Subtle Twist on the Monty Hall Problem---Does It Make a Difference? In the Monty Hall problem, when the host picks a door and reveals an goat, does it make any difference if he did not know which door the real car was behind, and he just happened to pick a door with a goat?

Ok, ill turn my comment into a full answer:
Let $C$ be the event where you choose the car and let $G$ be the event that Monty Hall shows the goat. Then the probability that the other door has a car given that monty shows you a goat is:
$P(\neg C|G)=\frac{P(\neg C)(P(G|\neg C)}{P(\neg C)(P(G|\neg C) + P(C)P(G|C)}=\frac{\frac{2}{3}\times\frac{1}{2}}{\frac{2}{3}\times\frac{1}{2} + \frac{1}{3}\times1} = \frac{1}{2}$, which is different from the original problem.
Intuitively, this is because Monty Hall is not really providing you with any additional information by showing you a goat (obviously he does if he reveals a car)
