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Yes, but that's kind of a weird scenario. Imagine you're taping episodes of this game show for broadcast. The contestant picks a door, and then Monty picks one

Yes, but that's kind of a weird scenario. Imagine you're taping episodes of this game show for broadcast. The contestant picks a door, and then Monty picks one of the two remaining doors at random. Monty has 1/3 chance of accidentally picking the car, and if he does, we're scrapping the entire episode. So to enforce your constraints, we have to scrap 1/3 of all episodes, we're only ever broadcasting the 2/3 of all episodes taped where Monty, by chance, didn't pick the door with the car behind it. Your numbers are correct, but that interpretation of the problem doesn't make a lot of sense if you think it through.

username90 7y ago
Move the scenario from Monty Hall to a street trickster, then this interpretation makes a lot more sense than the original one since a street trickster is unlikely to try to increase your chance of winning. As for the game show, nothing says that he needed to give this choice every time, it was just some random thing he did to make it more interesting not an important part of the show.
OscarCunningham 7y ago
We don't have to scrap the remaining episodes. Suppose that the situation is the Monty always picks randomly and in the situations where he reveals a car the episode just ends then. Then if when you go on the show Monty happens to reveal a goat should you switch? The answer is that it doesn't matter whether you do or not.
- henrikschroder 7y ago
  Yes, but that is not what you claimed. You claimed: > The point is that given that the door Monty opened had a goat behind it your probability of winning would then be 1/2. If the episode just ends when Monty picks the car, then I lose if he picks the car. I don't just vanish into thin air such that the universe erases my run. I was on the show, and I lost. And that means my odds of winning are still just 1/3, not 1/2 as you claimed. Yes, the game has changed such that it doesn't matter if I switch doors or not with this addition, but the underlying odds don't change.
  - username90 7y ago
    You are wrong here, the problem statement states "Monty Hall opened a door and it had a Goat", so the situation you describe where he selected a car didn't happen. Since he didn't select a car you now have 50% chance behind both doors. But you are right that you have 1/3 chance from the beginning, but the situation described in the problem is not at the beginning.
    - henrikschroder 7y ago
      No, that is not how the original Monty Hall problem was formulated. The original said: "the host, who knows what's behind the doors, opens another door, which has a goat." Your formulation of the problem changes it from the original Monty Hall problem to something more similar to the Tuesday Boy problem, where the trick is that you have pre-selected a bunch of outcomes and discarded others, changing the probabilities. And going back to the comments above this, it's not the case that Monty choosing a door by random is the thing that matters, it's the thing that in this version he chose a door by random and you discarded all the outcomes where Monty chose the car. It's not the case that people are confused by the problem because they interpret the problem like you did. People understand that a goat door is always eliminated as part of the game, and they still don't agree that switching is better.
  - OscarCunningham 7y ago
    I feel like you're fighting the meaning of the word given here. You might as well say that your probability of winning is tiny because you're unlikely be picked as a contestant, or even less likely than that because you might live somewhere than the US. The whole premise of the problem is that you are already on the show looking at an open door with a goat behind it.

username90 7y ago

Move the scenario from Monty Hall to a street trickster, then this interpretation makes a lot more sense than the original one since a street trickster is unlikely to try to increase your chance of winning. As for the game show, nothing says that he needed to give this choice every time, it was just some random thing he did to make it more interesting not an important part of the show.

OscarCunningham 7y ago

We don't have to scrap the remaining episodes. Suppose that the situation is the Monty always picks randomly and in the situations where he reveals a car the episode just ends then. Then if when you go on the show Monty happens to reveal a goat should you switch? The answer is that it doesn't matter whether you do or not.

henrikschroder 7y ago
Yes, but that is not what you claimed. You claimed: > The point is that given that the door Monty opened had a goat behind it your probability of winning would then be 1/2. If the episode just ends when Monty picks the car, then I lose if he picks the car. I don't just vanish into thin air such that the universe erases my run. I was on the show, and I lost. And that means my odds of winning are still just 1/3, not 1/2 as you claimed. Yes, the game has changed such that it doesn't matter if I switch doors or not with this addition, but the underlying odds don't change.
- username90 7y ago
  You are wrong here, the problem statement states "Monty Hall opened a door and it had a Goat", so the situation you describe where he selected a car didn't happen. Since he didn't select a car you now have 50% chance behind both doors. But you are right that you have 1/3 chance from the beginning, but the situation described in the problem is not at the beginning.
  - henrikschroder 7y ago
    No, that is not how the original Monty Hall problem was formulated. The original said: "the host, who knows what's behind the doors, opens another door, which has a goat." Your formulation of the problem changes it from the original Monty Hall problem to something more similar to the Tuesday Boy problem, where the trick is that you have pre-selected a bunch of outcomes and discarded others, changing the probabilities. And going back to the comments above this, it's not the case that Monty choosing a door by random is the thing that matters, it's the thing that in this version he chose a door by random and you discarded all the outcomes where Monty chose the car. It's not the case that people are confused by the problem because they interpret the problem like you did. People understand that a goat door is always eliminated as part of the game, and they still don't agree that switching is better.
- OscarCunningham 7y ago
  I feel like you're fighting the meaning of the word given here. You might as well say that your probability of winning is tiny because you're unlikely be picked as a contestant, or even less likely than that because you might live somewhere than the US. The whole premise of the problem is that you are already on the show looking at an open door with a goat behind it.