Since a lot of university mathematicians, engineers, and physicists famously got the answer wrong, I've classified The Monty Hall Problem as more of an "English parsing" problem instead of a math problem. From wikipedia[1], the problem as stated:
>"Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?"
The key phrase that everyone glosses over is: ", and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat."
If it was reworded as "the host who knows what's behind the doors and _always_ opens the door with a goat and _never_ opens the door with the car", the mathematicians would have seen the 2/3rds probability. The original wording with the nonchalant "say" in the middle of ", opens another door, _say_ No. 3, which has a goat" makes it seems like there was no link between the host's knowledge of what's behind the door and the opening of door #3 and therefore random. Therefore, the 50/50 probability seemed to be the correct answer.
> If it was reworded as "the host who knows what's behind the doors and _always_ opens the door with a goat and _never_ opens the door with the car", the mathematicians would have seen the 2/3rds probability.
There seems to be extensive examples of mathematicians encountering very clearly worded descriptions of the problem, but still failing to jump to the right answer.
> The original wording with the nonchalant "say" in the middle of ", opens another door, say No. 3, which has a goat." makes it seems like there was no link between the host's knowledge of what's behind the door and the opening of door #3 and therefore random. Therefore the 50/50 probability seemed to be the correct answer.
Yeah, but it's not 50/50. I think you're focusing on the wrong part there. The issue isn't the host's knowledge, it's the question of whether he always opens a door. If he always opens a door at random, and if in this particular run through he happens to open a door with a goat, then you will still double your odds if you switch. The host doesn't need to know what's behind the door for this to work!
What does need to be the case is that the host always opens a door (or chooses whether or not to open a door for reasons other than his knowledge of what's behind them). In a hypothetical variant where Monty only opens a door and offers you the chance to switch if he knows you picked a car, you should never switch. In a variant where he only does so if he knows you didn't, you always should. :)
>, and if in this particular run through he happens to open a door with a goat, then you will still double your odds if you switch.
Yes, I agree.
>The host doesn't need to know what's behind the door for this to work!
The host has to know it and always open the goat door to prevent mathematicians from making a repeated simulation in their minds where the host sometimes opens a door with a car behind it.
That would be over-reading and misinterpreting the MH wording but that's why the problem is so enduring -- it uses clever text to misdirect a lot of smart people away from the correct 2/3rds answer.
Weirdly enough, in the scenario where the host doesn't know which door has the goat and just happens to open the right door, your odds of winning the car if you switch actually does go back to 50-50.
In your program, you have to classify all cases where the host picks the prize on his second selection as 'passed', and put them aside. This affects the result, as in all those cases the subject had picked one of the wrong boxes, so you have biased the cases you actually consider towards those where the subject originally picked the right box.
In the case where he just opens random doors, it doesn't matter that you switch, your door and the other unopened have the same chance.
After all, it'd be the same if you chose the door to be opened. If that mattered, you should always mentally choose a door, choose another to randomly open, switch and make that your choice. That's absurd.
The whole thing is that be cause the host has to open a door with a goat, switching means you basically get to choose both other doors.
To be clear, as we are all in agreement that the correct answer is the same whether the question states the host has knowledge or not, there is no doubt here that the question, exactly as stated, is unambiguous and has one correct answer. If someone chooses to use faulty reasoning in solving it, they have made a mistake as surely as if they had made an arithmetical error in calculating the odds incorrectly. Mathematicians and philosophers, in particular, are supposed to know how to be rigorous.
I think this has wandered across the threshold into "if they were asked a different question, they would have given a different answer", which might be true, and, if tested experimentally, might throw some light on why this problem seems unreasonably baffling, though it would be hard to find sufficient suitable candidates for this experiment (maybe among schoolchildren who have shown they have a good grasp of the relevant statistics).
Personally, I would guess that people are led down the wrong path by causal intuition - i.e. "opening the second box makes no difference to where the prize is", or perhaps by an intuition that the choice to switch or not is as random a choice (i.e. over uniform odds) as their initial pick. I realize that at this point, these guesses have no more evidence for them than does the hypothesis that people are being misled by language clues.
I think if the mathematician (or anyone) gets to that point, he's solved the problem even if the answer is in the form of a probing question.
There might be confusion about whether or not the host "cheats" and always shows a goat, but if you're asking this question you get to the answer. The issue is that people don't get to this question. I think that's not a wording-misdirection thing, it's something about how people think.
> Yeah, but it's not 50/50. I think you're focusing on the wrong part there. The issue isn't the host's knowledge, it's the question of whether he always opens a door. If he always opens a door at random, and if in this particular run through he happens to open a door with a goat, then you will still double your odds if you switch. The host doesn't need to know what's behind the door for this to work!
