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> It’s important that Monty looked behind the doors before choosing which to open. This is often not explicitly stated when the problem is given. It is even no

by lqet 6y ago

> It’s important that Monty looked behind the doors before choosing which to open. This is often not explicitly stated when the problem is given. It is even not a 100% clear from the statement above. Monty always chooses a door with a goat. So: 1. You choose a door. 2. Prob that there is a car behind it: 1/3 3. Prob that the car is behind the two other doors: 2/3 4. If the car was behind the two other doors (which, remember, has p=2/3), Monty will choose the door without a car for you, and the door with a car will remain closed. In this case you are guaranteed to have the car if you switched. So with switching, the overall probability is 2/3. Without, its the original 1/3. If you did not understand that Monty always chooses a goat door, but the person giving you the problem does, or vice versa, then what usually happens is that both of you try to explain why your intuition is correct. Because most people don't talk formal probabilities, your explanations will be so vague that the other person will not realize your different understanding. You will discuss forever, you will both be right, and you will part ways with the strange feeling that maybe the other person was right, when all along you were talking about different problems. This is why this problem is so notorious.

mannykannot 6y ago
>> It’s important that Monty looked behind the doors before choosing which to open. > This is often not explicitly stated when the problem is given, which imho is the whole reason this problem has the reputation of being hard to understand. If that were the only difficulty, why have so many people continued to have trouble accepting it even after this misunderstanding has been cleared up, and even after the correct answer has been explained to them? According to Wikipedia, even Paul Erdős remained unconvinced until he was shown a computer simulation. I recall mention of an analysis of the responses to Vos Savant's Parade article, concluding that a majority disputing the result were aware of this constraint, and I will post a link if I can find it again (though if a majority did not explain their reasoning, it may not be possible to figure out what assumptions they made. Nevertheless, the question in my first paragraph still stands.)
- brmgb 6y ago
  What makes the three doors Monty Hall so counterintuitive is that people tend to correctly reason about the case where the second door is randomly opened and don't understand why it doesn't apply. I believe that this example makes understanding why people don't get it easier: you are looking for someone with one of your friend. You know they are in one of three rooms. Right before you can open the first one, your friend opens the second one and say: "not there". People assume the Monty Hall problem means that it's more likely your friend is in the third room and not the one you were going to open and think it's silly. And they are right to think that. What they don't get is that the case where your friend opened the correct door is part of the switching choice in the Monty Hall situation.
  - mannykannot 6y ago
    Your first paragraph looks very plausible, though I am not aware of any data saying it is a prevalent one. Are you suggesting that people still look at this problem as if the choice was random, even when they know it is not? (e.g., if they think the 'random' and 'chosen' cases are equivalent.) That is also plausible, but if it is common, that would imply that the phrasing of the problem is not the root cause of their misunderstanding of the optimal strategy. Personally, IIRC, my first reaction was to assume the second box opening was not random (perhaps only because the question is not phrased as being conditional on this act revealing a goat) but did not see how this gave any useful information. Here's another possible way of getting it wrong, regardless of the phrasing of the problem: assuming that, after the reveal (and whether one thinks of it as random or not), one is, as it were, starting over, except with a choice between two boxes rather than three, and no other information.
- cgriswald 6y ago
  I definitely don't think it is the only point of confusion, but I definitely think it is a big one. We discussed the problem in my discrete math class and the professor had stated the problem without the constraint. Many in the class, being far too young to even know who Monty Hall is, were confused, and several had their 'aha!' moment once that constraint was mentioned. > According to Wikipedia, even Paul Erdős remained unconvinced until he was shown a computer simulation. Interestingly, coding up my own simulation took me from understanding the problem to grokking it. I'd definitely suggest anyone who has programming ability but doesn't grok the problem to write up a simulation.
