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Comment by mikenew | Hacker News Reader

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.

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"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 (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.

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?

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.

And that they can't open the door chosen by the contestant! Also perhaps more subtly, that Monty picks at random when given the option. The original problem prompt is _highly_ underspecified (and makes assumptions that are counter-intuitive to how a game show host might behave).

The problem has posed by von Savant specified the host knew where the prize was. And claims the host reveals a non prize for. It is fairly obvious the host won't reveal the winning door, as it would end the problem. Not sure how the assumptions are fighter intuitive.

Of course Hall himself said maybe I open a door because I want you to switch.

You are not missing anything but you have gained 97 doors!

To get the subtlety here, you probably need to model it which is what I did. That's why I'm an IT bod these days, with a Civil Engineering degree and not a mathematician or statistician.

YOU aren't missing anything. But a lot of people can't give over "two doors means fifty fifty"

What you're missing here is that you already have enough intuition to translate it into a more obvious statistical problem, but most people come across this as blank slates, and under very different framing.

Even when relayed the right way and the wording isn't necessarily unclear or misleading per se, it is still framed in a particular way and context that pushes you to believe your choice is inconsequential, less about probability and math, and more for psychological games and dramatic effect in the context of a TV show.

In your example, the fact that the presenter opens all 98 doors and not just one is telling. If they only opened 1, and then told you to stick to your original or open one of the remaining 98 doors, that he picks for you, would you change?

This is what you're missing, I think. Most people don't even get to the point where they reframe the "would you change" question to "what are the chances my original choice was correct vs not". Most people get stuck on "He knows something; he clearly demonstrated very theatrically that he knew the car wouldn't be behind door No.2. And now he's asking me to swap my choice for door No.67. Why 67? Is he trying to trick me? Is this a psychological trick to get me to forfeit my door, which may be the right one? Should I trust him? After all, HOW IS MY DOOR ANY DIFFERENT THAN ALL THE OTHERS?"

(the fallacious thinking, obviously, lying in the last part here)

One thing you're missing: "The host of this game, who knows where the prize is, then opens one door without a prize"

Does the host _always_ do this, or does the host have discretion about _whether_or_not_ to do this? This has not been stated in any of the problem descriptions here, and I believe it's highly significant. If the host can choose whether or not to do this, then there is the potential for mindgames. If the host cannot choose whether or not to do this, that must be stated.

The Monty Hall problem absolutely becomes more obvious when the host's behavior is explicitly documented. And as this article points out, many key aspects of the host's behavior are implausible, which is a large part of what makes it counter-intuitive: https://ima.org.uk/4552/dont-switch-mathematicians-answer-mo...

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.

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.

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

I disagree with your conclusion that the odds remain 2/3 even if the host doesn’t know where the prize is. If that were the case, it would mean that if the same scenario is repeated many many times, then in 2/3 of the cases you would still pick the door with the prize with the same strategy. But what about the instances where the host picks the door with prize before you even get a chance to pick the other door? These cases happen with a probability of 2/3 * 1/2 = 1/3. Therefore, you only get to choose in 2/3 of the cases and in 1/2 of those cases you have already picked the correct door the first time.

Do you still have the opportunity to switch if he picks the winning door?

Even if you could, you already know that you're losing, unless he doesn't show you what's behind it.

If the game is still going, you know that he picked an empty door regardless of whether he knew he was doing it

Here, I wrote a simulation in 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

I'm not going to create an account on that site just for your thing but since you essentially haven't defined what "the host chooses randomly" means in this circumstance", I don't see how a print-out or whatever it gives would enlighten me.

Edit: plus your claim is senseless on it's face. Why would what the host knew about the door influence the probability of the contestant's choice being correct? The contestant picks first and the host doesn't influence the contestant.

Didn't realize that gitlab.com doesn't allow you to view things if you're not logged in. Here it is on github: https://github.com/michaelnew/monty_hall/blob/main/monty_hal...

