The day this was published in Parade magazine, I was sure that she was wrong.

Even though I was a computer programmer then (1990), I didn't own a computer. I drove across town to my parent's house to use my father's computer to prover her wrong.

I didn't even have to run the program. Just the act of writing the program made me realize that she was right. When I was writing the part where the host picks which door to open, I had a revelation.

Because in 2/3 scenarios you had to make the host choose the door without a car?

The host can never choose the door with the car, because then the game is over. Ergo the host has to choose the door with goat, every time.

this constraint was unclear to me in the original problem statement, which is why I thought she was wrong, too. It wasn't until somebody wrote in a letter saying they wrote a computer program that I understand what the rules were, and then it was obvious.

I wish the problem was stated more clearly to say, the host has to choose the door with goat.

> I wish the problem was stated more clearly to say, the host has to choose the door with goat.

Yes, that one point would make many people instantly realize the probability changed. Nevertheless, you'd think mathematicians and statisticians sending her offensive letters should've known better.

Right, and the host is only making a meaningful choice in the 2/3 scenarios where you didn't pick the car in the first place.

In this case, though, I don't believe it's just calculation vs. intuition. The reason the 100 or million door problem helped me understand was because it clarified an implicit rule of the game: that the door the host chooses to open is not random. If it were random, it would make for a very boring game ("Uhh, there's the prize, you win, I guess that's game over.") Once I understood that the host would only ever open "goat" doors, that's when I better understood the argument.

That was exactly what happened with me. I hadn't understood in the original proposal that the host's choice is not random. At some point during someone's offer of the '100 door' version, I realised they were operating with this extra piece of information I lacked.

Indeed, that is a crucial bit of information that is not made explicit in the original framing of the problem.

I think it's made pretty explicit. In this version: "Then, the host, who is well-aware of what’s going on behind the scenes, opens door #3, revealing one of the goats." The host has to open a door that doesn't show the car.

Yeah - I don't think that statement clarifies. The rule that the host will never reveal the prize should be made explicit.

You're quoting Priceonomic's 2021 description of the problem. This is NOT the 1990 Parade Magazine description of the problem that generated all the responses.

The 1990 Parade Magazine description is almost identical (and actually more explicit, since "say #3" is a removable parenthetical): "the host, who knows what’s behind the doors, opens another door, say #3, which has a goat" -- https://web.archive.org/web/20130121183432/http://marilynvos...

The host's knowledge is explicitly mentioned, and the only purpose this could have is that so he can use it to avoid giving the game away.

That is literally not explicit (where by "literally" I mean literally, not figuratively, and by "explicit", I mean explicit, not implicit). It is sort of hinted at, but it is not explicitly said that the host will mechanically reveal a door that has a goat. It is quite conceivable that the host picks a door randomly, or in fact that he picks the door with the car with a certain probability (saving the show quite some money).

> It is literally explicit in the 1990 rendition.

> You pick a door, say #1, and the host, who knows what’s behind the doors, opens another door, say #3, which has a goat

Remove the parenthetical example "say #3", and the parenthetical "who knows what's behind the doors" and that sentence reads: "...and the host opens another door which has a goat".

The "has a goat" is not a hypothetical example. It's a (both literal and explicit) declaration of the rules statement. The rules seem very clear that the host will open a door with a goat.

It matters whether they choose the door because it has a goat vs. they randomly choose a door that happens to have a goat.

It's not explicitly saying the former even when stating the result, because everything is written as a certainty in retrospect. You can say, "the roulette wheel stopped on 00," but it was still a random event when it happened.

nope, it's ambiguous as stated, if you're a literalist.

It's a tv show - a valid purpose is that he needs to stretch by 30 seconds before going to commercial. Another purpose is that the audience thinks seeing a goat is funny. Another purpose is to prove the show uses two different goats and doesn't do a switcheroo behind the scenes.

What the Parade article does NOT say is that the first door opened is ALWAYS not the contestant's choice and ALWAYS reveals a goat.

