5 ms·

If Monty revealed a goat, then goat-car-goat didn't happen and you just have car-goat-goat or goat-goat-car as the only two possibilities.

henrikschroder 7y ago
Yes, but you changed the original problem from "Monty reveals a goat" to "Monty revealed a goat". One little letter and the whole thing is a completely different problem.
- daveFNbuck 7y ago
  That doesn't change the problem. Either way, you're evaluating whether to switch after the reveal.
  - henrikschroder 7y ago
    ? Upthread I see that you are correctly writing about the different ways you can model Monty's behaviour, and how each different way changes the nature of the problem, the outcomes, and the odds of switching vs. staying. So I don't understand what you are arguing against right now? I'm saying that "Monty reveals a goat" is what happens in the correct interpretation of the problem, i.e. Monty knows where the car is, and always chooses a door with a goat in every single instance of game play. In this case switching gives you 2/3 odds of winning the car. And I'm saying that "Monty revealed a goat" is what happens in the incorrect interpretation of the problem, i.e. Monty does not know where the car is, and in this single instance of game play that you happen to be participating in, he happened to pick a door with a goat. And if you have gotten to this point, switching is 50/50. Just like you say. Do you understand my way of describing these two different models of Monty?
    - daveFNbuck 7y ago
      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

Yes, but you changed the original problem from "Monty reveals a goat" to "Monty revealed a goat". One little letter and the whole thing is a completely different problem.

daveFNbuck 7y ago
That doesn't change the problem. Either way, you're evaluating whether to switch after the reveal.
- henrikschroder 7y ago
  ? Upthread I see that you are correctly writing about the different ways you can model Monty's behaviour, and how each different way changes the nature of the problem, the outcomes, and the odds of switching vs. staying. So I don't understand what you are arguing against right now? I'm saying that "Monty reveals a goat" is what happens in the correct interpretation of the problem, i.e. Monty knows where the car is, and always chooses a door with a goat in every single instance of game play. In this case switching gives you 2/3 odds of winning the car. And I'm saying that "Monty revealed a goat" is what happens in the incorrect interpretation of the problem, i.e. Monty does not know where the car is, and in this single instance of game play that you happen to be participating in, he happened to pick a door with a goat. And if you have gotten to this point, switching is 50/50. Just like you say. Do you understand my way of describing these two different models of Monty?
  - daveFNbuck 7y ago
    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.