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How does it matter how he opened a goat door? If you allow the possibility of Monty opening a car door, you change the problem entirely, and you before stating
by joppy 6y ago
How does it matter how he opened a goat door? If you allow the possibility of Monty opening a car door, you change the problem entirely, and you before stating whether it changes or doesn't change the odds, you need to tell us your new interpretation of the problem.
- deleted 6y ago[deleted]
- dllthomas 6y agoImagine he picks one of the following ways: 1) picks randomly and opens the door, or 2) picks randomly and then, if that door has a car, switches. His application of knowledge is converting those cases that would have been thrown away in case 1 into victories for "switch". 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.
- xanathar 6y agoProblem-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