That he always opens a door that is not the contestant's choice is a necessary implication of saying that he opens "another" door. That he opens a door containing a goat is similarly implicated. It's not even an implication, really; it's just definitional in vernacular English. And in any event, any of the academics writing in would have been perfectly familiar with the semantic structure of such logic puzzles.

Also, most of the population of the country would have been familiar with the game show. Let's Make a Deal was one of the most popular programs on television. Even if they didn't particularly like it, people didn't have much of a TV program selection back then. And even if someone didn't regularly watch TV, it's likely they would have seen it and been familiar w/ the rules regardless simply because it was so widely watched around them. The vehemence of the respondents may even have been an effect of their familiarity, unable to recognize and accept that they'd missed a crucial analytical element after having watched the show so many times before.

What makes the puzzle counterintuitive, IMO, is that you're actually answering two different questions: the first before the host makes his selection, and the second (the only one that matters) after he makes his selection. Developing the habit of rigorously "updating your priors"--that is, recognizing when the available evidence has changed, requiring a reassessment--is an applied skill that even experts aren't particularly good at. (Nor was I the first time I read the problem.) The real-life narrative structure actually makes for a great logic puzzle as people are primed to miss the final constraint (his selection) for what it is. They're used to analyzing static situations, or a series of linked static situations--solve X correctly to solve Y. That makes the Monty Hall Problem peculiarly distinct from the typical type of logic puzzle that simply includes a misleading, irrelevant constraint or requires resolving a complex set of constraints.

> That he always opens a door that is not the contestant's choice is a necessary implication of saying that he opens "another" door

It really isn't. It's possible Monte Hall always opens the doors in reverse numerical order, so if you picked Door #3, that would get opened first. Or that if Door #3 contained the car, that would be revealed before any goats. The way the scenario was initially described by Ms. Vos Savant is not clear about that.

And no, not everybody knew Let's Make a Deal in great detail. It's always been a daytime tv show, meaning, it airs while most people are at work.

> > That he always opens a door that is not the contestant's choice is a necessary implication of saying that he opens "another" door

> It really isn't. It's possible Monte Hall always opens the doors in...

It really is. The "other" in "another" means, quite literally, "not the one already mentioned". And since the only door mentioned before that was the one the contestant had chosen, "he then opens another" not only implies but quite literally and explicitly says that this "another" door the host opens absolutely isn't, cannot be, the one the contestant chose. That's quite simply what "another" literally means.

The problem the reader originally submitted clearly implied both those things, at least clearly enough for Marylin

> and the host, who knows what’s behind the doors, opens another door ... which has a goat

The explanation Marylin originally wrote makes the second assumption explicitly clear to avoid any possibility of confusion

> Suppose there are a million doors, and you pick door #1. Then the host, who knows what’s behind the doors and will always avoid the one with the prize, opens them all except door #777,777. You’d switch to that door pretty fast, wouldn’t you?

People still wrote in claiming she was wrong.

> Since you seem to enjoy coming straight to the point, I’ll do the same. You blew it! Let me explain. If one door is shown to be a loser, that information changes the probability of either remaining choice, neither of which has any reason to be more likely, to 1/2. As a professional mathematician, I’m very concerned with the general public’s lack of mathematical skills. Please help by confessing your error and in the future being more careful.

She then explained the solution a second time with a table clearly illustrating the assumptions and the odds.

People continued to write in claiming she was wrong.

> I’m receiving thousands of letters, nearly all insisting that I’m wrong

> 92% are against my answer, and and of the letters from universities, 65% are against my answer.

Lack of clarity did not generate all the letters.

I run a small website with logic puzzles. I get thousands of emails. Some tell me the problem is misleading. Some tell me the answer is wrong.

The one thing I have learned is that once someone's mind is made up, it will never change.

The other thing I have learned, is that a good problem is succinct.

The smartest people in the room will ask clarifying questions to test out their assumptions before they even attempt an answer.

Yes.

The probability that car is behind the door NOT selected by the host is 1/2 - because the host had only two choices which are equally good for him.

Probability that car is behind the door initially selected by the user is 1/3 because that is what happens when you randomly choose one out of three.

