If you want to start a thread about the logic of the Blue Eye problem, I’m happy to engage. Instead, you started a thread about a “common mistake” in science and philosophy. And, rather than actually show that these mistakes are present in the canonical solution to the Blue Eye problem, rather than actually laying out your case, rather than quoting what I or anyone who’s endorsed the canonical solution has actually said (instead of making up a quote that is trivially fallacious), you just state that it “dawned” on you and then explore an intro-level logical fallacy.
Whatever prompted you to have your dawning, that thread is not about the logic of the Blue Eye problem. So far, this is the only thread in which that problem has been directly addressed.
This problem is pretty tedious, I think I must be doing it wrong.
[tab]Here are some equations:
A = C+G
B = H+D
B = I+F
C = G+E
D = H+J
E = G+I
F = J+D
H = I+J
A+B = C+E+F
A+C = B+H+I+F
A+G = B+H+I
Should be enough. Let’s try and solve for a variable in terms of J.
A+C = B+H+I+F
(C+G)+C = (I+F)+H+I+F
2(G+E)+G=2(I+F)+(I+J)
2(G+(G+I))+G = 3I+2F+J
5G+2I = 3I+2F+J
F = J+D
F = 2J+H
F = 3J+I
5G+2I = 3I+2(3J+I)+J
5G+2I = 5I+7J
5G = 3I+7J
A+G = B+H+I
G = (H+D)+H+I-(C+G)
2G = 2H+(H+J)+I-(G+E)
3G = 3H+J+I-(G+I)
4G = 3H+J
4G = 3(I+J)+J
4G = 3I+4J
G = (3I+4J)/4
5((3I+4J)/4) = 3I+7J
(15/4)I+5J = 3I+7J
(15/4)I-3I = 2J
(3/4)I = 2J
If J = 1 → I = 1.5
→ H = 2.5
D = 3.5
B = 6
F = 4.5
A+C = 14.5
A+G = 10
C-G=4.5
E = 4.5
G = 3
C= 7.5
A = 7…
Shit.[/tab]
Yep, definitely did it wrong