18 ms·

Making the Monty Hall problem weirder but obvious

196 points by dyno-might 6y ago

jashmenn 6y ago
Here's what worked for me: Imagine there are 1,000 doors and you pick 1. All other doors except 1 are opened and you're given the offer: keep the door you picked, or pick this other door. What are the chances you picked the right door (vs. this other door)? People seem to intuitively understand that having only one door unopened is a massive "hint" to where the prize is. (I learned this idea from Better Explained: https://betterexplained.com/articles/understanding-the-monty-hall-problem/ https://betterexplained.com/articles/understanding-the-monty...)
- deleted 6y ago
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- lordnacho 6y ago
  For someone who doesn't get it, the problem is then whether the correct extension is all the doors being opened. With 3 doors it's the same. For me the most sensible explanation requires you to know that a dud door is always opened, thus the probability from the 2/3 is the one you are switching to.
- thehappypm 6y ago
  I don’t think this helps me understand it. In the 1,000 doors problem, my odds of being right initially were something like 1/1000 and then it changes to something like 998/1000 or 999/1000 for switching, I can’t intuitively grasp exactly what the odds become of winning if I switch, I just know it’s high. Bringing it down to 3 doors doesn’t help me much — it’s still something like 1/2 or 1/3.
  - ragnese 6y ago
    You don't need to know the exact odds to understand that it's higher. I think that's the main takeaway of making it a 1,000 door problem. It makes it intuitive that the correct solution is to switch. The exact probability doesn't matter.
  - klmadfejno 6y ago
    probability of winning if you switch is 1 - (1 / n) aka (n-1)/n. probability of winning if you don't is 1 / n. By switching, you are simply betting that your original guess of 1/n was wrong.
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  - cestith 6y ago
    Try thinking of it this way. It may or may not be any more useful. I've seen different people come to understand the problem from different examples. 1. Observe that 3/3 = 1. Pedantic, yes, but good for frame of mind here. 2. Pick one of three doors. (1/3 odds) 3. Gain information that one of the three doors is a loser. 4. Note your odds on choosing the original door correctly are still 1/3. 5. Note that if you change doors, there are still 2/3 doors there to choose. 5. Note you're not going to switch to the known loser door, so if you change doors you know 100% which of the other 2/3 of doors to choose. The intuition usually is that you're down to two doors after the loser door is opened, but that's not the case. There are still three doors. The host has just told you that if you trade doors, you know which door to trade for. So trade for it. Note there's a newer version of "Let's Make a Deal", hosted by Wayne Brady, but there is no option to switch after a losing door has been shown in that version.
    - thehappypm 6y ago
      Step 4, kind of a mystery. Why are my odds still the same even though the situation is different?
      - Swenrekcah 6y ago
        Because it isn't different really. It is always a goat door that is opened, so you don't gain any information about your door by the opening of the goat door. I'm thinking of a number between 1 and 10, guess it. If I now tell you a number I promise is not the one I was thinking and not your number, you have no more information about if you were correct.
      - SamBam 6y ago
        Because it really centers on the initial premise: Monty will always open a goat door after your choice, no matter what. So, you make a totally random choice. That choice must be 1/3 right, right? Now the thing that you already knew would definitely happen happens: Monty opens a goat door. How can your odds suddenly jump to 1/2? Are you saying every single time you play the game, you always have a 1/2 chance of getting it right first time?
      - 6y ago
  - delecti 6y ago
    IMO what helps is to imagine slightly changing the order. First step is still that you pick a door. There's a 1/3 chance it has the car. Now you can either keep that single door (with a 1/3 chance of a car), or switch and get both of the other two doors (each with a 1/3 chance of the car, for a total of 2/3 chance). After you pick, I'll reveal all the goats.
  - bartc 6y ago
    Assume there are seven billion people on the planet. One of them knows the location of a specific hidden treasure. I know who it is and I ask you to guess who it is, but you have no possible way of knowing or even getting a hint about it. You pick some random person. I then bring in another stranger and tell you that the person who knows where the hidden treasure is is either the random person you chose or the one I brought in. At this point, there are only two possibilities: 1. You happened to randomly choose the right person on Earth and in my surprise, I had to pick some other random stranger to pretend they knew the secret. 2. You chose a total rando who has no idea what's going on and the person I brought in is in fact the one who knows where the treasure is
  - Closi 6y ago
    Try watching this video below, which explains the 1/100 analogy with a real world example: https://youtu.be/GPoPSNxV1D4?t=365 https://youtu.be/GPoPSNxV1D4?t=365 I've timestamped the relevant bit - but you should watch the full thing from the start, it's very entertaining :)
- kmm 6y ago
  That explanation never worked for me, because you can turn it around to the situation where Monty does not know where the car is. Say there are 1000 doors, and you pick door 429. On his way to open door 429, Monty stumbles, falls, and accidentally knocks open every door except door 128. If by some coincidence all opened doors happened to contain goats, you will have nothing to gain from switching. Very counter-intuitive, but just as true as the original problem. A possible intuition here is that Universes where your first pick was the door with the car, which initially were just 1 in a 1000 compared to Universes in which you picked a goat, will suddenly become massively overrepresented. After all, in these types of Universe Monty's Fall couldn't possibly have shown a car, whereas most of the other Universes will not survive to the next "round". Of course, if this happened in real life, Bayesian thinking would increase the likelihood of hypotheses such as, for example, "The door containing the car has a better lock" to such an extent that I would switch.
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  - tzs 6y ago
    You can't really turn it around, because Monty knowing and using his knowledge of where the car is to reveal only goats is what makes switching advantageous. In the case of the clumsy Monty of your example, it goes like this: 1. There is a 1/1000 chance door 429 has the car. 2a. If it has the car, then when Monty accidentally opens 998 doors no car will be revealed. This does not change the chances that 429 has that car, which remain 1/1000. 2b. If 429 does NOT have the car, then 998/999 times that Monty accidentally opens 998 doors, he will reveal a car, which presumably ends that game. There is only a 1/999 chance that he will not reveal the car and the game proceeds. 3. Thus, there are two cases where the game reaches the point of two remaining doors, with 998 revealed, the car is behind one of the two, and you have a chance to switch. 3a. Your door has the car, which happens 1/1000 games. 3b. Your door does not have the car, which happens 999/1000 x 1/999 games, or 1/1000 games. In other words, if the clumsy Monty version is played repeatedly, 998 out of 1000 games end without even getting too the point you get a chance to switch, and 2 get to where you get the chance. In those two, one has the car in your door, one not. There is no advantage to switching. In the case of the systematic Monty who knows where everything is and ALWAYS opens 998 goats, it goes like this: 1. There is a 1/1000 chance your door, 429, has the car. 2a. If it had the car, Monty opens 998 doors that do not have the car, leaving one door besides your yours. 2b. If your door did not have the car, it is one of the 999, and Monty systematically opens the 998 of those 999 that do not have the car. 3. You always reach the choice stage. You can either get there via 2a, which always results in the car being behind your door, or via 2b, which always results in the car being behind the other door. 3a. You get there via 2a in 1/1000 games. 3b. You get there via 2b in 999/1000 games. If you do not switch, you only if and only if you got there via 2a, so you only win 1/1000 games. If you always switch, you win if and only if you get there via 2b, so you win 999/1000 games.
- filoeleven 6y ago
  I was lucky enough to get this explanation from my high school physics teacher, who first presented the classic Monty Hall problem and then illustrated the changing of the odds by substituting all of the lockers in our school for the three doors. Switching gives a clear advantage. The rest of this post is an anecdote from the same class that this brought to mind, and is unrelated to the topic. Maybe we can say it shows how good teachers engage their students or something, but really it’s just a good yarn. We were learning about inelastic vs elastic collisions, and how an elastic collision has 2x the energy of an inelastic one. The teacher asked for a volunteer, and a bright-eyed student rose to the occasion. The teacher gave him some safety glasses and told him to lie down on the floor. The teacher took the inelastic ball and said, “Okay, I’m gonna drop this on your forehead now, ready?” PLONK. “Ow.” “Remember that feeling! This is the elastic one, and it has the same mass, so it should hurt twice as much.” PLONK. “Ow.” The teacher asked, “So, did the second one hurt more than the first?” The rest of us anticipated the experimental confirmation of what we’d just learned about. “...I couldn’t really tell the difference,” said the student. “Yeah,” said the teacher, “I knew you wouldn’t. I just wanted to see if you’d let me do it.”