No, if the host always opens a door, but has no clue what's behind them, you chances will not go up by switching. That's the central clue of the Monty Hall problem, which really helped me to grok why the odds go up by switching. By opening a door, a Monty Hall who knows where everything is, is giving you information in a way that a random choice cannot do.
Let's assume you always pick door 1. Then there are 6 equiprobable possible universes:
1. Prize behind door 1, Monty opens door 2
2. Prize behind door 1, Monty opens door 3
3. Prize behind door 2, Monty opens door 2
4. Prize behind door 2, Monty opens door 3
5. Prize behind door 3, Monty opens door 2
6. Prize behind door 3, Monty opens door 3
If no prize appears behind the door he opens, we're sure we're not in universes 3 and 6. That leaves behind 4 universes, in 2 of which we'll win the prize if we switch, so a 50/50 chance.
What's the difference between this version and the normal Monty Hall? In the usual version, universes 3 and 6 never existed in the first place, and universes 1 and 2 are the same. Whether he opens the 2nd or the 3rd door depends on the protocol he uses, but is irrelevant.
The 6 universes are not equiprobable. 1+2, 3+4 and 5+6 are equiprobable (1/3 each). 1 and 2 are equiprobable (1/6 each), 3 and 6 are impossible (0 each), 4 and 5 are equiprobable (1/3 each). You should switch in 2/3 of the cases.
I was incorrect to assert that you should still switch if the host opens the door at random. My bad. (A good reminder of how hard it is to reason about this type of stats I think...)
This might be a small difference, but is this an english parsing or reality parsing issue?
I mean, shouldn't these questions come up as you're mentally interrogating the scenario, and trying to win the car? What happens if the host opens a car door, do I win?
If the host opens doors randomly, I have a 2/3 chance of winning anyway. But, I didn't have enough information to know that when I picked a door first. By the time I do, I'm already back to 50-50 odds. I don't think people make either 2/3 game their mental model of the game.
IDK about non-mathematicians, but I taught a parlour-games-ish parent day class at a high school once. We solved this problem until the kids got it. Then they had to explain it to other kids. My hint was emphasising "every week he shows a goat." Then basically this program, "playing" over and over on blackboard and recording results. Either way, it takes effort before people think of the entire game (including the 2nd choice) as a single game mechanic. Almost everyone gets it if you make them play the host.
Maybe mathematicians are less prone to suspicion... Kids and adult non-mathematicians usually suspect the host of being up to something, so they are animating him in their mental model.
Anyway, I think there's something interesting about this problem, and how people have difficulty thinking about it. I wonder if the same mechanic, but with a differently decorated story would be easier for people? A robot host, for instance.
The host never opens a car door because the host knows where the car is. The host always opens a non-car door and gives the player a choice to make the situation interesting.
How do you know that? Depending on the exact wording of the problem, or even the actual television show, the host can totally open the car door. He just didn't, this time.
I agree that the wording of the problem is important, but I'm not sure if that's what was causing the mathematicians to get the wrong answer. I think I would have naturally assumed that the host was opening a door without the car in order to raise tension, and that he would therefore never open the door with the car.
It would be interesting to survey mathematicians, but I don't think there's enough of them who haven't already heard the problem.
I disagree that it is an English parsing problem. It is a logic parsing problem. Though, it would be great to poll people in other languages to try to verify if language or culture affect how people approach the problem.
It is a logic problem because people tend to look at one part of the problem instead of the whole. They see a door taken from the pool of choices, now seeing two doors, they jump to the idea that it is 50/50. If they approach it as a simulation, it becomes clearer.
A related question: if I have two children, and one is a boy, what is the probability the other is a boy?
Got your answer? Most people, just like in the Monty Hall problem, get it wrong. If you run the simulation of options, it is clear. The options are BB, GB, BG (the option GG does not exist). So the answer is 1/3. Even seeing this, many folks have a mental block and want to say that the independent act of the other child is a given gender is still 50/50, which, ignoring the information already given in the scenario, would be true.
Back to Monty Hall. Your choices are between C, G1, and G2. Your starting position can be any of those.
C -> G1 or G2 is shown
G1 -> G2 is shown
G2 -> G1 is shown
In two of those situations, the car has not yet been picked, so you have a 2/3 chance of picking the car if you switch your door choice after the goat reveal.
People tend to think of the last door choice as independent of their earlier information. Again, I think it is a human logic flaw, but I would love a study that showed different languages or cultures approaching this problem differently!
