Yeah, this problem is special because it's so familiar. If you ask a riddle about two trains moving at different speeds, people will treat it like a math problem and just assume they won't get it. The Monty Hall problem really makes people upset when they don't get it.
TFA is very well written, but in almost every reproduction of the Monty Hall problem you’ll find it very poorly written such that one would reasonably be led to believe the question is “if I have two doors what is the probability.”
Don’t believe me, go search for memes presenting the problem. It’s been on the internet for decades at this point.
The setup is near always to send a poor rewrite then link you to a story of Marilyn von Savant when you get 50-50. It’s an age old internet troll.
I think this story in its entirety says a lot about ignoring the importance of good technical writing… and how people on the internet love tricking others to feel “right.”
Of the cited hate mail Savant received, I have to wonder how many were setup with poor writing on the topic.
> Of the cited hate mail Savant received, I have to wonder how many were setup with poor writing on the topic.
Have you spent at least equal time considering, given the tenor of much of it, how many were instead willingly exposing their simple misogyny?
You might be a bit younger than me, but I remember this well, and I remember this being one of my first introductions into the idea that men who are wrong will lash out with misogyny if it is a woman who proves their ignorance. Even academics.
Interestingly I’ve correctly grow up more recently such that I instead see internet simps here rather than academic misogny as notable, seeing as the tables have largely socially turned since your (I assume by the username) birth year of 1875.
> Of the cited hate mail Savant received, I have to wonder how many were setup with poor writing on the topic.
The article includes the specific wording that Savant used in her column. It seems pretty clear.
> I think this story in its entirety says a lot about ignoring the importance of good technical writing… and how people on the internet love tricking others to feel “right.”
... or, since the column that provoked all hate mail was in 1990 and people saw it in print publications, perhaps it says very little about the internet. And given the higher effort of writing an actual letter to Savant then vs now, it says that people's desire to "correct" a woman is pretty high.
Was it poorly written? It was clear enough for you and many other people to write simulations, to do mathematical analysis, etc.
Or could it be the case that some very accessible true facts in probability are counter-intuitive, but even after we're convinced that we were mistaken in our intuition, people will prefer to blame the problem?
Note, I think people sharing this kind of thing _do_ tend to be smug, but of course if the first time they heard it they were also confused, they have no reason to be smug.
I think the meme I was originally sent as a part of my story was a shitty meme involving gold and silver balls, rather than this specific article from Pricenomics which painfully explains everything.
What I was trying to get at is there are poor reproductions used to trick people that go around and must be a big part of the meme.
> However, the odds of going through this weird game show setup are 2/3 as stated.
This was one I could never get my head around, so I simply listed all the possible scenarios, and counted the number of car vs. goat results. Sure enough, it worked out to 2/3 and my mind was blown.
> I, like many, stated “50-50” because the odds of getting one of two doors right is in fact 50-50.
Only if (in fact) it's an independent choice. It's not.
If the prizes had been juggled behind the curtain after one door had been opened, then you'd be right. But the "weird setup", which is the point of the puzzle, makes it clear.
Let's be real here: the point here for most of us is to think "oh it's 50:50", be proved wrong, enjoy being humbled, learn something new about how to interrogate a problem, and enjoy setting this puzzle to younger thinkers in the future.
Not to attack the (very closely-worded) description of the problem because you're upset to be wrong.
It's a probability puzzle and a character test for thinkers.
Most replications of the puzzle are incredibly bad such that one is led to the independent branch with little indication that there was an intent to examine the totality of circumstance.
I find the problem interesting for its historically poor examples floating on the internet and their application socially beyond a math puzzle.
Marva Watkins in Chicago, Illinois, writes:
What is your real last name?
Marilyn responds:
It's "vos Savant," which was my mother's maiden name and is my own legal name. (My mother's maiden name is just as real as my father's boyhood name.) My maternal grandfather was Giuseppe vos Savant. Coincidentally, he married a woman named Maria Savant, my maternal grandmother. Growing up, I never thought about "Savant" being a word, too. I'm sure that people with the name "Miller" don't think about it, either!
By the way, my mother changed her surname to my father's surname when she got married, which was traditional at the time. I use her name, and my brother uses his name.
> Imagine that you’re on a television game show and the host presents you with three closed doors. Behind one of them, sits a sparkling, brand-new Lincoln Continental; behind the other two, are smelly old goats. The host implores you to pick a door, and you select door #1. Then, the host, who is well-aware of what’s going on behind the scenes, opens door #3, revealing one of the goats.
> “Now,” he says, turning toward you, “do you want to keep door #1, or do you want to switch to door #2?”
There are some implied assumptions. You need to know that the car is randomly placed (or that your door choice was made at random), and that the host always has to show you an empty (ok, goat) door before offering you to switch.
Under those assumptions, the winning odds of the switching strategy equal the odds of your initial pick NOT being the car.
> You need to know that the car is randomly placed
You don't need to know this.
> and that the host always has to show you an empty (ok, goat) door before offering you to switch
You know this through the line, "Then, the host, who is well-aware of what’s going on behind the scenes." If you can't understand that a game show host won't spoil the game they're hosting, that's on you and a good opportunity to pick up common sense.
Just look at this thread, filled with people who got the wrong answer and want to prove that either the smartest person in the world was wrong or that the question is ambiguous.
Do you think that given the context of the puzzle – about a US game show, printed in a US paper that the question is being asked in many developing countries?
Or are you looking to find fault with something for the sake of finding fault?
33% of the time you pick the car and I show you a goat. You switch and lose 100% of the time.
66% of the time you pick a goat and I don’t show you a goat. Assuming you don’t switch because you believe this means you have an equal odd on your current door, you lose 100% of the time.
You've restated the problem (incorrectly) -- changed it.
There's still a goat you could show me. And it is a fact that you show me a goat. Nowhere in the problem does it suggest there is a chance you show me a goat.
I do honestly admire your dogged commitment, and I think the way you are committed shows up an important point about the article and the history of the problem.
Which is that one can quite clearly fairly argue the point, as you are doing, without resorting to misogynistic or patronising rudeness as so many did at the time!
You’ve incorrectly assumed that I would be showing you a goat in every case. But that is not included in the prompt. All you know is that you were shown a goat on one play. This is possible in the setup I’ve listed. You landed in that 33% case where you chose a car.
There is nothing in the prompt that says we are not playing my variant of the game.
Being told that you SAW a goat does not mean you would ALWAYS SEE a goat if the previous conditions had gone differently. And that is why you’re wrong :)
Again
> all you know is that he opened a goat door
> You’ve incorrectly assumed that I would be showing you a goat in every case. But that is not included in the prompt.
Yes. It is. It is a fixed part of the scenario. Monty opens a door and shows you a goat. He knows it is going to be a goat (he is "well-aware of what is going on behind the scenes"). He's showing you a goat as part of the problem which is: should you switch?
Again: think through what the problem actually SAYS:
> Imagine that you’re on a television game show and the host presents you with three closed doors. Behind one of them, sits a sparkling, brand-new Lincoln Continental; behind the other two, are smelly old goats. The host implores you to pick a door, and you select door #1. Then, the host, who is well-aware of what’s going on behind the scenes, opens door #3, revealing one of the goats.
> “Now,” he says, turning toward you, “do you want to keep door #1, or do you want to switch to door #2?”
No matter what: the goat Monty shows you is not a matter of chance. It is a fixed part of the problem. It's there in writing: he shows you a goat. Full stop.
> Imagine that you’re on a television game show and the host presents you with three closed doors. Behind one of them, sits a sparkling, brand-new Lincoln Continental; behind the other two, are smelly old goats. The host implores you to pick a door, and you select door #1. Then, the host, who is well-aware of what’s going on behind the scenes, opens door #3, revealing one of the goats.
You know nothing except you picked door 1 and then he decided to show door 3.
You do not know if he would have shown door 3 had you picked door 2! This is NOT in the prompt. You cannot assume that.
The experience of this contestant is fully and equally possible within my version of the game too!
> You do not know if he would have shown door 3 had you picked door 2! This is NOT in the prompt. You cannot assume that.
Monty isn't a contestant. Monty's action and outcome is part of the fixed description of the problem. He opens one of the doors and reveals a goat. Full stop.
If you think revealing one of the two goats (the other of which might or might not be behind your current selection) isn't information of value in assessing whether your chances in that scenario are improved by switching, I'd encourage you to consider why.
And now I really must leave it to someone else to help you, if they will. But I very much appreciate your politeness.
> Monty isn't a contestant. Monty's action and outcome is part of the fixed description of the problem. He opens one of the doors and reveals a goat. Full stop.
Incorrect and I invite you to cite the line of the prompt that says otherwise. This is the critical missing piece of information that makes it the commonly understood “Monty hall problem” and not the improperly stated prompt that we actually have.
“Monty did this” != “Monty would have done this regardless of your previous choices”.
Unless this is stated, you just have an incomplete problem. If you don’t know the mechanics by which Monty decides to share information, it does not give you probabilistic info.
Probability questions frankly often require you to make judgements about the probable starting conditions based on information given so far. You frequently need to recognize that your current situation is the result of prior processes that are or are not defined.
No, you're solving the wrong thing. This is a thought experiment, not a real life scenario with ambiguity.
In this thought experiment, MH will always open a door to a goat on purpose and always ask if you want to switch, because the entire point of the thought experiment is that people find the 2/3 odds if you switch unintuitive and baffling, leading to extremely long thread chains.
You're doing the equivalent of saying "why can't we wait and see if the cat meows" when talking about the Schrödinger's thought experiment. It can't because that's not the point of the exercise.
With all due respect, no. Please don’t jump in at the end of a long thread and bring it back to the beginning. We are not discussing “The Monty Hall Problem”. We are discussing the specifics of the actual original prompt with its failure to document all assumptions.
I am curious how you feel about the "Monty Fall"/"Monty Crawl" problems linked elsewhere in the thread.[0]
For my part I am somewhat sympathetic to jncfhnb's point. The exact phrasing that Vos Savant was asked did not specify the rules that the host was required to open a door, nor that it always must be a goat door. It simply says that the host has knowledge of what's behind the doors, and in this particular iteration of the game, he showed you a goat and asked you about switching.
That does not exclude a scenario where the host is a manipulative fellow, who chose to show you the goat only because he knows you are about to win, trying to convince you to lose. A contestant on the real show would surely worry about this possibility.
Of course, the people who wrote to disagree with Vos Savant almost never said "the problem is not fully stated", they said "it's 50/50 you fool", which isn't right. Additionally, since it wouldn't be a math problem at all if we let the host have agency, it is reasonable to assume the unspoken rule that he does not, leading to Marilyn's correct 2/3rds answer.
It doesn't make a difference what causes Monty to reveal a goat.
A goat behind a door the contestant did not choose is eliminated. That is all that matters. It could have been done by a Heath-Robinson machine, or a passing lonely shrubber, or elves.
Monty isn't a floating variable in the puzzle who makes choices. His choice is fixed. Which I imagine is why vos Savant adds the actually extraneous information that Monty is fully aware what is going on behind the scenes -- to underscore the concept that Monty isn't a variable.
The fact that a goat behind an unchosen door was revealed is what is crucial to the setup of the entire puzzle.
And that -- despite jncfhnb's protestations -- is information that makes the puzzle determinate.
> It doesn't make a difference what causes Monty to reveal a goat.
Oh, but it does! See the "Monty Fall" version of the problem, in which Monty accidentally trips and opens a door, which just happens to reveal a goat. In this variant there is no advantage gained by switching, because no more information was revealed about the remaining unopened door.
The information gain only happens in the original game because we know that Monty was forced to avoid the winning door in the (66% likely) case where we didn't already pick it.
They are two different problems. Map out all the scenarios exhaustively and you'll find the difference.
In both cases we were originally 1/3 chance of being right. That is not in dispute.
In the original (fully defined) "Monty Hall", Monty was going to show us a goat no matter what. It's part of the rules, he has to show a goat. So the fact that we see a goat behind the revealed door is no surprise, and no new information. But which of the two unchosen doors was the goat, is valuable information because in 2/3s of the scenarios Monty's hand was tied and he HAD to show that door to avoid revealing the remaining car.
In the "Monty Fall" problem, the fact that we see a goat at all is interesting information. This becomes more likely when we picked the car in the first place, because if we had initially picked the car, and a random other door is opened, it's 100% going to be a goat, whereas if we had picked the goat in the first place, we are only 50% likely to see a goat when a random other door is opened. Let's call the goats Alice and Bob to illustrate this point. We know we DID see a goat, but we don't know which of these equally probable scenarios led to that:
1. We picked the car and saw Alice
2. We picked the car and saw Bob
3. We picked Alice and saw Bob
4. We picked Bob and saw Alice
Notice how "we picked the car" originally had 1/3 odds but represents half the scenarios that remain possible, because there are two ways to see a goat with that start, while only one way to see a goat with the others.
This kind of brings the problem back around to similar territory as Bertrand's Box[0] where the fact that you drew a gold coin is already hinting to you that you're more likely on the "both gold" box than on the "half gold" box.
Since this bugged me all day, and I suspect you are the kind of person where it bugged you all day, too, here is a better description of the "fall"/"hall" distinction.
I think we can agree that these are the six possible, equally likely, configurations of the problem starting from me having chosen door 1. G1 here is "goat 1" and G2 is "goat 2". For each possible prize behind my chosen door, there are two possible configurations of the remaining prizes.
My Door | Door 2 | Door 3
Car | G1 | G2
Car | G2 | G1
G1 | Car | G2
G1 | G2 | Car
G2 | Car | G1
G2 | G1 | Car
With the "Monty Hall" problem, Monty uses his knowledge to always open a goat door. Thus we see the following revealed options and resulting 2/3s probability of switching succeeding. This is the classic version of the problem.
My Door | Door 2 | Door 3 | Monty Reveals | Switch Result
Car | G1 | G2 | Either | Lose
Car | G2 | G1 | Either | Lose
G1 | Car | G2 | G2 | Win
G1 | G2 | Car | G2 | Win
G2 | Car | G1 | G1 | Win
G2 | G1 | Car | G1 | Win
With "Monty Fall", the first thing that happens is a randomly chosen door, that isn't our own, reveals a goat. This is interesting. In the classic problem we were always going to see a goat next, because those are the rules Monty plays by. But in this case, the fact that we randomly found one wasn't guaranteed.
Essentially, you are blindfolded and throw a dart at the 2x6 grid of cells under the headers "door 2" and "door 3", and I tell you that the cell you've hit is a goat. What do you know about the row you hit being a switch-or-stay row? Well, half the possible goats you might've hit are in the first 2 scenarios where you should stay, and half the possible goats are in the last 4 scenarios where you should switch. So you're at 50/50. You don't have any new information to switch on.
My Door | Door 2 | Door 3
Car | G1(a) | G2(b)
Car | G2(c) | G1(d)
G1 | Car | G2(e)
G1 | G2(f) | Car
G2 | Car | G1(g)
G2 | G1(h) | Car
You are just as likely to be looking at (a), (b), (c), or (d) (so you should stay) as you are to be looking at (e), (f), (g), or (h) (so you should switch). It is 50/50 in this version of the problem.[footnote]
This may make it confusing going back to the original. I seem to have shown that both ways make sense but still, how is it different? Imagine it like Monty is doing a random dice roll for which door to open, and he simply juices the outcome by correcting it to the goat door when a car door is selected, since he can't reveal a car and spoil the game. Now we have these equally possible scenarios (a) through (l) for his fair dice roll...
