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> In this: "the host will open the door at random and in this example it happen to have a goat" vs "the host will never open a door with a car behind it".) If

> In this: "the host will open the door at random and in this example it happen to have a goat" vs "the host will never open a door with a car behind it".) If the host opens a door with a goat, then it doesn't matter whether or not it was intentional.

dllthomas 5y ago
Yes it does. Or more precisely, it matters whether the host could be counted on to do so reliably; the mechanism for that doesn't matter. There's a difference here, that our language obscures, between procedure and hypothetical.
- tpush 5y ago
  The intention of the host only matters if the contestant would have to choose the subsequent action (switching or not) before the hosts opens a door. If the host has revealed a goat door, and the contestant then has to decide what to do, the intentions of the host for having chosen the door are irrelevant.
  - dllthomas 5y ago
    (Noting again that the mechanism doesn't matter, what matters is the odds of various behavior by the host, but - I think reasonably - using "intentions of the host" as a proxy for that.) The intentions of the host do matter. Imagine the host picks the correct door by the following procedure: 1) picks an available door at random; 2) if that door has a goat, opens it; 3) if that door has the car, opens the other door. I hope you will agree that this is equivalent to the problem as originally intended - Monty can be relied on to reveal a goat, and exactly why doesn't matter. Breaking it down into equally likely cases, assuming the contestant picks door 3: A) The car is behind door 1, Monty picks door 1, Monty corrects. B) The car is behind door 1, Monty picks door 2 C) The car is behind door 2, Monty picks door 1 D) The car is behind door 2, Monty picks door 2, Monty corrects E) The car is behind door 3, Monty picks door 1 F) The car is behind door 3, Monty picks door 2 When Monty reveals the goat behind (say) door 2, we know we're in case A, B, or F. All remain equally likely, and switching wins in A and B. If Monty would not have corrected, then revealing the goat behind door 2 eliminates (the new) A as well, leaving us with only B and F, again equally likely. If all of this remains unconvincing, I encourage you to write a simple simulation of the problem.
    - tpush 5y ago
      So, thinking about it really hard and reading about it online: My comment was definitely wrong: If Monty could have opened a car door, but just didn't, then duh the probabilities for the car to be behind the doors are different than if Monty always opens a goat door. So in that way, the intentions of Monty, meaning how he chooses, definitely matter. But I think your example here doesn't show that? Are you trying to illustrate the Monty Fall variation?
      - dllthomas 5y ago
        Sorry it took me so long to get back to this; real life intervenes sometimes. I think what I was trying to do was frame the original Monty Hall problem as a variant of Monty Fall, in a way that (I hoped) makes it clear where Monty is doing work to convert some outcomes into other outcomes (and therefore producing different likelihoods).

dllthomas 5y ago

Yes it does. Or more precisely, it matters whether the host could be counted on to do so reliably; the mechanism for that doesn't matter. There's a difference here, that our language obscures, between procedure and hypothetical.

tpush 5y ago
The intention of the host only matters if the contestant would have to choose the subsequent action (switching or not) before the hosts opens a door. If the host has revealed a goat door, and the contestant then has to decide what to do, the intentions of the host for having chosen the door are irrelevant.
- dllthomas 5y ago
  (Noting again that the mechanism doesn't matter, what matters is the odds of various behavior by the host, but - I think reasonably - using "intentions of the host" as a proxy for that.) The intentions of the host do matter. Imagine the host picks the correct door by the following procedure: 1) picks an available door at random; 2) if that door has a goat, opens it; 3) if that door has the car, opens the other door. I hope you will agree that this is equivalent to the problem as originally intended - Monty can be relied on to reveal a goat, and exactly why doesn't matter. Breaking it down into equally likely cases, assuming the contestant picks door 3: A) The car is behind door 1, Monty picks door 1, Monty corrects. B) The car is behind door 1, Monty picks door 2 C) The car is behind door 2, Monty picks door 1 D) The car is behind door 2, Monty picks door 2, Monty corrects E) The car is behind door 3, Monty picks door 1 F) The car is behind door 3, Monty picks door 2 When Monty reveals the goat behind (say) door 2, we know we're in case A, B, or F. All remain equally likely, and switching wins in A and B. If Monty would not have corrected, then revealing the goat behind door 2 eliminates (the new) A as well, leaving us with only B and F, again equally likely. If all of this remains unconvincing, I encourage you to write a simple simulation of the problem.
  - tpush 5y ago
    So, thinking about it really hard and reading about it online: My comment was definitely wrong: If Monty could have opened a car door, but just didn't, then duh the probabilities for the car to be behind the doors are different than if Monty always opens a goat door. So in that way, the intentions of Monty, meaning how he chooses, definitely matter. But I think your example here doesn't show that? Are you trying to illustrate the Monty Fall variation?
    - dllthomas 5y ago
      Sorry it took me so long to get back to this; real life intervenes sometimes. I think what I was trying to do was frame the original Monty Hall problem as a variant of Monty Fall, in a way that (I hoped) makes it clear where Monty is doing work to convert some outcomes into other outcomes (and therefore producing different likelihoods).