So the setup of the scenario here is exactly the traditional Monty Hall problem. Except, before the game starts, you decide to cheat and open a door and it happens to be a goat. Before you can peek at the two other doors, the game begins, and you pick a door that's not the one you just peeked into (obviously, because you peeked a goat). Here it seems the probability to win a car is $\frac{1}{2}$, because you improved your original $\frac{1}{3}$ chance of winning a car by cheating.

Then the game host opens the same door you just peeked at before the game started (knowing it was a goat, like the traditional Monty Hall problem). Is it still a $\frac{2}{3}$ chance of winning the car if you switch, even though you already knew the informatiom beforehand? Or does it still remain $\frac{1}{2}$ because you already knew the information before, so you just screwed yourself out of the better $\frac{2}{3}$ probability by cheating?

If you originally picked a door with a goat, then the game host must pick the door you peeked at, because the other door contains the car. In this situation, it doesn't seem like anything changes because you already knew the door the game host opened beforehand by cheating.

If you originally picked a door with a car, the game host can open either other door, because they both have goats. If the game host opens the door you did *not* peek at and reveals a goat, then you know with $100\%$ probability that the door you originally picked contains the winning car. However, if the game host opens the door you peeked at, then it seems like the same situation in the paragraph I described before this one.