- snapetom 6y ago
  That's the explanation that helped me, and led me to realize something about this problem. The problem asks what is the best strategy. It doesn't ask what should you do at that moment. At that moment implies you should process the information available to you right then and there. The information available to you at that time - two doors, ignores prior information, which I think is the counterintuitive aspect that trips people up.
- Green_man 6y ago
  When I first heard about the problem, I struggled with the seemingly 50/50 chances. either you picked the right door and now switch and lose, or you picked a wrong door and switch and win. Switching seemed a zero sum game. The explanation that works best for me is that you were more likely to have picked a wrong door in the first place, so while the impacts are opposite equals, the likelihoods are not equivalent.
- notsuoh 6y ago
  Wow. I "understood" the reasoning behind the original and knew that was the right answer, but your anecdote just made it totally click. Of course if I choose one random locker there is a 1/1000 chance of getting a prize, and if Monty opens 998 other empty lockers, of course I should switch. That would make switching be the correct choice in 999/1000 times.
- NE2z2T9qi 6y ago
  This is the most intuitive explanation by far. I'd even say 1 million doors to really drive the point home. You choose a door with only 1 in a million odds of it being the door with a prize. Monty Hall know where the prize is and will only open the remaining doors he KNOWS doesn't have the prize. If he then opens up 999,998 doors without a prize behind them and asks if you want to keep your original door or switch, you'd obviously know that Monty's last remaining door must be the one with the prize.
twiceaday 6y ago
The opening of the door seems irrelevant and only serves to confuse. There will always be a goat door to open, who cares if it gets opened prior to the choice? The choice you are being given is "keep one door" or "choose both of the other doors." That is functionally the choice because Monty always opens a goat door. The chance of the other two doors containing a car is not affected.
- lcuff 6y ago
  I like this phraseology. How to make the answer intuitive is the objective, and underlining that you're choosing two doors, one of which is going to be wrong, really helps.
  - ju-st 6y ago
    Yes I finally understood it. At the beginning you are choosing one door vs "the others". If you could you would already chose "the others" because they have a 67% win probability but you are only allowed to chose one single door. Then he opens all the wrong doors of "the other doors". The "other doors" still have 67% probability and now only one door is remaining in the "other doors". Obviously that last remaining door now has 67%.
- basch 6y ago
  That's what I dont understand about all the other explanations trying to "simplify" the situation. Just ask people, would you prefer 1 door or 2 doors. I've never understood how expanding it to 100 or 1000 doors is simplier than asking "which is better odds 1/3rds or 2/3rds." I also dont believe the original intent of the question was ever meant to be ambiguous with regard to whether he had knowledge of the goat door or whether he chose at random. The intent was for him to have prior knowledge or impeccable luck, and the wordsmithing of the question came later as, in my opinion, a failed rebuttal to the simplicity of the question. The question might have been worded to not be immediately obvious, but it was not intended to have different correct outcomes depending on interpretation.
  - klmadfejno 6y ago
buildbot 6y ago
For anyone that doesn’t get it after this explanation or even the 1000 doors trick: Try visualizing how you’d pseudocode this game - it literally didn’t click for me until right now, and now it seems much more intuitive.
klmadfejno 6y ago
> It’s important that Monty looked behind the doors before choosing which to open. This is where people’s intuition usually fails. If he had chosen a door at random — in a way that he risked possibly exposing a car, then the situation would be different. (In that case, there’s no advantage or harm in switching.) This one took a second to rationalize. The reason it works is you have the same chance of winning overall (assuming you can safely choose the car if he opens the car by chance), it's just that the value of winning from switching vs. staying has been shifted into the probability of winning by default. The usual mental trick is to extend to 1,000,000 doors. If you pick one door, then are told that all of the alternatives except one are the correct answer, you should obviously reason that the door you didn't choose is the correct answer, unless you got the 1/1,000,000 guess. Odds of winning if you switch are 999,999 / 1,000,000 If the host instead opens 999,998 doors randomly, you have a 999,998 / 1,000,000 chance of winning by default. The remaining two doors have equal chance of winning, giving you the same total odds of 999,999 / 1,000,000 no matter which you choose. Which makes sense, because both situations are, more or less, being given 999,999 chances to guess the lucky door.
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tromp 6y ago
Most of the subtlety of the problem lies in this small sentence in the Side Notes: "He deliberately chooses to show you goats." This is not made as explicit in the standard formulation "the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat." Which could be read as "which happens to have a goat". It's the ambiguity in Monte's door opening strategy that leads to different answers.
- lcuff 6y ago
  One of the realities we face here is that different people have different brains which work differently. Thus different phraseology is going to help. Adding variant phrases such as "He will never show you a car", will probably help for some.
  - tromp 6y ago
    It's also possible that Monty only gives you the option to switch if you chose the car. Or only if you chose door No 1. All of which affect the answer. Therefore, it must be made fully explicit what Monty can and cannot do.
- basch 6y ago
  Whether or not he knew or whether it was luck, it doesnt exactly matter, although traditional stats knowledge would make you think it does. The important thing to understand is that the premise of the question says he will show you a goat. If you rerun the experiment 10,000 times, he will show you a goat 100% of the time, either through peaking, premonition, or consistent luck. The problem gets trickier because people start applying domain knowledge of stats, and treating it as a simulation with random events. The goat being chosen is not random, it is an event that occurs 100% of the time in the premise of the thought experiment. Thinking about the random chance of him choosing the car is outside the bounds of the axiom/postulate we start with. tldr: it doesnt matter how he opened a goat door, all that matters is that he did.
  - tromp 6y ago
    You said it well: "he will show you a goat." But the standard formulation just says Monte opened a door and it had a goat behind it. No explicit mention of intention. It matters not that he happened to do. It matters that he will.
c3534l 6y ago
I got out three playing cards and did the experiment myself over and over. That was many years ago and now whenever I see something on the Monty Hall problem it seems so obvious as if I can no longer even see why people think its unintuitive. The reason its unintuitive is that people don't really have experience with anything that works this way. I'm not sure that the linked explanation will be that helpful to people because of that, although it does make it plainly obvious that Monty Hall knows which door has the prize and he's telegraphing to you which door it might be by only opening doors which don't have it in there. I suppose if that's the missing part of the puzzle for you, then that will help.
- OJFord 6y ago
  I've always thought of it like the 'denominator' of probability for the second stage became 2/3 (vs. the usual/first stage 1). So when he reveals a goat door you know that door is certainly not a winner, 0 probability (of 2/3), and the switch door is certainly (if it were one of those) the winner, 1 probability (of 2/3). But I still used to manage to confuse myself thinking it's intuitively 'more likely' to be the original door 'now' that it isn't one of the others. Until I studied information theory at university and it really clicked - Monty's door choice has lower entropy than your initial pick! (But I acknowledge you can't say to the masses 'look look let's simplify this, if we just step back and take an information theoretic approach -')
- dingaling 6y ago