We have to think in terms of two different probabilities: a) That user selects the correct door initially and b) That door not selected by the host has the car.

But yes it is counter-intuitive. I can't quite point out which step in the flawed reasoning of no benefit from switching occurs. How do people come up with the incorrect result, and what are the flawed assumptions they make that cause them to arrive at the wrong answer? Where exactly does the logical error happen?

People come up with the incorrect result because with 2 doors remaining, it would initially appear that they both have equal chance of containing the car. Because it is unclear HOW we arrived at the final two doors. If we arrived there randomly, the probability is in fact 50-50. If there was some manipulation - i.e. if the host knows where the car is AND deliberately opens doors that do NOT contain the car - then the probability is not 50-50, and switching away from the initially-chosen door is advantageous.

I came up with a slightly different reasoning why you should switch: When the host is about to open a door there are 2 possibilities, he can either choose between 2 goats, OR he can choose between a car and a goat.

In more than 50% of the cases the host encounters the situation where he must choose between a car and and goat. Why because in > 50% of the cases the user's initial choice does NOT have car behind it, so the host must choose between a goat and a car in > 50% of the cases.

And of course he chooses the goat. So in > 50% of the cases the remaining door will have the car behind it,

So by switching, you have > 50% chance of winning the car. Is this the correct answer? It would seem to me your chance to win if you switch is 1/2 x 2/3. Is this correct?

What is counter-intuitive is why the action by the host should give us any information about which door has the car. And it does not do so always. It only gives us information in 2/3rds of the cases. If the host chooses between two goats that should not give us any information at all should it? But that is good enough for us since we play the odds, and switching seems to increase our odds.

But, now I have my doubts. This kind of probabilistic reasoning would seem to apply only if we repeat the experiment many times. But no contestant gets to play this game more than once. Probability theory does not tell us anything about what happens in a single experiment. So is it rational for the player to think they should switch if they can play this game only once?

How much should they be willing to pay for the option of being able to switch the door, if they know they can only play this game once?

I find it a little more intuitive in a visual way with the 100 doors. If the 98 doors that the host subsequently eliminates leaves one in the middle of the pack (e.g. #52), most rational people would look at that door and say "hmm its probably in there rather than in the one i picked, #1"

Except how many people, when presented with option of picking one of a hundred doors, picks Door #1?

Further, the problem with the 100-door (or 1000, or million, etc) rationalization is that it did not appear in the initial Parade Magazine article which generated all the negative responses. Sure, today you know the secret and can explain the problem well, but it isn't fair to judge the people whose only exposure to the issue was the first, original Parade Magazine article.

Eh, the formulation I like goes:

You pick a door.

The host runs up and knocks on one of the other doors, without opening it or saying anything. He just indicates a door.

Now you have the option to stick with your one original door, or to take both of the other two doors.

Which is better?

And then: how is that scenario different from the one where he opens the special door and reveals a goat?

I read this and thought that sometimes, the host would open the door with the car.

Marilyn had the same framing of the problem as everyone else.

Marilyn made her key assumptions explicit in the original explanation she gave.

92% of people still disagreed with her.

And most of them seemed fully aware that she had the guiness world record for the highest IQ in the world.

To paraphrase Paul Graham's recent article, If the smartest person in the world proposed an idea that sounded preposterous, I'd be very reluctant to say "You are wrong."

One of the sources of confusion with this was Monty Hall did not always behave the way the problem dictates. He sometimes would open the prize door right away, or he would reveal a goat that was not behind a door or etcetera.

One of the sources of confusion with this is Monty Hall did not always behave the way the problem dictates. He sometimes would open the prize door right away, or he would reveal a goat that was not behind a door or etcetera.

Can’t it be random if there are two goat doors the host has to choose from? But I see your point about it clarifying the problem.

* Most of all, it's the calculation vs intuition that I like. You can do the calculation, or run simulations and prove correctness.*

It puzzles me that so many smart people got it wrong. It’s so easy to check your working.