ogre_codes 6y ago
This explanation is IMO more intuitive than the linked one.
tasty_freeze 6y ago
It doesn't matter at all if Monty knows which doors are winners and losers. If Monty doesn't know, then sometimes the game will be ruined because he will expose the grand prize and then the game is moot. But if is simply lucky by showing the goat door vs he picked it with foreknowledge doesn't change the odds in any way.
- lqet 6y ago
  It does matter. If you are looking at a single instance where Monty got lucky and chose a goat door, then the probability that the door you have chosen has the car is 1/2, and switching doesn't change anything. This can easily be tested: just write a simulation that runs the experiment with Monty choosing a random door, and discard the instances where the game was "ruined" because he picked the car. In the remaining instances, both strategies will perform the same.
  - Taek 6y ago
    It matters because in the case where Monty does know to avoid the door, those games end up as wins for the person who selected to switch, as opposed to broken games. You are taking an empty result and replacing it with a strict win.
    - SamBam 6y ago
      I think lqet's point is that if it's just recounted as being thrown into a situation -- you've picked a door, and Monty opens a door, and it's a goat -- you don't know if you're in a world where Monty always picks a goat or one where he just happened to be lucky. In the second scenario, it doesn't matter if you switch.
      - couchand 6y ago
        Well in that situation it's still beneficial to switch, unless the probability of being in the first scenario is exactly zero.
jtsiskin 6y ago
Is it important the host looked behind the doors? If the host had just picked at random from the two doors, and it happened to show a goat, the odds would be the same.
- crooked-v 6y ago
  > If the host had just picked at random The overall point is that the host is not picking at random, and is thus affecting the outcome in a statistically reliable way.
  - fouronnes3 6y ago
    This is gonna have anthropic principle vibes but here we go. Either the host looks behind the door, or they don't and we just don't talk about games where they accidentally show you the car. The probabilities are the same. We're already conditioned on being in the "they show you a goat" universe.
    - samatman 6y ago
      No. If you're playing blackjack, and you hit on a 19, but you have x-ray vision and you know a 2 is at the top of the deck, you're not playing the same game as you would be if you didn't have that knowledge. Same with the Monty Hall problem. It is a critical distinction whether opening the door has a 0% chance, or a 33% chance, of showing you a car.
    - dllthomas 6y ago
      Simulate it. I thought as you think. I wrote a simulation to show I was right. I was wrong. I have had this interchange several times. Invariably it goes one of two ways. They have endless reasons why they have to be right and they don't need to write a goddamn simulation, or they tell me they wrote the simulation and they have learned they were wrong.
      - Arnavion 6y ago
        Specifically, it's the difference between var num_stays_wins = 0; var num_switches_wins = 0; for (var i = 0; i < 100000; i++) { const car_door = Math.floor(Math.random() * 3); const my_choice = Math.floor(Math.random() * 3); var montys_choice; do { montys_choice = Math.floor(Math.random() * 3); } while (montys_choice === my_choice); if (montys_choice === car_door) { i -= 1; continue; } if (my_choice === car_door) { num_stays_wins += 1; } else { num_switches_wins += 1; } } console.log(num_stays_wins, num_switches_wins); and var num_stays_wins = 0; var num_switches_wins = 0; for (var i = 0; i < 100000; i++) { const car_door = Math.floor(Math.random() * 3); const my_choice = Math.floor(Math.random() * 3); var montys_choice; do { montys_choice = Math.floor(Math.random() * 3); } while (montys_choice === my_choice || montys_choice === car_door); if (my_choice === car_door) { num_stays_wins += 1; } else { num_switches_wins += 1; } } console.log(num_stays_wins, num_switches_wins); (The difference is in the calculation of `montys_choice`.)
adrianmonk 6y ago
It's also not always emphasized that the host never picks the same door you did. The host's behavior is not done in isolation. It is a response to your choice.
Swizec 6y ago
Thank you this is the first explanation of Monty Hall that finally makes sense. Now it intuitively makes sense. I was always told Monty reveals new info and that changes probabilities. Never made sense. Your 2/3 vs 1/3 explanation makes it obvious.
