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This is an extremely important part of the problem that is always glossed over in explanations of the Monty hall problem, and it really bothers me that it's ign

This is an extremely important part of the problem that is always glossed over in explanations of the Monty hall problem, and it really bothers me that it's ignored because it is crucial to determine what is the "correct" decision. If the game show host is opening a door at random (i.e. it's possible that he opens the door with the car behind it OR the door you already picked) then the outcome is 50/50 whether you switch or not. But if the host knows what's behind the doors, and purposefully opens one of the remaining two with the goat, then you should switch doors to increase your odds to 66 percent. On the above article this is only briefly addressed and not explained. In many tellings it is not addressed at all.

dylan604 3y ago
Not "rigging" the decision, would be pointless in the game. Yes, it is failed to be mentioned, but if the host revealed the car, then what's the point?
tromp 3y ago
In this article it's addresses with this sentence > The show's host then reveals the contents of one of the two remaining boxes, but, importantly, always a "goat box." but it could be made slightly more explicit as: The show's host, who knows the location of the car, is then compelled to always reveal one of the two unselected boxes as containing a goat.
- andy800 3y ago
  As far as solving the problem in an academic sense, your explanation is clearer. However, one of the things that makes the whole thing fascinating is that the actual players are not told this specific information, that the host knows. If you watch the show religiously it might dawn on you, hey, he must not be picking the preview door randomly because he has never accidentally revealed the car -- but that's beyond most people's observation. In practice, the revelation of one door is likely to make the player even more committed to the original choice -- the odds (seemingly, not actually) just went from 33% to 50%, momentum is good, I'm feeling lucky, no way am I going to switch now!
  - DoctorOetker 3y ago
    > If you watch the show religiously it might dawn on you, hey, [...] You can't watch a hypothetical game show. Let's make a deal contains a myriad of different gameplay formats.
    - andy800 3y ago
      The "Big Deal" -- the game where the player chooses one of three doors -- was the finale of every episode of Let's Make a Deal under Monty Hall. Still is, in the Wayne Brady revival. Not hypothetical.
jjnoakes 3y ago
> If the game show host is opening a door at random (i.e. it's possible that he opens the door with the car behind it OR the door you already picked) then the outcome is 50/50 whether you switch or not. I don't think this is true. Assuming you mean that if the host opens my box, I switch to one if the other two at random with equal probability (even if he opened my box and it was the car), and assuming if the host doesn't open my box that I stay or switch to the other unopened box with equal probability (even if the host opened the box with the car), I think the odds are 33% that I win the car, and 66% that I don't.
- cman1444 3y ago
  Consider it framed this way: You pick door #3. The host says they're feeling generous so he is going to give you a free door. He blindly reaches into a bag with three marbles labeled 1,2, and 3. He pulls out the marble labeled 2. He opens door 2 and there's a goat behind it. Do you now switch to door 1? The answer is that it doesn't matter, it's 50/50.
  - jjnoakes 3y ago
    Right, in the world where the door he already opened was done randomly (i.e. it could have been a goat or it could have been a car, and it could have been the door you also chose), then if he happened to both (a) open one of the doors you didn't pick, and (b) he opened a goat door, then switching doesn't matter. But that's only one branch of the probability tree, and if you investigate all of them (assuming he chooses a door at random equally, might open yours, and might open a car), then it's not 50/50, it's 33/66 I think. Perhaps you just weren't clear enough that you think the 50/50 is limited to only a subset of the possible outcomes in your scenario.
    - kgwgk 3y ago
      What would "switching" even mean if the door that was opened was the door that you had picked and the car was there? What would be the point of switching to the remaining door if the door that was opened was not the one you picked and the car was there? (The probability of getting the car would be zero either way.) You're right that the subspace of scenarios where (a) and (b) happen is just part of the full space of things that could have happened - but it's the only part relevant for the problem. (We're explicitely told that (a) and (b) are true!)
travisjungroth 3y ago
It’s also helpful (but not essential) to realize that the choice of which door to open has two parts, a rule and a decision. If the participant is doing a switch or a stay strategy, the decision aspect has no impact on winning. The rule part is Monty can’t reveal the car. The decision comes up if the participant chooses the car first. Monty will have two doors to choose from to open. But if you have a stay strategy you’ll win for either choice, and if you have a switch strategy you’ll lose no matter what. So his decision doesn’t matter. This makes it very easy to simulate. It also might lead to people to understanding the solution. If you pick wrong first, you’ll see staying always loses and switching always wins. The probability of picking right first is 1/3, of picking wrong first 2/3. Since switch always works when you pick wrong and you pick wrong 2/3 times, go with that.