By "the host chooses randomly", I mean the host picks from the remaining two doors without any knowledge of whether or not the contestant chose the prize door, or which of the two remaining doors contains a prize. The host flips a coin and if it's heads, they choose the remaining door to the left. Tails and they choose the one to the right. Which means that in 1 of 3 games the host will accidentally reveal the prize.

Whether or not the host is acting with knowledge to filter out incorrect choices or they are just randomly revealing doors is what makes the difference between a 50/50 probability between the two remaining door, or a 1/3 vs 2/3 probability.

Tails and they choose the one to the right. Which means that in 1 of 3 games the host will accidentally reveal the prize.

And what happens then? I mean, I think people have said from the start, that this behavior is outside the specification of the problem - which is that the host opens a door and reveals nothing.

The situation is about only the situation where the host choose the door with nothing. The host "opens one door without a prize". The key detail is this, "opens one door without a prize", not "knows where the prize is". If the host doesn't know where the prize but still, by chance, "opens one door without a prize", then, in this situation the contestant's information remains the same and the odds remain the same.

I don't know what to tell you at this point. From the wikipedia article:

> Most people come to the conclusion that switching does not matter because there are two unopened doors and one car and that it is a 50/50 choice. This would be true if the host opens a door randomly, but that is not the case; the door opened depends on the player's initial choice, so the assumption of independence does not hold.

Random means there's a possibility the host opens the door containing the prize. The Monty Hall problem specifies that that is not a possible scenario, and that the host does not choose randomly.

> If the host doesn't know where the prize but still, by chance, "opens one door without a prize", then, in this situation the contestant's information remains the same and the odds remain the same.

The host cannon open a door by chance if there's no chance of opening the door with a prize behind it.

The host doesn't know exactly where the prize is, but he knows that the door he's opening is a dud; that's the information that he has and the audience doesn't.

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.

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!

> 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.

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.

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.

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?

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.

Think of it this way:

Each door has 1/3 chance of containing the prize, that never changes, so your choice has a 1/3 chance of being right and the unselected doors have a 2/3 chance. Again, these facts will never change, you'd surely agree.

So once one of those unselected doors is opened, the remaining door has a 2/3 chance because you now know which of the unselected doors you must chose to capture the 2/3 chance (because the opened door has been revealed to be empty). Those pair of doors still contain the 2/3 chance, you just know which to choose now.

The problem formulation doesn't really work for n<3, as you cannot safely open n-2 doors anymore. Of course you get semantical problems then.

But even then, I don't see your problem here. The invariant is that the chosen door has probability 1/n to contain the prize, and all other doors together have probability 1-1/n. This still holds for n=2: Both doors have a probability of 50% to contain the prize.

Also if you do not believe or grok the mathematics, you can just whip up a program to simulate the problem, and verify the probabilities empirically.

The door that's left over went through many trials of not being selected by the host.

The one you chose doesn't get those additional tests. If it's bad, that door had 98 chances where it could have been chosen to be opened by the host, and didn't get chosen. It's either a very lucky empty door, or the correct door.

You can take this all the way to the limiting case where you pick one out of infinite doors, where the probability of you getting it right is ~0. Then, you're presented with a second door, knowing that there's definitely a prize between the two. You're prior knowledge that there's no chance that it's behind the door you picked doesn't go away because you've been given a second door.

Hell, you can go even further, and say you pick between several doors known to be empty, then be presented with a second door and the knowledge that there's a winning door between the two. Your chances looking at the doors starting with two doors is still 50/50, unless you remember that your door is already empty

Or think about it in reverse. While you’re blindfolded the host opens 99/100 doors. While you’re still blindfolded he then gives you a door to choose from one of those 100. If you pick a different door than the one closed door, fine. If you pick the same door as the host, the host will choose at random a different door to close.

Now you’ve done all this and taken off your blindfold. If you guessed any door but the closed one, you chose wrong. The host asks you what you would like to do — stay with your choice or switch. So do you switch?

I actually meant to respond to the person below you who said that this explanation was bullshit but I’ll just leave this here.

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.

> 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.)

Here, I wrote some code to prove it: 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

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.

