know her own hat color if Logician 1 wore a red hat and Logician 3 wore a blue hat? There would still be both red and blue hats left in the pool.

#15

Because if 2 is wearing a blue hat, 1 would know her own hat colour, as both other hats would be blue.

So, if 3 is wearing a blue hat, and 1 doesn’t know her own hat colour, then 2 knows that 2’s hat is not blue, and is therefore red.

ETA: Here are the other three scenarios (there are only 7, as 1:B 2:B 3:B is not a valid scenario, there being only two blue hats):

1:R 2:B 3:B
1: I see two blue hats. Mine must be red. “Yes.” (Red)
2: There’s only one way that 1 can know what her own color is. “Yes.” (Blue)
3. There’s only one way that 1 can know what her own color is. “Yes.” (Blue)

1:R 2:R 3:B
1: I see one red and one blue hat, which leaves two red and one blue that mine can be. "I don’t know."
2: 3’s hat is blue. If my hat is blue, 1 would know her own hat colour. “Yes.” (Red).
3. 2 knows her own hat colour. She wouldn’t be able to figure it out if my hat was red. “Yes.” (Blue).

1:B 2:R 3:B
1: I see one red and one blue hat, which leaves two red and one blue that mine can be. "I don’t know."
2: I see two blue hats. Mine must be red. “Yes.” (Red).
3. 2 knows her own hat colour. She wouldn’t be able to figure it out if my hat was red. “Yes.” (Blue).

#16

The game is afoot!

#17

So Logician 3 knows her color because the other two Logicians couldn’t answer? Wow, that takes my brain for a walk.

Thank you!

#18

You’re welcome!

#19

What if they were shitty logicians?

#20

You’ve convinced me.

#21

No. The game is about hats. Not feet. Pay attention.

#22

There are seven possible combinations

However, if the hats were [r b b], 1 would know her hat instantly, and if the hats were [b r b], 2 would know her hat instantly. That leaves five possible combinations. Also, this means that if she sees one blue and one red, she knows her hat must be red. And, obviously, if she sees two blue hats she knows her hat is red.

There are five possible combinations, and three of them work out to 3’s hat being red. The two tricky ones left over are [r r r] and [r r b].

However, if the hats were [r r b], 2 would see 3’s blue hat and know her own hat can’t possibly be blue, or else 1 would answer red. [R r b] gets thrown out.

In all the cases where the first two answers are don’t know, 3’s hat must be red.

#23

I’m trying to sort this out as I got this wrong. In your scenario [ r r b], wouldn’t 1 and 2 be in the same situation? 1 would see a red and blue hat, and 2 would see a red and blue hat. From 2’s perspective 1 would only know her hat color if she saw two blue hats. 2 now knows that 1… oh… just got it. Thanks!

#24

These iterated theory-of-mind puzzles make for an excellent children’s book:

#26

This is easier to understand if you imagine person B and C, ask A " what color is my hat?"

and A answers to both " I see at least 1 red hat" which is equivalent to "I DNK my hat color"
since the first person A can only know their hat color if they see 2 Blue hats

Then pretend only C asks B “what color is my hat?”

and B answers “I see at least 1 Red hat” which is equivalent to "I DNK my hat color"
since B can know their hat color only if they see a Blue hat on C.
this is true whether or not person A has a red or blue hat on.

B really only needs to look at C’s hat to be able to deduce the color of B’s hat.
since A told them at least one of B and C’s hats is Red.

C can be blind.

#27

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