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So the author presents the Monty Hall problem this way (very explicitly saying that the host knows where the prize is an will not reveal it): > You are given a

So the author presents the Monty Hall problem this way (very explicitly saying that the host knows where the prize is an will not reveal it): > You are given a choice of three doors, behind one is a prize. You can choose one door. The host of this game, who knows where the prize is, then opens one door without a prize (again – they know where the prize is and deliberately choose one of the unchosen doors without a prize), and then ask if you want to change your choice to the other unopened door. If you change your choice your odds of winning go up from 1/3 to 2/3. which to me makes intuitive sense. Every other time I've encountered this it's usually presented in a very ambiguous way, where it isn't clear if the host has any information about what's behind the doors. If the host is choosing a door randomly and it doesn't happen to contain the prize, your odds don't improve if you switch your answer. ...I guess I just want to say that I think the Monty Hall problem is dumb. It's not some mind blowing truth of probabilities that human brains can't grasp; it's just a poorly worded problem that causes you to make incorrect assumptions. You could argue that if you don't know whether or not the host has any knowledge about where the prize is you might as well switch, because if they do you've improved your odds and if they don't you haven't made things worse. But still, the "problem" itself doesn't seem that interesting.

syki 5y ago
In the Monty Hall problem the host always knows where the prize is. It’s a mind blowing problem because usually a person’s intuition is wrong. For almost everyone they need to carefully analyze the problem in order to understand why their intuition is wrong. There are lots of such examples in math.
- davemp 5y ago
  Most peoples’ intuition is that the host wouldn’t always offer the choice unless it’s more likely that you’ve selected the correct door.
mtlogstdo 5y ago
I don't really understand how you think the game could work if the host didn't know where the prize is? They'd keep accidentally revealing the prize, and then asking you whether you want to switch to the other remaining door, at which point... what's the point? The game is over. I've not usually seen it explicitly stated like this that the host knows, but it's always seemed obvious to me that they must.
- seoaeu 5y ago
  There's an alternative version where the host opens a door at random. If it is a goat, they proceed the same as the normal version. But if it reveals the car, the host says, "Sorry, the car was behind door C. You lose." And a third version where the host only asks you to switch if you initially picked the car, and otherwise always ends the game early. The best part is that during the game you can't actually tell which rule set you are playing with!
HWR_14 5y ago
> you change your choice your odds of winning go up from 1/3 to 2/3 which to me makes intuitive sense Well, my guess is that's because you've fixed your intuitions either by prior exposure to the problem or education. Certainly I think it's intuitive - now. The point is most people's intuitions think it changes from 1/3 to 1/2, regardless of whether you switch or not. But hey, maybe Bayesian inference is intuitive to you without any math education. Meanwhile, when this was published in 1990 and thousands of people (10% with PhDs) wrote into the magazine that it was an error. So some people have a very strong contrary intuition.
gerdesj 5y ago
"which to me makes intuitive sense" Your intuition is either very good or complete bollocks, or at least worryingly odd 8) Monty Hall is a really clever problem and worth studying in some depth. Whenever I've encountered it, the rules are always given without ambiguity. Even so, it is very hard to get to the bottom of the probabilities. You can reason your way through it and possibly get to the right answer, unaided. In which case you are far brighter than the followers of a maths column that introduced this problem back in the day. Ah, let's get to the canonical source of knowledge: https://en.wikipedia.org/wiki/Monty_Hall_problem https://en.wikipedia.org/wiki/Monty_Hall_problem (para 1) Years (>30) ago I ran a sort of Monte Carlo simulation on this thing. It took a while to encode the rules in GWBASIC and luckily I had a copy of Peitgen's (both Beauty and Science of Fractals) and I was able to get roughly or reasonably random numbers to do the simulation by following his algorithm for generating them! After a while of running on my 286 I got a result. Then I spent a few hours fixing bugs. It took a few more iterations until I got an answer that looked right or at least was between 0 and 1 and didn't end up in syntax error 8) Dr Ian Stewart gives a great description of the problem in one of his books too.
