That has been part of the explicit problem ever since it was first presented back in 1975.
"Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, 'Do you want to pick door No. 2?' Is it to your advantage to switch your choice?"
- allways open a door
- allways open a door with a goat
- and not open a door at random
Often discussion of the solution reveals that this is not clear.
No, if both you and Monty pick a door at random, there’s 1/3 chance of a car behind each door. If Monty’s door reveals a goat, it’s 50/50 for the remaining two doors. It’s mandatory to specify Monty’s procedure precisely.
I won't respond further.
No it is not, unless it is explicitly stated that Monty knows where the car is and that he deliberately opens a door with a goat. Just look at the discussions in the comment here.