My Door | Door 2 | Door 3
Car | G1(a) | G2(b)
Car | G2(c) | G1(d)
G1 | Car(e) | G2(f)
G1 | G2(g) | Car(h)
G2 | Car(i) | G1(j)
G2 | G1(k) | Car(l)
Which he corrects, avoiding cars, to:
My Door | Door 2 | Door 3
Car | G1(a) | G2(b)
Car | G2(c) | G1(d)
G1 | Car | G2(f,e)
G1 | G2(g,h)| Car
G2 | Car | G1(i,j)
G2 | G1(k,l)| Car
Now we are back to the original game scenario where we see a goat no matter what. And we can see that 8 of the possible ways we might have arrived at seeing this goat come from "switch" rows while 4 come from "stay" rows.
[footnote] If this part of the explanation is bugging you, consider these related problems:
Problem 1. There are two opaque, externally identical bags, each containing 2 marbles. One bag contains 2 black marbles. The other bag contains 1 black marble and 1 white marble.
You choose a bag and draw a marble from it, without looking inside. The marble is black. What should you conclude are the odds that the remaining marble in the chosen bag is black?
Answer: .elbram kcalb dnoces a dnif ot ylekil sdriht-owt era ew oS .gab etihw-dna-kcalb eht morf si eno ylno dna ,gab kcalb-lla eht morf era elbram kcalb a werd ew hcihw ni soiranecs elbissop eht fo owT
Problem 2. There are three opaque, externally identical bags. One bag contains 2 black marbles. The other two bags each contain 1 black marble and 1 white marble.
Again you choose a bag and draw a marble from it, without looking inside. The marble is black. What should you conclude are the odds that the remaining marble in your chosen bag is black?
Answer: .tnecrep ytfif era elbram kcalb dnoces a gniward fo sddo ruO .gab kcalb-lla eht nesohc gnivah fo sddo ytfif-ytfif ta won era ew oS .sgab etihw-dna-kcalb tnereffid owt eht morf era owt dna ,gab kcalb-lla eht morf era elbram kcalb a werd ew hcihw ni soiranecs elbissop eht fo owT
You can connect Problem 2 to our random door opening and a goat being revealed, in the Monty Fall (with an F) problem.
I'm sorry what now? Stop second guessing the puzzle which is clearly stated.
I'm so, so done with this now that I am actually going to render inoperable my only HN account so I cannot possibly come back to this, or any other thread.
It is NOT STATED in THIS VERSION OF THE PROMPT that monte will always open a door and that it will always be a goat door.
It is simply stated that on one play of the game, Monte CHOSE to open a door and it was a goat. If this is truly all you know, you have learned nothing.
It is only useful information if monte explicitly opens a goat door for you. If he opens a door at random and gets goat your information gain goes away even if it’s the same door!
I’ve literally mapped it out for you else where. There’s only 4 outcomes. You can check yourself.
I got the problem wrong because I didn't know he never opens the door picked nor a door with a prize behind it. That's because I hadn't seen the show since I was about 5 years old, because that's about the time I stopped watching daytime TV.
That is also in the problem description. Again, it doesn't matter whether he does or doesn't ever do it (though no game show host would ruin the game).
It is the described problem that matters.
There was so, so much writing like this at the time, people saying "well, if...".
I could not grasp the probability maths (still shit at it, sorry Ms Von Savant) but I could write a program to experimentally demonstrate it, so I did.
In QBasic. Because that's what we had and we were happy.
The way the problem works started coming together in my head as I coded it. The code’s structure reflected the problem’s structure. I realized I previously misunderstood something about the problem statement. Later, I read someone saying many mathematicians got this problem wrong because of how it’s worded.
So, it may be challenging, not because probability was challenging, but because it was unclearly specified. That’s the cause of many software errors, too.
People who get it wrong are usually projecting some implied understanding onto it, which is why they get it wrong.
Which is the point of the problem. It's designed to reveal this tendency in analytical thinking leading to unexpected outcomes. You get it wrong, you're gobsmacked, you understand why, you gain some enlightenment from it. It's fun to be wrong in ways you later understand. (This is one of the thrills of programming: that moment during debugging when you shriek with joy: "YES, IT BROKE!")
The fact that it also exposes the way some overly confident people lash out with anger that exposes other biases is the point of the article, I think.
Yeah. I think my program code was ten doors or something.
It was years before I understood the actual lesson here -- both the probability lesson and also the way it hints at the fundamental truth that in life, binary choices may have been weighted in ways you don't understand, perhaps by the people asking you to make the choice.
Ha, I did the same thing after first reading about the problem. I understood the probabilities from her explanation, but it felt more indisputable after seeing the results play out after a thousand tries. Guess you could call it a "Monty" Carlo simulation.
I never understood this one. Not choosing a new door seems like an implied 50/50 choice to keep your original selection.
edit: I know it works out on paper, but if I put myself in the situation, I know that if I decided not to choose again that is a choice and I am re-choosing the same door.
Yes, although I wasn't very good at statistics in university (I found it too vague and non-math-y...), this one I never had much problems with; the fact that Monty is guaranteed to open a door with a goat and you already selected a door moved 1/3 to the door we both didn't select. You have the 1/3rd that you DID select the car, so the other he DIDN't select must have 2/3rd.
It is, strictly as written. The wording, both in the original article and the many rewrites, does not fully explain the problem as intended. The show host not only knows where the prize is, but actively chooses never to reveal the door where the prize is. The reveal is not random. The reveal is a tell where the prize isn't!
Worded more carefully, the answer is more intuitive, and this is probably why there was controversy surrounding it.
The naming is taken from an actual game show, which perhaps confusingly seems to have been a third variant on the same problem, using more steps and trick monetary incentives to make fore more exciting viewing.
The remaining doors are not random. The one you picked still has a 1/3 chance of winning, that doesn't change with opening another door. However switching doors is essentially the same as picking all the other doors at once, since the door with a goat is not opened, thus the 2/3.
Imagine the thing with 100 doors and imagine the host isn't opening any doors, but just telling you what's behind them and it becomes pretty obvious.
> Suppose you were new to the contest and were allowed to put money down on the choice at the point that there were two doors left.
That would be a completely different game. The goat doesn't get reshuffled. The doors that remain are not random and most importantly, you know which door you picked in the first round. It doesn't become a 50:50 chance just because there are two doors, as the goat isn't distributed over those two doors, but across all three.
Imagine 1000 doors. You pick one. In round two you are asked if you want to stay with that one door or pick all the other 999 doors at once. What do you do?
Case 1: Suppose there are n doors and Monty does not know what is behind any of the doors. You choose one door at random. Then Monty opens (n-2) of the remaining doors AT RANDOM until there is only one other door left. By chance none of the doors he opened had the car behind it. Then he asks you if you want to switch. Should you switch or not?
Case 2: Suppose there are n doors and Monty knows what is behind every one of the doors. You choose one door at random. Monty deliberately opens (n-2) of the remaining doors from the left to right, skipping the door with the car. Then he asks you if you want to switch. Should you switch or not?
Whether n=3 or n=100, it seems to me that it does not matter whether Monty Hall has complete knowledge or zero knowledge of the location of the car. You are required to make a choice under the condition where there is only one other unopened door and all the other doors did not reveal a car. The player's original choice was correct with probability 1/n and the probability of the complementary event must be (n-1)/n. The player's strategy of switching will result in a win with probability of (n-1)/n.
The intuitive way for me to understand the Monte Hall problem is to pretended there are 1000 doors. You pick 1. There’s a 1 in 1000 chance you get it right. The host then opens 998 doors that don’t have the prize. Do you keep your original or do you switch? Are the odds 50:50?
The most common unintended interpretation seems to be that Monty Hall is only coincidentally opening a door with a goat behind it. I don't quite understand how people arrive at that interpretation, but the 1,000-door variant should make it even more of a stretch to interpret it as Monty Hall coincidentally doing something so extremely improbable.
I think the reason it's so hard to understand is that we try to break the problem down into more digestible problems, and in this case, if you only consider the last stage of the game, it certainly seems like it should be 50:50 odds.
In other words, at the final round, you are presented with two doors. Ignore the fact that you already chose one, and assume you were starting at this point: you enter the game, see two doors, and need to pick the one you think hides a car. So intuitively, your odds of choosing which of two doors has the car are 1/2.
It's hard to really understand why it matters that the previous rounds occurred at all — why it matters how you narrowed the selection to these two doors. I don't think the 1000 door version makes that easier to understand, because the same thing is true there — if you come into a game, see two doors, and are asked to choose between them, it's very hard to understand why it matters whether in a previous round there were 2, 3, 10, or 1000 doors — there are only two now, when you make your choice.
That's especially tempting since we're used to probability problems often being set up where considering previous state is the sucker's answer. E.g. a fair coin has landed on heads three times in a row, what are the odds of it being tails on the next throw?
>It's hard to really understand why it matters that the previous rounds occurred at all
It's not hard at all. Humans are not perfectly rational beings. It's not purely about odds, it's about emotions and psychology. In a version where there's no previous selection and it's 50:50, than I pick one and live with it. In the canonical example, the previous selection means that the contestant has already staked their claim, and changing it to the loosing door would have a different emotional response than a simple 50:50 shot with no previous selection. There's a reason why Roulette shows you the last X spins and whether they're odd/even, red/black... because humans make the assumption that the last disconnected data point somehow impacts the current one.
You're also glossing over the fact that there was a 1:3 chance you picked the right door to begin with, and 2:3 chance the "other" door was right. The last round isn't 50:50, as you so claim, because there was prior information
This is the same reason why the 1000 door example helps explain things; because the math is fundamentally the same, yet significantly more imbalanced. We can also think about how we'd feel about our selection in the 1000 door version, which is likely significantly less confidence, and therefore more likelihood of switching. Whether it's 3 doors, or 1000, the math still say switching is optimal, and our psychology and emotions deal with the choice of switching different in each case.
Suppose after you make your initial choice, instead of opening a door, Monty simply asks if you'd like to switch to BOTH the other two doors, such that you win if the prize is behind EITHER of them.
That switch is intuitively a great deal, giving you 2/3 odds. The only way you can lose is in the 1/3rd case where you already picked a winner. The original scenario is equivalent to this, since by revealing a goat Monty is allowing you to pick "the best of" the two doors.
---
Often people who don't like this problem complain that actually Monty, with his knowledge of the winning door, may be trying to trick you into losing. Maybe he offers this trade conditionally based on whether you've already selected a winner. Of course, this wouldn't support the most common intuitive answer of "it's 50/50", either, making this a bad excuse. If we accept this behavior from Monty -- not stated in the problem -- it becomes a very uninteresting problem based on whether we are dealing with Evil Monty who only offers the trade when it's a bad one (stay wins 100% of the time), Friendly Monty who only offers the trade when it's a good one (switch wins 100% of the time), or somewhere in the middle. It has no analytical answer. So for this to be a solvable problem at all we must assume we are dealing with Fair Monty who offers the trade all the time, or at least, offers it at random times not based on the contestant's initial pick.
I do think this modified scenario makes the solution much clearer. At the same time, I also think that it feels like a significantly-enough different scenario from the original that saying the two scenarios have the same probability then becomes the non-intuitive part. The act of revealing what's behind the second door seems like it should change the probability from 1/3 (one door) vs. 2/3 (one of two doors) to 1/2 (one door of the remaining unopened two) vs. 1/2 (one door of the remaining unopened two, since one was "eliminated").
It's amazing how even seeing the probabilities written out, or running simulations, doesn't really make it easier to truly understand the result.
I think the important part of the problem is the unstated assumptions around it, and I suspect some people may see different assumptions than others.
If Monty always opens a door, and uses his knowledge of which door has the prize to ensure that the door he opens is always empty, then you should switch, because he's providing you with extra information about which door has the prize.
However, if Monty changes it up and sometimes doesn't open a door, or opens the door with the prize, the whole problem changes. If he only opens a door and offers you the chance to switch when you've picked the door with the prize, you should never switch of course. If he opens a random door and might reveal the prize (after which he obviously won't let you switch anymore), switching doesn't change your odds. I think a lot of people see it as one of those two situations.
Your formulation is once again ambiguous. Instead of clearly stating that Monty had to open 998 empty doors, you state that he did that. Which could mean he happened to do that.
Had to vs happened to makes a big difference.
Had he opened them at random and by (extremely small) chance they happened to be empty, the odds are quite different.
There's a variant where Monty trips and opens one of the doors by accident. Then it is 50/50 again (because he could have tripped into the car door). See also this video where cutting off a question in the middle changes the probability because of the information it provides: https://www.youtube.com/watch?v=bDZieLmya_I
You had a 1/1000 chance of picking the right door at once.
Monty had a 1/1000 chance of opening only empty doors.
He had a 998/1000 chance of opening a door to a car, but that didn't happen.
So it's either of the other two cases, with equal probabilities.
If he used his knowledge to never open a car, the 998/1000 case wouldn't exist, and there would be a 999/1000 chance that the door he left unopened had the car.
If Monty opens the door by chance there are 3 equally likely cases: There's a 1/3 chance you picked the car and Monty shows you a goat, 1/3 chance you picked a goat and Monty shows you the other goat, 1/3 chance you picked a goat and Monty shows you a car. So if Monty shows you a goat you have equal probability of being in one of the first two cases.
If Monty doesn't open the door by chance then he never shows you a car. So 2/3 of the time you picked a goat and Monty shows you the other goat.
Yup, the fact that you only have an advantage in switching when Monty "leaks information" in explicitly choosing _not_ to open a certain door as [1] pointed out, is likely the crux of what makes this unintuitive, since it is a very unusual prior.
It changes. Monty's behaviour influenced by his knowledge were the car is. He is leaking information. It is a probabilistic leak: if you first picked by a chance the only door with a car, then Monty is free to open any of remaining doors.
But if you had chosen a door with a goat, then Monty has no choice at all, he must open the only door with a goat that you didn't pick. It is a leak.
From other hand if Monty picked a door by random, he would not leak his secret, but he might open a door with the car accidentally.
> but he might open a door with the car accidentally.
Yes, but the fact that the problems never mention this possibility makes it pretty clear to me that, from the contestant's perspective, it is guaranteed that this will not happen. The original problem even mentions that Monty Hall knows what's behind the doors, which gives a clear idea of how this guarantee is implemented (versus, say, the contestant's memory being wiped and the game reset every time a car is revealed).
The language of the problem is still ambiguous, of course, because all human language is ambiguous. It could be that the car is a Hot Wheels car and the goat is actually a more valuable prize. We could quibble endlessly about the ambiguity of the problem statement, but I personally find the mathematical problem of the traditional intended interpretation more interesting.
I think GP is implying that you would reject 50:50 as absurd in this case (answer that rhetorical question with a "no"), and therefore should also reject it in the original. I agree that was not clear, though.
I actually like to back up even further and start with this problem: There are 1000 doors. You pick one. There's a 1 in 1000 chance you get it right. Now, do you want what's behind your door, or what's behind the other 999 doors? https://dynomight.net/2020/09/17/making-the-monty-hall-probl...
Do you really want 998 goats though? It must cost a fortune to transport them home from the studio, finding a space for them to live in, and feed them, etc.
The way I think about it, in general, is that Monty, by purposely not opening the correct door, is giving you a huge hint and thus changing the game which changes the odds.