> if I have two children, and (number) one is a boy, what is the probability the other is a boy?
The people who parse the sentence as the gender of the first child is given, will insist the odds are 50/50. Once you see the sample space, the sentence is unambigous.
I don't think it's the wording - for me the problem is that I get the intuition that, at the time just after the game show host opened a door with a goat behind it, the system can be in one of four states of equal probability:
* Closed door 1 chosen (incorrect)
* Closed door 2 chosen (incorrect)
* Closed door 1 chosen (correct)
* Closed door 2 chosen (correct)
... each with a probability of 1/4. This is probably simply because that's the kind of situation we're familiar with. And it's completely wrong in this case.
The actual state space at that point in time is (you can easily verify this by drawing the state transitions down on a piece of paper)
* Incorrect door 1 chosen, incorrect door 2 open (1/3, losing)
* Incorrect door 2 chosen, incorrect door 1 open (1/3, losing)
* Correct door chosen, incorrect door 1 open (1/6, winning)
* Correct door chosen, incorrect door 2 open (1/6, winning)
... and I think the reason that's not intuitive at first is that we're unfamiliar with thinking about a system transitioning through states with different probabilities: If you chose the incorrect door at first, the system state after the game show host intervenes is 100% determined (he will open the only one other incorrect door, and these states have 1/3 probability), but if you chose the correct door at first, he has a 50% choice and you will end up in one of two different states (1/3 * 1/2 = 1/6 probability).
So, in effect, you have four states of unequal probability after the game show host opens the door, and by switching doors you reverse the probabilities to 1/3 losing and 2/3 winning.
> The original wording with the nonchalant "say" in the middle of ", opens another door, _say_ No. 3, which has a goat" makes it seems like there was no link between the host's knowledge of what's behind the door and the opening of door #3 and therefore random. Therefore, the 50/50 probability seemed to be the correct answer.
If the host opened a door at random, then sometimes that door would contain the car. At which point, the entire exercise is meaningless—the host would in effect be asking "do you want to switch your door (which you know contains a goat) for another door (which you know contains a goat)?"
Asking the question at all when they just lost—especially given that the host just cost them the game is clearly not a desirable scenario for any game show.
So any interpretation of the problem which would allow for this scenario can be easily discarded as incorrect.
I would have thought that most times someone starts going down that road - of thinking that the host's choice had been a random selection from the two remaining doors - she would immediately see that it would be the wrong way to think about the problem because the prize would sometimes be revealed, a situation that does not occur in the question as posed, and which, when it does occur, renders the question of switching moot.
In other words, if one starts by assuming that the question implies a random choice, it quickly becomes apparent that the question would not be well-formed under that assumption. To be a well-formed question under that assumption, it would have to say something like "the host flips a coin to pick one of the closed boxes which, when opened, reveals a goat", and that is a different question, not the original one rephrased.
I think that is a bit far-fetched. In the context of a game show it would be strange to think that the host just makes a random choice. Also, why would there be the additional information that the host knows what's behind the doors if he just chooses randomly.
Since the game show stands to lose by you getting the car, it's not an unreasonable assumption that you would only be given the choice to switch had you picked the car in the first place, especially from someone who may know about game theory but may not be an expert on American TV shows.
Since a lot of university mathematicians, engineers, and physicists famously got the answer wrong
Did they? The wikipedia page you link says "After the problem appeared in Parade, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine, most of them claiming vos Savant was wrong (Tierney 1991)." And there's a story about Erdős and the number of stories about Erdős far exceed the number of stories that would be plausible about anyone, even Erdős.
Your rephrasing actually changes the question. It says that the scenario is run repeatedly, while the question is about a one-off event. By phrasing it as one of a series of events, you have told the subject how to go about answering the problem.
In an earlier reply, I suggested that the only way to interpret the question as posed is that the host is not making a random selection, and the question would have to be phrased differently to allow an interpretation in which the host's choice is random - e.g."the host flips a coin to pick one of the closed boxes which, when opened, reveals a goat".
I find that one can get a better intuitive grasp on many things when you take them to the extreme. In this case, suppose that instead of three doors, there are one hundred doors and behind 99 of them there are goats and behind 1 there's the car. After you select a door, say number 14. The host then tells you that it might be behind door number 14 but out of all the other doors he opens all but door 92, so you are left with two doors to choose from. I think the choice is quite obvious at this point. If not, then try a million doors...
As a high-schooler, this was my approach and it really locked in a strong understanding. You can really grok that with 1000 doors, for all intents and purposes, you basically had zero chance of getting it right on the first try, therefore you have basically a 100% chance of being right by switching.
