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The Monty Hall problem is actually a bit more subtle than this article makes out. The exact wording of the problem can change the answer. For example, in this v

The Monty Hall problem is actually a bit more subtle than this article makes out. The exact wording of the problem can change the answer. For example, in this version > There are two doors with goats and one with a car. You choose one door from the three. The host selects one of the doors with a goat from the remaining two doors, and opens it. Should you switch doors if given the chance? has a different answer than this one > There are two doors with goats and one with a car. You choose one door from the three. The host chooses one of the remaining two doors at random and opens it, showing a goat. Should you switch doors if given the chance? In the first example the hosts choice is either forced (if you chose a goat initially) or doesn't make a difference (if you chose a car). In the second example, the host could have opened a door with a car because he was choosing the doors at random. Given that he didn't, your estimate of the probability that you have chosen the car should increase - in fact, it should increase to 1/2, and switching doesn't make any difference. To me, this is the really interesting thing about the Monty Hall problem - the knowledge and intentions of the host are important in the correct solution to the problem! So there are two ways to come up with the answer "it doesn't matter if you switch" - 1. You understand the problem in the first sense above, and your intuition leads you to an incorrect answer. 2. You understand the problem in the second sense above, and you are correct that it doesn't matter whether you switch or not.

tromp 12y ago
And then there's this version > There are two doors with goats and one with a car. You choose one door from the three. The host may select one of the doors with a goat from the remaining two doors, and open it. Should you switch doors if given the chance? Now the probability can be anything from 0 to 1, depending on what makes the host tick... Considering the usual ambiguity in the problem formulation, the only safe reply is "It depends".
randomstring 12y ago
Which would explain why Erdos might be confounded by the answer. He was an expert at numbers, not psychology. Proofs are built with an adversarial mind set. Always assuming the worst possible case and proving that a theorem holds in all conditions. The "hand of god" (or in this case the game show host) doesn't move to make theorems easier to prove. That isn't the universe works!
jayvanguard 12y ago
This is exactly right. It drives me nuts that most explanations gloss over this. If you're a mathematician you don't tend to introduce any additional assumptions unless they are stated. Assuming the host has knowledge of what is behind all of the doors (or at least where the car is) changes everything. Some wordings of the question include it, others don't. Of course a non-mathematician might assume things about the game show host based on their knowledge of game show hosts but a mathematician is entirely correct to simply take the question at face value and not inject additional unstated assumptions. Otherwise what is preventing them from putting in other completely wacky assumptions like "the host wants you to win the car and knows you know about the monty hall problem and is using reverse psychology"?
____a 12y ago
My understanding is that in both situations as you described above you should switch. You are still in a new conditional state. Telling you that the host opened another door by chance and it is a goat removes the possibility that he opened the door with the car behind it by chance. You have eliminated one of the options and you have the 2/3 chance of winning by switching.
- throwaway283719 12y ago
  No. In the first case, the fact that the host shows you a goat has not put you in any new state of the world (the host was going to show you a goat regardless of what happened). In the second case, you now know that you are not in one of the states of the world where the host accidentally opened a door that had a car behind it. They could have opened a door with a car, but they didn't. That gives you some information - it stands to reason that since they didn't open a door with a car, it was probably a bit more likely that they would pick a door with a goat, which makes it more likely that there are two goats than that there is one car and one goat, which is equivalent to saying that it's a bit more likely that you have already chosen the car.
  - ____a 12y ago
    Okay, yes. Had to work it out on paper to convince myself that is works out to 1/2 in the second case. Thank you.
- deleted 12y ago
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Cushman 12y ago
It's definitely fascinating! Though I feel like attributing this to the intentions of the host is misleading, because it implies something mystical or quantum might be going on. It's more straightforward to just say it depends on the actions of the host. Specifically, the host in the first formulation never reveals a car, even if she wants to. The host in the second formulation reveals a car roughly 1/3 of the time, even if she'd rather not. Thus, intuitively, it makes sense that the former leaks information about the place of the car, and the latter doesn't. [0] Of course that's just a rewording of your point, based on my sense that "knowledge and intentions" are concepts that apply readily to game show hosts, but awkwardly to rule-following systems in general. [0] To the extent that revealing something can not leak information about it... Edit: To elucidate my quibble further, as a third formulation: "The host chooses one of the two remaining doors at random, which, by coincidence, is always a goat." I assert we should always switch: The intention of the host is the same as in the second case, but the behavior of the system, magically known to us, is that of the first. We make decisions based on that.
- dllthomas 12y ago
  Lots of quantum things are surely going on... But really, it has to do with the information content of the actions of the host.
  - Cushman 12y ago