When I first heard of this problem, I got it wrong too. Then I was told the answer, and to prove it to myself, it’s trivial to list the scenarios and simulate on paper. It’s still uncomfortable to think about sometimes, but I know how to show it’s true.

But all these people never thought to just do the working? Crazy!

This seems the most interesting aspect of the problem:

"However, the probability of winning by always switching is a logically distinct concept from the probability of winning by switching given that the player has picked door 1 and the host has opened door 3." [1]

[1] https://en.wikipedia.org/wiki/Monty_Hall_problem#Criticism_o...

Probability of an outcome always depends on what pool you're sampling from.

People pretty much always understand what the options are, but still get the analysis wrong. That is what I find interesting. Not the idea that there's some secret mechanism at play that alters the odds of a specific case. Even in that wonky "open the door on the right when possible" world, a contestant that always switches will still win 2/3 of the time.

I think people didn't check because the answer of 50% is so strikingly obvious (despite being wrong) that it seems like there's no work to check.

It's right depending on your assumptions. If the host has rules of only picking the right most door the contestant doesn't pick, and it happens to be a goat then the answer is 50%. Basically if the host is just as willing to show a car as show a goat and it is a goat which is shown, the answer is 50%.

There is a confirmation bias that people expect people who are aware of the Monty Hall Problem expect others to be wrong about it and attribute it to people not understanding. I think it's perfectly acceptable for someone to read the vague "the host, who is well-aware of what’s going on behind the scenes" and not assume that means he chooses to pick a goat. They're probably right, and the people saying 50% are reasoning along the lines of "there are two doors so 50/50".

For rigor, I think the host's rules need to be more strict: "The host knows what is behind the doors and will never reveal the car".

Or a 50% answer is correct if you lay out the principles of how the host acts given his knowledge.

Intention doesn’t matter. The problem states that the host opens another door and reveals a goat.

It doesn’t matter that the host meant to do that. Now that the information is revealed, the probabilities have changed.

Obviously if the host revealed the car, that would also affect the probabilities!

No. Intention does matter. If the host chooses the goat by chance ( with the possibility to choose the car instead ) then my chances to win are 50% ( switching or not ).

But we are talking about Monty Hall after all. So I could be wrong :P

Nope. If the host shows the goat _for any reason_, then the 2/3 chance that the doors you didn't choose contain the car now applies to the one unchosen door remaining -- and you should switch.

Staying is an equally good option as switching in this version if the host revealed a goat, since the host randomly revealing a goat now makes it more likely that your initial choice was correct.

P(initial choice is correct) = 1/3

P(host shows goat) = 2/3

P(host shows goat | initial choice is correct) = 1

and vice versa P(host shows goat | initial choice is wrong) = 1/2

Applying Bayes:

P(initial choice is correct | host shows goat) = P(host shows goat | initial choice is correct) * P(initial choice is correct) / P(host shows goat) = 1 * 1/3 / (2/3) = 1/2

So, initial choice now has 1/2 odds for being right. The host revealing a goat gave us information on whether he was playing the P=1/2 or the P=1 game.

Intuitively applying this with the 100 door version: the host getting lucky and revealing 98 goats in a row makes it fairly likely that the initial door was correct all along and they didn't get super lucky avoiding the car 98 times, as there's a good chance the contestant helped them by hiding the car.

Perfect case study right here. It’s trivial to simulate and y’all aren’t budging. In this case, simulating, I see 2/3 for switching after the host shows a goat on purpose, and 1/2 for switching when the host happens to have shown a goat on accident.

It does matter if he meant to or not, because if he chooses randomly from the remaining doors, the probability of him revealing a goat depends on whether you have already chosen the car. Whereas if you know he could only ever reveal a goat, then the fact that he revealed a goat tells you nothing about your current selection.

If you are interested in more details, Wikipedia lists this as the "Ignorant Monty" variant.

Whether Monte Hall is counter intuitive is a function of how the question is phrased.

When you phrase it in a way that underlines the mechanical nature of the host's decision, people get it right. When you phrase it in a way that suggests the host's choice is itself random, people get it wrong.

I think the first formulation primes people to think of it from the perspective of the host, which is the right perspective for this problem.

