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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
by xanathar 6y ago
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.
- deleted 6y ago[deleted]
- xanathar 6y agoIf you want to visualize them in "enumeration of cases". 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 Problem-2: 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 We know Monty didn't pick a car, that reduces it to: 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 Problem-2: 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 Or, 1/3 for Problem-1, 2/4 (=1/2) for Problem-2