    Just speaking for me, I'm not confident there's a "really" about this, so I'm not too concerned with finding it. I'm just trying to find the easiest way for our co-commenters to understand why the correct answer is, in fact, correct :)
    - dllthomas 12y ago
      Right. My point was just that the actions of the host, as far as we observe, can be the same in both cases.
      - Cushman 12y ago
Sandman 12y ago
Hmm.. I don't think there's a difference. In two out of three cases, one of the remaining two doors contains a car. If the host opens the one that does, ok, bummer, you lost. But if not, there's a good chance that the other door contains a car. Remember that you always had only one in three chances of choosing the right door, this doesn't change.
- jayvanguard 12y ago
  There is a difference. Write a simulation in your favorite language. You'll be forced to decide whether the door with the car ever gets revealed to the contestant. In those cases does the contestant get to switch and win the car? Or did you just throw those cases out in your code? If so, you made an assumption that isn't aways stated clearly in the problem.
  - skizm 12y ago
    But your question was if the host chooses a door at random and reveals a goat. Which means it doesn't matter if he had knowledge or not. You told us the outcome. If you were writing a computer simulation there would be no rgn for the host picking his door. The host would always open the door with the goat, same as if he knew since you already told us he opens the door with the goat.
    - jayvanguard 12y ago
      He does in this single instance, but in order to calculate the probably you need to consider all the possible cases. Does the case where opens the door with the car calculate in? Why or why not? Seriously, write the simulation to calculate the probabilities. It makes the assumptions very obvious.
      - skizm 12y ago
        You told us he opens the door with the goat. All cases consist of the host opening the door with a goat after you have picked.
ghusbands 12y ago
On top of this, if you're at all unsure about the wording, you can apply the meta-solution that it's best to switch, because the chance of it giving you the car is always at least 1/2 and has a non-zero probability of being 2/3. That gives an expected cross-universe probability of more than 1/2.
- jayvanguard 12y ago
  This is true but it isn't really a mathematical question then with a single correct answer. If I was in a culture where gameshow hosts behaved differently the question would have different priors and a different answer. What if all gameshow hosts shot contestants immediately after they won? Would you be better of switching? No, you'd be dead, so you didn't "win" the car. Completely silly example but it shows why this can't be taken seriously as a mathematical question with probabilities without injecting additional assumptions (unless of course it is stated clearly but then I bet few mathematicians would get tripped up).
deleted 12y ago
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- Philadelphia 12y ago
  No, he didn't. See the article linked in my other comment. He had many more options, including opening the contestant's door, offering money if the contestant abandoned the door, offering money to switch, etc.
Philadelphia 12y ago
My favorite overview of this is a New York Times article that asked Monty Hall himself about the problem [1]. I think I first saw it linked from from his biographical page on Wikipedia. [1] http://www.nytimes.com/1991/07/21/us/behind-monty-hall-s-doors-puzzle-debate-and-answer.html?src=pm&pagewanted=1 http://www.nytimes.com/1991/07/21/us/behind-monty-hall-s-doo...
epidemian 12y ago
Wow. As many others here, i was sure there wouldn't be a difference if Monty picked at random AND we ignored the cases where he picks the door with the car. But after reading "write the damn simulation!" too many times, i wrote one, being sure it would prove me right. Of course, i was wrong :D The simulation testing the four possible scenarios: #!/usr/bin/env ruby def run_games(monty_knows:, change_choice:) options = [:goat1, :goat2, :car] total_games = 0 wins = 0 100000.times do choice = options.sample monty_choice = (options - [choice, (:car if monty_knows)]).sample # Don't consider cases where Monty shows a car. next if monty_choice == :car if change_choice choice = (options - [monty_choice, choice]).first end total_games += 1 wins += 1 if choice == :car end wins / total_games.to_f end puts 'Monty knows & keep choice: %.2f' % run_games(monty_knows: true, change_choice: false) puts 'Monty knows & change choice: %.2f' % run_games(monty_knows: true, change_choice: true) puts 'Monty ignores & keep choice: %.2f' % run_games(monty_knows: false, change_choice: false) puts 'Monty ignores & change choice: %.2f' % run_games(monty_knows: false, change_choice: true) Sample output: Monty knows & keep choice: 0.33 Monty knows & change choice: 0.67 Monty ignores & keep choice: 0.50 Monty ignores & change choice: 0.50 So, in the case of Monty picking at random, even if we ignore the cases where he picks the car, your chances of winning do not improve by changing your first choice or not. After verifying that the simulation makes sense, i tried to figure out why it makes sense. In an information theory kinda way, it makes sense that if Monty picks at random then he is not providing any new information: you are just being shown what was behind a random door you didn't choose and you can't choose now. But that explanation is not intuitive to my brain. I found it more intuitive to actually draw a decision tree with the possibilities and their probabilities. The tree is actually very simple: With the Always Change Door strategy: car (1/3) -> LOSE (Monty always reveals goat, / we change and we always lose) first choice \ goat (1/2) -> WIN (we change to the \ / car door) goat (2/3) -> Monty choice \ car (1/2) -> IGNORE The probability of winning is 1/3 (2/3 * 1/2). With the Always Keep Door strategy: car (1/3) -> WIN (doesn't matter what Monty chooses) / first choice \ goat (1/2) -> LOSE (we keep our \ / goat-hiding door) goat (2/3) -> Monty choice \ car (1/2) -> IGNORE Same tree, but with win and lose cases inverted. The probability of winning is also 1/3. In conclusion, my intuition sucks. Thanks for showing me so!