  > I suppose if that's the missing part of the puzzle for you, then that will help. Of course it will help, because it changes the entire scenario from two people playing a game of chance to one person playing a game of deduction against secret knowledge. The odds don't change because of some quirk of the Universe, they change because one player is changing the parameters due to his knowledge. > I got out three playing cards and did the experiment myself over and over. How did you do that when you didn't know which cards were 'goats'?
choko 6y ago
I've had issues understanding the Monty Hall problem for years until now. What made it click was that, at the end of the article, it's explained that the doors are not opened by Monty at random. My previous reads about the problem did not disclose this, so I had made the assumption that the door opening was random. I've also never seen the show, for what it's worth.
- Sohcahtoa82 6y ago
  The problem explicitly says that he opens the door to reveal a goat, though. If he was opening the door at random, then he would have a 1/3 chance to reveal the car. Since he always reveals the goat, to me that implies that it's not random.
  - ghusbands 6y ago
    No, it states only what is happening this time. It does not say "always" or talk about what happens/happened for other contestants.
  - choko 6y ago
    An accurate and precise description of the problem is essential to understanding. Without explicitly stating that the door choice is non-random, I did not make the assumption.
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lqet 6y ago
> It’s important that Monty looked behind the doors before choosing which to open. This is often not explicitly stated when the problem is given. It is even not a 100% clear from the statement above. Monty always chooses a door with a goat. So: 1. You choose a door. 2. Prob that there is a car behind it: 1/3 3. Prob that the car is behind the two other doors: 2/3 4. If the car was behind the two other doors (which, remember, has p=2/3), Monty will choose the door without a car for you, and the door with a car will remain closed. In this case you are guaranteed to have the car if you switched. So with switching, the overall probability is 2/3. Without, its the original 1/3. If you did not understand that Monty always chooses a goat door, but the person giving you the problem does, or vice versa, then what usually happens is that both of you try to explain why your intuition is correct. Because most people don't talk formal probabilities, your explanations will be so vague that the other person will not realize your different understanding. You will discuss forever, you will both be right, and you will part ways with the strange feeling that maybe the other person was right, when all along you were talking about different problems. This is why this problem is so notorious.
- mannykannot 6y ago
  >> It’s important that Monty looked behind the doors before choosing which to open. > This is often not explicitly stated when the problem is given, which imho is the whole reason this problem has the reputation of being hard to understand. If that were the only difficulty, why have so many people continued to have trouble accepting it even after this misunderstanding has been cleared up, and even after the correct answer has been explained to them? According to Wikipedia, even Paul Erdős remained unconvinced until he was shown a computer simulation. I recall mention of an analysis of the responses to Vos Savant's Parade article, concluding that a majority disputing the result were aware of this constraint, and I will post a link if I can find it again (though if a majority did not explain their reasoning, it may not be possible to figure out what assumptions they made. Nevertheless, the question in my first paragraph still stands.)
  - brmgb 6y ago
    What makes the three doors Monty Hall so counterintuitive is that people tend to correctly reason about the case where the second door is randomly opened and don't understand why it doesn't apply. I believe that this example makes understanding why people don't get it easier: you are looking for someone with one of your friend. You know they are in one of three rooms. Right before you can open the first one, your friend opens the second one and say: "not there". People assume the Monty Hall problem means that it's more likely your friend is in the third room and not the one you were going to open and think it's silly. And they are right to think that. What they don't get is that the case where your friend opened the correct door is part of the switching choice in the Monty Hall situation.
oh_sigh 6y ago
I think the confusion comes from the oracle-like knowledge of Monty Hall - he knows what is behind each door and decides which door to open based on that knowledge. If Monty Hall wasn't an oracle, and just opened a random door after the guess, people's intuitions would be correct.
osipov 6y ago
The analysis of the Game 3 in the article is wrong. You should switch.
- nkurz 6y ago
  Your comment would be better if you could explain why you think it is wrong, rather than merely making an empty assertion. I think you are just misinterpreting the analysis. It says "Option B still gets you the car 90% of the time", and Option B entails switching. Do you read it differently?
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eithed 6y ago
If there are infinite number of doors, then does switching guarantee us the car? Also, Game 5: There are 2 doors. A car is randomly placed behind one, and goats behind the others. You pick one door. Monty looks behind the other doors. He chooses 0 of them with goats behind them, and opens them. You get two options: Option A: You get whatever is behind the door you picked. Option B: You get whatever is behind the other closed door. Should you switch?
- meatmanek 6y ago
  If Monty opens all infinity-minus-two doors except for your initial choice and the one with the car, then switching almost surely[1] gives you a car. 1. https://en.wikipedia.org/wiki/Almost_surely https://en.wikipedia.org/wiki/Almost_surely
  - eithed 6y ago
    It's counterintuitive, as magically the car appears behind the door that Monty chose. But tbf chances of it being behind the door you've chosen were 0 to begin with.
sduff 6y ago
What works for me is scaling the number of doors to 100 (or more). Now the initial guess is correct only 1% of the time, while switching is 99%. Simulated many more variations at https://simonduff.net/monty_hall/ https://simonduff.net/monty_hall/
Tomminn 6y ago
This seems overly complicated. Here's an answer on Quora which is a more straight forward deweirdification. https://www.quora.com/In-the-monty-hall-problem-how-does-opening-the-second-door-skew-the-probability-in-favor-of-the-initially-unchosen-door https://www.quora.com/In-the-monty-hall-problem-how-does-ope... Reprinted: Q: In the monty hall problem, how does opening the second door skew the probability in favor of the initially unchosen door? A: The Monty Hall problem is generally poorly described, in order to make the conclusion seem more surprising then it is. The actual Monty Hall game — as imagined by the people who are asking the question— is set up like this: In front of you are 3 doors, there is a goat behind two of them, and a car behind the other one. In *round 1* of the game, you select a *pair* of doors, from which *one* “goat containing door” will be *automatically eliminated from*, leaving only *one* door of the selected pair of doors in play, (and only *two* of initial *three* doors in play). In round 2 of the game, you guess which of the two remaining doors in play has the car. The choice is this: should you choose the remaining door from the pair selected in round 1, or should you choose the door which was not part of the selected pair in round 1? When phrased like this, the answer is fairly obvious: the pair of doors contains a car 2/3 of the time, whereas the non-paired door contains a car 1/3 of the time. The Monty Hall problem— as normally described— messes this all up by introducing a game show host. This is an agent who— seemingly by their own whim— changes the game you thought you were playing, and introduces round 1 of the game once you have guessed the door you initially think the car is behind. The rules this game show host agent are following are almost never described to a sufficient degree to ensure the game is equivalent to the game laid out above. And yet, the people asking this problem pretend that it is exactly equivalent when they ask you for an answer. It’s generally a poorly described problem, whose answer depends entirely on what kind of agent the game show host is. Don’t worry if it doesn’t make sense to you as it’s usually described. If you can understand why— in the two round game I describe above— it’s better to pick the door from the pair of doors, rather than the single door, you understand probability just fine.
millstone 6y ago
> But he doesn’t choose the door at random. He deliberately chooses to show you goats. Since this is always possible, it tells you nothing What happens if Monty does not ever choose at random. Say that Monty always opens the the highest possible door. If you choose door 1 and Monty reveals door 2, then switching (to 3) is 100% win. If you choose door 1 and Monty reveals door 3, then switching (to 2) is 50% win. I think it's a crucial unstated assumption that Monty does choose randomly among available goat doors.
- kgwgk 6y ago
  The assumption is that you don’t know how he picks the door when there are two goats. It’s a reasonable assumption when the problem statement doesn’t say anything about it. It could be random, it could be the highest, it could be the lowest, it could depend on the position of the stars. The point is that you don’t know so the answer cannot depend on it.