tedunangst 6y ago
What kind of game shows are people watching where the host opens a door and says, hey, there's the prize but too bad, now you can't win it? Of course he always picks a door with a goat.
- tshaddox 6y ago
  Yeah. That is literally the whole reason to use the game show analogy.
  - deleted 6y ago
    [deleted]
- ipsum2 6y ago
  Deal or No Deal does this - the contestant selects a briefcase, then opens up briefcases randomly. It's very similar to the Monty Hall problem, with some twists iirc.
- lone-commenter 6y ago
  The original game in Let's Make a Deal went exactly like you described. Monty Hall opened the remaining doors; no option to switch. Then there is the ambiguity as whether the host will always offer you to switch or just when he hasn't revealed the car.
- xorcist 6y ago
  Why not? Reveal all the choices the contestant didn't make, one at a time, slowly increasing the odds of a prize. This adds to the suspense. It's used all the time.
dlandis 6y ago
I get what you are saying, but if I look at it differently I still intuitively feel like it should be p=1/2 if they switch instead of 2/3. For example: - the bottom line is Monty will always eliminate one wrong choice - therefore when they switch, they are choosing between one wrong and one right door, every time, regardless of what came before. (p=1/2) (edit: thanks i see, no need for any more answers)
- ubercow13 6y ago
  It’s not regardless of what came before. That’s like saying, if there were two doors and Monty simply tells you which one the car is behind, there’s still a 50/50 chance of each door being right. No, there’s a 100% chance the one he told you is right. You learn real new information when Monty opens a door just like when he tells you something.
- monkeydreams 6y ago
  > therefore when they switch, they are choosing between one wrong and one right door, every time, regardless of what came before. (p=1/2) I get why you intuit this - I had a lot of trouble with it at first as well. Common sense is lying to you, however, because you are still dealing with the original odds (1/3 vs 2/3). Monty's choice is always to choose a Goat (probability 1.0), therefore his effect on the probabilities will always be with a weighting of 1.0. You end up with a probably spread of (1/3 * 1.0 vs 2/3 * 1.0). Therefore it is always in your interests to choose Monty's remaining door.
- SamBam 6y ago
  I think if you really read over the article, you'll see this is wrong. If not, tell us on which step you disagree with the author. A more common re-telling of the story: Monty has 1 million doors, with only one car behind one. When you pick a door, Monty will always open 999,998 other doors, all of which show goats, leaving one other door conspicuously closed. Do you still think you have 50% chance of being right by sticking with your choice?
  - oswald433 6y ago
    This got me closer to understanding. But, after 999,998 doors are opened, although the one other door is conspicuously closed, my door is also closed. That seems conspicuous, too. In that moment, there are 2 doors to pick from and so 50/50 makes sense. As some other comments have mentioned, I think this isn't a fully satisfying explanation because it's hard to reason about the difference between a general strategy vs. an in-the-moment choice; why does it matter that I've watched Monty narrow the choices down to 2 doors, vs. seeing 2 doors from the start?
godelski 6y ago
The problem is confusing because it appears that the choices are independent when they really are dependent. That's what confuses people and what needs to be adequately explained (and thus explicitly acknowledged).
ibiza 6y ago
> So with switching, the overall probability is 2/3. Without, its the original 1/3. Thank you for spelling this out so clearly. If explained in terms of strategies (always switch v. always stick), the probabilities should be crystal clear. What does the switcher have to do to win? Pick the wrong door first (2/3). What does the sticker have to do to win? Pick the right door first (1/3).
danielscrubs 6y ago
Your explanation was great! You should become a teacher. It really nails the problem at hand
conjectures 6y ago
This carries over to a general problem with wordy riddles and puzzles. They rest on a slew of assumptions and tacit knowledge. If the riddler fails to pluck out the essential parts, then there's a good chance the interaction will fail.
bart_spoon 6y ago
> Monty always chooses a door with a goat. Yes, this is the single most critical piece of information, and when people struggle to understand the Monty Hall problem, its almost always because this wasn't made clear. When it is, the problem becomes much more intuitive.

