3 ms·

Understanding the probability tree is fairly interesting though. If he opens a door and tells you he knows it's a goat, you double your likelihood of winning b

Understanding the probability tree is fairly interesting though. If he opens a door and tells you he knows it's a goat, you double your likelihood of winning by switching to the remaining door. If he opens a door randomly, and gets a goat, you don't modify your likelihood at all by switching. Saying 1 door or 2 doors doesn't mean you actually grasp the entirety of the problem.

basch 6y ago
>If he opens a door randomly, and gets a goat, you don't modify your likelihood at all by switching. At best, that was intended as a red herring to make the correct solution less obvious. It was not intended as an alternate correct answer. It's not how the game ever worked. The initial version of the question said he opens a goat door. Whether he knew it would be a goat door or not is slightly irrelevant, because your odds of being right the first time were 1/3. As far as the premise of the thought experiment, a goat door always gets opened, either through peaking, premonition, or consistent luck. >Saying 1 door or 2 doors doesn't mean you actually grasp the entirety of the problem. I disagree, and if you don't see it as that simple, you are falling for the trap that makes the question fun.
- klmadfejno 6y ago
  I don't have a particularly strong opinion on this, but if we accept that its interesting as a thought experiment and not a useful strat should one find themselves time traveled into the monte hall game show, I think it's worth exploring the very closely related variants of the thought experiment.
  - basch 6y ago
    If I time traveled into the show, and he opened a goat door, I would switch.
    - twiceaday 6y ago
      There is no better strategy than switching. Only under certain rules is there an equally good strategy in staying.
      - klmadfejno 6y ago
        Not if switching costs a dollar :)
  - boomboomsubban 6y ago
    Not important, and just barely relevant, but "Let's Make a Deal" was Monty Hall's gameshow and it's still being made.
majormajor 6y ago
> If he opens a door randomly, and gets a goat, you don't modify your likelihood at all by switching. Saying 1 door or 2 doors doesn't mean you actually grasp the entirety of the problem. What do you mean "opens a door randomly"? Is he picking from all three doors? Yours, and the two others? In that cases you get interesting but trivially-obvious-what-move-to-make scenarios like "he chose your door and showed you you were right" "he chose your door and showed you you were wrong" "he chose a different door which had the car"... Do you just mean the subset of "he chose randomly and happened to draw a goat out of one of the two you did not choose"? In which case switching isn't beneficial because you no longer are also capturing the cases that would otherwise be the "he chose randomly and opened the one with the car that you did not choose" that are included in the original "switch or not" decision because he always goes to a goat?
- basch 6y ago
  >Do you just mean the subset of "he chose randomly and happened to draw a goat out of one of the two you did not choose"? In which case switching isn't beneficial because Yes switching is. Switching is beneficial IF he shows you a goat out of the doors you didnt choose.
  - kgwgk 6y ago
    https://news.ycombinator.com/item?id=24713352 https://news.ycombinator.com/item?id=24713352
  - majormajor 6y ago
    If the host opens a door truly at random... - you choose door 1, goat -- host chooses door 1, goat --- switch: 1/2 chance of being right --- stay: you're wrong -- host chooses door 2, goat --- switch: you are right --- stay: you are wrong -- host chooses door 3, car --- switch: you are right --- stay: you are wrong - you choose door 2, goat -- host chooses door 1, goat --- switch: you are right --- stay: you're wrong -- host chooses door 2, goat --- switch: 1/2 chance of being right --- stay: you are wrong -- host chooses door 3, car --- switch: you are right --- stay: you are wrong - you choose door 3, car -- host chooses door 1, goat --- switch: you are wrong --- stay: you are right -- host chooses door 2, goat --- switch: you are wrong --- stay: you are right -- host chooses door 3, car --- switch: you are wrong --- stay: you are right so there are 6 times out of 9 where the host shows you a goat total, and you are right if you switch 3 out of those 6 times. if you stay each of those 6 times, you're right 2 out of the 6 times. there are 4 times the host shows you a goat that you didn't choose, and if you switch you are right 2 out of those times. but in this scenario, you are actually right 2 out of those 4 times if you stay, too. this seems opposite of what i expected. I guess if the host is choosing at random, and you conditionalize on "being shown a goat AND a door you didn't choose" then you're reducing that particular sample space down to be more heavy in the "you got it right originally" scenario, because the "you got it wrong originally" scenarios rule out half the goat doors from the hosts choices we're considering.