Taniwha 6y ago
So the strategy "always choose the other door" boils down to the following simulation, which kind of makes it obvious { int count = 0; int i; for (i = 0; i < 1000; i++) { int choice1 = rand()%3; int actual = rand()%3; if (choice1 != actual) count++; }
Angostura 6y ago
When explaining to friends, I found simply increasing the number of doors the most helpful way of explaining. There are 1,000 doors - you are passed to pick one. Monty removes all the doors except for your door and one other. There is money behind one of the doors. Do you stick, or change?
- andrewla 6y ago
  Of course you stick -- Monte obviously opened the other doors just to trick you into switching because he knows you got the right answer. Of course he can do this; you already know that 999 doors have goats, so whether or not you picked correctly he can open 998 doors to reveal goats. If you had picked a goat door, he would have just opened the door you picked or opened the door that has the money to show you that you got the wrong answer. The assumption that Monty Hall will always offer you a chance to switch is what is broken in the problem statement and the reason why so many people think that the correct answer is unintuitive.
stakkur 6y ago
What if you really just wanted one goat?
- kgwgk 6y ago
  https://xkcd.com/1282/ https://xkcd.com/1282/
  - stakkur 6y ago
    There's an XKCD for everything
phaemon 6y ago
As has been explained in the other comments, Door A has a 1/3 chance of being the right choice and Door C (the unopened door) has a 2/3 chance of being the right choice. All true... but if you toss a coin to choose whether to switch or not, the odds are 50:50...
- jjnoakes 6y ago
  Sure, but why is this noteworthy? Choosing between any two outcomes with a coin flip where one outcome is good X% of the time and the other is good 1-X% (the rest) of the time will always give you a 50/50 good outcome...
aflag 6y ago
What works for me is thinking: if you pick a goat, switching always wins you the game.
nkrisc 6y ago
If you were correct with your initial choice, switching is a loss. If you were wrong on your initial choice, switching is a win. Your initial choice is only correct 1/3 times, but your initial choice is wrong 2/3 times. It's pretty simple.
- oswald433 6y ago
  This worked for me. Avoiding the "and now you have 2 doors..." moment in your explanation I think made it easier to stay away from seeing it as a 50/50
hnracer 6y ago
If someone is struggling to understand it with three initial doors, instead start with a million initial doors and close all but two.
EGreg 6y ago
Actually, conditional probability is interesting https://en.m.wikipedia.org/wiki/Doomsday_argument https://en.m.wikipedia.org/wiki/Doomsday_argument Just the fact that you’re doing an experiment is already extra information!
LatteLazy 6y ago
If you swap, monty gives you the best outcome from the 2 doors you didn't originally pick. Better 2 doors than 1.
pathikrit 6y ago
When I first read Monty Hall problem, I actually thought it was blindingly obvious to switch. The fact the presenter had 2 doors to chose from and the one he opens is a goat _should_ increase the chance of the other one being not(goat) right?
- dingaling 6y ago
  Only if he's not obliged to open all the goat doors. That's where the "imagine there are 1000 doors" extrapolations fail. If there are 1000 doors and Monty can choose to open just one, then there's no point switching to one of the other 998. Those comparisons only work if he is obliged to open every goat door.
Myto 6y ago
Pretty much every description of the Monty Hall problem has the same flaw, and it is here also. The problem as given does not describe the general rules by which Monty operates. It describes only a single round of playing the game. Thefefore, Monty could be using the strategy of "if the player chose the car door, open a goat door and give the option to switch. Otherwise don't give the option to switch and the player wins the goat." In that case switching is a losing strategy.
- spawarotti 6y ago
  I would not say it is a flaw. I think it is a reasonable implicit assumption that there is only one round, unless explicitly stated otherwise.
  - Myto 6y ago
    It does not matter if there are many rounds or one, what matters is how Monty behaves. And that is given only for the current round, and not as a general rule.
- jjnoakes 6y ago
  That's no more a flaw than failing to explicitly say the car is valuable. I mean what if the car is a matchbox toy and the goats are worth more? That's not explicit either. Sometimes you have to use common sense, and I think every instance of the monty hall problem I've seen was sufficiently explicit (without being absurd), and the confusion was always around the math and probability and never around semantics or trickery.
  - Myto 6y ago
    Obviously the goal of the problem is to get the car. Pretending that the argument I have given is like making up something about a toy car just does not do anything. The fact is that the argument I have presented demonstrates that the problem as given is flawed and does not have a unique answer. Most people don't understand this and substitute the correct version of the problem in their mind, and then proceed to solve that by arguing about the probabilities. Of course the probabilities are what the problem is "supposed" to be about.
spawarotti 6y ago
Key observation: you lose when switching the door only if your first pick was winning. Your first pick had 1/3 chance of winning. So switching has 1/3 chance of losing. Thus it has 1 - 1/3 == 2/3 chance of winning.
- md224 6y ago
  Yeah, this is pretty close to how I see it. The key is making it clear that switching doors will always switch you from Lose to Win and vice versa. Switching doors is equivalent to switching outcomes. That's the critical fact, IMHO. Once that's accepted, the rest falls into place: there's a 2/3 chance that you picked a Losing door, so there's a 2/3 chance that you'll benefit from switching outcomes, and since you're guaranteed to switch outcomes if you switch doors, there's a 2/3 chance that you'll benefit from switching doors.
- the_af 6y ago
  Indeed, this is what made it click for me. Switching is equivalent to saying "I didn't get it right the first time", which is a good bet it has a probability of 2/3. Keeping the same door means "I bet I got it right the first time", which is only 1/3 probable!
emmelaich 6y ago
I think one way people get it wrong is by arguing from symmetry; if I choose 1 and 3 is shown to be a goat, then that's the same as choosing 2 and being shown 3 is a goat. So what's the diff? Which is wrong but seductive.
martincmartin 6y ago
Has anyone gone through old episodes of Let's Make A Deal, and looked at the success rate of those who change their guess, vs those who don't?
RickJWagner 6y ago
I'm just stunned anyone remembers Monte Hall. Kudos to the author.
nullc 6y ago
> Since you don’t care about goats, this makes no difference. Says you. I choose the goat!
anonymousiam 6y ago
Does (Did) Monty always make the offer, or did he only make the offer to switch sometimes? Perhaps he could have only made the offer if the contestant had chosen the winning door. This would change the odds considerably.
- justusthane 6y ago
  I don’t know about the game show, but the premise of the classic problem is that he always makes the offer.
m12k 6y ago
When you pick a door, there's three possible scenarios: A) You picked the car door B) You picked a goat door C) You picked the other goat door. All three are equally likely. But in B and C, the host only has one goat door available to open (because you already picked the other goat door) so since he has to show you a goat, he is effectively forced to tell you where the car is, since that must be the third door that has neither been picked nor opened yet - and switching doors is how you act on that information to get the car. It's only in scenario A that he gets a choice of which of the two doors to open. So there's a 2/3rds chance that he's telling you exactly where the car is, so you're better off acting on that info by switching.
jtsuken 6y ago
How about this? import numpy as np def play_monty_hall(rounds=100): car = np.random.randint(low=1,high=4,size=rounds) first_door = np.random.randint(low=1,high=4,size=rounds) switching_wins = car != first_door staying_wins = np.logical_not(switching_wins) print(sum(switching_wins)) print(sum(staying_wins))
- jtsuken 6y ago
  Result from 1,000,000 games: Switching wins: 667,299 Staying wins: 332,701
bumbledraven 6y ago
The kind of answer to this problem that I see most frequently uses logical reasoning. I don't think that's the best approach to start with, because it is abstract and disconnected from reality. It has a high chance of error. I think people would do better to write a computer simulation of the problem, or, for non-programmers, to design a game with dice that simulates the problem. When you execute the simulation, you pretty quickly realize what's going on. Then you can use logic or formal reasoning to put your intuition into words. A couple people here have said something similar (ctrl-f "simulation" on this page). According to https://www.mwsug.org/proceedings/2010/stats/MWSUG-2010-87.pdf https://www.mwsug.org/proceedings/2010/stats/MWSUG-2010-87.p... , even the great mathematician Paul Erdős wasn't convinced that switching was better until he saw a simulation: > Vazsonyi ran the program 100,000 times. Erdős watched the results of the simulation. The simulation results indicated that by switching, the odds of winning are indeed two out of three. Finally, he was grudgingly convinced that switching was better. He did not like it but seeing was believing. He could not argue with the results.
