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Yes, it does! The whole problem revolves around whether Monty Hall was giving you more information by his choice of which doors to open, or whether he was just

Yes, it does! The whole problem revolves around whether Monty Hall was giving you more information by his choice of which doors to open, or whether he was just opening doors randomly. If he was opening doors randomly then you're left with a one-in-two chance. But if he was giving you information by opening doors he knew were bad, you're left with two possibilities: the door you started with just happened to be the right door (a tiny chance), or (much more likely) the door you started with was not the right door - in which case Monty was forced to open every other door so as to avoid showing you the car, and he has eliminated all the non-car options.

mcphage 7y ago
What would happen if he opened the right door?
- rohit2412 7y ago
  The assumption is that this didn't happen. Hence you must remove these cases from your sample set while calculating probability
  - baddox 7y ago
    But that means that something knows which door was the winning one. Either Monty knew which one it was, or whatever mysterious Agent is sampling from all the possible outcomes knew (because the Agent only sampled the ones where Monty happened to open a non-winning door). Either way, you're given the choice between having the 1 door you initially chose, or both of the other doors.
    - leeoniya 7y ago
      > Either way, you're given the choice between having the 1 door you initially chose, or both of the other doors. i initially thought this too, but in both cases you're gonna see 2/3 doors. the thing that makes the difference is when he has the car (2/3 of the time), he has to tell you where it is. so the choice becomes between your unknown door and his 2/3 selectively chosen car door.
    - rohit2412 7y ago
      Monty, the game show host, knows.
  - dekhn 7y ago
    this is exactly what tripped me up when I first heard the problem and it took me a long time to understand the explanation because I didn't understand that Monty always opened the door that did not have the car behind it.
  - mcphage 7y ago
    But it’s from a game show. Like, an actual show. 1/3 of the time they can’t just stop filming and say "whoops let’s pretend that didn’t happen." That interpretation of the problem makes no sense whatsoever.
henrikschroder 7y ago
> 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

mcphage 7y ago

What would happen if he opened the right door?

rohit2412 7y ago
The assumption is that this didn't happen. Hence you must remove these cases from your sample set while calculating probability
- baddox 7y ago
  But that means that something knows which door was the winning one. Either Monty knew which one it was, or whatever mysterious Agent is sampling from all the possible outcomes knew (because the Agent only sampled the ones where Monty happened to open a non-winning door). Either way, you're given the choice between having the 1 door you initially chose, or both of the other doors.
  - leeoniya 7y ago
    > Either way, you're given the choice between having the 1 door you initially chose, or both of the other doors. i initially thought this too, but in both cases you're gonna see 2/3 doors. the thing that makes the difference is when he has the car (2/3 of the time), he has to tell you where it is. so the choice becomes between your unknown door and his 2/3 selectively chosen car door.
  - rohit2412 7y ago
    Monty, the game show host, knows.
- dekhn 7y ago
  this is exactly what tripped me up when I first heard the problem and it took me a long time to understand the explanation because I didn't understand that Monty always opened the door that did not have the car behind it.
- mcphage 7y ago
  But it’s from a game show. Like, an actual show. 1/3 of the time they can’t just stop filming and say "whoops let’s pretend that didn’t happen." That interpretation of the problem makes no sense whatsoever.

henrikschroder 7y ago

> 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