- mikenew 5y ago
  Maybe I'm just misunderstanding something then? I'm not trying to be dismissive or act like I think I have some special intuition here. It really does just seem straightforward. Imagine the problem this way: The host of a game show presents you with 100 doors, behind one of which is a prize. You pick one, and you know that your odds of having chosen the correct door are 1 in 100. The host, who knows where the prize is, then opens 98 other empty doors, leaving one. You know that the odds were 99 in 100 that the prize was behind one of those other doors, and the host just eliminated 98 of them. ...what am I missing here?
  - yata69420 5y ago
    I'm with you. Usually when it comes up in pop culture, it's not mentioned that the host can't open a door with a prize. That's the only reason it's confusing.
deleted 5y ago
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zeroxfe 5y ago
Wow... okay, I'm pretty sure you don't actually understand the Monty Hall problem.
- bhelkey 5y ago
  > I'm pretty sure you don't actually understand the Monty Hall problem What did you disagree with? Their understanding of the problem seems pretty solid to me.
WoahNoun 5y ago
>I've encountered this it's usually presented in a very ambiguous way, where it isn't clear if the host has any information about what's behind the doors. If the host is choosing a door randomly Why would the host reveal the car when the contestant has a chance to switch doors?
mhb 5y ago
Your intuition is wrong and it doesn't matter what the host knows before showing you the door that doesn't have the prize. Here's some explanation: https://betterexplained.com/articles/understanding-the-monty-hall-problem/ https://betterexplained.com/articles/understanding-the-monty...
- jodrellblank 5y ago
  How can they reliably show you the door that doesn't have the prize, if they don't know where the prize is? They can't, which means the host must know, which means it must matter what the host knows?
  - mhb 5y ago
    He's not reliably showing it. You're being presented with a situation in which the door shown doesn't have the prize.
  - mhb 5y ago
    An illuminating example uses a deck of cards. You're trying to get the ace of spades. You pick a card. I don't know anything but I turn over 50 of the 51 remaining cards and they're not the ace of spades. Do you want to switch your pick to the alternative unknown card?
  - galaxyLogic 5y ago
    To a casual observer it looks like they picked the door without price by chance.
  - deleted 5y ago
    [deleted]
- bhelkey 5y ago
  > Your intuition is wrong and it doesn't matter what the host knows before showing you the door that doesn't have the prize. This is not correct and your source doesn't support your point From your link: > Monty helps us by “filtering” the bad choices on the other side Monty knows which door holds the car and which holds the goat
LesZedCB 5y ago
best explanation i've heard is the game has 100 doors. you choose 1 of 100 possible doors. the host then opens 98 doors, all with nothing behind them. at that point it's much easier to see that you chances improve greatly by switching.
- RubyRidgeRandy 5y ago
  I always thought of this as a bullshit sleight-of-language problem. Realistically, it doesn't make a difference if you switch, because the chance of the prize being behind any door is 50%. The choice doesn't 'carry over' it's probability. You make a decision on one door out of 100. The odds are 1/100. Then all doors are eliminated besides one door and the one you already chose. the probability of the other door is 50%, BUT the probability of choosing the same door is 50% as well.
  - yayachiken 5y ago
    But the probability is not 50%. It is 99% to win if you switch and 1% if you stick with the same door. Imagine you repeat the process multiple times, but the contestant chooses to pick door 1 every time. Would you still assume that staying with door 1 every time would give you better odds, that is you would win 50% of the time if you always pick door 1 out of 100?
    - RubyRidgeRandy 5y ago
      Think about it this way: there are two doors as the starting condition, only one has the prize behind it. 50% odds. The host opens the wrong door, so only one remains. The prize MUST be behind the only remaining door. The host gives you the option to select that door or he can start a new game where there is only one door and one prize. Would anyone really argue that there is a difference in outcome here? But the logic I see is that people are saying you should switch because the probability in the first set is 50% even though its literally impossible in this scenario for it to be 50%. ----- I know mathematically it's wrong. But I hate this problem so much. I am convinced is a philosophical / semantical issue.