Think of it similarly to those people who play the same lotto #s all the time because they think eventually it has to hit. Well, no. It's a new game every time. If the lotto commission retired each combination that won each week, then sure. And this is essentially what Monty is doing. He's retiring some doors and changing the odds.
No explanation worked for me. I did a truth table, and said, "...huh."
You don't need to model all three door choices. Just say you pick Door A, not Door 1. The first door you didn't pick is B, and C is the other door. Then map the prize behind each of 3 doors, and whether you stay or change your answer.
Then count the number of successes for change versus stay. Spoilers: It's 3/6 vs 2/6.
My experience was I coded a simulation in Python and got 50/50.
... this revealed a major flaw in my intuition that helped me to grasp it. I had subconsciously been assuming Monty sometimes opens the door revealing the prize instead of always opening a goat door. When I coded that into the simulation, of course it broke (because the player was now playing foolishly, looking at the prize and intentionally choosing the closed door that definitely still had a goat behind it).
I drew a decision tree once to show all possible states, with the 1/3 and 2/3 probabilities along each edge, which finally made it click for the coworkers I was trying to convince.
In the tree it becomes blindingly obvious that the probabilities are.
I never understood why the 1000 door version was more intuitive. Here's how I understand it.
There's 1/3 chance of picking the car. 1/3 of the time you pick the car, switch, and lose. 2/3 of the time you pick a goat, switch, and win. Why? Because 2/3 of the time you picked a goat, Monty shows you the other goat, so if you switch you definitely get the car.
I never liked the 10^x other doors version either. My thought was, what say me and a friend play a the same time, and pick the same initial door? If I NEVER change, and he ALWAYS does... I'm still going to win 1/3 of the time, and if Monty always shows the goat, ONE OF US has to win, so if I'm 1/3 by not switching, he HAS to be 2/3 by switching.
When first confronted with this, I wrote a simulation in GW-Basic on a PC with an orange monochrome monitor and ran it a bunch of times to convince myself switching was the right thing to do, statistically.
I came up with another way to visualize it that made more sense than increasing the number of doors, but it requires that one accept the fact that your chances of winning are 1/3 if you don't switch, and you know Monty always shows a losing door.
Did the host have to open 998 doors, or could he have chosen to just let you open your selected door? If he's "out to get you" then you get very little information here, because he could have just let you open your door. The only reason he'd open the other doors is to try to get you to switch, which means that it's pretty close to 50-50.
> 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?
Without additional assumptions this original wording is not clear, it could be that the host always shows a goat, then one should switch, or the host shows something at random and we just look at the cases where he shows a goat, then it doesn't matter.
> Without additional assumptions this original wording is not clear
The only problem you are being asked to solve is the one described only in the words of the problem.
That's actually what this is all about. This is a learning exercise, not just about probability, but about comprehension of what the problem as described exactly says.
Incorrect. The host doesn’t need to choose randomly. His purposeful behavior could also be random. The only way you know the switching to be optimal is if it specifically says he picks a goat door to open.
If it says he opens a door, and it has a goat, then you’ve learned nothing about the remaining two.
It doesn't matter. Because the problem explains what happens: he opens a door and reveals a goat.
That is crucial information. There's now only one goat and one car left. But the choices have not been shuffled: you're definitely still pointing at your original choice.
What were the odds that your original choice is a goat? Two in three.
If you picked a goat, what is guaranteed to be behind the other door? A car.
So what are the odds that behind the other door is a car? Two in three.
Incorrect. You have no way to know if he would only show you a goat if you picked the car.
If that were the case then switching is a guaranteed loss. Simply knowing that he opened a door and showed a goat does not mean he would have done this regardless of your choice.
You cannot assume that your context would apply from all starting conditions!
That’s why this problem is kind of bad. It does not describe the behavior of the host. It describes the perspective of a contestant halfway through the game.
Dumb example. Host flips a coin 9 times in a row and lands heads each time. If you believe this to have happened by chance then it’s probably rigged because that’s insane and it’ll likely be heads again.
If it’s ALWAYS heads then you haven’t learned anything.
It's actually counter-intuitive. I was going to argue on your side, and then I wrote up a quick program that proved me wrong[0]. Let's go through the scenarios, with Goat A, Goat B, and Car C. In the scenario where Monty picks a door purposefully, always selecting a goat, the scenarios are:
You picked A, Monty showed you B, and you switch to get C.
You picked B, Monty showed you A, and you switch to get C.
You picked C, Monty showed you either A or B, and you switch to get the other goat.
So a 2 / 3 chance of getting the car if you always switch.
If Monty is choosing randomly, we have the following scenarios:
Initial Choice | Monty's choice | Remaining Door
A | B | C
A | C | B
B | A | C
B | C | A
C | A | B
C | B | A
But we know in the problem statement that Monty hall showed us a goat, so we can eliminate possibilities 2 and 4 to get:
Initial Choice | Monty's choice | Remaining Door
A | B | C
B | A | C
C | A | B
C | B | A
Whether you switch or not, you have a 50/50 chance.
I'm not great with probabilities, but the major difference I can see is that in the first scenario, if you pick the car, Monty will either show you the goat A or B with equal probability as a part of the same 1/3 scenario. So you have really:
1/3: You picked A, Monty showed you B, and you switch to get C.
1/3: You picked B, Monty showed you A, and you switch to get C.
1/6: You picked C, Monty showed you A, and you switch to get B.
1/6: You picked C, Monty showed you B, and you switch to get A.
But in the second scenario, each of those options is actually 1/4, because he was choosing randomly. Most importantly, each option was a 1/6, but two options where you selected a goat were eliminated because those were ones where Monty selected the car.
You confused yourself. There are four possible outcomes at the end, but they are not equally likely. So not 50/50.
If Monty always shows a goat, it is undeniably a 2/3 chance to win on switch.
The nuance here being discussed is whether you can assume Monty would have shown you the goat had you chosen a different initial door just because he showed you one this time. If you don’t know that, then you don’t learn anything.
Edit: actually I misread. You just chose to ignore the cases where Monty revealed a car. Which is correct although most people chalk that up as a win or loss.
I'm covering the problem statement. I'm ignoring the cases where Monty revealed a car because it didn't happen. You can't chalk it up as a win or a loss, because it didn't happen.
It's always "You choose a door, the host reveals a goat behind another door. Do you switch?" The scenario does not include him revealing the car.
If he intentionally chooses a goat, switching gets the car 2 times out of 3.
If he chose randomly and just happened to choose a goat, switching doesn't matter. It's 50/50.
Of course, these aren't the only scenarios. As others have mentioned, if Monty can decide whether to reveal what's behind a door, he can act maliciously and only reveal a goat if you've already selected a car, and switching will always make you lose. Without knowing these constraints, a single answer isn't knowable, as you mentioned before.
Yes you’re correct. You’re just the first person I’ve ever seen cover the true random Monty variety who doesn’t force the situation where he shows a car in as a win or loss.
Total aside here, but this article is a good example of the problem, so I just want to point it out.
People go digging, usually pretty hard, to find insults to use as capital in whatever point they are trying to make. In this case, yes, she received a lot of criticism and ad-hominem attacks for her correct answer. But she wasn't really criticized for being a woman, despite it being repeated multiple times in the article.
Look at the given list of people who wrote her angry letters and what they wrote:
Ph.D
Ph.D
Ph.D
Ph.D
Ph.D
Non-doctorate professor?
Don Rivers from Sunriver Oregon who made a sexist remark.
Why is a comment from Don Rivers given so much weight? Who the fuck is Don Rivers from the town of 1,400 people in central Oregon? Man or woman, if you receive 10,000 hate letters from the general population, you can surely find whatever criticism you want if you dig hard enough. But why pay attention or give any weight to these absolute nobodies?
> But she wasn't really criticized for being a woman
Nobody is saying she was criticised for being a woman, though. What they are observing is that it was the basis of and the tenor of attacks on her character, intelligence, certainty etc.
It was nasty and it really happened, is all I'm saying. Are we better? I hope so. But I don't know that I think so.
Exactly. Would the tone of the letters been different if the column was said to be have written by a man? Likely they would be different.
This is what makes discussions about discrimination so hard- it can be very subtle. And it can include situations like this one, where we don’t have a test environment where we can compare responses to the article written by a woman to the one written by a man.
As stated, the answer is indeterminate. To demonstrate, consider these scenarios.
- Savant's Monty: The host knows where the car is, and wants to prolong the game. Switching wins 2/3 of the time.
- Ignorant Monty: The host has no idea where the car is. Revealing the goat was luck and your odds remain 50-50.
- Malicious Monty: The host knows where the car is, and wants you to lose. The fact that the host didn't reveal the car means that you have the right door. You'll have no chance of winning if you switch.
Nowhere in the problem are Monty's knowledge and motivation stated. When explained, most go along with the assumptions that Savant's argument introduces. When the actual Monty was asked, it turned out that he was usually ignorant. And the malicious scenario demonstrates how different the answer can be.
The fact that all three answers are consistent with the statement of the problem means that it is indeterminate.
I mean, the original run ended and the first revival also went bankrupt before I was born, but my understanding of the show is that the host always presented the three doors [at the end, to the show's winner, similar to the prize puzzle at the end of Wheel of Fortune] and always eliminated one of the losers.
Fun Facts which I think I'm remembering correctly: it wasn't always a goat, they would have other random animals behind the doors as well and because of the laws regulating game shows, the goats were actually a prize, for which you would receive "the cash equivalent" of the goat, if you picked that door. But because the goat was the prize, they couldn't force you to take the cash-equivalent of the goat, if you really wanted to keep the goat. Which was a big headache the handful of times it happened, because the goats were a rental.
So you made an assertion about what happened 100% of the time on the show, when you don't even remember it???
I went and re-read the interview with him that I'd read decades ago. It turns out that we're both wrong. The game was more complicated. He not only knew where the prize was, he was offering people money about whether or not to switch. He knew the answer, and was trying to manipulate the psychology of the person in front of him for the sake of entertainment. Whether he was on the side of the contestant depended on his mood. His quote about the actual game was:
*My only advice is, if you can get me to offer you $5,000 not to open the door, take the money and go home."
And if there's actual show where this experiment was done dozens of times, surely there's tons of data to confirm the correct approach here. Did anyone ever document all of that?
So long as you describe a Malicious Monty that might not have shown you the door, you could also have a Benevolent Monty who won't show you the door if you've gotten it right.
Though I'm not sure I've ever considered the Malicious Monty, only the other two.
It’s still wrong. He needs to declare (or at least he needs to consistently) open an empty box; not any random box or a box of his choosing at his whim. If he opens a random box; you have not learned anything about the keys GIVEN he opened an empty box
This paper adds that in as an assumption after the prompt, which I’m pretty sure is not the original prompt
If I understand correctly, you’re saying Monte’s intention (randomly picking an empty box vs purposely picking an empty box) is effecting the odds that the box in hand has keys?
Also, do you have any evidence that this isn’t the original?
> If I understand correctly, you’re saying Monte’s intention (randomly picking an empty box vs purposely picking an empty box) is effecting the odds that the box in hand has keys?
Sort of. If you’re saying the next box is chosen at random then there are 2 of 6 possible end games in which the key is chosen; 2 of 6 in which you pick an empty box and can switch for the keys; and 2 of 6 that both of the remaining are empty.
Since the prompt says that you did in fact open an empty box, that removes the 2 where you open the keys. So it’s 50/50.
When you know for a fact that the keys will never be chosen, the probability of picking an empty box when you chose an empty box goes from 50% to 100%. Meaning it now occupies twice the probability space. That’s now it’s 2/3 chance of winning.
You truly learn nothing if Monte randomly opens one of the doors and it is not the keys.
> Also, do you have any evidence that this isn’t the original?
The probability of A given B is the probability of A and B divided by the probability of observing B. And the probability of observing B depends on counterfactuals of various sorts. "What would happen if...?" And that's where intention comes in.
In this case B is "Monty opens an empty box". The probability of the event B depends on Monty's knowledge and intent. If Monty knows where the prize is, and always avoids it, then Monty always opens an empty box. Probability 1. If Monty is clueless, then Monty opens an empty box with probability 2/3. And if Monty is knowledgeable and malicious, then Monty opens an empty box with probability 1/3.
Event A is that you have found the prize and Monty found an empty box. We're assuming that this is the probability that you initially found the prize, and so has probability 1/3. And so we get that Savant's Monty leaves you with odds (1/3)/1 = 1/3 of having the prize, ignorant Monty leaves you with odds (1/3)/(2/3) = 1/2 of having the prize, and malicious Monty leaves you with odds (1/3)/(1/3) = 1 of having the prize.
I find it absurd that I've never looked at it this way and recognized the fourth possibility. HELPFUL Monty knows the answer, and is giving you every chance. So if you had the prize, helpful Monty would show you that you're a winner, otherwise helpful Monty will give you another chance. What helpful Monty changes is the probability of A and B. If you had the prize, you would have been shown it. Therefore the probability of A and B is 0, and you really, really want to take Monty's hint and switch.
The statement of the problem is the guarantee, just like the statement of the problem is the guarantee that there is always 1 car and 2 goats, that you always get to initially choose 1 door, etc.
You're right. The problem is stated multiple times in the article. As is usual, most of the statements do not address knowledge. But Marilyn's own statement did. However she did not address motivation.
Monty himself claims that his motivation varied depending on his mood. The actual game had more complications. And his statement about the real game was, "My only advice is, if you can get me to offer you $5,000 not to open the door, take the money and go home."
Ignorant Monty can't exist. Monty is the host of a game show, not a contestant. He's not trying to win a prize, and if he opens a prize door, it's a production failure. Whether he knows or not, somebody told him which door to open, and that door does not have the prize behind it.
Malicious Monty, however, can exist. He only ever opens a second door when you've picked correctly, but when you haven't simply accepts your choice.
The problem was stated correctly and clearly, though, as long as you assume that Monty is not a person, has no motivation at all, and as part of a math problem simply wants to reveal a goat after you've made a choice and has the means to do it. That's Savant's Monty: the Monty that doesn't add significant features to the question.
Ignorant Monty absolutely can exist. The person standing in front of the audience is not necessarily the person who arranged the prize. And if he's not, he knows nothing more than the audience.
I was, however, wrong that Monty was ignorant. By his own statement, he always knew the answer. But whether or not he was on your side depended on his mood. And he applied various forms of pressure to get you to make the right, or wrong, decision.
Therefore the problem was not stated correctly. A correct statement needs to not just state what you witness, but the counterfactuals about possibilities that you didn't witness.
I swear this problem only ever causes confusion because the original wording is ambiguous about what the host's motivations are. Once you realize the host is trying to get you to lose and knows what's behind each door and must open a door, then it's clearer that you should switch.
The host’s motives don’t matter at all. If the first choice is a goat, then the host does not have any choice about the door to open, so their intent is irrelevant. If the first choice is a goat, then the outcome is the same regardless of the door the host chooses.
The host does not really have any agency here.
The rules are intended to confuse people and trick them into thinking both remaining doors have the same probability. I can see how this would favour the house on average because once people committed and chose a door, they are invested and would tend not to change their mind. But the host’s actions have no effect. And even then, this might be an intuition, but does not provide any understanding as to what exactly happens.
I'm not sure if you would call it 'agency' but the host is following it's own specific door-picking policy - a policy which will never result in a car being revealed. It is precisely this policy and no other which makes the contestant switch the correct move. But the original wording doesn't say that - it just says the host opens a door and it happens to have a goat behind it.