Edit: Sorry, didn't realize that gitlab.com requires you to be logged in to view things (that's dumb). Here it is on github: https://github.com/michaelnew/monty_hall/blob/main/monty_hal...

That is my premise exactly. If the host chooses at random, then there is indeed a 1/3 chance of choosing the door that contains the prize. In which case the game is discarded (or restarted, if you like). If that is the behavior of the host, then in the event that the host randomly chooses a door without the prize (which should happen 2/3 of the time), then the choice between the original door and the remaining door is 50/50. If the host never chooses the door that contains the prize (and therefore the game never has to be restarted), then the host is not choosing randomly.

It's stated pretty clearly in the wikipedia article. The 2/3 probability is only true when these three conditions are met:

1. The host must always open a door that was not picked by the contestant.

2. The host must always open a door to reveal a goat and never the car.

3. The host must always offer the chance to switch between the originally chosen door and the remaining closed door.

Ah, you’ve invented (and implemented!) a very strange variant indeed. The contestant chooses a door. If they choose right, they win. Otherwise the host effectively flips a coin and, with 50% probability, the game starts over. Otherwise the contestant loses.

I’m not convinced this has anything to do with the host’s _knowledge_ per se.

For example, consider a different version. The contestant chooses a door. The host then peeks behind a door the door contestant didn’t choose (without showing anyone else what’s there) at random. If the prize isn’t there, the host reveals that there is no prize behind the door. Otherwise the host forgets they peeked and repeats the peek-and-choose step until they randomly get an empty door. Once an empty door has been revealed, the contestant may choose to switch. In this version, the host has no secret knowledge, but the 1/3 vs 2/3 probabilities still happen.

> The contestant chooses a door. If they choose right, they win. Otherwise the host effectively flips a coin and, with 50% probability, the game starts over. Otherwise the contestant loses.

That isn't what happening in the first simulation. The game doesn't end if the contestant chooses the prize door initially. The contestant simply chooses a door. The host randomly chooses between the remaining doors. If they reveal the prize, the game is restarted. If they reveal an empty door, then the game is logged as either "the original door was correct" or "the remaining door was correct". This is the definition of "the host chooses randomly".

> The host then peeks behind a door the door contestant didn’t choose (without showing anyone else what’s there) at random. If the prize isn’t there, the host reveals that there is no prize behind the door. Otherwise the host forgets they peeked and repeats the peek-and-choose step until they randomly get an empty door.

Picking random doors until an empty one is found absolutely the same thing as randomly choosing amongst the available doors that don't contain a prize. You can implement any function you like here. Either the host always opens a door that does not contain the prize, or the host has a 1/3 chance of revealing the prize and the game is restarted. There is simply no other possibility.

> Picking random doors until an empty one is found absolutely the same thing as randomly choosing amongst the available doors that don't contain a prize. You can implement any function you like here. Either the host always opens a door that does not contain the prize, or the host has a 1/3 chance of revealing the prize and the game is restarted. There is simply no other possibility.

Not entirely. The issue is what, exactly, gets restarted.

Version A: contestant chooses a door. Then the host picks a door and, if the host finds the prize, restarts the whole game if they picked the door with the prize. There is a 1/3 chance the contestant finds the prize. There is a 1/3 chance the contestant does not find the prize and and host does not find the prize. And there is a 1/3 chance neither one finds the prize. So 1/3 of the time the game restarts and, conditioned on a completer run, the contestant’s first guess was right with 50% probability.

Version B: contestant chooses a door. Then the host picks a door and, if the host finds the prize, restarts just this step and picks another door. There is a 1/3 chance the contestant finds the prize. If this happens, the host won’t find the prize. If the contestant’s first guess was wrong, there is a 1/2 chance the host finds the prize on any given try, but the host keeps trying so the host will not find the prize when the loop is done. So, conditioned on a completed run, the contestant’s first guess is correct 1/3 of the time.

See the difference?

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.

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.

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.

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.

>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?

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.

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"

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...

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?

He's not reliably showing it. You're being presented with a situation in which the door shown doesn't have the prize.

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?

To a casual observer it looks like they picked the door without price by chance.

> 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

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

> 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.

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