joe_the_user 5y ago
If the host is choosing a door randomly and it doesn't happen to contain the prize, your odds don't improve if you switch your answer. I'm not sure what you mean here. If you pick 1 door in 3, the chance you've picked the door with the prize is 1/3 and so the chance the prize is behind another of the 2 remaining door is 2/3. If one of the doors you didn't pick is opened and reveals no prize and the situation is as-described, the chance of the prize being behind one of the non-picked doors remains the same, 2/3. And since you know now which of these door doesn't have the prize, the chance of the final unpicked door having the prize is now 2/3. The host has to pick a door with no prize for the described situation to happen but if the situation happens, your knowledge doesn't depend on the host's knowledge. Maybe there's some world where the host picks at random and so 1/3 of the time, the door with the prize open and you then know with 100% certainty where the prize (though whether anyone gets a prize is ambiguous, to say the least). But that revealed-prize situation isn't what the problem describes. Edit: The only way that host knowledge matters is if the host can choose whether or not to offer this particular deal based on whether the contestant picked the correct door originally. If we assume some optimal game theoretic counter-play on the part of host, then I would guess the host's action give no information and so there's no reason to switch. But that's a bit far from your comment.
- mikenew 5y ago
  Whether or not the host knows where the prize is absolutely makes a difference. There's a reason why it's stated twice in TFA, very unambiguously, that the host is intentionally opening a door they know does not contain the prize. In the case of a host that doesn't know where the prize is, your odds don't improve by switching. Think of it slightly differently: Imagine you pick a door, and then the host, who has no idea what's behind the doors, also picks one. Then the remaining door opens and reveals that there's nothing behind it. There's really no difference between your choice and the hosts.
  - joe_the_user 5y ago
    Think of it slightly differently: Imagine you pick a door, and then the host, who has no idea what's behind the doors, also picks one. Then the remaining door opens and reveals that there's nothing behind it. Yes, if the host were to pick a door, not tell you which it was or reveal anything else and then offer to switch your choice for their choice, there would be no difference in the odds of each choice. That just happens not to be the statement of the situation. The statement of situation is: you pick a door. The host picks a door, other than what you picked, then opens the door and reveals there's nothing there. This excludes the situation of the host picking the prize or the host picking the same door as you. How that exclusion happened doesn't matter. At that point, you get to choose your original pick or switch to the remaining door. At that point, what the host knew before irrelevant. You had a 1/3 chance of picking the door with the prize before regardless of the host, and now you can make a choice that gives every other possibility and so a 2/3 chance of getting the prize. Edit: I think that the problem is stated as "the host knows" because that means the host is guaranteed to open a door without the prize. It's not that the host's knowledge matters to the strategy, it's that host's knowledge makes it certain the situation will happen as described. See mtlogstdo's comment. https://news.ycombinator.com/item?id=29846498 https://news.ycombinator.com/item?id=29846498
chrismcb 5y ago
It isn't the Monty Hall problem if the host doesn't know where the prize is. The problem is generally stated, up funny, the host knows where the prize is. The issue isn't poorly worded, it is that a lot of people can't understand how two doors don't represent half and half. You are correct, in that if that just randomly selects one of the two doors, occasionally he will reveal the prize, changing to odds. Which is why it is usually started that the host knows and selects she empty door.
tzs 5y ago
