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(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. |
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?