Problem-1: "You pick a door. Then Monty opens a door he knows having the goat. Calculate probability of winning if staying."

Problem-2: "You pick a door. Monty opens a door at random, which just happens to be a goat. Calculate probability of winning if staying."

Both are conditional probability P(A|B) (that is, probability of A happening, under the assumption B has happened).

   - A is probability of picking the correct door at the first try (or switching if you prefer)
   - B is the probability that Monty picked a door with a goat

P(A|B) is defined as: P(A|B) = P(A and B) / P(B)

In both problems P(A) is 1/3.

In Problem-1, P(B) is 1 because Monty knows and it's not a random event and P(A and B) = P(A), so P(A|B) = P(A)*P(B)/P(B) = P(A) = 1/3.

In Problem-2, P(A) is 1/3, P(B) is 2/3, P(A and B) is 1/3 (if you pick a car, Monty is guaranteed to pick a goat), so P(A|B) = 1/3 / 2/3 = 1/2.