> When you phrase it in a way that suggests the host's choice is itself random, people get it wrong.

In other words, they still get it right; they get it right for the separate question that that phrasing implies.

If the host's choice is random, so that when you initially picked wrong it's equally probable that the host open the door with the car and then say "sorry, looks like you lost" (which is what I assumed when I first heard this problem, not being familiar with the show), then even if the host happened to open the door with a goat and give you a chance to switch, it doesn't matter if you take it or not. People are correct that, for that question, the probabilities are one in two for both of the remaining doors.

Can you help me understand - in my view the choice to switch doors or keep the same door is irrelevant, because even if you keep the same door you're making a choice that is now 2/3 of the right answer. If you switch or keep, you're still choosing from two doors that contain a car and a goat. The other door is no longer relevant and doesn't affect the new state at all.

It's your perspective (narrowing the choice down) that changes, not the position of the car. Like getting a run of red in roulette and thinking the next one has gotta be black.

When you first pick a door, you have a 1/3 chance of it being the right door. There's a 2/3 chance of it being behind a door you didn't pick.

When the host then opens a door, there's still a 2/3 chance that it is behind one of the doors you didn't pick. However, there's now only one door in this set, so there's 2/3 chance that it's behind _that_ door.

To look at another way, imagine if the host didn't reveal the content of the door, but gave you the option to switch to BOTH of the other doors instead of your door. Your odds of winning clearly go up, as you now have two chances to win (and all are of equal probability). That's equivalent to what's happening here. By showing you the losing door of those two, he doesn't change anything - there's still twice the chance it was behind one of the doors you didn't pick compared to the one you did, and by switching you win if it was behind either of them.

I've seen this explained ten different ways but this is the way I'm going to explain it from now on. thank you.

I wrote out a hopefully helpful intution down below [0]. Here's a terse part of it:

> Or just imagine monty hall with infinite doors. I'm thinking of a number between 1 and infinity (secretly, it's 19083412039102388171230123). You pick a number, and then I'll narrow down your choice to two options. Do you think you just happened to pick my number, or do you switch?

You're right that it's absolutely the narrowing down of the choice that's the key. If you happen to have not picked the prize door on your first try (more likely than not), the narrowing down will be your door and the prize door.

[0] https://news.ycombinator.com/item?id=27092614

This is the best explanation I've read thus far.

Ok. But you picked a very very very very small number between 1 and infinity. That can’t be random.

That it isn't random is precisely the point, isn't it?

The only difference that the "very very very very small number" makes is that the odds aren't infinity to one but two to one. Still winning odds.

Tons of common distributions have infinite support, but the details don't matter for this bit of intuition. If you want to formalize it in a simpler way, just consider it as an arbitrarily large finite number of doors with a uniform distribution.

Look at it from actual door, first guess door pairs -- which are the random uncorrelated events:

1,1 - switch loses (host may have shown door #2 or #3)

1,2 - switch wins (host showed door #3)

1,3 - switch wins (host showed door #2)

2,1 - switch wins (host showed door #3)

2,2 - switch loses (host may have shown door #1 or #3)

2,3 - switch wins (host showed door #1)

3,1 - switch wins (host showed door #2)

3,2 - switch wins (host showed door #1)

3,3 - switch loses (host may have shown door #1 or #2)

The choice of which door to show you is not a random event and is highly dependent upon the events which preceded it, because the host cannot show you the door it is actually behind, or the door that you picked.

incredibly succinct, thanks.

Here's another thought experiment.

Forget the opening of the door.

Start with picking a door at random, you have a 1/3 chance of having picked the car. On that I think we all agree.

Now let's say that the host offers to let you switch from the door you picked, to the other _two_ doors combined. He hasn't opened any doors, they are all closed, you're allowed to stick with your initial guess of one door, or switch to a combined guess of the other two doors.

If that's the case it should be fairly obvious that you have a 2/3 chance of getting the car by switching to the combined 2 doors.

Now that you've switched, would it really make a difference to your odds if the host opens one of your doors to reveal a goat? Would that lower your probably to 1/2 or would it remain 2/3?

