A hat puzzle


#1

Originally published at: http://boingboing.net/2017/04/05/a-hat-puzzle.html


#2

What is the color of the third logician’s hat?

Trick question. Obviously the well-mannered third logician, unlike their uncouth companions, did the proper thing when walking into a building while wearing a hat - they removed it, and therefore knew for certain what color their hat was.


#3

I got this one immediately (for the wrong reason). The third logician’s answer is actually unnecessary to solve the puzzle.

Logic involved:
The first logician looks at hats 2 and 3. Because she does not immediately answer “Red,” the answer cannot be 1:R 2:B 3:B. Therefore, at least one of the hats 2 and 3 are red.

The second logician looks at hats 1 and 3. If 3 is blue and 2 is blue, then 1 would have answered “Yes.” If 3 is blue and 2 is red, 1 would have answered “No.” So, if 3 is blue, then either way, 2 would know what colour her own hat is. Since she doesn’t, 3 is red.

Since the third logician has been observing the other two and knows this, she is able to answer “Yes” without looking.

My initial solution was that 1’s answer rules out both 2 and 3 wearing blue; 2’s answer rules out 3 wearing blue, and 3’s ability to answer depends on her separate perceptions of 1 and 2’s hats, meaning that all three of them are wearing red. But that was wrong, because the puzzle was already solved before 3 had a chance to contribute.


#4

But there is only a one in ten chance that the conditions allowing the third logician to answer with certainty will arise. They must have visited a lot of bars.


#5

I get 60%, not 10%.

As far as I’ve been able to determine, all three questions would be answered the same no matter what the first two hats are. The only thing that matters to the riddle is that the third hat is red, which has a 3/5 (60%) chance.

How are you calculating one-in-ten?


#6

I hate these puzzles. I have never managed to solve a single one of these things. And I’m really smart!


#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.