What you're not understanding is that they could just add easily incorrectly deduce their eye colour. — flannel jesus
No they won't. Let's take the example with 3 blue, 3 brown, and 1 green.
Each blue's reasoning is:
A1. Green sees blue
A2. Therefore, if I don't see blue then I must be blue
A3. Therefore, if I see one blue and he leaves on the first day then I must not be blue
A4. Therefore, if I see one blue and he doesn't leave on the first day then I must be blue
A3. Therefore, if I see two blue and they leave on the second day then I must not be blue
A4. Therefore, if I see two blue and they don't leave on the second day then I must be blue
A3. Therefore, if I see three blue and they leave on the third day then I must not be blue
A4. Therefore, if I see three blue and they don't leave on the third day then I must be blue
B1. Green sees brown
B2. Therefore, if I don't see brown then I must be brown
B3. Therefore, if I see one brown and he leaves on the first day then I must not be brown
B4. Therefore, if I see one brown and he doesn't leave on the first day then I must be brown
B3. Therefore, if I see two brown and they leave on the second day then I must not be brown
B4. Therefore, if I see two brown and they don't leave on the second day then I must be brown
B5. Therefore, if I see three brown and they leave on the third day then I must not be brown
B6. Therefore, if I see three brown and they don't leave on the third day then I must be brown
Given that each blue sees 2 blue and 3 brown, they can rule out some of these premises:
A1. Green sees blue
A2. Therefore, if I don't see blue then I must be blue
A3. Therefore, if I see one blue and he leaves on the first day then I must not be blue
A4. Therefore, if I see one blue and he doesn't leave on the first day then I must be blue
A3. Therefore, if I see two blue and they leave on the second day then I must not be blue
A4. Therefore, if I see two blue and they don't leave on the second day then I must be blue
A3. Therefore, if I see three blue and they leave on the third day then I must not be blue
A4. Therefore, if I see three blue and they don't leave on the third day then I must be blue
B1. Green sees brown
B2. Therefore, if I don't see brown then I must be brown
B3. Therefore, if I see one brown and he leaves on the first day then I must not be brown
B4. Therefore, if I see one brown and he doesn't leave on the first day then I must be brown
B3. Therefore, if I see two brown and they leave on the second day then I must not be brown
B4. Therefore, if I see two brown and they don't leave on the second day then I must be brown
B5. Therefore, if I see three brown and they leave on the third day then I must not be brown
B6. Therefore, if I see three brown and they don't leave on the third day then I must be brown
Then, come the third day, they can rule out one more:
A1. Green sees blue
A2. Therefore, if I don't see blue then I must be blue
A3. Therefore, if I see one blue and he leaves on the first day then I must not be blue
A4. Therefore, if I see one blue and he doesn't leave on the first day then I must be blue
A3. Therefore, if I see two blue and they leave on the second day then I must not be blue
A4. Therefore, if I see two blue and they don't leave on the second day then I must be blue
A3. Therefore, if I see three blue and they leave on the third day then I must not be blue
A4. Therefore, if I see three blue and they don't leave on the third day then I must be blue
B1. Green sees brown
B2. Therefore, if I don't see brown then I must be brown
B3. Therefore, if I see one brown and he leaves on the first day then I must not be brown
B4. Therefore, if I see one brown and he doesn't leave on the first day then I must be brown
B3. Therefore, if I see two brown and they leave on the second day then I must not be brown
B4. Therefore, if I see two brown and they don't leave on the second day then I must be brown
B5. Therefore, if I see three brown and they leave on the third day then I must not be brown
B6. Therefore, if I see three brown and they don't leave on the third day then I must be brown
Given that one of the As and one of the Bs must obtain, and given that only A4 is left of the As, blue knows on the third day that A4 must obtain and so that they are blue.
The complimentary set of arguments will have all browns deducing that they are brown on the third day.
All without green having to say anything.