It might help to have a contrasting example to clarify the key details.

Let's say we buy four scratch-off tickets. The clerk selling the tickets assures us that one of the four is a winner. You choose two tickets, and I take the remaining two.

I scratch one of my tickets, revealing that it is a dud. At this point, I suggest we should swap your two tickets for my two tickets. You start to object, but I point out that this is just like the Monty Hall problem.

When we made our initial choice, there was a 2/4 chance the winning ticket was in the set you picked, and a 2/4 chance that it was in the set that I picked. The fact that I revealed a ticket didn't change that. So you should be willing to trade, as both sets of tickets are equally likely to contain the winner.

You refuse, obviously. The difference is that whether there's two goats behind his doors or one, Monty will always choose a goat. He provides no information about the door you've already chosen, as the probability of him choosing a goat is unchanged based on whether you picked correct or not. I, on the other hand, could have revealed the winning ticket, but the odds of me doing so would be lower if you already had it in your possession. So, my dud updates the probability of all remaining tickets.

Intuitively it makes sense it would be half, because there was a 33% chance of winning but now it's 50/50 because 3-1 = 2, but when you read it logically it makes sense.

Yours is the first explanation I 'got', thanks!

Glad it helped, I just came up with it and it made it click more for me.

The other door is not irrelevant.

When you first picked a door, there was a 33% chance it was right, and a 67% chance one of the other two doors was right.

Once the other door got opened, it is still a 67% chance that the other two doors is right, but you now know which one of those two it would be - the one which wasn’t opened.

Thats what I'm stuck on, I suppose.

Imagine the entire situation is reversed. The Reverse Monty Hall problem.

I'm given a choice between two doors, once of which contains a car and one contains a goat. I have to choose, 1 or 2.

Then the game show host reveals that there was also a third door, which contained a goat, which is no longer relevant and never was relevant. I'm then also asked to choose a door (which is also irrelevant since the problem is backwards and the supposed aim is to get the car).

Even if that last step repeats 1000 times with 1000 doors and 1 car, each removing a goat-door, the only relevant choice is still the first one as the host appears to be adding new information, but it's always irrelevant information as a new choice is always made at the end.

In your reverse problem, I don't see how the new door can ever be relevant, as you know the car is behind 1 or 2, so you're only ever interested in picking between those 2.

In the original problem, its presence _is_ relevant, as the car could be behind doors 1, 2, or 3. Say you pick door 1 - there's a 1/3 chance you are right.

The host is then left with doors 2 and 3. We know there is a 2/3 chance the car is behind _one_ of these doors. When the presenter reveals a goat (say in door 2), he is reveling information about this set of doors - there is still a 2/3 chance that the car is behind one of the doors in this set, but there's only one door left we don't know anything about (3). There is therefore a 2/3 chance that it is behind _this_ door.

I think it boils down to the fact that time isn’t reversible, so the “reverse problem” isn’t like the “forward” version.

Let’s take some new problems. Suppose you have 100 doors and no switching. Your probability is 1/100, even if the host later opens a goat door, so in that sense the new information is irrelevant. But if we “reverse it” and the host opens the door first and you guess second, your probability improves to 1/99. So now suddenly the same information is relevant. Two things to observe here, one is that the forward and reverse problems are different, the other is that the relevance or irrelevance of the information depends on the direction of time. If you learn the information before you act it is relevant, afterward it is irrelevant.

One way to think about Monty Hall is you’re deciding which of these games to play. If you will stick with your first decision, you are sorta turning it into the toy problem above, where you decide the door first and then the goat information is irrelevant. Vs if you will switch, the goat door is opened before you decide, which is relevant.

Another way to think about it is with two contestants. Let’s say I pick the door initially, then someone opens the goat door, and finally you decide whether to switch. In this scenario, you don’t have self-preference bias to stick with “my” original door. In fact, my decision was the irrelevant information. It doesn’t matter at all what door I picked, what matters is whether you pick the right door, and involving me at all is a kind of misdirection to anchor you to the 1/3 probability.