basch 6y ago

>If he opens a door randomly, and gets a goat, you don't modify your likelihood at all by switching. At best, that was intended as a red herring to make the correct solution less obvious. It was not intended as an alternate correct answer. It's not how the game ever worked. The initial version of the question said he opens a goat door. Whether he knew it would be a goat door or not is slightly irrelevant, because your odds of being right the first time were 1/3. As far as the premise of the thought experiment, a goat door always gets opened, either through peaking, premonition, or consistent luck. >Saying 1 door or 2 doors doesn't mean you actually grasp the entirety of the problem. I disagree, and if you don't see it as that simple, you are falling for the trap that makes the question fun.

klmadfejno 6y ago
I don't have a particularly strong opinion on this, but if we accept that its interesting as a thought experiment and not a useful strat should one find themselves time traveled into the monte hall game show, I think it's worth exploring the very closely related variants of the thought experiment.
- basch 6y ago
  If I time traveled into the show, and he opened a goat door, I would switch.
  - twiceaday 6y ago
    There is no better strategy than switching. Only under certain rules is there an equally good strategy in staying.
    - klmadfejno 6y ago
      Not if switching costs a dollar :)
- boomboomsubban 6y ago
  Not important, and just barely relevant, but "Let's Make a Deal" was Monty Hall's gameshow and it's still being made.

majormajor 6y ago

> If he opens a door randomly, and gets a goat, you don't modify your likelihood at all by switching. Saying 1 door or 2 doors doesn't mean you actually grasp the entirety of the problem. What do you mean "opens a door randomly"? Is he picking from all three doors? Yours, and the two others? In that cases you get interesting but trivially-obvious-what-move-to-make scenarios like "he chose your door and showed you you were right" "he chose your door and showed you you were wrong" "he chose a different door which had the car"... Do you just mean the subset of "he chose randomly and happened to draw a goat out of one of the two you did not choose"? In which case switching isn't beneficial because you no longer are also capturing the cases that would otherwise be the "he chose randomly and opened the one with the car that you did not choose" that are included in the original "switch or not" decision because he always goes to a goat?

basch 6y ago
>Do you just mean the subset of "he chose randomly and happened to draw a goat out of one of the two you did not choose"? In which case switching isn't beneficial because Yes switching is. Switching is beneficial IF he shows you a goat out of the doors you didnt choose.
- kgwgk 6y ago
  https://news.ycombinator.com/item?id=24713352 https://news.ycombinator.com/item?id=24713352
- majormajor 6y ago
  If the host opens a door truly at random... - you choose door 1, goat -- host chooses door 1, goat --- switch: 1/2 chance of being right --- stay: you're wrong -- host chooses door 2, goat --- switch: you are right --- stay: you are wrong -- host chooses door 3, car --- switch: you are right --- stay: you are wrong - you choose door 2, goat -- host chooses door 1, goat --- switch: you are right --- stay: you're wrong -- host chooses door 2, goat --- switch: 1/2 chance of being right --- stay: you are wrong -- host chooses door 3, car --- switch: you are right --- stay: you are wrong - you choose door 3, car -- host chooses door 1, goat --- switch: you are wrong --- stay: you are right -- host chooses door 2, goat --- switch: you are wrong --- stay: you are right -- host chooses door 3, car --- switch: you are wrong --- stay: you are right so there are 6 times out of 9 where the host shows you a goat total, and you are right if you switch 3 out of those 6 times. if you stay each of those 6 times, you're right 2 out of the 6 times. there are 4 times the host shows you a goat that you didn't choose, and if you switch you are right 2 out of those times. but in this scenario, you are actually right 2 out of those 4 times if you stay, too. this seems opposite of what i expected. I guess if the host is choosing at random, and you conditionalize on "being shown a goat AND a door you didn't choose" then you're reducing that particular sample space down to be more heavy in the "you got it right originally" scenario, because the "you got it wrong originally" scenarios rule out half the goat doors from the hosts choices we're considering.