kgwgk 6y ago
For the “The host knowledge and deliberate choice are irrelevant, if he opened a random door nothing would change.” crowd: If he’s going to open a random door he may just leave it to you (because you don’t know anything). The problem becomes: 0) you’re presented three doors, there is a car behind one of them 1) you pick one door 2) you choose one of the other doors and open it: there is a goat Now, you have the choice between keeping your initial door and switching to the other unopened door.
- the_af 6y ago
  Even in this scenario, you should switch too. Failing to switch is equivalent to saying "I chose correctly the first time", which has 1/3 probability of being right.
  - kgwgk 6y ago
    No, in this sceneario it has a 1/2 probability of being right. P(I chose correctly the first time | all doors are closed) = 1/3 but P(I chose correctly the first time | I opened one of the other doors at random and there was a goat) = 1/2 Without loss of generality, we can say that I picked door A and opened door B findind a goat. P(car@A | goat@B) = P(car@A and goat@B) / P(goat@B) = P(goat@B | car@A) P(car@A) / P(goat@B) = 1 x 1/3 / 2/3 = 1/2 In case it's not clear where the formulas above come from: P(car@A and goat@B) = P(car@A) P(goat@B | car@A) is equal to P(goat@B and car@A) = P(goat@B) P(car@A | goat@B)
    - the_af 6y ago
      Hmmm. Makes sense. I iterated a simulation of 10000 cases, discarding every case where the second random door reveals a car (because this breaks the game; at this point there is nothing to guess anymore) and it gives a probability of winning of only 0.42 if you switch. So it's neither 1/2 nor 2/3. Curious!
rahimnathwani 6y ago
Assuming you understand the rule (that Monty will always open a door you didn't choose, and that door always has a donkey behind it), then one way to understand the game is: You can choose to have: - The most valuable prize that's behind door A, or - The most valuable prize that's behind doors B or C When you look at it this way, it's obvious that you would rather have the most valuable prize from the 'other' doors, and the way to do that is to switch. Then you don't need to think about whether the probabilities change once Monty opens the door with the donkey. If you need convincing, here's a simple python script based on the above: https://gist.github.com/rahimnathwani/2b6ca328a74b37b952c75d14b6b34a93 https://gist.github.com/rahimnathwani/2b6ca328a74b37b952c75d... If you're still thinking about whether the probabilities change, consider whether 'Monty opens a door with a donkey behind it' is new information. It's not because he always does that. And the two doors you picked are fungible/identical except for physical position, as both are in the set of doors you didn't pick. So which one he opens is irrelevant.
nobaelazum 6y ago
Some explanations of this problem are far too complicated. This is how I explain it. 1/3 of the time, you guess correctly on your initial guess. If you switch, you'd be wrong. 2/3 of the time, you guess incorrectly on your initial guess. If you switch, you're right. So when switching, the expecting outcome 1/3 of the time is 0, but the expected outcome 2/3 of the time 1. (1/3)0 + (2/3)1 = 2/3
- kgwgk 6y ago
  You have solved a different problem. You didn’t include in your a analysis the “he opens one of the other doors and there is a goat” bit. The original problem is more complicated.
Dryondristica 6y ago
This problem is good because it illustrates a good way to think through any counter-intuitive problem - increase the numbers to amplify the effect. I remember being taught a similar thing in physics class, to make the weight stupidly heavy or the charge stupidly large, and then see what happens. This kind of thinking can be applied in a lot of different areas.
- theelous3 6y ago
  An even easier way to achieve this is to just say there are 100 doors, you pick one, he opens 98 goat doors and asks you to if you want to switch. I have yet to meet someone who didn't get it from that. No need to have this complex evolving ruleset and so on.
trashtester 6y ago
3 Traits can cause a person to get the Monty Hall problem wrong: 1) Stubborness: "I've made my pick, and I'm sticking to it, options will just make me double down!" 2) Suspiciousness: "I'm faced with a man in a suit, looking like a salesperson, and he's trying to make me change my mind. I think he is up to no good!" 3) Stupidity (relative to those that get it right): "I overestimate my ability at statistics, and I think there is an even chance of a car between each door after one goat has been revealed!" And for those who get it right, not from luck, but from understanding the statistics, there is always the Two Envelopes problem that is very similar, yet so much harder.
tpoacher 6y ago
I had given a very similar answer here: https://stats.stackexchange.com/questions/41208/the-sleeping-beauty-paradox/410602#410602 https://stats.stackexchange.com/questions/41208/the-sleeping... in an attempt to draw a parallel to the Sleeping Beauty paradox.
steerablesafe 6y ago
Monty hall is a conditional probability problem. Conditional probability problems are typically not intuitive and require very precise definition of the condition. Rule of thumb: don't meddle with the condition. An algebraic, Bayes-theorem solution to the problem: The relevant events: A_x = "contestant first picks door x" B_y = "Monty opens door y, revealing a goat" C_z = "price is behind door z" We want to calculate P(C_z|A_x & B_y) for certain combinations of x,y and z. I assume x=1, y=2 for the following calculations (A = A_1, B = B_2). Assumptions: P(C_z) = 1/3, the price can be behind any door with equal probabilities A and C are independent, the contestant has no prior knowledge of the placement of the price P(B|A&C_3)=1, that is Monty opens door 2 with probability 1 if the contestant first opened door 1 and the price is behind door 3, Monty deliberately picks the door with the goat, very important! P(B|A&C_2)=0, Monty never opens the door with the price. P(B|A&C_1)=1/2, Monty equally randomly picks between two doors when he can. Now substitute into all the probabilities: P(C_3|A & B) //probability for winning when switching = P(A & B|C_3)*P(C_3) / P(A & B) = P(B|A & C_3)*P(A|C_3)*P(C_3) /(P(B|A)*P(A)) // P(A|C_3) = P(A) due to independence = P(B|A & C_3)*P(C_3) /P(B|A) //expand denominator = P(B|A & C_3)*P(C_3) /( P(B|A & C_1)*P(C_1) + P(B|A & C_2)*P(C_2) + P(B|A & C_3)*P(C_3) ) // use P(C_1) = P(C_2) = P(C_3) = 1/3 = P(B|A & C_3) /( P(B|A & C_1) + P(B|A & C_2) + P(B|A & C_3) ) // substitute all our assumptions above = 1 / (1 + 0 + 1/2) = 2/3 We can see that the condition B is not trivial and requires precise knowledge of Monty's strategy. Meddling with this condition results in different outcomes.
317070 6y ago
There is another important assumption: what if Monty knows which door has a car in advance and: 1) if you pick the car, Monty opens up other doors as in the regular problem 2) if you do not pick a car, Monty just opens your door and says "unfortunate pick, contestant". If Monty is an adversarial agent (which is not an unfair assumption in television land), then your strategy changes. So an important assumption to make for the Monty Hall problem to work, is that Monty announced what he will do after you pick your door _before_ you pick your door. Only in that case do you have your counterfactuals correct and in that case it is indeed better to swap doors. It is a hidden assumption in most probability theory type of answers.
- andrewla 6y ago
  This is at the heart of why so many people find this unintuitive. People who know the right answer are assuming that this is a repeated game, where every time Monty Hall does the same thing. Every instance given in the linked article includes the assumption that you will always have a choice. If you encounter this situation, how can you possibly know whether you would be offered the choice had you picked the wrong one?
klausjensen 6y ago
...I am not a very smart man, and after reading this - it still did not click why the correct choice is to switch. So I went on youtube.. Watching the first 3 mins of this 5 min video made it click: https://www.youtube.com/watch?v=4Lb-6rxZxx0 https://www.youtube.com/watch?v=4Lb-6rxZxx0
tracerbulletx 6y ago
Just using more doors (both for the total, and the number he opens) is probably enough to make it clear. Like 1000 doors. If I choose one at random I am pretty unlikely to be right on that first choice. But then he opens 998 of the unchosen doors, knowing he won't open the door with the car, I would think oh if it was originally in any of the 999 unchosen doors it is in the one he didn't open.
vanusa 6y ago
But that was the beauty of the original MH problem: It's so everyday and normal seeming -- and while too cryptic or opaque -- just opaque and non-obvious enough to make it, even for many smart and well educated people -- an exquisitely slippery trap to fall in.
arh68 6y ago
The only "surprising" concept in Monty Hall is that a game show host would do anything at all to help a contestant. I don't know if this is just an Americanism, but the default assumption is this guy's trying to trick you out of your money. To explain Monty is helping you is, at all times, a stretch of the imagination.