Think from the perspective of your first choice. You had a 66.66% chance of choosing the wrong door.

So now I just eliminate one of the doors you didn't choose. There are two left, including the one you chose first, which only had a 33.33% chance of being correct. Nothing else has changed about the problem.

What if we started with a thousand doors? You chose 1, with a 999/1000 chance of being wrong.

Of the remaining doors, I open 499 doors. The prize is behind one of the remaining 500 that you did not choose the first time. Do you want to have a chance to pick from one of the 500?

I like expanding it out to 1000 doors. Even though the logic is the same between 3 and 1000 doors, the chances are more intuitive with more doors.

I think the part that really throws the brain off with the 3 door version is that the host opening a door seems like another random event but it's actually dependent on the state of the game at that point. Our natural intuition of the chances is for a version of the game where you pick a door, then the host randomly picks (and does not open as you also haven't) a different door, and then you have to choose if you want to switch to the door that neither of you picked.

> The other door is no longer relevant and doesn't affect the new state at all.

There is no new state, that's the thing. By picking one door you split these in two groups: one group with 1/3 to win, the other with 2/3 to win. When the host reveals the door with a goat behind it from the group that is 2/3 to win, it doesn't change the fact that that group had 2/3 to win to begin with.

I mean: that's how I see it.

The trick is that the host does not pick a door to open at random... If they did there would be a 1/3 probability they opened your door. If they didn't open yours under that circumstance, there would be a 50/50 chance you already picked the good door.

But it's not random... Monty will never open the door you chose. That's what messes up the intuitive probabilities.

Not only will he not open the door you choose, he won't open the door in front of the car either. I think that's the key point that confuses people. He knows where the car is and he always exposes a goat instead.

It's the presenter sharing knowledge with you, by opening a door that isn't the one you choose nor one that has the prize that changes the odds. And it doesn't help you if you stay with the same door because he never opens your door anyway, so it only adds knowledge about the other doors.

Maybe someone can help me with this -- what if for round two, you flipped a coin: heads you pick the door you picked previously, tails you pick the other door. Does that change things at all, since you're "switching" each time?

--

Edited -- Actually @haberman's comment that "By opening a door, Monty lets you cover 100% of the "everything else" makes the most sense to me as I think about this. Imagine there is no Monty hall, just three doors, and I say "you can choose one door and you win if there's a car behind it, or choose two doors and see if there's a car behind it." Clearly you're better off choosing two doors. And that's in fact, functionally what happens by switching after a door is eliminated.

This shows how much of our intelligence is crystalized (external). We look intelligent because we're reusing the solutions found by others. As soon as we can't do that it becomes apparent that the intelligence of a single person is not so great. It took us millennia to get where we are.

I was trying to think this through and realized it would be faster and more accurate to just write up a little script. Pretty sure this is accurate:

https://replit.com/@NateWildermuth/Monty-Hall-Problem

Thinking about this was a lot easier once I could look at the code. It seems like what's happening is that when changing doors, you are actually changing to two doors rather than one, giving you 2/3 odds?

I don’t think you need the 100 or 1,000,000 doors thing.

If you decide before not to switch then your goal is to pick the car which you have a 1/3 chance to do.

If you decide before to switch then your goal is to pick a goat which you have a 2/3 chance to do.

My favorite thing is when somebody says, “what if I want to win the goat?” Because you instantly know they’re a contrarian for its own sake.

The thing that bothers me about this problem is that (as far as I recall) it's not asserted that the host always gives the player a chance to switch. If the host sometimes gives people a choice and sometimes doesn't, based on unknown criteria, then all bets are off.

In the probability puzzle, the chance to switch is part of the basis of the puzzle. It is always offered.

In the actual game show, the rules were more flexible. The OP article mentions this. For example, the host could "offer the contestant cash NOT to switch." That introduces a more psychological poker-like aspect, but that example still allows for switching and doesn't change the underlying probabilities.

Yeah, for instance if the host only offers the switch when you've picked the car, obviously you shouldn't switch. That said, this does not seem to be the thread the respondents were pulling at...