tromp 12y ago

And then there's this version > There are two doors with goats and one with a car. You choose one door from the three. The host may select one of the doors with a goat from the remaining two doors, and open it. Should you switch doors if given the chance? Now the probability can be anything from 0 to 1, depending on what makes the host tick... Considering the usual ambiguity in the problem formulation, the only safe reply is "It depends".

randomstring 12y ago

Which would explain why Erdos might be confounded by the answer. He was an expert at numbers, not psychology. Proofs are built with an adversarial mind set. Always assuming the worst possible case and proving that a theorem holds in all conditions. The "hand of god" (or in this case the game show host) doesn't move to make theorems easier to prove. That isn't the universe works!

jayvanguard 12y ago

This is exactly right. It drives me nuts that most explanations gloss over this. If you're a mathematician you don't tend to introduce any additional assumptions unless they are stated. Assuming the host has knowledge of what is behind all of the doors (or at least where the car is) changes everything. Some wordings of the question include it, others don't. Of course a non-mathematician might assume things about the game show host based on their knowledge of game show hosts but a mathematician is entirely correct to simply take the question at face value and not inject additional unstated assumptions. Otherwise what is preventing them from putting in other completely wacky assumptions like "the host wants you to win the car and knows you know about the monty hall problem and is using reverse psychology"?

____a 12y ago

My understanding is that in both situations as you described above you should switch. You are still in a new conditional state. Telling you that the host opened another door by chance and it is a goat removes the possibility that he opened the door with the car behind it by chance. You have eliminated one of the options and you have the 2/3 chance of winning by switching.

throwaway283719 12y ago
No. In the first case, the fact that the host shows you a goat has not put you in any new state of the world (the host was going to show you a goat regardless of what happened). In the second case, you now know that you are not in one of the states of the world where the host accidentally opened a door that had a car behind it. They could have opened a door with a car, but they didn't. That gives you some information - it stands to reason that since they didn't open a door with a car, it was probably a bit more likely that they would pick a door with a goat, which makes it more likely that there are two goats than that there is one car and one goat, which is equivalent to saying that it's a bit more likely that you have already chosen the car.
- ____a 12y ago
  Okay, yes. Had to work it out on paper to convince myself that is works out to 1/2 in the second case. Thank you.
deleted 12y ago
[deleted]

Cushman 12y ago

It's definitely fascinating! Though I feel like attributing this to the intentions of the host is misleading, because it implies something mystical or quantum might be going on. It's more straightforward to just say it depends on the actions of the host. Specifically, the host in the first formulation never reveals a car, even if she wants to. The host in the second formulation reveals a car roughly 1/3 of the time, even if she'd rather not. Thus, intuitively, it makes sense that the former leaks information about the place of the car, and the latter doesn't. [0] Of course that's just a rewording of your point, based on my sense that "knowledge and intentions" are concepts that apply readily to game show hosts, but awkwardly to rule-following systems in general. [0] To the extent that revealing something can not leak information about it... Edit: To elucidate my quibble further, as a third formulation: "The host chooses one of the two remaining doors at random, which, by coincidence, is always a goat." I assert we should always switch: The intention of the host is the same as in the second case, but the behavior of the system, magically known to us, is that of the first. We make decisions based on that.

dllthomas 12y ago
Lots of quantum things are surely going on... But really, it has to do with the information content of the actions of the host.
- Cushman 12y ago
  Just speaking for me, I'm not confident there's a "really" about this, so I'm not too concerned with finding it. I'm just trying to find the easiest way for our co-commenters to understand why the correct answer is, in fact, correct :)
  - dllthomas 12y ago
    Right. My point was just that the actions of the host, as far as we observe, can be the same in both cases.
    - Cushman 12y ago

Sandman 12y ago

Hmm.. I don't think there's a difference. In two out of three cases, one of the remaining two doors contains a car. If the host opens the one that does, ok, bummer, you lost. But if not, there's a good chance that the other door contains a car. Remember that you always had only one in three chances of choosing the right door, this doesn't change.