196 points by dyno-might 6y ago

Here's what worked for me: Imagine there are 1,000 doors and you pick 1. All other doors except 1 are opened and you're given the offer: keep the door you picked, or pick this other door. What are the chances you picked the right door (vs. this other door)? People seem to intuitively understand that having only one door unopened is a massive "hint" to where the prize is. (I learned this idea from Better Explained: https://betterexplained.com/articles/understanding-the-monty-hall-problem/ https://betterexplained.com/articles/understanding-the-monty...)

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lordnacho 6y ago
For someone who doesn't get it, the problem is then whether the correct extension is all the doors being opened. With 3 doors it's the same. For me the most sensible explanation requires you to know that a dud door is always opened, thus the probability from the 2/3 is the one you are switching to.
thehappypm 6y ago
I don’t think this helps me understand it. In the 1,000 doors problem, my odds of being right initially were something like 1/1000 and then it changes to something like 998/1000 or 999/1000 for switching, I can’t intuitively grasp exactly what the odds become of winning if I switch, I just know it’s high. Bringing it down to 3 doors doesn’t help me much — it’s still something like 1/2 or 1/3.
- ragnese 6y ago
  You don't need to know the exact odds to understand that it's higher. I think that's the main takeaway of making it a 1,000 door problem. It makes it intuitive that the correct solution is to switch. The exact probability doesn't matter.
- klmadfejno 6y ago
  probability of winning if you switch is 1 - (1 / n) aka (n-1)/n. probability of winning if you don't is 1 / n. By switching, you are simply betting that your original guess of 1/n was wrong.
- deleted 6y ago
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- cestith 6y ago
  Try thinking of it this way. It may or may not be any more useful. I've seen different people come to understand the problem from different examples. 1. Observe that 3/3 = 1. Pedantic, yes, but good for frame of mind here. 2. Pick one of three doors. (1/3 odds) 3. Gain information that one of the three doors is a loser. 4. Note your odds on choosing the original door correctly are still 1/3. 5. Note that if you change doors, there are still 2/3 doors there to choose. 5. Note you're not going to switch to the known loser door, so if you change doors you know 100% which of the other 2/3 of doors to choose. The intuition usually is that you're down to two doors after the loser door is opened, but that's not the case. There are still three doors. The host has just told you that if you trade doors, you know which door to trade for. So trade for it. Note there's a newer version of "Let's Make a Deal", hosted by Wayne Brady, but there is no option to switch after a losing door has been shown in that version.
  - thehappypm 6y ago
    Step 4, kind of a mystery. Why are my odds still the same even though the situation is different?
    - Swenrekcah 6y ago
      Because it isn't different really. It is always a goat door that is opened, so you don't gain any information about your door by the opening of the goat door. I'm thinking of a number between 1 and 10, guess it. If I now tell you a number I promise is not the one I was thinking and not your number, you have no more information about if you were correct.
    - SamBam 6y ago
      Because it really centers on the initial premise: Monty will always open a goat door after your choice, no matter what. So, you make a totally random choice. That choice must be 1/3 right, right? Now the thing that you already knew would definitely happen happens: Monty opens a goat door. How can your odds suddenly jump to 1/2? Are you saying every single time you play the game, you always have a 1/2 chance of getting it right first time?
    - 6y ago
- delecti 6y ago
  IMO what helps is to imagine slightly changing the order. First step is still that you pick a door. There's a 1/3 chance it has the car. Now you can either keep that single door (with a 1/3 chance of a car), or switch and get both of the other two doors (each with a 1/3 chance of the car, for a total of 2/3 chance). After you pick, I'll reveal all the goats.
- bartc 6y ago
  Assume there are seven billion people on the planet. One of them knows the location of a specific hidden treasure. I know who it is and I ask you to guess who it is, but you have no possible way of knowing or even getting a hint about it. You pick some random person. I then bring in another stranger and tell you that the person who knows where the hidden treasure is is either the random person you chose or the one I brought in. At this point, there are only two possibilities: 1. You happened to randomly choose the right person on Earth and in my surprise, I had to pick some other random stranger to pretend they knew the secret. 2. You chose a total rando who has no idea what's going on and the person I brought in is in fact the one who knows where the treasure is
- Closi 6y ago
  Try watching this video below, which explains the 1/100 analogy with a real world example: https://youtu.be/GPoPSNxV1D4?t=365 https://youtu.be/GPoPSNxV1D4?t=365 I've timestamped the relevant bit - but you should watch the full thing from the start, it's very entertaining :)
kmm 6y ago
That explanation never worked for me, because you can turn it around to the situation where Monty does not know where the car is. Say there are 1000 doors, and you pick door 429. On his way to open door 429, Monty stumbles, falls, and accidentally knocks open every door except door 128. If by some coincidence all opened doors happened to contain goats, you will have nothing to gain from switching. Very counter-intuitive, but just as true as the original problem. A possible intuition here is that Universes where your first pick was the door with the car, which initially were just 1 in a 1000 compared to Universes in which you picked a goat, will suddenly become massively overrepresented. After all, in these types of Universe Monty's Fall couldn't possibly have shown a car, whereas most of the other Universes will not survive to the next "round". Of course, if this happened in real life, Bayesian thinking would increase the likelihood of hypotheses such as, for example, "The door containing the car has a better lock" to such an extent that I would switch.
- deleted 6y ago
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- tzs 6y ago
  You can't really turn it around, because Monty knowing and using his knowledge of where the car is to reveal only goats is what makes switching advantageous. In the case of the clumsy Monty of your example, it goes like this: 1. There is a 1/1000 chance door 429 has the car. 2a. If it has the car, then when Monty accidentally opens 998 doors no car will be revealed. This does not change the chances that 429 has that car, which remain 1/1000. 2b. If 429 does NOT have the car, then 998/999 times that Monty accidentally opens 998 doors, he will reveal a car, which presumably ends that game. There is only a 1/999 chance that he will not reveal the car and the game proceeds. 3. Thus, there are two cases where the game reaches the point of two remaining doors, with 998 revealed, the car is behind one of the two, and you have a chance to switch. 3a. Your door has the car, which happens 1/1000 games. 3b. Your door does not have the car, which happens 999/1000 x 1/999 games, or 1/1000 games. In other words, if the clumsy Monty version is played repeatedly, 998 out of 1000 games end without even getting too the point you get a chance to switch, and 2 get to where you get the chance. In those two, one has the car in your door, one not. There is no advantage to switching. In the case of the systematic Monty who knows where everything is and ALWAYS opens 998 goats, it goes like this: 1. There is a 1/1000 chance your door, 429, has the car. 2a. If it had the car, Monty opens 998 doors that do not have the car, leaving one door besides your yours. 2b. If your door did not have the car, it is one of the 999, and Monty systematically opens the 998 of those 999 that do not have the car. 3. You always reach the choice stage. You can either get there via 2a, which always results in the car being behind your door, or via 2b, which always results in the car being behind the other door. 3a. You get there via 2a in 1/1000 games. 3b. You get there via 2b in 999/1000 games. If you do not switch, you only if and only if you got there via 2a, so you only win 1/1000 games. If you always switch, you win if and only if you get there via 2b, so you win 999/1000 games.
filoeleven 6y ago
I was lucky enough to get this explanation from my high school physics teacher, who first presented the classic Monty Hall problem and then illustrated the changing of the odds by substituting all of the lockers in our school for the three doors. Switching gives a clear advantage. The rest of this post is an anecdote from the same class that this brought to mind, and is unrelated to the topic. Maybe we can say it shows how good teachers engage their students or something, but really it’s just a good yarn. We were learning about inelastic vs elastic collisions, and how an elastic collision has 2x the energy of an inelastic one. The teacher asked for a volunteer, and a bright-eyed student rose to the occasion. The teacher gave him some safety glasses and told him to lie down on the floor. The teacher took the inelastic ball and said, “Okay, I’m gonna drop this on your forehead now, ready?” PLONK. “Ow.” “Remember that feeling! This is the elastic one, and it has the same mass, so it should hurt twice as much.” PLONK. “Ow.” The teacher asked, “So, did the second one hurt more than the first?” The rest of us anticipated the experimental confirmation of what we’d just learned about. “...I couldn’t really tell the difference,” said the student. “Yeah,” said the teacher, “I knew you wouldn’t. I just wanted to see if you’d let me do it.”
snapetom 6y ago
That's the explanation that helped me, and led me to realize something about this problem. The problem asks what is the best strategy. It doesn't ask what should you do at that moment. At that moment implies you should process the information available to you right then and there. The information available to you at that time - two doors, ignores prior information, which I think is the counterintuitive aspect that trips people up.
Green_man 6y ago
When I first heard about the problem, I struggled with the seemingly 50/50 chances. either you picked the right door and now switch and lose, or you picked a wrong door and switch and win. Switching seemed a zero sum game. The explanation that works best for me is that you were more likely to have picked a wrong door in the first place, so while the impacts are opposite equals, the likelihoods are not equivalent.
notsuoh 6y ago
Wow. I "understood" the reasoning behind the original and knew that was the right answer, but your anecdote just made it totally click. Of course if I choose one random locker there is a 1/1000 chance of getting a prize, and if Monty opens 998 other empty lockers, of course I should switch. That would make switching be the correct choice in 999/1000 times.
NE2z2T9qi 6y ago
This is the most intuitive explanation by far. I'd even say 1 million doors to really drive the point home. You choose a door with only 1 in a million odds of it being the door with a prize. Monty Hall know where the prize is and will only open the remaining doors he KNOWS doesn't have the prize. If he then opens up 999,998 doors without a prize behind them and asks if you want to keep your original door or switch, you'd obviously know that Monty's last remaining door must be the one with the prize.

twiceaday 6y ago

The opening of the door seems irrelevant and only serves to confuse. There will always be a goat door to open, who cares if it gets opened prior to the choice? The choice you are being given is "keep one door" or "choose both of the other doors." That is functionally the choice because Monty always opens a goat door. The chance of the other two doors containing a car is not affected.

lcuff 6y ago
I like this phraseology. How to make the answer intuitive is the objective, and underlining that you're choosing two doors, one of which is going to be wrong, really helps.
- ju-st 6y ago
  Yes I finally understood it. At the beginning you are choosing one door vs "the others". If you could you would already chose "the others" because they have a 67% win probability but you are only allowed to chose one single door. Then he opens all the wrong doors of "the other doors". The "other doors" still have 67% probability and now only one door is remaining in the "other doors". Obviously that last remaining door now has 67%.
basch 6y ago
That's what I dont understand about all the other explanations trying to "simplify" the situation. Just ask people, would you prefer 1 door or 2 doors. I've never understood how expanding it to 100 or 1000 doors is simplier than asking "which is better odds 1/3rds or 2/3rds." I also dont believe the original intent of the question was ever meant to be ambiguous with regard to whether he had knowledge of the goat door or whether he chose at random. The intent was for him to have prior knowledge or impeccable luck, and the wordsmithing of the question came later as, in my opinion, a failed rebuttal to the simplicity of the question. The question might have been worded to not be immediately obvious, but it was not intended to have different correct outcomes depending on interpretation.
- klmadfejno 6y ago

buildbot 6y ago

For anyone that doesn’t get it after this explanation or even the 1000 doors trick: Try visualizing how you’d pseudocode this game - it literally didn’t click for me until right now, and now it seems much more intuitive.