Judging by the hazing Marilyn vos Savant received for her resolution of the Monty Hall problem, it seems like Twitter's preferred style of commentary isn't all that new.
No. I am not a mathematician but I got it once I understood that this is not drawing coloured balls randomly from a bag (which is how all my school probability problems seemed to go).
That is:
- the setup of the system matters.
- The state of the system at the point of the decision to switch matters.
- The choices don’t get re-randomised.
So the probabilities assigned to the original choice (and the remaining alternative) still count.
If the host had closed a curtain over the stage and randomised the remaining doors, then it would be 50:50.
But he didn’t. So you’re still in the probabilities of the original choice.
One of the goats has been removed. The car and one goat remain: you know this for sure.
You are being offered a door knowing that behind it must, necessarily, be the opposite of your original choice, and the probabilities have not been reset.
If you originally picked the goat, that door absolutely has a car behind it. And there's a 2/3 chance you picked the goat originally. So by inference there's a 2/3 chance the door has a car behind it. You should switch.
I didn’t get it until I had written a simulation to see it for myself though!
I think I have a simple explanation. There's 1/3 chance of picking the car. 1/3 of the time you pick the car, switch and lose. 2/3 of the time you pick a goat, switch and win. Why? Because 2/3 of the time you picked a goat, Monty shows you the other goat, so if you switch you definitely get the car.
Yes but in fact Monty always shows you a goat. There's always a goat for him to choose regardless of your choice, he knows where it is, and he's not going show you the car or the goat you already picked.
Showing you the goat is the event that, as you say, guarantees that the other door has the opposite of your original choice behind it. Because nobody closed the curtain to shuffle the choices.
One of the things I think people struggle with -- and I struggle with -- is that probability isn't about hypothesising about a single event that happened and how it might have happened. It's about encapsulating all the possible ways a single specified scenario can play out in a single expression.
Monty shows you a goat this time, but this means Monty always shows you a goat. There's no scenario where he is unable to show you a goat. Just like you always only pick one door. And there's always only two goats and one car.
Based on other comments, it seems the people that refuse to accept the answer that it's better to switch and not 50/50 are nitpicking how the game is described.
The idea is supposed to be that Monty always reveals a goat. He does not pick a door at random to reveal before offering the choice to switch. If you pick door A, and the car is in door B, he will reveal door C and ask if you want to switch. If you pick A, the car is in C, then he will open door B and ask if you want to switch.
Expanding it to 100 doors with Monty opening 98 of them and then offering you the opportunity to switch should make it very intuitive. Let's say you pick door 7, and Monty then opens every door except your door and say, door 48, all revealing goats, then it should make you think "Wow...very odd that he didn't reveal door 48...there must be a reason he skipped that specific door".
Make it 1,000,000 doors. You choose door 1. Monty opens every door except number 423,901, and all the doors he opened had goats. You're tired and hungry because all you've done for 4 days is watch Monty open doors to goats, but you should certainly be thinking it's odd that he skipped that one specific door. You should probably switch, though maybe take a nap before claiming your car.
In addition to the "100 door" formulation that helped me a lot, the way I rationalise it is that:
* When you choose your first door, the chance the car is behind that door is 1/3, for obvious reasons.
* When Monty opens the goat door, nothing actually changes that is relevant to your odds. He is always going to open a goat door, whether your first choice is a car or a goat, so it's basically irrelevant. He hasn't move the car, he hasn't given you any information you didn't already know when you made your first choice. So the odds of your first choice being correct can't have changed. They are still 1/3.
* The only possible outcomes are that your first-chosen door has the car, or the other remaining door has the car. The probability if your first-chosen door having the car is (still) 1/3, and the probabilities of all possible outcomes must add up to 1, so the probability of the other door having the car are 1 - 1/3 = 2/3.
I recall reading a lot of the replies and one always struck me - a (I believe) teenager wrote a program that proved her right.
That person wrote a program and proved her right, while all these stuffy old men got just downright testy with her - one even getting misogynistic with here. It was gross to read about.
But good on the teenager for actually sciencing it.
Arguably the most fascinating thing about the Monty Hall problem is that although this solution is extremely widely known in academic circles, the game is still played today on Let's Make a Deal! in the exact same way: three doors, pick one, a Zonk is revealed behind another, switch or keep?
(Although there is one variable I don't know the answer to: I don't know if they always give you the offer to switch. Sometimes I believe it's a different offer, "Keep the door or take this money out of my hand". The host being able to choose which offer is made shifts the odds a bit, because the host may be choosing the offer based on whether you've picked the right door initially, and may be making that choice in a biased or randomized way. There's hidden data there we don't have as observers).
There's the Monty Hall Problem and then there's the Second Order Monty Hall Problem Problem.
That is, did the wording as it was originally printed unambiguously define the problem to be solved, or was it ambiguous enough that some people who got it wrong got it wrong for the right reason?
Like many others, I completely missed that when Monty opens a door to show you a goat, he always shows you a goat. I think under close inspection, the problem doesn't actually allow you to interpret otherwise. But, should a problem like this depend on parsing and identifying the ambiguity, or should the problem be defined with explicit details about the door opening?
Given how many otherwise extremely smart people have misunderstood the problem, I am still unsure whether this was an intentional choice in framing the question. I got it wrong and thought Vos Savant was wrong until I read one of the replies in her follow-up article (yes I read this when it first came out!) said they wrote a computer program and reproduced her results. The program implemention worked because they correctly modelled Monty's behavior. I think when people end up writing the program out the ambiguity becomes more obvious.
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 #1, and the host, who knows what’s behind the doors, opens another door, say #3, which has a goat. He says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice of doors?
Craig F. Whitaker
Columbia, Maryland
Yes; you should switch. The first door has a 1/3 chance of winning, but the second door has a 2/3 chance. Here’s a good way to visualize what happened. 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?
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Yes, the question as posed by a reader was ambiguous but Marilyn made a very reasonable interpretation based on what someone at that time would probably tend to know about game shows. And she even gave an explanation (using more doors) that I find tends to work with folks combined with the explicit assumption that the host will always pick goats.
Yes, I'm literally talking about this sentence: "host, who knows what’s behind the doors, opens another door, say #3, which has a goat."
I parsed that as "50% of the time, monty opens another door and it has a car and you win immediately, and 50% of the time, monty opens another door and it has a goat". In retrospect I think my brain just sort of pictured that and proceeded to assume there was no reason to switch, and it wasn't until I read some answers (not her explanation) that I understood I was wrong (and I went through a lot of anger, like many of the commenters, and thought she was very wrong before that).
If there had been exactly one more sentence saying "monty always opens a door showing a goat", I'm pretty sure I would have recognized that. BTW I was familiar with the game (I liked Jeopardy better, as Let's Make a Deal was a fairly dull show) and I don't think they had a game that was identical to the problem as described.
> 50% of the time, monty opens another door and it has a car and you win immediately
Why would you think you won? You didn't choose that one with the car and you aren't offered the opportunity to switch to the open one, only the closed one. The only interpretation which makes sense is that he always shows you a goat.
I disagree. If Monty always reveals a goat, there is no tension. You know exactly what is going to happen. Maybe it could be used a way to pad the running time of the show, but it would not add tension. "Coming up next: Monty reveals what's behind one of the other doors!" seems like something a game show would do. Whereas "Coming up next: Monty reveals a goat!" does not.
Yeah, you make a good point. I can sort of see it both ways, I think it would depend on how the show handled it. Leaving it across a commercial break then yeah it probably does make more sense the door is random.
counterintuitively, it doesn't actually matter whether monty knew there was a goat before he opened the door; what matters is that you observed the goat, so you know you're not in one of the possible worlds where he opened the car door
If you want the result where there is a lower conditional probability of your originally selected door having the prize versus the remaining unopened door, then it completely does matter to have the setup be "Monty always opens a door with a goat" instead of "Monty opens a door at random, and in this particular case, it happened to have a goat behind it".
edit: i ran a monty carlo simulation¹ and i was doing the math wrong. it really does matter if monty knows or not. here's the simulation where he knows:
In [15]: non_censored_trials = got_car_trials = 0
In [16]: for trial in range(100_000):
...: car_door = random.randrange(3) # the other two doors have goats
...: your_door = random.randrange(3)
...: monty_door = random.choice(list({0, 1, 2} - {your_door, car_door}))
...: if monty_door == car_door:
...: print("Monty showed the car, never mind")
...: continue
...: non_censored_trials += 1
...: your_new_choice = next(iter({0, 1, 2} - {your_door, monty_door})) # you change your choice
...: if your_new_choice == car_door:
...: print(f"You got the car because you changed from {your_door} to {your_new_choice}")
...: got_car_trials += 1
...: else:
...: print(f"Too bad you changed; you should have stuck with {your_door}")
(...output omitted...)
In [17]: got_car_trials / non_censored_trials
Out[17]: 0.66921
so in ⅔ of the cases, switching doors gets you the car. by contrast, if monty didn't know which door would reveal a car, it's only ½ of the cases:
In [21]: non_censored_trials = got_car_trials = 0
In [22]: for trial in range(100_000):
...: car_door = random.randrange(3) # the other two doors have goats
...: your_door = random.randrange(3)
...: monty_door = random.choice(list({0, 1, 2} - {your_door}))
...: if monty_door == car_door:
...: print("Monty showed the car, never mind")
...: continue
...: non_censored_trials += 1
...: your_new_choice = next(iter({0, 1, 2} - {your_door, monty_door})) # you change your choice
...: if your_new_choice == car_door:
...: print(f"You got the car because you changed from {your_door} to {your_new_choice}")
...: got_car_trials += 1
...: else:
...: print(f"Too bad you changed; you should have stuck with {your_door}")
(...output omitted...)
In [23]: got_car_trials / non_censored_trials
Out[23]: 0.49987257709086
so if monty picked the goat door on purpose, you do gain by switching. but if he just got lucky, you don't
I dunno about anybody else, but to me teaching people how to convert word problems into code/simulations and how to interpret the results is one of the most important things a country that wants to be wealthy and powerful in the future should be doing.
I've always admired folks who can do these sorts of things in their head, while it takes me a bunch of time to inspect the code and convince myself it's an accurate representation of the word problem.
in this case I couldn't verify it against an actual experiment, which i think is good practice for newly programmed simulations, but when I saw that removing car_door from monty's choices made the probability go from 50% to 67%, i was reasonably sure that i hadn't fucked up the code, just the stuff i did in my head
One way to think about is, suppose you switch -- why not switch back?
In the random-open case, you really know nothing new about either of the closed doors. If you can talk yourself into switching, you could make an equally good argument for switching back.
In the Monty-knows-and-always-shows-goat case, you have gained information about one of the closed doors. You haven't gained any information about your initial pick door. But the other remaining door, you know there's a 2/3rds chance that Monty was forced to avoid it so as not to reveal the car. Only in the 1/3rd case where you were already on the car does Monty have freedom to open either door willy nilly.
You can tell because only trials where you chose a goat in the first round can be "censored". Your first pick is still twice as likely to be a goat as a car, but half of the times you do choose a goat on round 1 you won't get a chance to switch. Whereas if your first pick was a car, the game is guaranteed to complete (and switching is guaranteed to lose).
But if the car door is picked, there's no further game to play. Surely the only interesting thing to ask is, conditioned on seeing a goat, what's the probability the third door contains a car. The cases of observing a car are irrelevant since that's not the scenario. I'm still not convinced it's any different whether Monty Hall knows or not so long as the goat door is opened.
Edit: having written this, thinking about the 100 door case. If 98 random doors open (that aren't the one I picked) then the fact they all contain goats is pretty suggestive that I have the car, or at least 50/50. I'm not convinced by the code example though.
your first paragraph is what i thought before running the simulation, but it turned out to be wrong. if my simulation isn't convincing to you, try writing one that is
Yes, it's exactly 50/50. It's not hard to work out the details, as I did in another thread elsewhere on this article. Here goes:
-There's a 1/100 chance my door is a car, in which case it doesn't matter which of the others stays closed, it will always be a goat. The game will proceed, and switching will lose.
-There's a 99/100 chance my door is a goat, in which case the car is behind some other door. Choosing 98 out of 99 doors to open at random is the same as choosing 1 out of 99 doors to leave closed at random. So the chance that the car stays hidden in this case (so that switching will win) is 1/99, and 98/99 that the game ends early because the car is revealed.
-Adding it up, the game ends without the chance to make a choice 99/100 * 98/99 = 98/100 of the time. Of the remaining 2%, 1/100 comes from the first case (switching loses) and 99/100 * 1/99 = 1/100 comes from the second case (switching wins). The strategies are equally effective.
> the fact they all contain goats is pretty suggestive that I have the car, or at least 50/50
That's exactly the point. If he doesn't know, then it's exactly 50/50 and there is no reason to switch. If he does know, then it's 1/NUM_DOORS versus NUM_DOORS-1/NUM_DOORS, so you'd be crazy not to switch.
The point is, if he picks at random, in the 100 door case, the vast majority of the time he will open the car door while opening those 98 doors. The case where you get to pick again would be exceedingly rare. Conversely, if he only opens goat doors, you will get your second pick 100% of the time.
I believe the confusion with "ignoring the car picked cases" comes due to thinking that they are additional games to the original ones, instead of being part of those that constitute the 2/3 in which the player starts picking wrong.
So you thought that eliminating them you were left with the original 1/3 vs 2/3, when you actually removed half of the 2/3.
I don't think your interpretation of the sentence is sensible. The sentence mentions that the host knows what's behind the doors. So, if he is allowed to open the door with the car, the problem would become insoluble and would just be about speculating on the host's personality. And it definitely doesn't support the conclusion that the probabilities become 50/50.
If the format of the game allows him to show the car to the player, how could his personality not enter into it? In every case where the player picks a goat-door, the host will be presented the option to either reveal the car or the goat. I mean, one can imagine various complicated scenarios in which the host might reveal the car exactly 50% of the time in such cases, but none seem like they can be reasonably arrived at.
it turns out that there is in fact no way that his personality could not enter into it, but that was not obvious to me until i did the simulation. even if he chose to reveal the car exactly 50% of the time in such cases, that would be a result of his personality, wouldn't it?
Not logical. It doesn't follow from the assumption that the host would not pose the question "Hey, here's the car, wanna switch?" that the contestant would automatically lose. You could just as well speculate that if the host opened the door to the car the contestant would automatically win the car.
There is no randomness in this sentence. The door has a goat, not a car.
Then again, you might read it as an example outcome. But even if the door was chosen randomly, it would not justify the 50%/50% answer.
Independently of that, assuming that the host chooses the door with the car and the goat with 50% probability each is the same mistake that confused so many, you cannot always assume uniform probability only because there are 2 options.
(I wanted to point out the flaws in your thinking because this has also been a confusing problem for me)
I think you are correct about my thinking: at the time, I assumed it was an example outcome and my brain turned off at that point and simply assumed the probs were still 50-50. In some sense I was both intellectually wrong (in my assumption about it being an example) and intellectually lazy (in my failure to work through the implications of my assumption).
Personally the whole thing taught me is one true hallmark of intelligence is the ability to eliminate unnecessary ambiguity and find the "right" answer.
This was an interesting discussion when I read about it long ago.
I don't think it's obvious that Marilyn's interpretation is the correct one. Two possible interpretations could be equally valid.
In law we have the "rule of the last antecedent" that says descriptive clauses modify the nearest antecedent noun.