jayvanguard 12y ago
There is a difference. Write a simulation in your favorite language. You'll be forced to decide whether the door with the car ever gets revealed to the contestant. In those cases does the contestant get to switch and win the car? Or did you just throw those cases out in your code? If so, you made an assumption that isn't aways stated clearly in the problem.
- skizm 12y ago
  But your question was if the host chooses a door at random and reveals a goat. Which means it doesn't matter if he had knowledge or not. You told us the outcome. If you were writing a computer simulation there would be no rgn for the host picking his door. The host would always open the door with the goat, same as if he knew since you already told us he opens the door with the goat.
  - jayvanguard 12y ago
    He does in this single instance, but in order to calculate the probably you need to consider all the possible cases. Does the case where opens the door with the car calculate in? Why or why not? Seriously, write the simulation to calculate the probabilities. It makes the assumptions very obvious.
    - skizm 12y ago
      You told us he opens the door with the goat. All cases consist of the host opening the door with a goat after you have picked.

ghusbands 12y ago

On top of this, if you're at all unsure about the wording, you can apply the meta-solution that it's best to switch, because the chance of it giving you the car is always at least 1/2 and has a non-zero probability of being 2/3. That gives an expected cross-universe probability of more than 1/2.

jayvanguard 12y ago
This is true but it isn't really a mathematical question then with a single correct answer. If I was in a culture where gameshow hosts behaved differently the question would have different priors and a different answer. What if all gameshow hosts shot contestants immediately after they won? Would you be better of switching? No, you'd be dead, so you didn't "win" the car. Completely silly example but it shows why this can't be taken seriously as a mathematical question with probabilities without injecting additional assumptions (unless of course it is stated clearly but then I bet few mathematicians would get tripped up).

deleted 12y ago

[deleted]

Philadelphia 12y ago
No, he didn't. See the article linked in my other comment. He had many more options, including opening the contestant's door, offering money if the contestant abandoned the door, offering money to switch, etc.

Philadelphia 12y ago

My favorite overview of this is a New York Times article that asked Monty Hall himself about the problem [1]. I think I first saw it linked from from his biographical page on Wikipedia. [1] http://www.nytimes.com/1991/07/21/us/behind-monty-hall-s-doors-puzzle-debate-and-answer.html?src=pm&pagewanted=1 http://www.nytimes.com/1991/07/21/us/behind-monty-hall-s-doo...

epidemian 12y ago

Wow. As many others here, i was sure there wouldn't be a difference if Monty picked at random AND we ignored the cases where he picks the door with the car. But after reading "write the damn simulation!" too many times, i wrote one, being sure it would prove me right. Of course, i was wrong :D The simulation testing the four possible scenarios: #!/usr/bin/env ruby def run_games(monty_knows:, change_choice:) options = [:goat1, :goat2, :car] total_games = 0 wins = 0 100000.times do choice = options.sample monty_choice = (options - [choice, (:car if monty_knows)]).sample # Don't consider cases where Monty shows a car. next if monty_choice == :car if change_choice choice = (options - [monty_choice, choice]).first end total_games += 1 wins += 1 if choice == :car end wins / total_games.to_f end puts 'Monty knows & keep choice: %.2f' % run_games(monty_knows: true, change_choice: false) puts 'Monty knows & change choice: %.2f' % run_games(monty_knows: true, change_choice: true) puts 'Monty ignores & keep choice: %.2f' % run_games(monty_knows: false, change_choice: false) puts 'Monty ignores & change choice: %.2f' % run_games(monty_knows: false, change_choice: true) Sample output: Monty knows & keep choice: 0.33 Monty knows & change choice: 0.67 Monty ignores & keep choice: 0.50 Monty ignores & change choice: 0.50 So, in the case of Monty picking at random, even if we ignore the cases where he picks the car, your chances of winning do not improve by changing your first choice or not. After verifying that the simulation makes sense, i tried to figure out why it makes sense. In an information theory kinda way, it makes sense that if Monty picks at random then he is not providing any new information: you are just being shown what was behind a random door you didn't choose and you can't choose now. But that explanation is not intuitive to my brain. I found it more intuitive to actually draw a decision tree with the possibilities and their probabilities. The tree is actually very simple: With the Always Change Door strategy: car (1/3) -> LOSE (Monty always reveals goat, / we change and we always lose) first choice \ goat (1/2) -> WIN (we change to the \ / car door) goat (2/3) -> Monty choice \ car (1/2) -> IGNORE The probability of winning is 1/3 (2/3 * 1/2). With the Always Keep Door strategy: car (1/3) -> WIN (doesn't matter what Monty chooses) / first choice \ goat (1/2) -> LOSE (we keep our \ / goat-hiding door) goat (2/3) -> Monty choice \ car (1/2) -> IGNORE Same tree, but with win and lose cases inverted. The probability of winning is also 1/3. In conclusion, my intuition sucks. Thanks for showing me so!