That statement of the Monty Hall problem is still not quite correct. It needs to also include that you are told beforehand that the host will be opening a door and giving you a chance to switch. If the host is not bound to make the offer, they could do something like only make the offer if you have picked the right door. Switching would then always lose. People would learn that is the case, and no one would switch which would not be as interesting for the audience. To avoid this the host could give the option to switch half the time when the player picks an empty door and all the time when they pick the prize door. On average under this approach the player wins 1/3 of the time regardless of their choice.
goto11 5y ago
Yeah, the scenario is often stated ambiguously. The important points: 1) The prizes are placed behind the doors before the game begins. 2) The host will open a door regardless of whether the contestant have selected the door with the prize or not. 3) The door to open is always selected from the remaining doors. The host will never open the door the contestant selected. 4) The door is not selected at random. The host knows where the prize is, and will only open a door without a price. The counter-intuitive part is not the probabilities, it is the rules the host operates. Some of the rules (2,3) depends on the host not knowing or ignoring where the prize is, while some (4) depend on the host using the knowledge of where the prize is.
- galaxyLogic 5y ago
  The trick is that it looks like the host is doing the same thing as you, choosing one of the remaining doors. That trick us into thinking the host is operating under the same rules as the guest, it is like a game and in a game typically all players have to follow the same rules. So it is natural to assume this is so. The situation would be perhaps easier to understand correctly if the host didn't open a door, but told you "I can guarantee to you the car is NOT behind this door"
amluto 5y ago
> If the host is choosing a door randomly and it doesn't happen to contain the prize, your odds don't improve if you switch your answer. No. If the host chooses a door randomly and it doesn’t happen to contain the prize, the outcome is exactly the same as if the host does know where the prize is. Look up “conditional probability” for how to calculate this. The host’s knowledge has no effect other than to prevent the awkward outcome in which the host opens the door with the prize. Even if the host did open the door with the prize on occasion, the outcomes would still be the same if the rule was that the whole game starts over if the host accidentally reveals the prize. (Well, the game would take longer to finish, but that has no effect on the probability of winning.)
- mikenew 5y ago
  Here, I wrote some code to prove it: https://gitlab.com/mikenew/monty_hall/-/blob/main/monty_hall.nim https://gitlab.com/mikenew/monty_hall/-/blob/main/monty_hall... If you have nim installed you can run it yourself with `nim c -r monty_hall.nim`. I ran 1 million iterations, and here are the results: Running the simulation 1000000 times where the host chooses doors randomly 500193 # of times the contestant's originally chosen door contained the prize 499807 # of times the remaining door contained the prize Running the simulation 1000000 times where the host intentionally chooses empty doors 334432 # of times the contestan'ts originally chosen door contained the prize 665568 # of times the remaining door contained the prize
  - amluto 5y ago
    I can’t read your code because I’m not logged in. But you pretty clearly have a bug. If the host does not know which door has a prize and opens a door at random, there is a 1/3 chance the host opens the door with the prize, and the game stops (or otherwise deserves its own logged outcome). And there’s no conceivable way that the host’s knowledge, strategy, or anything else could make the probability that the contestant’s originally chosen door contains a prize be anything other than 1/3. The contestant originally guesses with clean slate and has a 1/3 chance of guessing right, full stop.
    - 5y ago
Tarsul 5y ago
Thank you. The articles about this problem constantly neglect those ambiguities of the problem. I'm not arguing that people are good with probabilites (many are not), I'm arguing that the Monthy Hall problem has been overstated as a problem about understanding probabilities when in reality it's more a "understanding/articulating the premise correctly"-problem. Which is why I'm of the opinion that article writers should leave Monthy Hall alone in the future and write about different things. Or maybe someone should make a case study where the participants have to play a Monty Hall game and afterwards answer in a questionnaire why they chose whatever they did. Which would probably depend A LOT on how well the experiment would've been explained at the start, but you could test if the participants understood it in the questionnaire. That way you could finally have a more sophisticated answer about whether this problem has any merit regarding behavioral biases.