klmadfejno 6y ago

> It’s important that Monty looked behind the doors before choosing which to open. This is where people’s intuition usually fails. If he had chosen a door at random — in a way that he risked possibly exposing a car, then the situation would be different. (In that case, there’s no advantage or harm in switching.) This one took a second to rationalize. The reason it works is you have the same chance of winning overall (assuming you can safely choose the car if he opens the car by chance), it's just that the value of winning from switching vs. staying has been shifted into the probability of winning by default. The usual mental trick is to extend to 1,000,000 doors. If you pick one door, then are told that all of the alternatives except one are the correct answer, you should obviously reason that the door you didn't choose is the correct answer, unless you got the 1/1,000,000 guess. Odds of winning if you switch are 999,999 / 1,000,000 If the host instead opens 999,998 doors randomly, you have a 999,998 / 1,000,000 chance of winning by default. The remaining two doors have equal chance of winning, giving you the same total odds of 999,999 / 1,000,000 no matter which you choose. Which makes sense, because both situations are, more or less, being given 999,999 chances to guess the lucky door.

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tromp 6y ago

Most of the subtlety of the problem lies in this small sentence in the Side Notes: "He deliberately chooses to show you goats." This is not made as explicit in the standard formulation "the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat." Which could be read as "which happens to have a goat". It's the ambiguity in Monte's door opening strategy that leads to different answers.

lcuff 6y ago
One of the realities we face here is that different people have different brains which work differently. Thus different phraseology is going to help. Adding variant phrases such as "He will never show you a car", will probably help for some.
- tromp 6y ago
  It's also possible that Monty only gives you the option to switch if you chose the car. Or only if you chose door No 1. All of which affect the answer. Therefore, it must be made fully explicit what Monty can and cannot do.
basch 6y ago
Whether or not he knew or whether it was luck, it doesnt exactly matter, although traditional stats knowledge would make you think it does. The important thing to understand is that the premise of the question says he will show you a goat. If you rerun the experiment 10,000 times, he will show you a goat 100% of the time, either through peaking, premonition, or consistent luck. The problem gets trickier because people start applying domain knowledge of stats, and treating it as a simulation with random events. The goat being chosen is not random, it is an event that occurs 100% of the time in the premise of the thought experiment. Thinking about the random chance of him choosing the car is outside the bounds of the axiom/postulate we start with. tldr: it doesnt matter how he opened a goat door, all that matters is that he did.
- tromp 6y ago
  You said it well: "he will show you a goat." But the standard formulation just says Monte opened a door and it had a goat behind it. No explicit mention of intention. It matters not that he happened to do. It matters that he will.

c3534l 6y ago

I got out three playing cards and did the experiment myself over and over. That was many years ago and now whenever I see something on the Monty Hall problem it seems so obvious as if I can no longer even see why people think its unintuitive. The reason its unintuitive is that people don't really have experience with anything that works this way. I'm not sure that the linked explanation will be that helpful to people because of that, although it does make it plainly obvious that Monty Hall knows which door has the prize and he's telegraphing to you which door it might be by only opening doors which don't have it in there. I suppose if that's the missing part of the puzzle for you, then that will help.

OJFord 6y ago
I've always thought of it like the 'denominator' of probability for the second stage became 2/3 (vs. the usual/first stage 1). So when he reveals a goat door you know that door is certainly not a winner, 0 probability (of 2/3), and the switch door is certainly (if it were one of those) the winner, 1 probability (of 2/3). But I still used to manage to confuse myself thinking it's intuitively 'more likely' to be the original door 'now' that it isn't one of the others. Until I studied information theory at university and it really clicked - Monty's door choice has lower entropy than your initial pick! (But I acknowledge you can't say to the masses 'look look let's simplify this, if we just step back and take an information theoretic approach -')
dingaling 6y ago
> I suppose if that's the missing part of the puzzle for you, then that will help. Of course it will help, because it changes the entire scenario from two people playing a game of chance to one person playing a game of deduction against secret knowledge. The odds don't change because of some quirk of the Universe, they change because one player is changing the parameters due to his knowledge. > I got out three playing cards and did the experiment myself over and over. How did you do that when you didn't know which cards were 'goats'?

choko 6y ago

I've had issues understanding the Monty Hall problem for years until now. What made it click was that, at the end of the article, it's explained that the doors are not opened by Monty at random. My previous reads about the problem did not disclose this, so I had made the assumption that the door opening was random. I've also never seen the show, for what it's worth.

Sohcahtoa82 6y ago
The problem explicitly says that he opens the door to reveal a goat, though. If he was opening the door at random, then he would have a 1/3 chance to reveal the car. Since he always reveals the goat, to me that implies that it's not random.
- ghusbands 6y ago
  No, it states only what is happening this time. It does not say "always" or talk about what happens/happened for other contestants.
- choko 6y ago
  An accurate and precise description of the problem is essential to understanding. Without explicitly stating that the door choice is non-random, I did not make the assumption.

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lqet 6y ago

> It’s important that Monty looked behind the doors before choosing which to open. This is often not explicitly stated when the problem is given. It is even not a 100% clear from the statement above. Monty always chooses a door with a goat. So: 1. You choose a door. 2. Prob that there is a car behind it: 1/3 3. Prob that the car is behind the two other doors: 2/3 4. If the car was behind the two other doors (which, remember, has p=2/3), Monty will choose the door without a car for you, and the door with a car will remain closed. In this case you are guaranteed to have the car if you switched. So with switching, the overall probability is 2/3. Without, its the original 1/3. If you did not understand that Monty always chooses a goat door, but the person giving you the problem does, or vice versa, then what usually happens is that both of you try to explain why your intuition is correct. Because most people don't talk formal probabilities, your explanations will be so vague that the other person will not realize your different understanding. You will discuss forever, you will both be right, and you will part ways with the strange feeling that maybe the other person was right, when all along you were talking about different problems. This is why this problem is so notorious.

mannykannot 6y ago
>> It’s important that Monty looked behind the doors before choosing which to open. > This is often not explicitly stated when the problem is given, which imho is the whole reason this problem has the reputation of being hard to understand. If that were the only difficulty, why have so many people continued to have trouble accepting it even after this misunderstanding has been cleared up, and even after the correct answer has been explained to them? According to Wikipedia, even Paul Erdős remained unconvinced until he was shown a computer simulation. I recall mention of an analysis of the responses to Vos Savant's Parade article, concluding that a majority disputing the result were aware of this constraint, and I will post a link if I can find it again (though if a majority did not explain their reasoning, it may not be possible to figure out what assumptions they made. Nevertheless, the question in my first paragraph still stands.)
- brmgb 6y ago
  What makes the three doors Monty Hall so counterintuitive is that people tend to correctly reason about the case where the second door is randomly opened and don't understand why it doesn't apply. I believe that this example makes understanding why people don't get it easier: you are looking for someone with one of your friend. You know they are in one of three rooms. Right before you can open the first one, your friend opens the second one and say: "not there". People assume the Monty Hall problem means that it's more likely your friend is in the third room and not the one you were going to open and think it's silly. And they are right to think that. What they don't get is that the case where your friend opened the correct door is part of the switching choice in the Monty Hall situation.

oh_sigh 6y ago

I think the confusion comes from the oracle-like knowledge of Monty Hall - he knows what is behind each door and decides which door to open based on that knowledge. If Monty Hall wasn't an oracle, and just opened a random door after the guess, people's intuitions would be correct.

osipov 6y ago

The analysis of the Game 3 in the article is wrong. You should switch.

nkurz 6y ago
Your comment would be better if you could explain why you think it is wrong, rather than merely making an empty assertion. I think you are just misinterpreting the analysis. It says "Option B still gets you the car 90% of the time", and Option B entails switching. Do you read it differently?