by lqet 6y ago

mannykannot 6y ago

>> It’s important that Monty looked behind the doors before choosing which to open. > This is often not explicitly stated when the problem is given, which imho is the whole reason this problem has the reputation of being hard to understand. If that were the only difficulty, why have so many people continued to have trouble accepting it even after this misunderstanding has been cleared up, and even after the correct answer has been explained to them? According to Wikipedia, even Paul Erdős remained unconvinced until he was shown a computer simulation. I recall mention of an analysis of the responses to Vos Savant's Parade article, concluding that a majority disputing the result were aware of this constraint, and I will post a link if I can find it again (though if a majority did not explain their reasoning, it may not be possible to figure out what assumptions they made. Nevertheless, the question in my first paragraph still stands.)

brmgb 6y ago
What makes the three doors Monty Hall so counterintuitive is that people tend to correctly reason about the case where the second door is randomly opened and don't understand why it doesn't apply. I believe that this example makes understanding why people don't get it easier: you are looking for someone with one of your friend. You know they are in one of three rooms. Right before you can open the first one, your friend opens the second one and say: "not there". People assume the Monty Hall problem means that it's more likely your friend is in the third room and not the one you were going to open and think it's silly. And they are right to think that. What they don't get is that the case where your friend opened the correct door is part of the switching choice in the Monty Hall situation.
- mannykannot 6y ago
  Your first paragraph looks very plausible, though I am not aware of any data saying it is a prevalent one. Are you suggesting that people still look at this problem as if the choice was random, even when they know it is not? (e.g., if they think the 'random' and 'chosen' cases are equivalent.) That is also plausible, but if it is common, that would imply that the phrasing of the problem is not the root cause of their misunderstanding of the optimal strategy. Personally, IIRC, my first reaction was to assume the second box opening was not random (perhaps only because the question is not phrased as being conditional on this act revealing a goat) but did not see how this gave any useful information. Here's another possible way of getting it wrong, regardless of the phrasing of the problem: assuming that, after the reveal (and whether one thinks of it as random or not), one is, as it were, starting over, except with a choice between two boxes rather than three, and no other information.
cgriswald 6y ago
I definitely don't think it is the only point of confusion, but I definitely think it is a big one. We discussed the problem in my discrete math class and the professor had stated the problem without the constraint. Many in the class, being far too young to even know who Monty Hall is, were confused, and several had their 'aha!' moment once that constraint was mentioned. > According to Wikipedia, even Paul Erdős remained unconvinced until he was shown a computer simulation. Interestingly, coding up my own simulation took me from understanding the problem to grokking it. I'd definitely suggest anyone who has programming ability but doesn't grok the problem to write up a simulation.

ogre_codes 6y ago

This explanation is IMO more intuitive than the linked one.

tasty_freeze 6y ago

It doesn't matter at all if Monty knows which doors are winners and losers. If Monty doesn't know, then sometimes the game will be ruined because he will expose the grand prize and then the game is moot. But if is simply lucky by showing the goat door vs he picked it with foreknowledge doesn't change the odds in any way.

lqet 6y ago
It does matter. If you are looking at a single instance where Monty got lucky and chose a goat door, then the probability that the door you have chosen has the car is 1/2, and switching doesn't change anything. This can easily be tested: just write a simulation that runs the experiment with Monty choosing a random door, and discard the instances where the game was "ruined" because he picked the car. In the remaining instances, both strategies will perform the same.
- Taek 6y ago
  It matters because in the case where Monty does know to avoid the door, those games end up as wins for the person who selected to switch, as opposed to broken games. You are taking an empty result and replacing it with a strict win.
  - SamBam 6y ago
    I think lqet's point is that if it's just recounted as being thrown into a situation -- you've picked a door, and Monty opens a door, and it's a goat -- you don't know if you're in a world where Monty always picks a goat or one where he just happened to be lucky. In the second scenario, it doesn't matter if you switch.
    - couchand 6y ago
      Well in that situation it's still beneficial to switch, unless the probability of being in the first scenario is exactly zero.

jtsiskin 6y ago

Is it important the host looked behind the doors? If the host had just picked at random from the two doors, and it happened to show a goat, the odds would be the same.

crooked-v 6y ago
> If the host had just picked at random The overall point is that the host is not picking at random, and is thus affecting the outcome in a statistically reliable way.
- fouronnes3 6y ago
  This is gonna have anthropic principle vibes but here we go. Either the host looks behind the door, or they don't and we just don't talk about games where they accidentally show you the car. The probabilities are the same. We're already conditioned on being in the "they show you a goat" universe.
  - samatman 6y ago
    No. If you're playing blackjack, and you hit on a 19, but you have x-ray vision and you know a 2 is at the top of the deck, you're not playing the same game as you would be if you didn't have that knowledge. Same with the Monty Hall problem. It is a critical distinction whether opening the door has a 0% chance, or a 33% chance, of showing you a car.
  - dllthomas 6y ago
    Simulate it. I thought as you think. I wrote a simulation to show I was right. I was wrong. I have had this interchange several times. Invariably it goes one of two ways. They have endless reasons why they have to be right and they don't need to write a goddamn simulation, or they tell me they wrote the simulation and they have learned they were wrong.
    - Arnavion 6y ago
      Specifically, it's the difference between var num_stays_wins = 0; var num_switches_wins = 0; for (var i = 0; i < 100000; i++) { const car_door = Math.floor(Math.random() * 3); const my_choice = Math.floor(Math.random() * 3); var montys_choice; do { montys_choice = Math.floor(Math.random() * 3); } while (montys_choice === my_choice); if (montys_choice === car_door) { i -= 1; continue; } if (my_choice === car_door) { num_stays_wins += 1; } else { num_switches_wins += 1; } } console.log(num_stays_wins, num_switches_wins); and var num_stays_wins = 0; var num_switches_wins = 0; for (var i = 0; i < 100000; i++) { const car_door = Math.floor(Math.random() * 3); const my_choice = Math.floor(Math.random() * 3); var montys_choice; do { montys_choice = Math.floor(Math.random() * 3); } while (montys_choice === my_choice || montys_choice === car_door); if (my_choice === car_door) { num_stays_wins += 1; } else { num_switches_wins += 1; } } console.log(num_stays_wins, num_switches_wins); (The difference is in the calculation of `montys_choice`.)

adrianmonk 6y ago

It's also not always emphasized that the host never picks the same door you did. The host's behavior is not done in isolation. It is a response to your choice.