syki 5y ago

In the Monty Hall problem the host always knows where the prize is. It’s a mind blowing problem because usually a person’s intuition is wrong. For almost everyone they need to carefully analyze the problem in order to understand why their intuition is wrong. There are lots of such examples in math.

davemp 5y ago
Most peoples’ intuition is that the host wouldn’t always offer the choice unless it’s more likely that you’ve selected the correct door.

mtlogstdo 5y ago

I don't really understand how you think the game could work if the host didn't know where the prize is? They'd keep accidentally revealing the prize, and then asking you whether you want to switch to the other remaining door, at which point... what's the point? The game is over. I've not usually seen it explicitly stated like this that the host knows, but it's always seemed obvious to me that they must.

seoaeu 5y ago
There's an alternative version where the host opens a door at random. If it is a goat, they proceed the same as the normal version. But if it reveals the car, the host says, "Sorry, the car was behind door C. You lose." And a third version where the host only asks you to switch if you initially picked the car, and otherwise always ends the game early. The best part is that during the game you can't actually tell which rule set you are playing with!

HWR_14 5y ago

> you change your choice your odds of winning go up from 1/3 to 2/3 which to me makes intuitive sense Well, my guess is that's because you've fixed your intuitions either by prior exposure to the problem or education. Certainly I think it's intuitive - now. The point is most people's intuitions think it changes from 1/3 to 1/2, regardless of whether you switch or not. But hey, maybe Bayesian inference is intuitive to you without any math education. Meanwhile, when this was published in 1990 and thousands of people (10% with PhDs) wrote into the magazine that it was an error. So some people have a very strong contrary intuition.

gerdesj 5y ago

"which to me makes intuitive sense" Your intuition is either very good or complete bollocks, or at least worryingly odd 8) Monty Hall is a really clever problem and worth studying in some depth. Whenever I've encountered it, the rules are always given without ambiguity. Even so, it is very hard to get to the bottom of the probabilities. You can reason your way through it and possibly get to the right answer, unaided. In which case you are far brighter than the followers of a maths column that introduced this problem back in the day. Ah, let's get to the canonical source of knowledge: https://en.wikipedia.org/wiki/Monty_Hall_problem https://en.wikipedia.org/wiki/Monty_Hall_problem (para 1) Years (>30) ago I ran a sort of Monte Carlo simulation on this thing. It took a while to encode the rules in GWBASIC and luckily I had a copy of Peitgen's (both Beauty and Science of Fractals) and I was able to get roughly or reasonably random numbers to do the simulation by following his algorithm for generating them! After a while of running on my 286 I got a result. Then I spent a few hours fixing bugs. It took a few more iterations until I got an answer that looked right or at least was between 0 and 1 and didn't end up in syntax error 8) Dr Ian Stewart gives a great description of the problem in one of his books too.

mikenew 5y ago
Maybe I'm just misunderstanding something then? I'm not trying to be dismissive or act like I think I have some special intuition here. It really does just seem straightforward. Imagine the problem this way: The host of a game show presents you with 100 doors, behind one of which is a prize. You pick one, and you know that your odds of having chosen the correct door are 1 in 100. The host, who knows where the prize is, then opens 98 other empty doors, leaving one. You know that the odds were 99 in 100 that the prize was behind one of those other doors, and the host just eliminated 98 of them. ...what am I missing here?
- yata69420 5y ago
  I'm with you. Usually when it comes up in pop culture, it's not mentioned that the host can't open a door with a prize. That's the only reason it's confusing.

deleted 5y ago

[deleted]

zeroxfe 5y ago

Wow... okay, I'm pretty sure you don't actually understand the Monty Hall problem.

bhelkey 5y ago
> I'm pretty sure you don't actually understand the Monty Hall problem What did you disagree with? Their understanding of the problem seems pretty solid to me.

WoahNoun 5y ago

>I've encountered this it's usually presented in a very ambiguous way, where it isn't clear if the host has any information about what's behind the doors. If the host is choosing a door randomly Why would the host reveal the car when the contestant has a chance to switch doors?

mhb 5y ago

Your intuition is wrong and it doesn't matter what the host knows before showing you the door that doesn't have the prize. Here's some explanation: https://betterexplained.com/articles/understanding-the-monty-hall-problem/ https://betterexplained.com/articles/understanding-the-monty...

jodrellblank 5y ago
How can they reliably show you the door that doesn't have the prize, if they don't know where the prize is? They can't, which means the host must know, which means it must matter what the host knows?
- mhb 5y ago
  He's not reliably showing it. You're being presented with a situation in which the door shown doesn't have the prize.
- mhb 5y ago
  An illuminating example uses a deck of cards. You're trying to get the ace of spades. You pick a card. I don't know anything but I turn over 50 of the 51 remaining cards and they're not the ace of spades. Do you want to switch your pick to the alternative unknown card?
- galaxyLogic 5y ago
  To a casual observer it looks like they picked the door without price by chance.
- deleted 5y ago
  [deleted]
bhelkey 5y ago
> Your intuition is wrong and it doesn't matter what the host knows before showing you the door that doesn't have the prize. This is not correct and your source doesn't support your point From your link: > Monty helps us by “filtering” the bad choices on the other side Monty knows which door holds the car and which holds the goat

LesZedCB 5y ago

best explanation i've heard is the game has 100 doors. you choose 1 of 100 possible doors. the host then opens 98 doors, all with nothing behind them. at that point it's much easier to see that you chances improve greatly by switching.

