The alternate puzzle description could even explicitly state that none of the three sages have any eyesight issues (though that is implied in hat puzzles unless an eyesight issue is mentioned, like color blindness in this puzzle). ]]>

1. All sages know one of them is colorblind and none of them knows who the colorblind sage is.

2. All sages know one of them is colorblind and only the colorblind sage knows he/she is colorblind.

3. All sages know one of them is colorblind and only the noncolorblind sages know who he/she is.

4. All sages know one of them is colorblind and all sages know who the colorblind sage is.

5. Something else.

It is hard to imagine a real-life sage who does not know whether he/she is colorblind. It is also hard to imagine that when facing a sage-test like this (and knowing that one of them is truly colorblind), the sages will not ask each other who he/she is in advance. So, I think 4 (above) is the default starting situation, right?

]]>The hat distributions could be as follows:

a) All three sages get red hats.

b) Some two sages get red hats, and the remaining sage gets a green hat.

c) One of the sages gets a red hat, and the remaining two sages get green hats. In this scenario, note that the two green hats would be worn by the two sages who are NOT color blind. If the color blind sage and a non color blind sage wore green hats, the remaining sage (who is not color blind) would have guessed that his/her own hat can only be red, and this announcement would have occurred within the first three responses (which didn’t happen).

Let R = Red, G = Green.

If ALICE was COLOR-BLIND, then the hat distributions (in the order of Alice, Bob, Carol) could be one of the following. Cases like G, R, G and G, G, R have been left out as explained in note c above.

R, R, R

G, R, R

R, G, R

R, R, G

R, G, G

If their hats were G, R, R, then non color blind Bob saying “I don’t know my color” would indicate to non color blind Carol that color-blind Alice and Carol are not both wearing Green (otherwise, Bob would have known that his own color is Red). Since Carol can see that Alice is wearing Green, Carol would know that her own hat color is Red. Since Carol also says “I don’t know my hat color”, color-blind Alice would know now that the hat distribution is NOT G, R, R. In all other distributions, note that Alice has a Red hat. So, in Alice’s second announcement (4th announcement overall), Alice would have announced her own hat color as Red. Since Alice again says that she doesn’t know her own color, we can conclude that Alice is NOT the color-blind sage.

If CAROL was COLOR-BLIND, then the hat distributions (in the order of Alice, Bob, Carol) could be one of the following. Cases like G, R, G and R, G, G have been left out as explained in note c above.

R, R, R

G, R, R

R, G, R

R, R, G

G, G, R

If their hats were R, R, G, then non color blind Alice saying “I don’t know my color” would indicate to non color blind Bob that Bob and color-blind Carol are not both wearing Green (otherwise, Alice would have known that her own color is Red). Since Bob can see that Carol is wearing Green, Bob would know that his own hat color is Red. Since Bob also says “I don’t know my hat color”, color-blind Carol would know that the hat distribution is NOT R, R, G. In all other distributions, note that Carol has a Red hat. So, color-blind Carol would have announced her own hat color as Red. Since Carol says that she doesn’t know her own color, we can conclude that Carol is NOT the color-blind sage.

Hence, Bob is the color-blind sage. Since color-blind Bob speaks second, he does not have enough information at that point to conclude anything. No one has definitive information to conclude his/her own hat color during the first 4 responses.

]]>