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> If he was opening doors randomly then you're left with a one-in-two chance. No. Or, well, it depends on what happens if Monty opens a door with the car. If I

> If he was opening doors randomly then you're left with a one-in-two chance. No. Or, well, it depends on what happens if Monty opens a door with the car. If I lose if he does that, then switching doors doesn't change the probability. It's easy to prove. Let's say that I always pick door #1, and Monty always picks door #2. Then we get the following probabilities: car-goat-goat: I stay, I win. I switch, I lose. goat-car-goat: I stay, I lose. I switch, I lose. goat-goat-car: I stay, I lose. I switch, I win. Staying: 1/3 chance of winning. Switching: 1/3 chance of winning.

OscarCunningham 7y ago
The point is that given that the door Monty opened had a goat behind it your probability of winning would then be 1/2.
- henrikschroder 7y ago
  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.
daveFNbuck 7y ago
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.
mlazos 7y ago
Yes this makes sense to me. The way I thought about it is if you win if he picks the door with the car, then you’re in essence getting two choices. So the total probability of winning is 1/3 (probability of Monty picking car) + 2/3 * 1/2 (prob of monty picking goat and you picking car). The reason I commented is I can compute the total probability of winning in your scenario by saying Monty has 2/3 chance of picking a goat and I have 1/2 chance of picking a car after that. The result then is 2/3 * 1/2 = 1/3

OscarCunningham 7y ago

The point is that given that the door Monty opened had a goat behind it your probability of winning would then be 1/2.

henrikschroder 7y ago
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.

daveFNbuck 7y ago

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.

mlazos 7y ago

Yes this makes sense to me. The way I thought about it is if you win if he picks the door with the car, then you’re in essence getting two choices. So the total probability of winning is 1/3 (probability of Monty picking car) + 2/3 * 1/2 (prob of monty picking goat and you picking car). The reason I commented is I can compute the total probability of winning in your scenario by saying Monty has 2/3 chance of picking a goat and I have 1/2 chance of picking a car after that. The result then is 2/3 * 1/2 = 1/3