Question about Monty Hall if you already knew a bad door beforehand 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.
If you pick the car first, as you say, the host has 50% chance of confirming you have the car, and 50% of giving you no new information.

If you pick the unknown goat first, the host gives you no new information.

Ultimately the host has a $\frac{3}{4}$ chance of revealing the goat you already knew about.

If you're picking between those two randomly, then in $\frac{1}{4}$ of all cases, you pick the car, then the host reveals the second goat, so you win the car. In the other $\frac{3}{4}$ of cases, you're back to the $\frac{2}{3}$ - $\frac{1}{3}$ situation, since in $\frac{2}{3}$ of those $\frac{3}{4}$, you had the goat to begin with, and in the other $\frac{1}{3}$ you had the car.

$$P(\mathrm{car}|\mathrm{unknown\,goat\,revealed}) = 1$$

$$P(\mathrm{car}|\mathrm{known\,goat\,revealed}) = \frac{P(\mathrm{car}\cap\mathrm{known\,goat\,revealed})}{P(\mathrm{known\,goat\,revealed})} = \frac{1/4}{3/4} = \frac{1}{3}$$

You can expect to get the car $\frac{3}{4}$ of the time by switching if the goat you knew about is revealed and staying if not.

Of course, if you pick randomly between the two unpicked doors, you're a fool, because picking the door you know to be a goat forces Monty to show you the other goat and hence brand new car :)
No, the probability is now $1/2$ not $2/3$ because for it to be $2/3$ with the switching strategy you need a $2/3$ chance of originally chosing the wrong door, but by cheating you lowered that to $1/2$