Under this interpretation "which has a goat" modifies the exemplary door "say #3". Same as saying host opens a door, for example door #3 containing a goat. It could have been say door #2 containing a car.
Marilyn's interpretation is that "which has a goat" modifies "host opens another door." Same as saying the door opened by the host always has a goat.
Language is inherently ambiguous. I don't think either interpretation is unreasonable...
I assume that at the time of writing, most readers would have been familiar with the actual game show and the fact that the host always opens a door with a goat?
As Monty Hall pointed out in interviews, he was not obligated to offer the opportunity to switch and sometimes did not. So in this case knowledge of the actual show may have contributed to confusion.
(But vos Savant stated in later columns that most of the critical responses she received assumed that the host was obligated to offer the switch, so they were genuinely confused by the paradox.)
> You pick a door and the host, who knows what’s behind the doors, opens exactly one of the two remaining doors according to the following algorithm: if both doors contain goats, he picks a door at random, else he picks the door he knows to contain the goat.
Your re-framing is actually not quite right either (or at least, not complete). Key point: if the host picks the car, the game essentially re-sets.
Imagine the extreme scenario with 999 goats and 1 car. You select the first door, the host opens 998 doors at random, leaving your selected door and one other. Two scenarios are now possible:
1. There was a car in the 998 doors that got opened. Tough luck, you lose the game automatically now (game resets).
2. There was not a car in the 998 doors. In this scenario, you are still advised to switch.
Edit - I guess I misremembered a simulation I once did about it.
As I explained in my reply to your comment on mine, this analysis doesn't work out, even in this very-lopsided case. If the game resets when a car is revealed, then games where you first choice happened to be a car will always complete, and games where your first choice was a goat will almost always reset. On average, of 1000 games, your first pick will be a goat 999 times, and of those, 998 will be reset because MH opens the car and 1 will complete and give you the car if you switch, whereas in 1 in 1000 you will pick the car first, in which case the game is certain to complete but you will lose the car if you switch. Therefore, most games will reset, but of the ones that complete, there is no advantage to switching.
Vos Savant mentions this, and she's kept track of who did and didn't understand the conditions:
> And a very small percentage of readers feel convinced that the furor is resulting from people not realizing that the host is opening a losing door on purpose. (But they haven’t read my mail! The great majority of people understand the conditions perfectly.)
Nice find! I have wondered this. It's easy to say after you have come around that you had first misunderstood the scenario and done the math right in that context, even if in truth you (implicitly) understood the scenario but just neglected to account for the constraints.
What astonishes me is not that people interpret the scenario differently, but that they are so quick to conclude that others are doing the math wrong and feel superior or even angry, rather than thinking about whether you might be talking at cross purposes about different scenarios.
So I see the second order problem a bit differently: it's not "was the statement unambiguous", it's about what you do with the (possibility of) ambiguity: do you assume you can't be wrong and therefore others must be, or do you try to figure out whether you were solving different problems?
There appears to be problems that make it easy for people to assume the worst of others.
And on a side note, I discovered about 6 months ago that mythbusters actually did this experiment. At the end of the day, few things work better than seeing something happen in real time.
There's the Monty Hall problem and then there's the "can I get away with falling back on claiming the question was somehow wrong after embarrassing myself by getting the Monty Hall problem incorrect" problem. No one who claimed it was unclear or ambiguous did so before getting it wrong.
It's really tiresome to have to always pretend to assume good faith when it's really clear what's going on.
That interpretation requires believing that the host, who knows what’s behind the doors, would sometimes open the door revealing the automobile and ask the contestant if they want to switch to that door. It doesn't really make sense, so that should tip off readers that they've misunderstood the show format.
Marilyn even made this explicit in her answer, writing that the host "knows what's behind the doors and will always avoid the one with the prize" [0]. So readers were arguing her answer was wrong even given this understanding of how the show works.
> That interpretation requires believing that the host, who knows what’s behind the doors
The host has to always open the door, too. Else-- if your strategy is always to switch when shown a goat, the host can choose to only show you a goat if you have already picked the car (which would cause you to lose 100%). Or various mixed strategies.
This is one of those problems that is hard to grasp at baseline, but a little harder to grasp because the situation isn't fully specified. As a result it tends to create a lot of controversy, like the one about an airplane on a treadmill.
The explanation in the article is fine as far as it goes, but I think it's much more helpful to emphasize one key fact, which is that MH has to show a goat, and can't open your first choice door, even if it's a goat. That means that if you did pick a goat the first time around, he can only have revealed the other goat, in which case switching will necessarily give you the car. That happens with probability 2/3.
This is implicit in the "100 doors, MH opens 98 of them" construction discussed in the article but making it explicit makes the idea quite clear and intuitive, even with just 3 doors.
I'm moderately sympathetic to the idea that this constraint is not immediately obvious from how the scenario is described, and could technically be ambiguous in some versions. I do think most people will agree that it makes sense to infer this constraint once you think about it. Obviously if he opened the door that you initially picked and it was a goat, you would know to switch to one of the other two (and have 1/2 chance of finding the car). (Or, more absurdly, he could just open the door with the car and you could choose it directly). Or you could imagine a game show where MH can choose to open a door or not, in which case the problem is not well-posed, because you would need to know how he makes the choice.
There are probably a lot of people at this point who have never watched one of these types of game shows. Maybe if they thought about it, they'd realize that MH probably will always show a goat but it wasn't explicit in the initial question that Marilyn answered.
Certainly, that constraint is important to the final solution.
I guess my point is that in the inevitable arguments, whether they are in advice columns or HN comment threads or around dinner tables, sometimes seem to their participants to be about how math works, when they're really about the problem description. Sometimes the disagreement amounts to "I agree that those constraints make sense and I wasn't properly accounting for them" and other times it amounts to "I was actually picturing a scenario where MH might have shown you a car". Either is ok, and there might be room for more discussion after getting there, but it feels like a big waste for either of these to get confused with a fruitless argument where both sides think the other is just doing the math wrong.
No, he wouldn't always open a door -- sometimes he'd offer a trade; like "you can take $1,000 cash from me now, or have your door".
I wonder if anyone has gone through every episode of "Let's Make a Deal" where this scenario occurs and look at the empirical distribution of switch/not-switch.
Actually it does matter. If you lose automatically when a car is shown, that will happen in 1/3 of games: none of the times your first pick was a car (0 * 1/3 = 0), half of the times your first pick was a goat (1/2 * 2/3 = 1/3). The remaining 2/3 are equally split between the times you chose a car first (which are guaranteed to complete) and the times you chose a goat first and MH randomly chose the other goat rather than the car.
(Note: this analysis does depend on the constraint that MH doesn't open your first pick door! Edit: actually the answer also becomes equal odds if he can pick any door, regardless of whether it was your first pick or the car. In that case, 1/3 of games will not complete, and the rest will be 50/50 between the unopened doors.)
Monty never shows the car. That was stated in the original reply from Marilyn, and it should be obvious to anyone who's watched a game show. Why on earth would monty open a door, show you the prize (which you now can't get) and then ask you if you want to switch your door?
No one would design a game show like that, so it means the question only has sense if interpreted as "Monty always opens a goat door".
> The explanation in the article is fine as far as it goes, but I think it's much more helpful to emphasize one key fact, which is that MH has to show a goat, and can't open your first choice door, even if it's a goat.
The second part I agree with, but not the first part; because it sounds like you're saying MH is required by the rules to reveal a goat. But the problem works fine if he always opens a remaining door at random and it happens to be a goat.
I'd say it as: once you had chosen a door, Monte was required to open some other door. And the door he happened to open, for whatever reason, had a goat.
This has been addressed several times in the comments on this post, and the answer is no, it does matter that he has to select a goat. If he might have selected a car and just got lucky in selecting the goat, then all of the following had equal probability:
1. you picked car first, he picked the first goat
2. you picked car first, he picked the other goat
3. you picked first goat first, he picked the car
4. you picked first goat first, he picked the other goat
5. you picked other goat first, he picked the car
6. you picked other goat first, he picked the first goat
You happen to know that he picked a goat, so you can eliminate 3 and 5, but the remaining four possibilities are still equally likely. Half of them let you win by switching. This is also the "MH stumbles into a door on accident" variant that others have mentioned.
Another way of getting intuition about your version is to realize that, if he might have shown you a car, then the fact that he has opened a goat is evidence against the hypothesis that your first pick was a goat, because half of the times you choose a goat he would open a car, whereas if you chose a car he would have to show you a goat every time. It eliminates some of the ways things could have turned out if you started on a goat without eliminating any of the ways things could have turned out if you started on a car, so the former is relatively less likely in retrospect than it was before MH opened the goat. More obvious in the 100-door case: if your first pick was a goat, then the fact that he hasn't shown you a car after opening 98 doors at random (or equivalently leaving one door closed at random) either means that you were very lucky in the first pick and got the can (in which case it's easy to open 98 doors without a car) or he got very lucky in his random choices. Most games played in this way (98%) would end with MH revealing the car, and the fact that you don't happen to be in one of them only tells you that one or the other of you got very lucky. The probabilities work out that the two options are equally likely.
I remember reading Marilyn vos Savant every week as a teenager. At the time I half suspected that she wasn't real--just a character made by a team or something. [This was before the internet, so I couldn't just ask Google.]
I'm happy that she's real and as smart as portrayed. But what really impressed me is how calmly she received all the negative feedback and personal insults. No matter what, she never lashed back. I have a feeling that's authentic too.
330 comments
[ 3.2 ms ] story [ 396 ms ] threadThe Time Everyone “Corrected” the World’s Smartest Woman (2015) - https://news.ycombinator.com/item?id=17686279 - Aug 2018 (112 comments)
Marilyn vos Savant and the Monty Hall Problem - https://news.ycombinator.com/item?id=9078660 - Feb 2015 (142 comments)
Game Show Problem (1990) - https://news.ycombinator.com/item?id=6386289 - Sept 2013 (31 comments)
Don’t believe me, go search for memes presenting the problem. It’s been on the internet for decades at this point.
The setup is near always to send a poor rewrite then link you to a story of Marilyn von Savant when you get 50-50. It’s an age old internet troll.
I think this story in its entirety says a lot about ignoring the importance of good technical writing… and how people on the internet love tricking others to feel “right.”
Of the cited hate mail Savant received, I have to wonder how many were setup with poor writing on the topic.
Have you spent at least equal time considering, given the tenor of much of it, how many were instead willingly exposing their simple misogyny?
You might be a bit younger than me, but I remember this well, and I remember this being one of my first introductions into the idea that men who are wrong will lash out with misogyny if it is a woman who proves their ignorance. Even academics.
The article includes the specific wording that Savant used in her column. It seems pretty clear.
> I think this story in its entirety says a lot about ignoring the importance of good technical writing… and how people on the internet love tricking others to feel “right.”
... or, since the column that provoked all hate mail was in 1990 and people saw it in print publications, perhaps it says very little about the internet. And given the higher effort of writing an actual letter to Savant then vs now, it says that people's desire to "correct" a woman is pretty high.
I guess it all depends on your perspective. I see a lot more people on the internet piling on with nitpicks to avoid feeling "wrong."
Note, I think people sharing this kind of thing _do_ tend to be smug, but of course if the first time they heard it they were also confused, they have no reason to be smug.
What I was trying to get at is there are poor reproductions used to trick people that go around and must be a big part of the meme.
This was one I could never get my head around, so I simply listed all the possible scenarios, and counted the number of car vs. goat results. Sure enough, it worked out to 2/3 and my mind was blown.
Only if (in fact) it's an independent choice. It's not.
If the prizes had been juggled behind the curtain after one door had been opened, then you'd be right. But the "weird setup", which is the point of the puzzle, makes it clear.
Let's be real here: the point here for most of us is to think "oh it's 50:50", be proved wrong, enjoy being humbled, learn something new about how to interrogate a problem, and enjoy setting this puzzle to younger thinkers in the future.
Not to attack the (very closely-worded) description of the problem because you're upset to be wrong.
It's a probability puzzle and a character test for thinkers.
I find the problem interesting for its historically poor examples floating on the internet and their application socially beyond a math puzzle.
That’s my view and point in commenting.
If you didn't notice, "Savant" means smart.
So is it just a coincidence? Karma? The pressure of carrying such a name drove her to smartness?
Marilyn vos Savant was born Marilyn Mach on August 11, 1946, in St. Louis, Missouri, to parents Joseph Mach and Marina vos Savant.
... so only 50% coincidence.
It reflects some broad understandings of the 1980s and what we valued.
Marva Watkins in Chicago, Illinois, writes: What is your real last name?
Marilyn responds: It's "vos Savant," which was my mother's maiden name and is my own legal name. (My mother's maiden name is just as real as my father's boyhood name.) My maternal grandfather was Giuseppe vos Savant. Coincidentally, he married a woman named Maria Savant, my maternal grandmother. Growing up, I never thought about "Savant" being a word, too. I'm sure that people with the name "Miller" don't think about it, either! By the way, my mother changed her surname to my father's surname when she got married, which was traditional at the time. I use her name, and my brother uses his name.
https://parade.com/418148/marilynvossavant/marilyns-surname/
> “Now,” he says, turning toward you, “do you want to keep door #1, or do you want to switch to door #2?”
There are some implied assumptions. You need to know that the car is randomly placed (or that your door choice was made at random), and that the host always has to show you an empty (ok, goat) door before offering you to switch.
Under those assumptions, the winning odds of the switching strategy equal the odds of your initial pick NOT being the car.
You don't need to know this.
> and that the host always has to show you an empty (ok, goat) door before offering you to switch
You know this through the line, "Then, the host, who is well-aware of what’s going on behind the scenes." If you can't understand that a game show host won't spoil the game they're hosting, that's on you and a good opportunity to pick up common sense.
Because he did it this time?
Amazed this nitpicking is still going on; it's illustrative.
This puzzle is like a shibboleth for nitpickers.
More like a shibboleth for sore losers.
Just look at this thread, filled with people who got the wrong answer and want to prove that either the smartest person in the world was wrong or that the question is ambiguous.
Or are you looking to find fault with something for the sake of finding fault?
The shibboleth for nitpickers title is locked up.
I instead love when people get uppity about others being wrong about Monty hall while still being wrong about Monty hall!
It matters what he did in the problem as it is described. Because that is what you're solving.
It could be that he always opens a goat door meaning you should switch.
It could be that he only opens a goat door if you picked the car and you should not.
It could be he only opens a goat door if you picked a goat and you should.
Probability cannot determine which of these is any more likely than the other. You have learned nothing.
This is precisely where you are wrong.
We play this game 1000 times.
33% of the time you pick the car and I show you a goat. You switch and lose 100% of the time.
66% of the time you pick a goat and I don’t show you a goat. Assuming you don’t switch because you believe this means you have an equal odd on your current door, you lose 100% of the time.
Congrats. You have lost every single round.
There's still a goat you could show me. And it is a fact that you show me a goat. Nowhere in the problem does it suggest there is a chance you show me a goat.
I do honestly admire your dogged commitment, and I think the way you are committed shows up an important point about the article and the history of the problem.
Which is that one can quite clearly fairly argue the point, as you are doing, without resorting to misogynistic or patronising rudeness as so many did at the time!
But you're still wrong. :-)
And I'm going to leave it here.
You’ve incorrectly assumed that I would be showing you a goat in every case. But that is not included in the prompt. All you know is that you were shown a goat on one play. This is possible in the setup I’ve listed. You landed in that 33% case where you chose a car.
