“If Student A saw two blue hats, then he would’ve been sure that he had a red hat. Student B knew that Student A passed because he either saw two red hats or one red hat and one blue hat. ” So now it’s B’s turn, and he knows that if C has a blue hat, and A has a red hat, his own hat must be red. (If it were blue, A would have known his own hat color) If C has a red hat and A has a red hat, B’s hat could still be red or blue. (Same if C has a red hat and A has a blue hat.) He passes, and this means C’s hat is red. So C knows his own hat color without looking. That was confusing….but I think it works.

A will pass because he/she saw 2 red hats but couldn’t

confirm if his/her hat is blue. And with that knowledge, student

B will only have to confirm that student A is not wearing a blue

hat and he

can be 100% positive that his hat is red and make a guess.

Therefore, 3 red hat wouldn’t be possible and that leaves you

with the answer.

The only way A could guess is if she saw two blue hats. She didn’t guess, so B knows that his hat and C’s are not both blue.

When B looked, if he’d seen a blue hat on C, he would have known his hat was red – again, because he knows that his and C’s hats are not both blue. He passed, so he must’ve seen a red hat on C. That was all C needed to know.

A could have seen anything else. B+R or R+R.

for B, then if C was blue, then obviously B could not be blue because A did not see two blue hats, therefore C could not be blue otherwise B would be wearing red.

Then C knows their hat is red.