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I was arguing against reveals vs revealed because I thought you were making a different point. I understand what you meant about that difference now. I'm still

I was arguing against reveals vs revealed because I thought you were making a different point. I understand what you meant about that difference now. I'm still not sure what the difference between these models has to do with my point about goat-car-goat. That can't happen in either model when we say that I'm picking door 1 and Monty is picking door 2.

henrikschroder 7y ago
Ok, right, the point I was arguing against the parent poster to my initial post in this thread is that he said that "Monty picks a door at random" implicitly means that we have to discard all the outcomes where Monty picked the car. But I'm saying that since that is a wrong interpretation of the original problem, any other wrong interpretation of the problem is equally valid. For example, if Monty picks the car, it's game over and the contestant never gets to choose. Or, if Monty picks the car, the show goes "oops", closes the door, shuffles the prizes, and lets Monty pick again until he picks a goat. There's not enough information in "Monty picks a door at random" to infer a single interpretation of the problem. "Monty picks a door at random AND he picked a goat". That's enough information to get down to your universe where goat-car-goat could never happen. Nitpicky semantics, sure. But to roll back to the original, original question: I don't think people get the Monty Hall problem wrong because they're interpreting it as "Monty picks a door at random and he picked a goat". They interpret the problem correctly, i.e. "Monty knows where the car is and always picks a goat", and they still get it wrong.
- daveFNbuck 7y ago
  Either of your interpretations of Monty picking a door at random lead to a conditional probability of 1/2 that your chosen door has the car given that he revealed a goat. I'm not saying that goat-car-goat could never happen. I'm saying that when you're evaluating what to do after Monty Hall has revealed a goat, the one piece of information you know is that goat-car-goat didn't happen. I guess you're analyzing from the point of view of someone who's trying to figure out their odds of winning with different strategies before playing. I agree with you that most people don't get the problem wrong because of interpretation. I'd go even further and say that many people who think they understand it don't get it right because of the interpretation. Most arguments I see for the correct answer don't use the assumption that Monty knows which door has the goat, and therefore work equally well under a model of Monty picking the door at random. They're just picking different ways to model the problem and assuming a uniform distribution. You get the answer right if you picked a model where the distribution is uniform under the assumptions in the problem. The real issue seems to be much simpler. Probability is not intuitive. Most people don't have the tools to figure out problems like this reliably, and those that do tend not to apply their tools when they think they see a simple obvious answer.
  - henrikschroder 7y ago
    > Either of your interpretations of Monty picking a door at random lead to a conditional probability of 1/2 that your chosen door has the car given that he revealed a goat. No, if Monty picks a door by random, but the show shuffles the doors until he gets it "right" and picks a goat, that is exactly the same scenario as if Monty knows where the goats are and picks one straight away, which means my odds of winning when switching is 2/3. So that is a scenario where "Monty picks a door by random" doesn't lead to the odds changing to 50/50.
    - daveFNbuck 7y ago
      If by shuffle you mean that Monty just gets to keep trying until he picks a goat, you're right. If instead you mean that they randomize all three doors and start over, the probability is 1/2 because half the time that the player has a goat you start over and get to pick again, which perfectly cancels out the double probability of having a goat. This can be analyzed pretty easily using the three possibilities you outlined earlier. Suppose the player gets door 1 and Monty always picks door 2. Each time they shuffle, you get one of the following with equal probability: A. car-goat-goat B. goat-car-goat C. goat-goat-car A means the player wins, B means we re-shuffle, C means the player loses. It's pretty easy to see that A and B are equally likely outcomes, so they become 50/50 odds after all the re-shuffling is done.

henrikschroder 7y ago

Ok, right, the point I was arguing against the parent poster to my initial post in this thread is that he said that "Monty picks a door at random" implicitly means that we have to discard all the outcomes where Monty picked the car. But I'm saying that since that is a wrong interpretation of the original problem, any other wrong interpretation of the problem is equally valid. For example, if Monty picks the car, it's game over and the contestant never gets to choose. Or, if Monty picks the car, the show goes "oops", closes the door, shuffles the prizes, and lets Monty pick again until he picks a goat. There's not enough information in "Monty picks a door at random" to infer a single interpretation of the problem. "Monty picks a door at random AND he picked a goat". That's enough information to get down to your universe where goat-car-goat could never happen. Nitpicky semantics, sure. But to roll back to the original, original question: I don't think people get the Monty Hall problem wrong because they're interpreting it as "Monty picks a door at random and he picked a goat". They interpret the problem correctly, i.e. "Monty knows where the car is and always picks a goat", and they still get it wrong.

daveFNbuck 7y ago
Either of your interpretations of Monty picking a door at random lead to a conditional probability of 1/2 that your chosen door has the car given that he revealed a goat. I'm not saying that goat-car-goat could never happen. I'm saying that when you're evaluating what to do after Monty Hall has revealed a goat, the one piece of information you know is that goat-car-goat didn't happen. I guess you're analyzing from the point of view of someone who's trying to figure out their odds of winning with different strategies before playing. I agree with you that most people don't get the problem wrong because of interpretation. I'd go even further and say that many people who think they understand it don't get it right because of the interpretation. Most arguments I see for the correct answer don't use the assumption that Monty knows which door has the goat, and therefore work equally well under a model of Monty picking the door at random. They're just picking different ways to model the problem and assuming a uniform distribution. You get the answer right if you picked a model where the distribution is uniform under the assumptions in the problem. The real issue seems to be much simpler. Probability is not intuitive. Most people don't have the tools to figure out problems like this reliably, and those that do tend not to apply their tools when they think they see a simple obvious answer.
- henrikschroder 7y ago
  > Either of your interpretations of Monty picking a door at random lead to a conditional probability of 1/2 that your chosen door has the car given that he revealed a goat. No, if Monty picks a door by random, but the show shuffles the doors until he gets it "right" and picks a goat, that is exactly the same scenario as if Monty knows where the goats are and picks one straight away, which means my odds of winning when switching is 2/3. So that is a scenario where "Monty picks a door by random" doesn't lead to the odds changing to 50/50.
  - daveFNbuck 7y ago
    If by shuffle you mean that Monty just gets to keep trying until he picks a goat, you're right. If instead you mean that they randomize all three doors and start over, the probability is 1/2 because half the time that the player has a goat you start over and get to pick again, which perfectly cancels out the double probability of having a goat. This can be analyzed pretty easily using the three possibilities you outlined earlier. Suppose the player gets door 1 and Monty always picks door 2. Each time they shuffle, you get one of the following with equal probability: A. car-goat-goat B. goat-car-goat C. goat-goat-car A means the player wins, B means we re-shuffle, C means the player loses. It's pretty easy to see that A and B are equally likely outcomes, so they become 50/50 odds after all the re-shuffling is done.