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I am curious how you feel about the "Monty Fall"/"Monty Crawl" problems linked elsewhere in the thread.[0] For my part I am somewhat sympathetic to jncfhnb's p

I am curious how you feel about the "Monty Fall"/"Monty Crawl" problems linked elsewhere in the thread.[0] For my part I am somewhat sympathetic to jncfhnb's point. The exact phrasing that Vos Savant was asked did not specify the rules that the host was required to open a door, nor that it always must be a goat door. It simply says that the host has knowledge of what's behind the doors, and in this particular iteration of the game, he showed you a goat and asked you about switching. That does not exclude a scenario where the host is a manipulative fellow, who chose to show you the goat only because he knows you are about to win, trying to convince you to lose. A contestant on the real show would surely worry about this possibility. Of course, the people who wrote to disagree with Vos Savant almost never said "the problem is not fully stated", they said "it's 50/50 you fool", which isn't right. Additionally, since it wouldn't be a math problem at all if we let the host have agency, it is reasonable to assume the unspoken rule that he does not, leading to Marilyn's correct 2/3rds answer. [0]https://web.archive.org/web/20230706235720/https://probability.ca/jeff/writing/montyfall.pdf https://web.archive.org/web/20230706235720/https://probabili...

bemusedthrow75 3y ago
It doesn't make a difference what causes Monty to reveal a goat. A goat behind a door the contestant did not choose is eliminated. That is all that matters. It could have been done by a Heath-Robinson machine, or a passing lonely shrubber, or elves. Monty isn't a floating variable in the puzzle who makes choices. His choice is fixed. Which I imagine is why vos Savant adds the actually extraneous information that Monty is fully aware what is going on behind the scenes -- to underscore the concept that Monty isn't a variable. The fact that a goat behind an unchosen door was revealed is what is crucial to the setup of the entire puzzle. And that -- despite jncfhnb's protestations -- is information that makes the puzzle determinate.
- rspeele 3y ago
  > It doesn't make a difference what causes Monty to reveal a goat. Oh, but it does! See the "Monty Fall" version of the problem, in which Monty accidentally trips and opens a door, which just happens to reveal a goat. In this variant there is no advantage gained by switching, because no more information was revealed about the remaining unopened door. The information gain only happens in the original game because we know that Monty was forced to avoid the winning door in the (66% likely) case where we didn't already pick it.
  - deleted 3y ago
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  - bemusedthrow75 3y ago
    Nope. But I am going to leave this to someone else to explain. I'm tired out now.
    - jncfhnb 3y ago
      Suppose Car is C. The only possible options are (equally likely) You choose A and get shown B You choose B and get shown A You choose C and get shown A You choose C and get shown B 4 options. You lose 50% of the time. The 2 options where you would normally get the 2/3 odds are explicitly ignored when you are told Monty chosen randomly and randomly got a goat. You choose A and get shown C or you choose B and get shown C are forbidden. 50/50 probability on the dot. If Monty chooses with intention then the options where you choose goat and are shown goat double in probability so you get back to 2/3
    - rspeele 3y ago
      They are two different problems. Map out all the scenarios exhaustively and you'll find the difference. In both cases we were originally 1/3 chance of being right. That is not in dispute. In the original (fully defined) "Monty Hall", Monty was going to show us a goat no matter what. It's part of the rules, he has to show a goat. So the fact that we see a goat behind the revealed door is no surprise, and no new information. But which of the two unchosen doors was the goat, is valuable information because in 2/3s of the scenarios Monty's hand was tied and he HAD to show that door to avoid revealing the remaining car. In the "Monty Fall" problem, the fact that we see a goat at all is interesting information. This becomes more likely when we picked the car in the first place, because if we had initially picked the car, and a random other door is opened, it's 100% going to be a goat, whereas if we had picked the goat in the first place, we are only 50% likely to see a goat when a random other door is opened. Let's call the goats Alice and Bob to illustrate this point. We know we DID see a goat, but we don't know which of these equally probable scenarios led to that: 1. We picked the car and saw Alice 2. We picked the car and saw Bob 3. We picked Alice and saw Bob 4. We picked Bob and saw Alice Notice how "we picked the car" originally had 1/3 odds but represents half the scenarios that remain possible, because there are two ways to see a goat with that start, while only one way to see a goat with the others. This kind of brings the problem back around to similar territory as Bertrand's Box[0] where the fact that you drew a gold coin is already hinting to you that you're more likely on the "both gold" box than on the "half gold" box. [0]https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox
- jncfhnb 3y ago
  Noooo It is not the fact that a goat is behind the door. It is the fact that a goat door would always have been opened! These two facts are not the same!
  - deleted 3y ago
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  - bemusedthrow75 3y ago
    I'm sorry what now? Stop second guessing the puzzle which is clearly stated. I'm so, so done with this now that I am actually going to render inoperable my only HN account so I cannot possibly come back to this, or any other thread.
    - jncfhnb 3y ago
      It is NOT STATED in THIS VERSION OF THE PROMPT that monte will always open a door and that it will always be a goat door. It is simply stated that on one play of the game, Monte CHOSE to open a door and it was a goat. If this is truly all you know, you have learned nothing. It is only useful information if monte explicitly opens a goat door for you. If he opens a door at random and gets goat your information gain goes away even if it’s the same door! I’ve literally mapped it out for you else where. There’s only 4 outcomes. You can check yourself. 2:2