Swizec 6y ago

Thank you this is the first explanation of Monty Hall that finally makes sense. Now it intuitively makes sense. I was always told Monty reveals new info and that changes probabilities. Never made sense. Your 2/3 vs 1/3 explanation makes it obvious.

tedunangst 6y ago

What kind of game shows are people watching where the host opens a door and says, hey, there's the prize but too bad, now you can't win it? Of course he always picks a door with a goat.

tshaddox 6y ago
Yeah. That is literally the whole reason to use the game show analogy.
- deleted 6y ago
  [deleted]
ipsum2 6y ago
Deal or No Deal does this - the contestant selects a briefcase, then opens up briefcases randomly. It's very similar to the Monty Hall problem, with some twists iirc.
lone-commenter 6y ago
The original game in Let's Make a Deal went exactly like you described. Monty Hall opened the remaining doors; no option to switch. Then there is the ambiguity as whether the host will always offer you to switch or just when he hasn't revealed the car.
xorcist 6y ago
Why not? Reveal all the choices the contestant didn't make, one at a time, slowly increasing the odds of a prize. This adds to the suspense. It's used all the time.

dlandis 6y ago

I get what you are saying, but if I look at it differently I still intuitively feel like it should be p=1/2 if they switch instead of 2/3. For example: - the bottom line is Monty will always eliminate one wrong choice - therefore when they switch, they are choosing between one wrong and one right door, every time, regardless of what came before. (p=1/2) (edit: thanks i see, no need for any more answers)

ubercow13 6y ago
It’s not regardless of what came before. That’s like saying, if there were two doors and Monty simply tells you which one the car is behind, there’s still a 50/50 chance of each door being right. No, there’s a 100% chance the one he told you is right. You learn real new information when Monty opens a door just like when he tells you something.
monkeydreams 6y ago
> therefore when they switch, they are choosing between one wrong and one right door, every time, regardless of what came before. (p=1/2) I get why you intuit this - I had a lot of trouble with it at first as well. Common sense is lying to you, however, because you are still dealing with the original odds (1/3 vs 2/3). Monty's choice is always to choose a Goat (probability 1.0), therefore his effect on the probabilities will always be with a weighting of 1.0. You end up with a probably spread of (1/3 * 1.0 vs 2/3 * 1.0). Therefore it is always in your interests to choose Monty's remaining door.
SamBam 6y ago
I think if you really read over the article, you'll see this is wrong. If not, tell us on which step you disagree with the author. A more common re-telling of the story: Monty has 1 million doors, with only one car behind one. When you pick a door, Monty will always open 999,998 other doors, all of which show goats, leaving one other door conspicuously closed. Do you still think you have 50% chance of being right by sticking with your choice?
- oswald433 6y ago
  This got me closer to understanding. But, after 999,998 doors are opened, although the one other door is conspicuously closed, my door is also closed. That seems conspicuous, too. In that moment, there are 2 doors to pick from and so 50/50 makes sense. As some other comments have mentioned, I think this isn't a fully satisfying explanation because it's hard to reason about the difference between a general strategy vs. an in-the-moment choice; why does it matter that I've watched Monty narrow the choices down to 2 doors, vs. seeing 2 doors from the start?

godelski 6y ago

The problem is confusing because it appears that the choices are independent when they really are dependent. That's what confuses people and what needs to be adequately explained (and thus explicitly acknowledged).

ibiza 6y ago

> So with switching, the overall probability is 2/3. Without, its the original 1/3. Thank you for spelling this out so clearly. If explained in terms of strategies (always switch v. always stick), the probabilities should be crystal clear. What does the switcher have to do to win? Pick the wrong door first (2/3). What does the sticker have to do to win? Pick the right door first (1/3).

danielscrubs 6y ago

Your explanation was great! You should become a teacher. It really nails the problem at hand

conjectures 6y ago

This carries over to a general problem with wordy riddles and puzzles. They rest on a slew of assumptions and tacit knowledge. If the riddler fails to pluck out the essential parts, then there's a good chance the interaction will fail.

bart_spoon 6y ago

> Monty always chooses a door with a goat. Yes, this is the single most critical piece of information, and when people struggle to understand the Monty Hall problem, its almost always because this wasn't made clear. When it is, the problem becomes much more intuitive.