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The hand-wavy way I've always thought about the Monty Hall problem is as follows: When the contestant picked a box he was making a random choice between three

The hand-wavy way I've always thought about the Monty Hall problem is as follows: When the contestant picked a box he was making a random choice between three boxes and had a probability of 1/3 of picking the car. The probability that one of the other two boxes was the car is thus 2/3. When Monty Hall opens a box it's still the case that the probability of the car being in the other two is 2/3 but the contestant can now eliminate the open box and knows that the probability that the car is in the other box is 2/3. Thus it's worth switching.

mattdeboard 11y ago
This is by far the most simple, concise explanation of the Monty Hall problem I've ever read: https://www.reddit.com/r/videos/comments/1bcuo8/this_has_been_explained_to_me_over_and_over_again/c95w7h9 https://www.reddit.com/r/videos/comments/1bcuo8/this_has_bee... > Basically, you probably got a goat the first time, so switch. It's so concise that I use more words to re-post it every time Monty Hall comes up than are in the explanation itself.
- adevine 11y ago
  The other phrasing of the problem that I've seen help people is this (I think was also originally from Marilyn Vos Savant): Suppose there are a thousand doors, and only one is a car. Monty opens 998 of them to reveal goats, leaving door #628 still hidden. Do you switch? That helps people understand that their original probability of picking the car the first time was very small, and didn't change because of what Monty did.
  - darkmighty 11y ago
    Most paradoxes vanish, become obvious this way (when you increase the numbers). The brothers paradox mentioned: 'I have two children. At least one is a boy.'/'I have two children. At least one is a boy born on Tuesday.'; 'What is the probability I have two boys?' drastically changes when you change to 'I have two children. At least one is a boy born at 7h:35m:42s.' -- it becomes clear that the succeeding information exactly specifies the boy as it becomes specific, in the limit turning into 'I have two children. The first is a boy.' (so the probability goes from 1/3 to 1/2).
  - thaumasiotes 11y ago
    I prefer this one: You pick a door. Monty opens another door, revealing a goat. You now get the option to switch from your door to the other two doors, keeping the car if it's behind either one. Whether Monty opens the door or not doesn't mean much.
    - baddox 11y ago
      An even simpler version, which is equivalent: Monty Hall says that you can either choose 1 door, or 2 doors. It can't get more obvious than that.
      - dllthomas 11y ago
        That is only equivalent if Monty (as intended but usually not stated clearly) always picks a door without a car using knowledge of where the car is. If, instead, Monty picks randomly from the other two doors and just happens not to have revealed the car this time, odds are the same whether you stick with your door or switch. Even in the case where your restatement is mathematically equivalent, it's obviously not at all equivalent as a puzzle because what is being requested by the puzzle is seeing that the original situation is mathematically equivalent to that one.
proksoup 11y ago
Monty chooses a door? I thought he selected it at random. That was always the confusing unstated assumption for me, that makes me think the problem is only confusing because of how the wording usually de-emphasizes that distinction. If he chooses a door, you can benefit from the secret information he reveals sometimes by doing so. If the door was chosen at random, he is not adding any information, so you can't act on it?
- maxerickson 11y ago
  He never reveals a car when he opens the door.
  - malisper 11y ago
    That actually doesn't matter. What does matter is if he knows he is going to reveal a door that is empty. If he does know you should switch. If he reveals a door arbitrarily, and it just so happens to be empty, it doesn't matter if you switch. Edit: Since I have been downvoted, here is a citation: http://probability.ca/jeff/writing/montyfall.pdf http://probability.ca/jeff/writing/montyfall.pdf (see Monty Fall problem)
    - maxerickson 11y ago
      I agree that my phrasing is not superiorally precise, but the meaning is clear enough, the choice of the door isn't arbitrary.
    - kgwgk 11y ago
      To be fully pedantic, what matters is not "if he knows he is going to reveal a door that is empty". Maybe he has no idea and he just reads his lines off a teleprompter. What matters is if you know that he is going to reveal a door that is empty, because then you can reason according to that information.
    - baddox 11y ago
      The notion that he always randomly opens a door without knowing what's behind it but the door is always a losing door is fairly bizarre and doesn't make much sense when you try to analyze probability in the context of repeated experiments. Something must make him always choose a losing door. Either he knows what's behind the doors, or there's some external force ("fate," "God," or whatever) that causes the door to always be a losing door. Regardless, the problem as stated has Monty Hall open a losing door. Thus it's clear that you should switch after Monty Hall does so, and a very simple computer program can show that you tend to win by switching. Another way to phrase the problem is that you choose one door, then Monty Hall (without revealing anything) gives you the option of taking the door you chose, or taking both the other doors. That's an equivalent problem (assuming the goat has zero utility), and it makes it very obvious that you should switch.
zcarter 11y ago
I like to explain it with a wager: pick a star from the night's sky. If you pick the star I'm thinking of, I'll give you a dollar. After you've made your selection, I'll give you the opportunity to switch to one other star, guaranteeing that either your star or this one alternative will be the star I had in mind. Do you want to switch?
dllthomas 11y ago
Your argument isn't really wrong as far as it goes, but it doesn't explain why it matters whether Monty deliberately opens a door without a car or picks randomly and just happened to open a car-less door this time.