This has always fascinated me. I can understand the reasoning behind it, but still don’t completely “get” it. It seems so counter intuitive. I ended up writing code to simulate it too. Glad it’s not just me that need that extra level of proof.
Edit: I very much appreciate the willingness to try and explain this below. Many have tried before; not least my poor despairing A-level maths teacher back in the day.
It's much clearer if instead of 3 doors, you use a big number, let's say 100 doors.
So you pick one of the 100 doors, then the host closes 98 other doors that are guaranteed to be wrong. Now, you can chose between that one door you started with, or go with the one door that's left after the host removed 98 wrong doors. Intuitively, which one is more likely to be right? The one you randomly picked out of 100, or the one left after 98 doors were discarded?
Thanks and yes it does make more sense when posed like that. I still (as something I consider a personal failing) end up intuiting a probability of 50%. It remains fascinating to me how different people’s brains perceive the world differently. At least I tell myself that when struggling through subjects I find tougher :)
I've seen this explanation a million times, and I don't think it's very good. Here is one that sounds almost as plausible yet yields the opposite result:
Divide the doors into two sets. The first set is the door you chose and the door that was opened; the second set is the remaining door. There is a 2/3 chance the correct door is in the first set and a 1/3 chance the correct door is in the second set. You are then told that one of the doors in the first set is not the correct door ...
I find this line of thought much more intuitive: there is a 2/3 chance your original guess was wrong. If your original guess was wrong, the other door (the one that is not the one you chose and is not the one that was opened) is the right door. So there is a 2/3 chance the other door is the right door.
Divide the doors into two sets. The first set is the door you chose and the door that was opened; the second set is the remaining door. There is a 2/3 chance the correct door is in the first set and a 1/3 chance the correct door is in the second set.
The error is assuming that the first set has a 2/3 probability of winning just because it contains 2 out of 3 doors while the way the set is constructed ensures that one of the doors, the door that is opened, never is the winning door. Including or excluding this door from a set must not change the probability you assign to this set containing the winning door, i.e. you can freely move this door between two sets or drop it altogether without affecting probabilities.
Your first arguement assumes that you can choose the door you pick initially and the door with the goat that I will open once you heve picked your first door before I have opened that second door. That is not even remotely plausible.
It's far easier to grasp when you consider that Monty opening one of the other 2 doors makes no difference to his offer (since he'll always open a door with a goat behind it anyway), i.e. he could also say "Do you want to stick with your door, or switch to both these other 2 doors?" without the unnecessary misdirection of opening the goat door beforehand.
When I was but a wee...teenager...I was introduced to the monty-hall problem.
It seemed so counter-intuitive, I can't say I wrote code to simulate it.
No.
Instead, I remember I got my dad and I said: "all right, here's what we're going to do. We're going to actually do this thing 100 times, and we're going to document the results on paper. Then I'm going to count up the results, cause this is bullshit!"
The host knows which door the car is behind. Therefore if he makes decisions based on that knowledge those decisions can give you information about the location of the car. If there is some action that he is more likely to perform when the car is behind door 1 than when it isn't, and he performs that action, then you should increase the probability you assign to the car being behind door 1.
In particular if you've chosen door 2 then the host is more likely to reveal door 3 if the car is behind door 1 than if it isn't.
It clicked with me when somebody explained it with playing cards but they used the entire deck.
Spread out an entire deck of 52 cards, face down, on a table. Guess which card is the ace of spades. Monty then flips over to reveal 50 of the remaining 51 cards (none of which are the ace of spades) and you are given a choice to stick with your original guess or switch to the other face-down card.
There are many interesting variations of this, some that give the opposite behavior, if you ever want to trick someone who already knows the Monty Hall problem.
The only way to win if we don’t switch is to have chosen the winning door (switch==false, chosen==winning); the only way to win if we switch is not to have chosen the winning door (switch==true, chosen!=winning).
Whether you should switch or not is highly dependent on what you know of the host. In the classical problem, you should switch. But there are hosts with which one should never switch.
>"Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?"
The key phrase that everyone glosses over is: ", and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat."
If it was reworded as "the host who knows what's behind the doors and _always_ opens the door with a goat and _never_ opens the door with the car", the mathematicians would have seen the 2/3rds probability. The original wording with the nonchalant "say" in the middle of ", opens another door, _say_ No. 3, which has a goat" makes it seems like there was no link between the host's knowledge of what's behind the door and the opening of door #3 and therefore random. Therefore, the 50/50 probability seemed to be the correct answer.
[1] https://en.wikipedia.org/wiki/Monty_Hall_problem