5 ms·
Whether or not he knew or whether it was luck, it doesnt exactly matter, although traditional stats knowledge would make you think it does. The important thing
by basch 6y ago
Whether or not he knew or whether it was luck, it doesnt exactly matter, although traditional stats knowledge would make you think it does.
The important thing to understand is that the premise of the question says he will show you a goat. If you rerun the experiment 10,000 times, he will show you a goat 100% of the time, either through peaking, premonition, or consistent luck.
The problem gets trickier because people start applying domain knowledge of stats, and treating it as a simulation with random events. The goat being chosen is not random, it is an event that occurs 100% of the time in the premise of the thought experiment. Thinking about the random chance of him choosing the car is outside the bounds of the axiom/postulate we start with.
tldr: it doesnt matter how he opened a goat door, all that matters is that he did.
- tromp 6y agoYou said it well: "he will show you a goat." But the standard formulation just says Monte opened a door and it had a goat behind it. No explicit mention of intention. It matters not that he happened to do. It matters that he will.
- spullara 6y agoIt doesn't, as long as he always shows a goat.
- roywiggins 6y agoWhat if he didn't know and just happened to open the goat door? Should you still switch?
- pxx 6y agoIt doesn't matter if you switch in that case. See https://news.ycombinator.com/item?id=24714612 https://news.ycombinator.com/item?id=24714612
- basch 6y agoSimulations where he opens a goat door are a subset of all simulations. The thought experiment is only concerned with the subset.
- dllthomas 6y agoBut... it does matter how he opened a goat door. Specifically, whether there was a chance that he might not have. I originally thought it couldn't matter. I wrote a simulation to demonstrate how right I was. I was wrong. I encourage duplicating the experience.
- joppy 6y agoHow 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.