bemusedthrow75 3y ago

It doesn't make a difference what causes Monty to reveal a goat. A goat behind a door the contestant did not choose is eliminated. That is all that matters. It could have been done by a Heath-Robinson machine, or a passing lonely shrubber, or elves. Monty isn't a floating variable in the puzzle who makes choices. His choice is fixed. Which I imagine is why vos Savant adds the actually extraneous information that Monty is fully aware what is going on behind the scenes -- to underscore the concept that Monty isn't a variable. The fact that a goat behind an unchosen door was revealed is what is crucial to the setup of the entire puzzle. And that -- despite jncfhnb's protestations -- is information that makes the puzzle determinate.

rspeele 3y ago
> It doesn't make a difference what causes Monty to reveal a goat. Oh, but it does! See the "Monty Fall" version of the problem, in which Monty accidentally trips and opens a door, which just happens to reveal a goat. In this variant there is no advantage gained by switching, because no more information was revealed about the remaining unopened door. The information gain only happens in the original game because we know that Monty was forced to avoid the winning door in the (66% likely) case where we didn't already pick it.
- deleted 3y ago
  [deleted]
- bemusedthrow75 3y ago
  Nope. But I am going to leave this to someone else to explain. I'm tired out now.
  - jncfhnb 3y ago
    Suppose Car is C. The only possible options are (equally likely) You choose A and get shown B You choose B and get shown A You choose C and get shown A You choose C and get shown B 4 options. You lose 50% of the time. The 2 options where you would normally get the 2/3 odds are explicitly ignored when you are told Monty chosen randomly and randomly got a goat. You choose A and get shown C or you choose B and get shown C are forbidden. 50/50 probability on the dot. If Monty chooses with intention then the options where you choose goat and are shown goat double in probability so you get back to 2/3
  - rspeele 3y ago
    They are two different problems. Map out all the scenarios exhaustively and you'll find the difference. In both cases we were originally 1/3 chance of being right. That is not in dispute. In the original (fully defined) "Monty Hall", Monty was going to show us a goat no matter what. It's part of the rules, he has to show a goat. So the fact that we see a goat behind the revealed door is no surprise, and no new information. But which of the two unchosen doors was the goat, is valuable information because in 2/3s of the scenarios Monty's hand was tied and he HAD to show that door to avoid revealing the remaining car. In the "Monty Fall" problem, the fact that we see a goat at all is interesting information. This becomes more likely when we picked the car in the first place, because if we had initially picked the car, and a random other door is opened, it's 100% going to be a goat, whereas if we had picked the goat in the first place, we are only 50% likely to see a goat when a random other door is opened. Let's call the goats Alice and Bob to illustrate this point. We know we DID see a goat, but we don't know which of these equally probable scenarios led to that: 1. We picked the car and saw Alice 2. We picked the car and saw Bob 3. We picked Alice and saw Bob 4. We picked Bob and saw Alice Notice how "we picked the car" originally had 1/3 odds but represents half the scenarios that remain possible, because there are two ways to see a goat with that start, while only one way to see a goat with the others. This kind of brings the problem back around to similar territory as Bertrand's Box[0] where the fact that you drew a gold coin is already hinting to you that you're more likely on the "both gold" box than on the "half gold" box. [0]https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox
jncfhnb 3y ago
Noooo It is not the fact that a goat is behind the door. It is the fact that a goat door would always have been opened! These two facts are not the same!
- deleted 3y ago
  [deleted]
- bemusedthrow75 3y ago
  I'm sorry what now? Stop second guessing the puzzle which is clearly stated. I'm so, so done with this now that I am actually going to render inoperable my only HN account so I cannot possibly come back to this, or any other thread.
  - jncfhnb 3y ago
    It is NOT STATED in THIS VERSION OF THE PROMPT that monte will always open a door and that it will always be a goat door. It is simply stated that on one play of the game, Monte CHOSE to open a door and it was a goat. If this is truly all you know, you have learned nothing. It is only useful information if monte explicitly opens a goat door for you. If he opens a door at random and gets goat your information gain goes away even if it’s the same door! I’ve literally mapped it out for you else where. There’s only 4 outcomes. You can check yourself. 2:2