RubyRidgeRandy 5y ago
I always thought of this as a bullshit sleight-of-language problem. Realistically, it doesn't make a difference if you switch, because the chance of the prize being behind any door is 50%. The choice doesn't 'carry over' it's probability. You make a decision on one door out of 100. The odds are 1/100. Then all doors are eliminated besides one door and the one you already chose. the probability of the other door is 50%, BUT the probability of choosing the same door is 50% as well.
- yayachiken 5y ago
  But the probability is not 50%. It is 99% to win if you switch and 1% if you stick with the same door. Imagine you repeat the process multiple times, but the contestant chooses to pick door 1 every time. Would you still assume that staying with door 1 every time would give you better odds, that is you would win 50% of the time if you always pick door 1 out of 100?
  - RubyRidgeRandy 5y ago
    Think about it this way: there are two doors as the starting condition, only one has the prize behind it. 50% odds. The host opens the wrong door, so only one remains. The prize MUST be behind the only remaining door. The host gives you the option to select that door or he can start a new game where there is only one door and one prize. Would anyone really argue that there is a difference in outcome here? But the logic I see is that people are saying you should switch because the probability in the first set is 50% even though its literally impossible in this scenario for it to be 50%. ----- I know mathematically it's wrong. But I hate this problem so much. I am convinced is a philosophical / semantical issue.

joe_the_user 5y ago

If the host is choosing a door randomly and it doesn't happen to contain the prize, your odds don't improve if you switch your answer. I'm not sure what you mean here. If you pick 1 door in 3, the chance you've picked the door with the prize is 1/3 and so the chance the prize is behind another of the 2 remaining door is 2/3. If one of the doors you didn't pick is opened and reveals no prize and the situation is as-described, the chance of the prize being behind one of the non-picked doors remains the same, 2/3. And since you know now which of these door doesn't have the prize, the chance of the final unpicked door having the prize is now 2/3. The host has to pick a door with no prize for the described situation to happen but if the situation happens, your knowledge doesn't depend on the host's knowledge. Maybe there's some world where the host picks at random and so 1/3 of the time, the door with the prize open and you then know with 100% certainty where the prize (though whether anyone gets a prize is ambiguous, to say the least). But that revealed-prize situation isn't what the problem describes. Edit: The only way that host knowledge matters is if the host can choose whether or not to offer this particular deal based on whether the contestant picked the correct door originally. If we assume some optimal game theoretic counter-play on the part of host, then I would guess the host's action give no information and so there's no reason to switch. But that's a bit far from your comment.

mikenew 5y ago
Whether or not the host knows where the prize is absolutely makes a difference. There's a reason why it's stated twice in TFA, very unambiguously, that the host is intentionally opening a door they know does not contain the prize. In the case of a host that doesn't know where the prize is, your odds don't improve by switching. Think of it slightly differently: Imagine you pick a door, and then the host, who has no idea what's behind the doors, also picks one. Then the remaining door opens and reveals that there's nothing behind it. There's really no difference between your choice and the hosts.
- joe_the_user 5y ago
  Think of it slightly differently: Imagine you pick a door, and then the host, who has no idea what's behind the doors, also picks one. Then the remaining door opens and reveals that there's nothing behind it. Yes, if the host were to pick a door, not tell you which it was or reveal anything else and then offer to switch your choice for their choice, there would be no difference in the odds of each choice. That just happens not to be the statement of the situation. The statement of situation is: you pick a door. The host picks a door, other than what you picked, then opens the door and reveals there's nothing there. This excludes the situation of the host picking the prize or the host picking the same door as you. How that exclusion happened doesn't matter. At that point, you get to choose your original pick or switch to the remaining door. At that point, what the host knew before irrelevant. You had a 1/3 chance of picking the door with the prize before regardless of the host, and now you can make a choice that gives every other possibility and so a 2/3 chance of getting the prize. Edit: I think that the problem is stated as "the host knows" because that means the host is guaranteed to open a door without the prize. It's not that the host's knowledge matters to the strategy, it's that host's knowledge makes it certain the situation will happen as described. See mtlogstdo's comment. https://news.ycombinator.com/item?id=29846498 https://news.ycombinator.com/item?id=29846498

chrismcb 5y ago

It isn't the Monty Hall problem if the host doesn't know where the prize is. The problem is generally stated, up funny, the host knows where the prize is. The issue isn't poorly worded, it is that a lot of people can't understand how two doors don't represent half and half. You are correct, in that if that just randomly selects one of the two doors, occasionally he will reveal the prize, changing to odds. Which is why it is usually started that the host knows and selects she empty door.