There is nothing in the prompt that says we are not playing my variant of the game.
Being told that you SAW a goat does not mean you would ALWAYS SEE a goat if the previous conditions had gone differently. And that is why you’re wrong :)
Again > all you know is that he opened a goat door
Yes. It is. It is a fixed part of the scenario. Monty opens a door and shows you a goat. He knows it is going to be a goat (he is "well-aware of what is going on behind the scenes"). He's showing you a goat as part of the problem which is: should you switch?
Again: think through what the problem actually SAYS:
> Imagine that you’re on a television game show and the host presents you with three closed doors. Behind one of them, sits a sparkling, brand-new Lincoln Continental; behind the other two, are smelly old goats. The host implores you to pick a door, and you select door #1. Then, the host, who is well-aware of what’s going on behind the scenes, opens door #3, revealing one of the goats.
> “Now,” he says, turning toward you, “do you want to keep door #1, or do you want to switch to door #2?”
No matter what: the goat Monty shows you is not a matter of chance. It is a fixed part of the problem. It's there in writing: he shows you a goat. Full stop.
> Imagine that you’re on a television game show and the host presents you with three closed doors. Behind one of them, sits a sparkling, brand-new Lincoln Continental; behind the other two, are smelly old goats. The host implores you to pick a door, and you select door #1. Then, the host, who is well-aware of what’s going on behind the scenes, opens door #3, revealing one of the goats.
You know nothing except you picked door 1 and then he decided to show door 3.
You do not know if he would have shown door 3 had you picked door 2! This is NOT in the prompt. You cannot assume that.
The experience of this contestant is fully and equally possible within my version of the game too!
Monty isn't a contestant. Monty's action and outcome is part of the fixed description of the problem. He opens one of the doors and reveals a goat. Full stop.
If you think revealing one of the two goats (the other of which might or might not be behind your current selection) isn't information of value in assessing whether your chances in that scenario are improved by switching, I'd encourage you to consider why.
And now I really must leave it to someone else to help you, if they will. But I very much appreciate your politeness.
Incorrect and I invite you to cite the line of the prompt that says otherwise. This is the critical missing piece of information that makes it the commonly understood “Monty hall problem” and not the improperly stated prompt that we actually have.
“Monty did this” != “Monty would have done this regardless of your previous choices”.
Unless this is stated, you just have an incomplete problem. If you don’t know the mechanics by which Monty decides to share information, it does not give you probabilistic info.
Probability questions frankly often require you to make judgements about the probable starting conditions based on information given so far. You frequently need to recognize that your current situation is the result of prior processes that are or are not defined.
In this thought experiment, MH will always open a door to a goat on purpose and always ask if you want to switch, because the entire point of the thought experiment is that people find the 2/3 odds if you switch unintuitive and baffling, leading to extremely long thread chains.
You're doing the equivalent of saying "why can't we wait and see if the cat meows" when talking about the Schrödinger's thought experiment. It can't because that's not the point of the exercise.
For my part I am somewhat sympathetic to jncfhnb's point. The exact phrasing that Vos Savant was asked did not specify the rules that the host was required to open a door, nor that it always must be a goat door. It simply says that the host has knowledge of what's behind the doors, and in this particular iteration of the game, he showed you a goat and asked you about switching.
That does not exclude a scenario where the host is a manipulative fellow, who chose to show you the goat only because he knows you are about to win, trying to convince you to lose. A contestant on the real show would surely worry about this possibility.
Of course, the people who wrote to disagree with Vos Savant almost never said "the problem is not fully stated", they said "it's 50/50 you fool", which isn't right. Additionally, since it wouldn't be a math problem at all if we let the host have agency, it is reasonable to assume the unspoken rule that he does not, leading to Marilyn's correct 2/3rds answer.
[0]https://web.archive.org/web/20230706235720/https://probabili...
A goat behind a door the contestant did not choose is eliminated. That is all that matters. It could have been done by a Heath-Robinson machine, or a passing lonely shrubber, or elves.
Monty isn't a floating variable in the puzzle who makes choices. His choice is fixed. Which I imagine is why vos Savant adds the actually extraneous information that Monty is fully aware what is going on behind the scenes -- to underscore the concept that Monty isn't a variable.
The fact that a goat behind an unchosen door was revealed is what is crucial to the setup of the entire puzzle.
And that -- despite jncfhnb's protestations -- is information that makes the puzzle determinate.
Oh, but it does! See the "Monty Fall" version of the problem, in which Monty accidentally trips and opens a door, which just happens to reveal a goat. In this variant there is no advantage gained by switching, because no more information was revealed about the remaining unopened door.
The information gain only happens in the original game because we know that Monty was forced to avoid the winning door in the (66% likely) case where we didn't already pick it.
But I am going to leave this to someone else to explain. I'm tired out now.
The only possible options are (equally likely)
You choose A and get shown B You choose B and get shown A You choose C and get shown A You choose C and get shown B
4 options. You lose 50% of the time.
The 2 options where you would normally get the 2/3 odds are explicitly ignored when you are told Monty chosen randomly and randomly got a goat.
You choose A and get shown C or you choose B and get shown C are forbidden.
50/50 probability on the dot.
If Monty chooses with intention then the options where you choose goat and are shown goat double in probability so you get back to 2/3
In both cases we were originally 1/3 chance of being right. That is not in dispute.
In the original (fully defined) "Monty Hall", Monty was going to show us a goat no matter what. It's part of the rules, he has to show a goat. So the fact that we see a goat behind the revealed door is no surprise, and no new information. But which of the two unchosen doors was the goat, is valuable information because in 2/3s of the scenarios Monty's hand was tied and he HAD to show that door to avoid revealing the remaining car.
In the "Monty Fall" problem, the fact that we see a goat at all is interesting information. This becomes more likely when we picked the car in the first place, because if we had initially picked the car, and a random other door is opened, it's 100% going to be a goat, whereas if we had picked the goat in the first place, we are only 50% likely to see a goat when a random other door is opened. Let's call the goats Alice and Bob to illustrate this point. We know we DID see a goat, but we don't know which of these equally probable scenarios led to that:
1. We picked the car and saw Alice
2. We picked the car and saw Bob
3. We picked Alice and saw Bob
4. We picked Bob and saw Alice
Notice how "we picked the car" originally had 1/3 odds but represents half the scenarios that remain possible, because there are two ways to see a goat with that start, while only one way to see a goat with the others.
This kind of brings the problem back around to similar territory as Bertrand's Box[0] where the fact that you drew a gold coin is already hinting to you that you're more likely on the "both gold" box than on the "half gold" box.
[0]https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox
I think we can agree that these are the six possible, equally likely, configurations of the problem starting from me having chosen door 1. G1 here is "goat 1" and G2 is "goat 2". For each possible prize behind my chosen door, there are two possible configurations of the remaining prizes.
With the "Monty Hall" problem, Monty uses his knowledge to always open a goat door. Thus we see the following revealed options and resulting 2/3s probability of switching succeeding. This is the classic version of the problem. With "Monty Fall", the first thing that happens is a randomly chosen door, that isn't our own, reveals a goat. This is interesting. In the classic problem we were always going to see a goat next, because those are the rules Monty plays by. But in this case, the fact that we randomly found one wasn't guaranteed.Essentially, you are blindfolded and throw a dart at the 2x6 grid of cells under the headers "door 2" and "door 3", and I tell you that the cell you've hit is a goat. What do you know about the row you hit being a switch-or-stay row? Well, half the possible goats you might've hit are in the first 2 scenarios where you should stay, and half the possible goats are in the last 4 scenarios where you should switch. So you're at 50/50. You don't have any new information to switch on.
You are just as likely to be looking at (a), (b), (c), or (d) (so you should stay) as you are to be looking at (e), (f), (g), or (h) (so you should switch). It is 50/50 in this version of the problem.[footnote]This may make it confusing going back to the original. I seem to have shown that both ways make sense but still, how is it different? Imagine it like Monty is doing a random dice roll for which door to open, and he simply juices the outcome by correcting it to the goat door when a car door is selected, since he can't reveal a car and spoil the game. Now we have these equally possible scenarios (a) through (l) for his fair dice roll...
Which he corrects, avoiding cars, to: Now we are back to the original game scenario where we see a goat no matter what. And we can see that 8 of the possible ways we might have arrived at seeing this goat come from "switch" rows while 4 come from "stay" rows.Problem 1. There are two opaque, externally identical bags, each containing 2 marbles. One bag contains 2 black marbles. The other bag contains 1 black marble and 1 white marble.
You choose a bag and draw a marble from it, without looking inside. The marble is black. What should you conclude are the odds that the remaining marble in the chosen bag is black?
Answer: .elbram kcalb dnoces a dnif ot ylekil sdriht-owt era ew oS .gab etihw-dna-kcalb eht morf si eno ylno dna ,gab kcalb-lla eht morf era elbram kcalb a werd ew hcihw ni soiranecs elbissop eht fo owT
Problem 2. There are three opaque, externally identical bags. One bag contains 2 black marbles. The other two bags each contain 1 black marble and 1 white marble.
Again you choose a bag and draw a marble from it, without looking inside. The marble is black. What should you conclude are the odds that the remaining marble in your chosen bag is black?
Answer: .tnecrep ytfif era elbram kcalb dnoces a gniward fo sddo ruO .gab kcalb-lla eht nesohc gnivah fo sddo ytfif-ytfif ta won era ew oS .sgab etihw-dna-kcalb tnereffid owt eht morf era owt dna ,gab kcalb-lla eht morf era elbram kcalb a werd ew hcihw ni soiranecs elbissop eht fo owT
You can connect Problem 2 to our random door opening and a goat being revealed, in the Monty Fall (with an F) problem.
It is not the fact that a goat is behind the door. It is the fact that a goat door would always have been opened! These two facts are not the same!
I'm so, so done with this now that I am actually going to render inoperable my only HN account so I cannot possibly come back to this, or any other thread.
It is simply stated that on one play of the game, Monte CHOSE to open a door and it was a goat. If this is truly all you know, you have learned nothing.
It is only useful information if monte explicitly opens a goat door for you. If he opens a door at random and gets goat your information gain goes away even if it’s the same door!
I’ve literally mapped it out for you else where. There’s only 4 outcomes. You can check yourself.
2:2
Only if there are three doors - or rather, only if the host eliminates all but two doors, one of which is the one you chose.
If there are, say, four doors, and the host eliminates one, your odds of success on switching are not 3/4.
It is the described problem that matters.
There was so, so much writing like this at the time, people saying "well, if...".
I could not grasp the probability maths (still shit at it, sorry Ms Von Savant) but I could write a program to experimentally demonstrate it, so I did.
In QBasic. Because that's what we had and we were happy.
The way the problem works started coming together in my head as I coded it. The code’s structure reflected the problem’s structure. I realized I previously misunderstood something about the problem statement. Later, I read someone saying many mathematicians got this problem wrong because of how it’s worded.
So, it may be challenging, not because probability was challenging, but because it was unclearly specified. That’s the cause of many software errors, too.
People who get it wrong are usually projecting some implied understanding onto it, which is why they get it wrong.
Which is the point of the problem. It's designed to reveal this tendency in analytical thinking leading to unexpected outcomes. You get it wrong, you're gobsmacked, you understand why, you gain some enlightenment from it. It's fun to be wrong in ways you later understand. (This is one of the thrills of programming: that moment during debugging when you shriek with joy: "YES, IT BROKE!")
The fact that it also exposes the way some overly confident people lash out with anger that exposes other biases is the point of the article, I think.
It was years before I understood the actual lesson here -- both the probability lesson and also the way it hints at the fundamental truth that in life, binary choices may have been weighted in ways you don't understand, perhaps by the people asking you to make the choice.
Turns out, instead, she's the G.O.A.T.
My character hinges on that understanding now.
edit: I know it works out on paper, but if I put myself in the situation, I know that if I decided not to choose again that is a choice and I am re-choosing the same door.
I found working through it to be really rewarding, you might too.
Spoiler alert: consider how Monty's choices are constrained by what he knows, and how the player can capitalize on that.
Worded more carefully, the answer is more intuitive, and this is probably why there was controversy surrounding it.
The naming is taken from an actual game show, which perhaps confusingly seems to have been a third variant on the same problem, using more steps and trick monetary incentives to make fore more exciting viewing.
Imagine the thing with 100 doors and imagine the host isn't opening any doors, but just telling you what's behind them and it becomes pretty obvious.
If there are only two choices remaining, I guess I don't understand the assertion that there is still a 1/3 chance of winning.
Suppose you were new to the contest and were allowed to put money down on the choice at the point that there were two doors left.
Then you repeated the bet every new contestant.
Would one door would be more profitable than the other?
That would be a completely different game. The goat doesn't get reshuffled. The doors that remain are not random and most importantly, you know which door you picked in the first round. It doesn't become a 50:50 chance just because there are two doors, as the goat isn't distributed over those two doors, but across all three.
Imagine 1000 doors. You pick one. In round two you are asked if you want to stay with that one door or pick all the other 999 doors at once. What do you do?
Case 2: Suppose there are n doors and Monty knows what is behind every one of the doors. You choose one door at random. Monty deliberately opens (n-2) of the remaining doors from the left to right, skipping the door with the car. Then he asks you if you want to switch. Should you switch or not?
Whether n=3 or n=100, it seems to me that it does not matter whether Monty Hall has complete knowledge or zero knowledge of the location of the car. You are required to make a choice under the condition where there is only one other unopened door and all the other doors did not reveal a car. The player's original choice was correct with probability 1/n and the probability of the complementary event must be (n-1)/n. The player's strategy of switching will result in a win with probability of (n-1)/n.
It's not 50/50 because the choice wasn't random.
The problem stated with 3 doors is hard, with 5 is still not quite there for many. 100, 100, 10_000_000... becomes almost absurd.
In other words, at the final round, you are presented with two doors. Ignore the fact that you already chose one, and assume you were starting at this point: you enter the game, see two doors, and need to pick the one you think hides a car. So intuitively, your odds of choosing which of two doors has the car are 1/2.
It's hard to really understand why it matters that the previous rounds occurred at all — why it matters how you narrowed the selection to these two doors. I don't think the 1000 door version makes that easier to understand, because the same thing is true there — if you come into a game, see two doors, and are asked to choose between them, it's very hard to understand why it matters whether in a previous round there were 2, 3, 10, or 1000 doors — there are only two now, when you make your choice.
It's not hard at all. Humans are not perfectly rational beings. It's not purely about odds, it's about emotions and psychology. In a version where there's no previous selection and it's 50:50, than I pick one and live with it. In the canonical example, the previous selection means that the contestant has already staked their claim, and changing it to the loosing door would have a different emotional response than a simple 50:50 shot with no previous selection. There's a reason why Roulette shows you the last X spins and whether they're odd/even, red/black... because humans make the assumption that the last disconnected data point somehow impacts the current one.
You're also glossing over the fact that there was a 1:3 chance you picked the right door to begin with, and 2:3 chance the "other" door was right. The last round isn't 50:50, as you so claim, because there was prior information
This is the same reason why the 1000 door example helps explain things; because the math is fundamentally the same, yet significantly more imbalanced. We can also think about how we'd feel about our selection in the 1000 door version, which is likely significantly less confidence, and therefore more likelihood of switching. Whether it's 3 doors, or 1000, the math still say switching is optimal, and our psychology and emotions deal with the choice of switching different in each case.