dylan604 3y ago

Not "rigging" the decision, would be pointless in the game. Yes, it is failed to be mentioned, but if the host revealed the car, then what's the point?

tromp 3y ago

In this article it's addresses with this sentence > The show's host then reveals the contents of one of the two remaining boxes, but, importantly, always a "goat box." but it could be made slightly more explicit as: The show's host, who knows the location of the car, is then compelled to always reveal one of the two unselected boxes as containing a goat.

andy800 3y ago
As far as solving the problem in an academic sense, your explanation is clearer. However, one of the things that makes the whole thing fascinating is that the actual players are not told this specific information, that the host knows. If you watch the show religiously it might dawn on you, hey, he must not be picking the preview door randomly because he has never accidentally revealed the car -- but that's beyond most people's observation. In practice, the revelation of one door is likely to make the player even more committed to the original choice -- the odds (seemingly, not actually) just went from 33% to 50%, momentum is good, I'm feeling lucky, no way am I going to switch now!
- DoctorOetker 3y ago
  > If you watch the show religiously it might dawn on you, hey, [...] You can't watch a hypothetical game show. Let's make a deal contains a myriad of different gameplay formats.
  - andy800 3y ago
    The "Big Deal" -- the game where the player chooses one of three doors -- was the finale of every episode of Let's Make a Deal under Monty Hall. Still is, in the Wayne Brady revival. Not hypothetical.

jjnoakes 3y ago

> If the game show host is opening a door at random (i.e. it's possible that he opens the door with the car behind it OR the door you already picked) then the outcome is 50/50 whether you switch or not. I don't think this is true. Assuming you mean that if the host opens my box, I switch to one if the other two at random with equal probability (even if he opened my box and it was the car), and assuming if the host doesn't open my box that I stay or switch to the other unopened box with equal probability (even if the host opened the box with the car), I think the odds are 33% that I win the car, and 66% that I don't.

cman1444 3y ago
Consider it framed this way: You pick door #3. The host says they're feeling generous so he is going to give you a free door. He blindly reaches into a bag with three marbles labeled 1,2, and 3. He pulls out the marble labeled 2. He opens door 2 and there's a goat behind it. Do you now switch to door 1? The answer is that it doesn't matter, it's 50/50.
- jjnoakes 3y ago
  Right, in the world where the door he already opened was done randomly (i.e. it could have been a goat or it could have been a car, and it could have been the door you also chose), then if he happened to both (a) open one of the doors you didn't pick, and (b) he opened a goat door, then switching doesn't matter. But that's only one branch of the probability tree, and if you investigate all of them (assuming he chooses a door at random equally, might open yours, and might open a car), then it's not 50/50, it's 33/66 I think. Perhaps you just weren't clear enough that you think the 50/50 is limited to only a subset of the possible outcomes in your scenario.
  - kgwgk 3y ago
    What would "switching" even mean if the door that was opened was the door that you had picked and the car was there? What would be the point of switching to the remaining door if the door that was opened was not the one you picked and the car was there? (The probability of getting the car would be zero either way.) You're right that the subspace of scenarios where (a) and (b) happen is just part of the full space of things that could have happened - but it's the only part relevant for the problem. (We're explicitely told that (a) and (b) are true!)

travisjungroth 3y ago

It’s also helpful (but not essential) to realize that the choice of which door to open has two parts, a rule and a decision. If the participant is doing a switch or a stay strategy, the decision aspect has no impact on winning. The rule part is Monty can’t reveal the car. The decision comes up if the participant chooses the car first. Monty will have two doors to choose from to open. But if you have a stay strategy you’ll win for either choice, and if you have a switch strategy you’ll lose no matter what. So his decision doesn’t matter. This makes it very easy to simulate. It also might lead to people to understanding the solution. If you pick wrong first, you’ll see staying always loses and switching always wins. The probability of picking right first is 1/3, of picking wrong first 2/3. Since switch always works when you pick wrong and you pick wrong 2/3 times, go with that.