tzs 5y ago

That statement of the Monty Hall problem is still not quite correct. It needs to also include that you are told beforehand that the host will be opening a door and giving you a chance to switch. If the host is not bound to make the offer, they could do something like only make the offer if you have picked the right door. Switching would then always lose. People would learn that is the case, and no one would switch which would not be as interesting for the audience. To avoid this the host could give the option to switch half the time when the player picks an empty door and all the time when they pick the prize door. On average under this approach the player wins 1/3 of the time regardless of their choice.

goto11 5y ago

Yeah, the scenario is often stated ambiguously. The important points: 1) The prizes are placed behind the doors before the game begins. 2) The host will open a door regardless of whether the contestant have selected the door with the prize or not. 3) The door to open is always selected from the remaining doors. The host will never open the door the contestant selected. 4) The door is not selected at random. The host knows where the prize is, and will only open a door without a price. The counter-intuitive part is not the probabilities, it is the rules the host operates. Some of the rules (2,3) depends on the host not knowing or ignoring where the prize is, while some (4) depend on the host using the knowledge of where the prize is.

galaxyLogic 5y ago
The trick is that it looks like the host is doing the same thing as you, choosing one of the remaining doors. That trick us into thinking the host is operating under the same rules as the guest, it is like a game and in a game typically all players have to follow the same rules. So it is natural to assume this is so. The situation would be perhaps easier to understand correctly if the host didn't open a door, but told you "I can guarantee to you the car is NOT behind this door"

amluto 5y ago

> If the host is choosing a door randomly and it doesn't happen to contain the prize, your odds don't improve if you switch your answer. No. If the host chooses a door randomly and it doesn’t happen to contain the prize, the outcome is exactly the same as if the host does know where the prize is. Look up “conditional probability” for how to calculate this. The host’s knowledge has no effect other than to prevent the awkward outcome in which the host opens the door with the prize. Even if the host did open the door with the prize on occasion, the outcomes would still be the same if the rule was that the whole game starts over if the host accidentally reveals the prize. (Well, the game would take longer to finish, but that has no effect on the probability of winning.)

mikenew 5y ago
Here, I wrote some code to prove it: https://gitlab.com/mikenew/monty_hall/-/blob/main/monty_hall.nim https://gitlab.com/mikenew/monty_hall/-/blob/main/monty_hall... If you have nim installed you can run it yourself with `nim c -r monty_hall.nim`. I ran 1 million iterations, and here are the results: Running the simulation 1000000 times where the host chooses doors randomly 500193 # of times the contestant's originally chosen door contained the prize 499807 # of times the remaining door contained the prize Running the simulation 1000000 times where the host intentionally chooses empty doors 334432 # of times the contestan'ts originally chosen door contained the prize 665568 # of times the remaining door contained the prize
- amluto 5y ago
  I can’t read your code because I’m not logged in. But you pretty clearly have a bug. If the host does not know which door has a prize and opens a door at random, there is a 1/3 chance the host opens the door with the prize, and the game stops (or otherwise deserves its own logged outcome). And there’s no conceivable way that the host’s knowledge, strategy, or anything else could make the probability that the contestant’s originally chosen door contains a prize be anything other than 1/3. The contestant originally guesses with clean slate and has a 1/3 chance of guessing right, full stop.
  - 5y ago

Tarsul 5y ago

Thank you. The articles about this problem constantly neglect those ambiguities of the problem. I'm not arguing that people are good with probabilites (many are not), I'm arguing that the Monthy Hall problem has been overstated as a problem about understanding probabilities when in reality it's more a "understanding/articulating the premise correctly"-problem. Which is why I'm of the opinion that article writers should leave Monthy Hall alone in the future and write about different things. Or maybe someone should make a case study where the participants have to play a Monty Hall game and afterwards answer in a questionnaire why they chose whatever they did. Which would probably depend A LOT on how well the experiment would've been explained at the start, but you could test if the participants understood it in the questionnaire. That way you could finally have a more sophisticated answer about whether this problem has any merit regarding behavioral biases.