Suppose after you make your initial choice, instead of opening a door, Monty simply asks if you'd like to switch to BOTH the other two doors, such that you win if the prize is behind EITHER of them.
That switch is intuitively a great deal, giving you 2/3 odds. The only way you can lose is in the 1/3rd case where you already picked a winner. The original scenario is equivalent to this, since by revealing a goat Monty is allowing you to pick "the best of" the two doors.
---
Often people who don't like this problem complain that actually Monty, with his knowledge of the winning door, may be trying to trick you into losing. Maybe he offers this trade conditionally based on whether you've already selected a winner. Of course, this wouldn't support the most common intuitive answer of "it's 50/50", either, making this a bad excuse. If we accept this behavior from Monty -- not stated in the problem -- it becomes a very uninteresting problem based on whether we are dealing with Evil Monty who only offers the trade when it's a bad one (stay wins 100% of the time), Friendly Monty who only offers the trade when it's a good one (switch wins 100% of the time), or somewhere in the middle. It has no analytical answer. So for this to be a solvable problem at all we must assume we are dealing with Fair Monty who offers the trade all the time, or at least, offers it at random times not based on the contestant's initial pick.
It's amazing how even seeing the probabilities written out, or running simulations, doesn't really make it easier to truly understand the result.
If Monty always opens a door, and uses his knowledge of which door has the prize to ensure that the door he opens is always empty, then you should switch, because he's providing you with extra information about which door has the prize.
However, if Monty changes it up and sometimes doesn't open a door, or opens the door with the prize, the whole problem changes. If he only opens a door and offers you the chance to switch when you've picked the door with the prize, you should never switch of course. If he opens a random door and might reveal the prize (after which he obviously won't let you switch anymore), switching doesn't change your odds. I think a lot of people see it as one of those two situations.
Had to vs happened to makes a big difference.
Had he opened them at random and by (extremely small) chance they happened to be empty, the odds are quite different.
Does it really change anything?
You had a 1/1000 chance of picking the right door at once.
Monty had a 1/1000 chance of opening only empty doors.
He had a 998/1000 chance of opening a door to a car, but that didn't happen.
So it's either of the other two cases, with equal probabilities.
If he used his knowledge to never open a car, the 998/1000 case wouldn't exist, and there would be a 999/1000 chance that the door he left unopened had the car.
If Monty opens the door by chance there are 3 equally likely cases: There's a 1/3 chance you picked the car and Monty shows you a goat, 1/3 chance you picked a goat and Monty shows you the other goat, 1/3 chance you picked a goat and Monty shows you a car. So if Monty shows you a goat you have equal probability of being in one of the first two cases.
If Monty doesn't open the door by chance then he never shows you a car. So 2/3 of the time you picked a goat and Monty shows you the other goat.
[1] https://news.ycombinator.com/item?id=39514463
But if you had chosen a door with a goat, then Monty has no choice at all, he must open the only door with a goat that you didn't pick. It is a leak.
From other hand if Monty picked a door by random, he would not leak his secret, but he might open a door with the car accidentally.
Yes, but the fact that the problems never mention this possibility makes it pretty clear to me that, from the contestant's perspective, it is guaranteed that this will not happen. The original problem even mentions that Monty Hall knows what's behind the doors, which gives a clear idea of how this guarantee is implemented (versus, say, the contestant's memory being wiped and the game reset every time a car is revealed).
The language of the problem is still ambiguous, of course, because all human language is ambiguous. It could be that the car is a Hot Wheels car and the goat is actually a more valuable prize. We could quibble endlessly about the ambiguity of the problem statement, but I personally find the mathematical problem of the traditional intended interpretation more interesting.
That still leads to the wrong answer, because he is retiring one of the losers, leaving a 50:50 chance of getting a prize.
The act of retiring leads to the wrong answer, so it does not aid in understanding.
You don't need to model all three door choices. Just say you pick Door A, not Door 1. The first door you didn't pick is B, and C is the other door. Then map the prize behind each of 3 doors, and whether you stay or change your answer.
Then count the number of successes for change versus stay. Spoilers: It's 3/6 vs 2/6.
... this revealed a major flaw in my intuition that helped me to grasp it. I had subconsciously been assuming Monty sometimes opens the door revealing the prize instead of always opening a goat door. When I coded that into the simulation, of course it broke (because the player was now playing foolishly, looking at the prize and intentionally choosing the closed door that definitely still had a goat behind it).
In the tree it becomes blindingly obvious that the probabilities are.
There's 1/3 chance of picking the car. 1/3 of the time you pick the car, switch, and lose. 2/3 of the time you pick a goat, switch, and win. Why? Because 2/3 of the time you picked a goat, Monty shows you the other goat, so if you switch you definitely get the car.
I never liked the 10^x other doors version either. My thought was, what say me and a friend play a the same time, and pick the same initial door? If I NEVER change, and he ALWAYS does... I'm still going to win 1/3 of the time, and if Monty always shows the goat, ONE OF US has to win, so if I'm 1/3 by not switching, he HAS to be 2/3 by switching.
I came up with another way to visualize it that made more sense than increasing the number of doors, but it requires that one accept the fact that your chances of winning are 1/3 if you don't switch, and you know Monty always shows a losing door.
Without additional assumptions this original wording is not clear, it could be that the host always shows a goat, then one should switch, or the host shows something at random and we just look at the cases where he shows a goat, then it doesn't matter.
The only problem you are being asked to solve is the one described only in the words of the problem.
That's actually what this is all about. This is a learning exercise, not just about probability, but about comprehension of what the problem as described exactly says.
That's what makes it so enormously valuable.
That’s what the sentence says: “and the host, who knows what's behind the doors, opens another door […] which has a goat”.
> or the host shows something at random and we just look at the cases where he shows a goat
Nothing says that the host opens a door at random. Quite the contrary, we know that the host has a perfect knowledge of the situation.
If it says he opens a door, and it has a goat, then you’ve learned nothing about the remaining two.
It doesn't matter. Because the problem explains what happens: he opens a door and reveals a goat.
That is crucial information. There's now only one goat and one car left. But the choices have not been shuffled: you're definitely still pointing at your original choice.
What were the odds that your original choice is a goat? Two in three.
If you picked a goat, what is guaranteed to be behind the other door? A car.
So what are the odds that behind the other door is a car? Two in three.
You should switch.
If that were the case then switching is a guaranteed loss. Simply knowing that he opened a door and showed a goat does not mean he would have done this regardless of your choice.
Not a probability problem without this info.
What do you mean?
It's specified in the scenario. He shows you a goat. It's right there. This isn't a variable. It's a fact.
Your only job is to work out whether, given the scenario described, it makes sense to switch. Given all your possible choices.
That’s why this problem is kind of bad. It does not describe the behavior of the host. It describes the perspective of a contestant halfway through the game.
Dumb example. Host flips a coin 9 times in a row and lands heads each time. If you believe this to have happened by chance then it’s probably rigged because that’s insane and it’ll likely be heads again.
If it’s ALWAYS heads then you haven’t learned anything.
If Monty is choosing randomly, we have the following scenarios:
But we know in the problem statement that Monty hall showed us a goat, so we can eliminate possibilities 2 and 4 to get: Whether you switch or not, you have a 50/50 chance.I'm not great with probabilities, but the major difference I can see is that in the first scenario, if you pick the car, Monty will either show you the goat A or B with equal probability as a part of the same 1/3 scenario. So you have really:
But in the second scenario, each of those options is actually 1/4, because he was choosing randomly. Most importantly, each option was a 1/6, but two options where you selected a goat were eliminated because those were ones where Monty selected the car.[0]: https://gist.github.com/Taywee/2ba202b1bf7af40293ecffb01c2ab...
If Monty always shows a goat, it is undeniably a 2/3 chance to win on switch.
The nuance here being discussed is whether you can assume Monty would have shown you the goat had you chosen a different initial door just because he showed you one this time. If you don’t know that, then you don’t learn anything.
Edit: actually I misread. You just chose to ignore the cases where Monty revealed a car. Which is correct although most people chalk that up as a win or loss.
It's always "You choose a door, the host reveals a goat behind another door. Do you switch?" The scenario does not include him revealing the car.
If he intentionally chooses a goat, switching gets the car 2 times out of 3.
If he chose randomly and just happened to choose a goat, switching doesn't matter. It's 50/50.
Of course, these aren't the only scenarios. As others have mentioned, if Monty can decide whether to reveal what's behind a door, he can act maliciously and only reveal a goat if you've already selected a car, and switching will always make you lose. Without knowing these constraints, a single answer isn't knowable, as you mentioned before.
People go digging, usually pretty hard, to find insults to use as capital in whatever point they are trying to make. In this case, yes, she received a lot of criticism and ad-hominem attacks for her correct answer. But she wasn't really criticized for being a woman, despite it being repeated multiple times in the article.
Look at the given list of people who wrote her angry letters and what they wrote:
Why is a comment from Don Rivers given so much weight? Who the fuck is Don Rivers from the town of 1,400 people in central Oregon? Man or woman, if you receive 10,000 hate letters from the general population, you can surely find whatever criticism you want if you dig hard enough. But why pay attention or give any weight to these absolute nobodies?Nobody is saying she was criticised for being a woman, though. What they are observing is that it was the basis of and the tenor of attacks on her character, intelligence, certainty etc.
It was nasty and it really happened, is all I'm saying. Are we better? I hope so. But I don't know that I think so.
This is what makes discussions about discrimination so hard- it can be very subtle. And it can include situations like this one, where we don’t have a test environment where we can compare responses to the article written by a woman to the one written by a man.
FWIW, Sunriver is a resort community outside of Bend. It's not the middle of nowhere, it's filled with tech industry vacationers and wealthy retirees.
- Savant's Monty: The host knows where the car is, and wants to prolong the game. Switching wins 2/3 of the time.
- Ignorant Monty: The host has no idea where the car is. Revealing the goat was luck and your odds remain 50-50.
- Malicious Monty: The host knows where the car is, and wants you to lose. The fact that the host didn't reveal the car means that you have the right door. You'll have no chance of winning if you switch.
Nowhere in the problem are Monty's knowledge and motivation stated. When explained, most go along with the assumptions that Savant's argument introduces. When the actual Monty was asked, it turned out that he was usually ignorant. And the malicious scenario demonstrates how different the answer can be.
The fact that all three answers are consistent with the statement of the problem means that it is indeterminate.
There was sufficient evidence for Savant's Monty -- the car was revealed 0% of the time, not 33 or 66% of the time.
Fun Facts which I think I'm remembering correctly: it wasn't always a goat, they would have other random animals behind the doors as well and because of the laws regulating game shows, the goats were actually a prize, for which you would receive "the cash equivalent" of the goat, if you picked that door. But because the goat was the prize, they couldn't force you to take the cash-equivalent of the goat, if you really wanted to keep the goat. Which was a big headache the handful of times it happened, because the goats were a rental.
I went and re-read the interview with him that I'd read decades ago. It turns out that we're both wrong. The game was more complicated. He not only knew where the prize was, he was offering people money about whether or not to switch. He knew the answer, and was trying to manipulate the psychology of the person in front of him for the sake of entertainment. Whether he was on the side of the contestant depended on his mood. His quote about the actual game was:
*My only advice is, if you can get me to offer you $5,000 not to open the door, take the money and go home."
Though I'm not sure I've ever considered the Malicious Monty, only the other two.
"and the host, who knows what’s behind the doors, opens another door [..] which has a goat."
The question is clear: The host (1) knows what is behind each door and (2) always shows a goat. It's clearly a determinate problem.
Well you do have that guarantee -- as it is stated in the problem. It's clear that no matter what door you open, you will be shown a goat.
https://www.jstor.org/stable/2683689
On the last line of the problem he opens an empty box.
This paper adds that in as an assumption after the prompt, which I’m pretty sure is not the original prompt
Also, do you have any evidence that this isn’t the original?
Sort of. If you’re saying the next box is chosen at random then there are 2 of 6 possible end games in which the key is chosen; 2 of 6 in which you pick an empty box and can switch for the keys; and 2 of 6 that both of the remaining are empty.
Since the prompt says that you did in fact open an empty box, that removes the 2 where you open the keys. So it’s 50/50.
When you know for a fact that the keys will never be chosen, the probability of picking an empty box when you chose an empty box goes from 50% to 100%. Meaning it now occupies twice the probability space. That’s now it’s 2/3 chance of winning.
You truly learn nothing if Monte randomly opens one of the doors and it is not the keys.
> Also, do you have any evidence that this isn’t the original?
The quote in the article?
The probability of A given B is the probability of A and B divided by the probability of observing B. And the probability of observing B depends on counterfactuals of various sorts. "What would happen if...?" And that's where intention comes in.
In this case B is "Monty opens an empty box". The probability of the event B depends on Monty's knowledge and intent. If Monty knows where the prize is, and always avoids it, then Monty always opens an empty box. Probability 1. If Monty is clueless, then Monty opens an empty box with probability 2/3. And if Monty is knowledgeable and malicious, then Monty opens an empty box with probability 1/3.
Event A is that you have found the prize and Monty found an empty box. We're assuming that this is the probability that you initially found the prize, and so has probability 1/3. And so we get that Savant's Monty leaves you with odds (1/3)/1 = 1/3 of having the prize, ignorant Monty leaves you with odds (1/3)/(2/3) = 1/2 of having the prize, and malicious Monty leaves you with odds (1/3)/(1/3) = 1 of having the prize.
I find it absurd that I've never looked at it this way and recognized the fourth possibility. HELPFUL Monty knows the answer, and is giving you every chance. So if you had the prize, helpful Monty would show you that you're a winner, otherwise helpful Monty will give you another chance. What helpful Monty changes is the probability of A and B. If you had the prize, you would have been shown it. Therefore the probability of A and B is 0, and you really, really want to take Monty's hint and switch.
It states you opened a door and then Monty opened a door.
It does not state that Monty would have opened a door if you had picked a different one.
If this is not communicated you don’t gain any knowledge from his reveal.
The typical proper Monty hall formula states this assumption clearly that he will always open a goat door. The original one does not state this.
You're right. The problem is stated multiple times in the article. As is usual, most of the statements do not address knowledge. But Marilyn's own statement did. However she did not address motivation.
Monty himself claims that his motivation varied depending on his mood. The actual game had more complications. And his statement about the real game was, "My only advice is, if you can get me to offer you $5,000 not to open the door, take the money and go home."
You can find the article I got that quote from at https://www.nytimes.com/1991/07/21/us/behind-monty-hall-s-do....
Malicious Monty, however, can exist. He only ever opens a second door when you've picked correctly, but when you haven't simply accepts your choice.
The problem was stated correctly and clearly, though, as long as you assume that Monty is not a person, has no motivation at all, and as part of a math problem simply wants to reveal a goat after you've made a choice and has the means to do it. That's Savant's Monty: the Monty that doesn't add significant features to the question.
I was, however, wrong that Monty was ignorant. By his own statement, he always knew the answer. But whether or not he was on your side depended on his mood. And he applied various forms of pressure to get you to make the right, or wrong, decision.
Therefore the problem was not stated correctly. A correct statement needs to not just state what you witness, but the counterfactuals about possibilities that you didn't witness.
The host does not really have any agency here.
The rules are intended to confuse people and trick them into thinking both remaining doors have the same probability. I can see how this would favour the house on average because once people committed and chose a door, they are invested and would tend not to change their mind. But the host’s actions have no effect. And even then, this might be an intuition, but does not provide any understanding as to what exactly happens.
