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.
> unless it is explicitly stated that Monty knows where the car is and that he deliberately opens a door with a goat.
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?"
There are three doors. You have picked one, leaving two other doors. It absolutely explicitly says he opens one door. The only options are a goat or a car. If it was a car, you would have lost already and so there is no problem. If it was random, you still get the same information (what is behind one of the unpicked doors).
> If it was random, you still get the same information
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.
This is goalpost moving that has nothing to do with the original point. If people misunderstand the conditions of the problem, that has nothing to do with intuitions about probability.
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.