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
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.
The complete cases for the choices is:
Problem-1:
1) You pick goat#1, Monty opens goat#2
2) You pick goat#2, Monty opens goat#1
3) You pick car, Monty chooses a goat of his liking
1) You pick goat#1, Monty opens goat#2
2) You pick goat#2, Monty opens goat#1
3) You pick car, Monty opens goat#1
4) You pick car, Monty opens goat#2
5) You pick goat#1, Monty opens car
6) You pick goat#2, Monty opens car
Problem-1:
1) You pick goat#1, Monty opens goat#2
2) You pick goat#2, Monty opens goat#1
3) You pick car, Monty chooses a goat of his liking
1) You pick goat#1, Monty opens goat#2
2) You pick goat#2, Monty opens goat#1
3) You pick car, Monty opens goat#1
4) You pick car, Monty opens goat#2
1) picks randomly and opens the door, or
2) picks randomly and then, if that door has a car, switches.
The counts bear this out. If you simulate it, 1/3 of all games go to 'stick' regardless. Either 1/3 or 2/3 go to 'switch', and either 1/3 or 0/3 (respectively) end up with Monty revealing the car.