I completely get all the explanations but it just feels too weird.
That is:
- the setup of the system matters.
- The state of the system at the point of the decision to switch matters.
- The choices don’t get re-randomised.
So the probabilities assigned to the original choice (and the remaining alternative) still count.
If the host had closed a curtain over the stage and randomised the remaining doors, then it would be 50:50.
But he didn’t. So you’re still in the probabilities of the original choice.
One of the goats has been removed. The car and one goat remain: you know this for sure.
You are being offered a door knowing that behind it must, necessarily, be the opposite of your original choice, and the probabilities have not been reset.
If you originally picked the goat, that door absolutely has a car behind it. And there's a 2/3 chance you picked the goat originally. So by inference there's a 2/3 chance the door has a car behind it. You should switch.
I didn’t get it until I had written a simulation to see it for myself though!
Showing you the goat is the event that, as you say, guarantees that the other door has the opposite of your original choice behind it. Because nobody closed the curtain to shuffle the choices.
One of the things I think people struggle with -- and I struggle with -- is that probability isn't about hypothesising about a single event that happened and how it might have happened. It's about encapsulating all the possible ways a single specified scenario can play out in a single expression.
Monty shows you a goat this time, but this means Monty always shows you a goat. There's no scenario where he is unable to show you a goat. Just like you always only pick one door. And there's always only two goats and one car.
(Full marks for username choice)
The idea is supposed to be that Monty always reveals a goat. He does not pick a door at random to reveal before offering the choice to switch. If you pick door A, and the car is in door B, he will reveal door C and ask if you want to switch. If you pick A, the car is in C, then he will open door B and ask if you want to switch.
Expanding it to 100 doors with Monty opening 98 of them and then offering you the opportunity to switch should make it very intuitive. Let's say you pick door 7, and Monty then opens every door except your door and say, door 48, all revealing goats, then it should make you think "Wow...very odd that he didn't reveal door 48...there must be a reason he skipped that specific door".
Make it 1,000,000 doors. You choose door 1. Monty opens every door except number 423,901, and all the doors he opened had goats. You're tired and hungry because all you've done for 4 days is watch Monty open doors to goats, but you should certainly be thinking it's odd that he skipped that one specific door. You should probably switch, though maybe take a nap before claiming your car.
* When you choose your first door, the chance the car is behind that door is 1/3, for obvious reasons.
* When Monty opens the goat door, nothing actually changes that is relevant to your odds. He is always going to open a goat door, whether your first choice is a car or a goat, so it's basically irrelevant. He hasn't move the car, he hasn't given you any information you didn't already know when you made your first choice. So the odds of your first choice being correct can't have changed. They are still 1/3.
* The only possible outcomes are that your first-chosen door has the car, or the other remaining door has the car. The probability if your first-chosen door having the car is (still) 1/3, and the probabilities of all possible outcomes must add up to 1, so the probability of the other door having the car are 1 - 1/3 = 2/3.
That person wrote a program and proved her right, while all these stuffy old men got just downright testy with her - one even getting misogynistic with here. It was gross to read about.
But good on the teenager for actually sciencing it.
(Although there is one variable I don't know the answer to: I don't know if they always give you the offer to switch. Sometimes I believe it's a different offer, "Keep the door or take this money out of my hand". The host being able to choose which offer is made shifts the odds a bit, because the host may be choosing the offer based on whether you've picked the right door initially, and may be making that choice in a biased or randomized way. There's hidden data there we don't have as observers).
That is, did the wording as it was originally printed unambiguously define the problem to be solved, or was it ambiguous enough that some people who got it wrong got it wrong for the right reason?
Like many others, I completely missed that when Monty opens a door to show you a goat, he always shows you a goat. I think under close inspection, the problem doesn't actually allow you to interpret otherwise. But, should a problem like this depend on parsing and identifying the ambiguity, or should the problem be defined with explicit details about the door opening?
Given how many otherwise extremely smart people have misunderstood the problem, I am still unsure whether this was an intentional choice in framing the question. I got it wrong and thought Vos Savant was wrong until I read one of the replies in her follow-up article (yes I read this when it first came out!) said they wrote a computer program and reproduced her results. The program implemention worked because they correctly modelled Monty's behavior. I think when people end up writing the program out the ambiguity becomes more obvious.
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 #1, and the host, who knows what’s behind the doors, opens another door, say #3, which has a goat. He says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice of doors? Craig F. Whitaker Columbia, Maryland
Yes; you should switch. The first door has a 1/3 chance of winning, but the second door has a 2/3 chance. Here’s a good way to visualize what happened. 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?
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Yes, the question as posed by a reader was ambiguous but Marilyn made a very reasonable interpretation based on what someone at that time would probably tend to know about game shows. And she even gave an explanation (using more doors) that I find tends to work with folks combined with the explicit assumption that the host will always pick goats.
I parsed that as "50% of the time, monty opens another door and it has a car and you win immediately, and 50% of the time, monty opens another door and it has a goat". In retrospect I think my brain just sort of pictured that and proceeded to assume there was no reason to switch, and it wasn't until I read some answers (not her explanation) that I understood I was wrong (and I went through a lot of anger, like many of the commenters, and thought she was very wrong before that).
If there had been exactly one more sentence saying "monty always opens a door showing a goat", I'm pretty sure I would have recognized that. BTW I was familiar with the game (I liked Jeopardy better, as Let's Make a Deal was a fairly dull show) and I don't think they had a game that was identical to the problem as described.
Why would you think you won? You didn't choose that one with the car and you aren't offered the opportunity to switch to the open one, only the closed one. The only interpretation which makes sense is that he always shows you a goat.
edit: this is wrong, dekhn was right, see below
edit: i ran a monty carlo simulation¹ and i was doing the math wrong. it really does matter if monty knows or not. here's the simulation where he knows:
so in ⅔ of the cases, switching doors gets you the car. by contrast, if monty didn't know which door would reveal a car, it's only ½ of the cases: so if monty picked the goat door on purpose, you do gain by switching. but if he just got lucky, you don't______
¹ thank you, manoj
I've always admired folks who can do these sorts of things in their head, while it takes me a bunch of time to inspect the code and convince myself it's an accurate representation of the word problem.
In the random-open case, you really know nothing new about either of the closed doors. If you can talk yourself into switching, you could make an equally good argument for switching back.
In the Monty-knows-and-always-shows-goat case, you have gained information about one of the closed doors. You haven't gained any information about your initial pick door. But the other remaining door, you know there's a 2/3rds chance that Monty was forced to avoid it so as not to reveal the car. Only in the 1/3rd case where you were already on the car does Monty have freedom to open either door willy nilly.
Edit: having written this, thinking about the 100 door case. If 98 random doors open (that aren't the one I picked) then the fact they all contain goats is pretty suggestive that I have the car, or at least 50/50. I'm not convinced by the code example though.
-There's a 1/100 chance my door is a car, in which case it doesn't matter which of the others stays closed, it will always be a goat. The game will proceed, and switching will lose.
-There's a 99/100 chance my door is a goat, in which case the car is behind some other door. Choosing 98 out of 99 doors to open at random is the same as choosing 1 out of 99 doors to leave closed at random. So the chance that the car stays hidden in this case (so that switching will win) is 1/99, and 98/99 that the game ends early because the car is revealed.
-Adding it up, the game ends without the chance to make a choice 99/100 * 98/99 = 98/100 of the time. Of the remaining 2%, 1/100 comes from the first case (switching loses) and 99/100 * 1/99 = 1/100 comes from the second case (switching wins). The strategies are equally effective.
That's exactly the point. If he doesn't know, then it's exactly 50/50 and there is no reason to switch. If he does know, then it's 1/NUM_DOORS versus NUM_DOORS-1/NUM_DOORS, so you'd be crazy not to switch.
The point is, if he picks at random, in the 100 door case, the vast majority of the time he will open the car door while opening those 98 doors. The case where you get to pick again would be exceedingly rare. Conversely, if he only opens goat doors, you will get your second pick 100% of the time.
So you thought that eliminating them you were left with the original 1/3 vs 2/3, when you actually removed half of the 2/3.
There is no randomness in this sentence. The door has a goat, not a car.
Then again, you might read it as an example outcome. But even if the door was chosen randomly, it would not justify the 50%/50% answer.
Independently of that, assuming that the host chooses the door with the car and the goat with 50% probability each is the same mistake that confused so many, you cannot always assume uniform probability only because there are 2 options.
(I wanted to point out the flaws in your thinking because this has also been a confusing problem for me)
Personally the whole thing taught me is one true hallmark of intelligence is the ability to eliminate unnecessary ambiguity and find the "right" answer.
I don't think it's obvious that Marilyn's interpretation is the correct one. Two possible interpretations could be equally valid.
In law we have the "rule of the last antecedent" that says descriptive clauses modify the nearest antecedent noun.
Under this interpretation "which has a goat" modifies the exemplary door "say #3". Same as saying host opens a door, for example door #3 containing a goat. It could have been say door #2 containing a car.
Marilyn's interpretation is that "which has a goat" modifies "host opens another door." Same as saying the door opened by the host always has a goat.
Language is inherently ambiguous. I don't think either interpretation is unreasonable...
https://en.wikipedia.org/wiki/Monty_Hall#Monty_Hall_Problem
(But vos Savant stated in later columns that most of the critical responses she received assumed that the host was obligated to offer the switch, so they were genuinely confused by the paradox.)
The problem statement is just a mess:
> 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.
even tiny edits like:
> You pick a door, say #1, and the host, who knows what’s behind the doors, opens another door which has a goat, say #3.
or even
> You pick a door, say #1, and the host, who knows what’s behind the doors, opens another door, say #3, which he knows has a goat.
make it clear that the last phrase is describing an invariant of the problem, not an artifact of the illustrative hypothetical.
> You pick a door and the host, who knows what’s behind the doors, opens exactly one of the two remaining doors according to the following algorithm: if both doors contain goats, he picks a door at random, else he picks the door he knows to contain the goat.
Imagine the extreme scenario with 999 goats and 1 car. You select the first door, the host opens 998 doors at random, leaving your selected door and one other. Two scenarios are now possible:
1. There was a car in the 998 doors that got opened. Tough luck, you lose the game automatically now (game resets).
2. There was not a car in the 998 doors. In this scenario, you are still advised to switch.
Edit - I guess I misremembered a simulation I once did about it.
A nice goat fetches up to $1000 to according to a quick google.
https://www.youtube.com/watch?v=AWvefaN8USk
Enjoy.
Or look at this case analysis for the three element case. https://news.ycombinator.com/item?id=24714612
You can get intuition by understanding that the cases where the game is ruined are only the cases where you didn't choose the car.
> And a very small percentage of readers feel convinced that the furor is resulting from people not realizing that the host is opening a losing door on purpose. (But they haven’t read my mail! The great majority of people understand the conditions perfectly.)
https://web.archive.org/web/20181118225305/http://marilynvos...
So I see the second order problem a bit differently: it's not "was the statement unambiguous", it's about what you do with the (possibility of) ambiguity: do you assume you can't be wrong and therefore others must be, or do you try to figure out whether you were solving different problems?
https://blog.xkcd.com/2008/09/09/the-goddamn-airplane-on-the...
There appears to be problems that make it easy for people to assume the worst of others.
And on a side note, I discovered about 6 months ago that mythbusters actually did this experiment. At the end of the day, few things work better than seeing something happen in real time.
There's the Monty Hall problem and then there's the "can I get away with falling back on claiming the question was somehow wrong after embarrassing myself by getting the Monty Hall problem incorrect" problem. No one who claimed it was unclear or ambiguous did so before getting it wrong.
It's really tiresome to have to always pretend to assume good faith when it's really clear what's going on.
Marilyn even made this explicit in her answer, writing that the host "knows what's behind the doors and will always avoid the one with the prize" [0]. So readers were arguing her answer was wrong even given this understanding of how the show works.
[0] https://www.newspapers.com/article/the-missoulian-the-monty-...
The host has to always open the door, too. Else-- if your strategy is always to switch when shown a goat, the host can choose to only show you a goat if you have already picked the car (which would cause you to lose 100%). Or various mixed strategies.
This is one of those problems that is hard to grasp at baseline, but a little harder to grasp because the situation isn't fully specified. As a result it tends to create a lot of controversy, like the one about an airplane on a treadmill.
How so? The problem is typically presented as a single trial, with nothing that clarifies the intent of the host.
This is implicit in the "100 doors, MH opens 98 of them" construction discussed in the article but making it explicit makes the idea quite clear and intuitive, even with just 3 doors.
I'm moderately sympathetic to the idea that this constraint is not immediately obvious from how the scenario is described, and could technically be ambiguous in some versions. I do think most people will agree that it makes sense to infer this constraint once you think about it. Obviously if he opened the door that you initially picked and it was a goat, you would know to switch to one of the other two (and have 1/2 chance of finding the car). (Or, more absurdly, he could just open the door with the car and you could choose it directly). Or you could imagine a game show where MH can choose to open a door or not, in which case the problem is not well-posed, because you would need to know how he makes the choice.
Certainly, that constraint is important to the final solution.
I wonder if anyone has gone through every episode of "Let's Make a Deal" where this scenario occurs and look at the empirical distribution of switch/not-switch.
(Note: this analysis does depend on the constraint that MH doesn't open your first pick door! Edit: actually the answer also becomes equal odds if he can pick any door, regardless of whether it was your first pick or the car. In that case, 1/3 of games will not complete, and the rest will be 50/50 between the unopened doors.)
No one would design a game show like that, so it means the question only has sense if interpreted as "Monty always opens a goat door".
The second part I agree with, but not the first part; because it sounds like you're saying MH is required by the rules to reveal a goat. But the problem works fine if he always opens a remaining door at random and it happens to be a goat.
I'd say it as: once you had chosen a door, Monte was required to open some other door. And the door he happened to open, for whatever reason, had a goat.
Wouldn't that work?
1. you picked car first, he picked the first goat
2. you picked car first, he picked the other goat
3. you picked first goat first, he picked the car
4. you picked first goat first, he picked the other goat
5. you picked other goat first, he picked the car
6. you picked other goat first, he picked the first goat
You happen to know that he picked a goat, so you can eliminate 3 and 5, but the remaining four possibilities are still equally likely. Half of them let you win by switching. This is also the "MH stumbles into a door on accident" variant that others have mentioned.
Another way of getting intuition about your version is to realize that, if he might have shown you a car, then the fact that he has opened a goat is evidence against the hypothesis that your first pick was a goat, because half of the times you choose a goat he would open a car, whereas if you chose a car he would have to show you a goat every time. It eliminates some of the ways things could have turned out if you started on a goat without eliminating any of the ways things could have turned out if you started on a car, so the former is relatively less likely in retrospect than it was before MH opened the goat. More obvious in the 100-door case: if your first pick was a goat, then the fact that he hasn't shown you a car after opening 98 doors at random (or equivalently leaving one door closed at random) either means that you were very lucky in the first pick and got the can (in which case it's easy to open 98 doors without a car) or he got very lucky in his random choices. Most games played in this way (98%) would end with MH revealing the car, and the fact that you don't happen to be in one of them only tells you that one or the other of you got very lucky. The probabilities work out that the two options are equally likely.
I'm happy that she's real and as smart as portrayed. But what really impressed me is how calmly she received all the negative feedback and personal insults. No matter what, she never lashed back. I have a feeling that's authentic too.