mattdeboard 11y ago

This is by far the most simple, concise explanation of the Monty Hall problem I've ever read: https://www.reddit.com/r/videos/comments/1bcuo8/this_has_been_explained_to_me_over_and_over_again/c95w7h9 https://www.reddit.com/r/videos/comments/1bcuo8/this_has_bee... > Basically, you probably got a goat the first time, so switch. It's so concise that I use more words to re-post it every time Monty Hall comes up than are in the explanation itself.

adevine 11y ago
The other phrasing of the problem that I've seen help people is this (I think was also originally from Marilyn Vos Savant): Suppose there are a thousand doors, and only one is a car. Monty opens 998 of them to reveal goats, leaving door #628 still hidden. Do you switch? That helps people understand that their original probability of picking the car the first time was very small, and didn't change because of what Monty did.
- darkmighty 11y ago
  Most paradoxes vanish, become obvious this way (when you increase the numbers). The brothers paradox mentioned: 'I have two children. At least one is a boy.'/'I have two children. At least one is a boy born on Tuesday.'; 'What is the probability I have two boys?' drastically changes when you change to 'I have two children. At least one is a boy born at 7h:35m:42s.' -- it becomes clear that the succeeding information exactly specifies the boy as it becomes specific, in the limit turning into 'I have two children. The first is a boy.' (so the probability goes from 1/3 to 1/2).
- thaumasiotes 11y ago
  I prefer this one: You pick a door. Monty opens another door, revealing a goat. You now get the option to switch from your door to the other two doors, keeping the car if it's behind either one. Whether Monty opens the door or not doesn't mean much.
  - baddox 11y ago
    An even simpler version, which is equivalent: Monty Hall says that you can either choose 1 door, or 2 doors. It can't get more obvious than that.
    - dllthomas 11y ago
      That is only equivalent if Monty (as intended but usually not stated clearly) always picks a door without a car using knowledge of where the car is. If, instead, Monty picks randomly from the other two doors and just happens not to have revealed the car this time, odds are the same whether you stick with your door or switch. Even in the case where your restatement is mathematically equivalent, it's obviously not at all equivalent as a puzzle because what is being requested by the puzzle is seeing that the original situation is mathematically equivalent to that one.

proksoup 11y ago

Monty chooses a door? I thought he selected it at random. That was always the confusing unstated assumption for me, that makes me think the problem is only confusing because of how the wording usually de-emphasizes that distinction. If he chooses a door, you can benefit from the secret information he reveals sometimes by doing so. If the door was chosen at random, he is not adding any information, so you can't act on it?

maxerickson 11y ago
He never reveals a car when he opens the door.
- malisper 11y ago
  That actually doesn't matter. What does matter is if he knows he is going to reveal a door that is empty. If he does know you should switch. If he reveals a door arbitrarily, and it just so happens to be empty, it doesn't matter if you switch. Edit: Since I have been downvoted, here is a citation: http://probability.ca/jeff/writing/montyfall.pdf http://probability.ca/jeff/writing/montyfall.pdf (see Monty Fall problem)
  - maxerickson 11y ago
    I agree that my phrasing is not superiorally precise, but the meaning is clear enough, the choice of the door isn't arbitrary.
  - kgwgk 11y ago
    To be fully pedantic, what matters is not "if he knows he is going to reveal a door that is empty". Maybe he has no idea and he just reads his lines off a teleprompter. What matters is if you know that he is going to reveal a door that is empty, because then you can reason according to that information.
  - baddox 11y ago
    The notion that he always randomly opens a door without knowing what's behind it but the door is always a losing door is fairly bizarre and doesn't make much sense when you try to analyze probability in the context of repeated experiments. Something must make him always choose a losing door. Either he knows what's behind the doors, or there's some external force ("fate," "God," or whatever) that causes the door to always be a losing door. Regardless, the problem as stated has Monty Hall open a losing door. Thus it's clear that you should switch after Monty Hall does so, and a very simple computer program can show that you tend to win by switching. Another way to phrase the problem is that you choose one door, then Monty Hall (without revealing anything) gives you the option of taking the door you chose, or taking both the other doors. That's an equivalent problem (assuming the goat has zero utility), and it makes it very obvious that you should switch.

zcarter 11y ago

I like to explain it with a wager: pick a star from the night's sky. If you pick the star I'm thinking of, I'll give you a dollar. After you've made your selection, I'll give you the opportunity to switch to one other star, guaranteeing that either your star or this one alternative will be the star I had in mind. Do you want to switch?

dllthomas 11y ago

Your argument isn't really wrong as far as it goes, but it doesn't explain why it matters whether Monty deliberately opens a door without a car or picks randomly and just happened to open a car-less door this time.