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eithed 6y ago

If there are infinite number of doors, then does switching guarantee us the car? Also, Game 5: There are 2 doors. A car is randomly placed behind one, and goats behind the others. You pick one door. Monty looks behind the other doors. He chooses 0 of them with goats behind them, and opens them. You get two options: Option A: You get whatever is behind the door you picked. Option B: You get whatever is behind the other closed door. Should you switch?

meatmanek 6y ago
If Monty opens all infinity-minus-two doors except for your initial choice and the one with the car, then switching almost surely[1] gives you a car. 1. https://en.wikipedia.org/wiki/Almost_surely https://en.wikipedia.org/wiki/Almost_surely
- eithed 6y ago
  It's counterintuitive, as magically the car appears behind the door that Monty chose. But tbf chances of it being behind the door you've chosen were 0 to begin with.

sduff 6y ago

What works for me is scaling the number of doors to 100 (or more). Now the initial guess is correct only 1% of the time, while switching is 99%. Simulated many more variations at https://simonduff.net/monty_hall/ https://simonduff.net/monty_hall/

Tomminn 6y ago

This seems overly complicated. Here's an answer on Quora which is a more straight forward deweirdification. https://www.quora.com/In-the-monty-hall-problem-how-does-opening-the-second-door-skew-the-probability-in-favor-of-the-initially-unchosen-door https://www.quora.com/In-the-monty-hall-problem-how-does-ope... Reprinted: Q: In the monty hall problem, how does opening the second door skew the probability in favor of the initially unchosen door? A: The Monty Hall problem is generally poorly described, in order to make the conclusion seem more surprising then it is. The actual Monty Hall game — as imagined by the people who are asking the question— is set up like this: In front of you are 3 doors, there is a goat behind two of them, and a car behind the other one. In *round 1* of the game, you select a *pair* of doors, from which *one* “goat containing door” will be *automatically eliminated from*, leaving only *one* door of the selected pair of doors in play, (and only *two* of initial *three* doors in play). In round 2 of the game, you guess which of the two remaining doors in play has the car. The choice is this: should you choose the remaining door from the pair selected in round 1, or should you choose the door which was not part of the selected pair in round 1? When phrased like this, the answer is fairly obvious: the pair of doors contains a car 2/3 of the time, whereas the non-paired door contains a car 1/3 of the time. The Monty Hall problem— as normally described— messes this all up by introducing a game show host. This is an agent who— seemingly by their own whim— changes the game you thought you were playing, and introduces round 1 of the game once you have guessed the door you initially think the car is behind. The rules this game show host agent are following are almost never described to a sufficient degree to ensure the game is equivalent to the game laid out above. And yet, the people asking this problem pretend that it is exactly equivalent when they ask you for an answer. It’s generally a poorly described problem, whose answer depends entirely on what kind of agent the game show host is. Don’t worry if it doesn’t make sense to you as it’s usually described. If you can understand why— in the two round game I describe above— it’s better to pick the door from the pair of doors, rather than the single door, you understand probability just fine.

millstone 6y ago

> But he doesn’t choose the door at random. He deliberately chooses to show you goats. Since this is always possible, it tells you nothing What happens if Monty does not ever choose at random. Say that Monty always opens the the highest possible door. If you choose door 1 and Monty reveals door 2, then switching (to 3) is 100% win. If you choose door 1 and Monty reveals door 3, then switching (to 2) is 50% win. I think it's a crucial unstated assumption that Monty does choose randomly among available goat doors.

kgwgk 6y ago
The assumption is that you don’t know how he picks the door when there are two goats. It’s a reasonable assumption when the problem statement doesn’t say anything about it. It could be random, it could be the highest, it could be the lowest, it could depend on the position of the stars. The point is that you don’t know so the answer cannot depend on it.

Taniwha 6y ago

So the strategy "always choose the other door" boils down to the following simulation, which kind of makes it obvious { int count = 0; int i; for (i = 0; i < 1000; i++) { int choice1 = rand()%3; int actual = rand()%3; if (choice1 != actual) count++; }

Angostura 6y ago

When explaining to friends, I found simply increasing the number of doors the most helpful way of explaining. There are 1,000 doors - you are passed to pick one. Monty removes all the doors except for your door and one other. There is money behind one of the doors. Do you stick, or change?

andrewla 6y ago
Of course you stick -- Monte obviously opened the other doors just to trick you into switching because he knows you got the right answer. Of course he can do this; you already know that 999 doors have goats, so whether or not you picked correctly he can open 998 doors to reveal goats. If you had picked a goat door, he would have just opened the door you picked or opened the door that has the money to show you that you got the wrong answer. The assumption that Monty Hall will always offer you a chance to switch is what is broken in the problem statement and the reason why so many people think that the correct answer is unintuitive.

stakkur 6y ago

What if you really just wanted one goat?

kgwgk 6y ago
https://xkcd.com/1282/ https://xkcd.com/1282/
- stakkur 6y ago
  There's an XKCD for everything

phaemon 6y ago

As has been explained in the other comments, Door A has a 1/3 chance of being the right choice and Door C (the unopened door) has a 2/3 chance of being the right choice. All true... but if you toss a coin to choose whether to switch or not, the odds are 50:50...

jjnoakes 6y ago
Sure, but why is this noteworthy? Choosing between any two outcomes with a coin flip where one outcome is good X% of the time and the other is good 1-X% (the rest) of the time will always give you a 50/50 good outcome...

aflag 6y ago

What works for me is thinking: if you pick a goat, switching always wins you the game.

nkrisc 6y ago

If you were correct with your initial choice, switching is a loss. If you were wrong on your initial choice, switching is a win. Your initial choice is only correct 1/3 times, but your initial choice is wrong 2/3 times. It's pretty simple.

oswald433 6y ago
This worked for me. Avoiding the "and now you have 2 doors..." moment in your explanation I think made it easier to stay away from seeing it as a 50/50

hnracer 6y ago

If someone is struggling to understand it with three initial doors, instead start with a million initial doors and close all but two.

EGreg 6y ago

Actually, conditional probability is interesting https://en.m.wikipedia.org/wiki/Doomsday_argument https://en.m.wikipedia.org/wiki/Doomsday_argument Just the fact that you’re doing an experiment is already extra information!

LatteLazy 6y ago

If you swap, monty gives you the best outcome from the 2 doors you didn't originally pick. Better 2 doors than 1.

pathikrit 6y ago

When I first read Monty Hall problem, I actually thought it was blindingly obvious to switch. The fact the presenter had 2 doors to chose from and the one he opens is a goat _should_ increase the chance of the other one being not(goat) right?

dingaling 6y ago
Only if he's not obliged to open all the goat doors. That's where the "imagine there are 1000 doors" extrapolations fail. If there are 1000 doors and Monty can choose to open just one, then there's no point switching to one of the other 998. Those comparisons only work if he is obliged to open every goat door.

Myto 6y ago

Pretty much every description of the Monty Hall problem has the same flaw, and it is here also. The problem as given does not describe the general rules by which Monty operates. It describes only a single round of playing the game. Thefefore, Monty could be using the strategy of "if the player chose the car door, open a goat door and give the option to switch. Otherwise don't give the option to switch and the player wins the goat." In that case switching is a losing strategy.

spawarotti 6y ago
I would not say it is a flaw. I think it is a reasonable implicit assumption that there is only one round, unless explicitly stated otherwise.
- Myto 6y ago
  It does not matter if there are many rounds or one, what matters is how Monty behaves. And that is given only for the current round, and not as a general rule.
jjnoakes 6y ago
That's no more a flaw than failing to explicitly say the car is valuable. I mean what if the car is a matchbox toy and the goats are worth more? That's not explicit either. Sometimes you have to use common sense, and I think every instance of the monty hall problem I've seen was sufficiently explicit (without being absurd), and the confusion was always around the math and probability and never around semantics or trickery.
- Myto 6y ago
  Obviously the goal of the problem is to get the car. Pretending that the argument I have given is like making up something about a toy car just does not do anything. The fact is that the argument I have presented demonstrates that the problem as given is flawed and does not have a unique answer. Most people don't understand this and substitute the correct version of the problem in their mind, and then proceed to solve that by arguing about the probabilities. Of course the probabilities are what the problem is "supposed" to be about.

spawarotti 6y ago

Key observation: you lose when switching the door only if your first pick was winning. Your first pick had 1/3 chance of winning. So switching has 1/3 chance of losing. Thus it has 1 - 1/3 == 2/3 chance of winning.

md224 6y ago
Yeah, this is pretty close to how I see it. The key is making it clear that switching doors will always switch you from Lose to Win and vice versa. Switching doors is equivalent to switching outcomes. That's the critical fact, IMHO. Once that's accepted, the rest falls into place: there's a 2/3 chance that you picked a Losing door, so there's a 2/3 chance that you'll benefit from switching outcomes, and since you're guaranteed to switch outcomes if you switch doors, there's a 2/3 chance that you'll benefit from switching doors.
the_af 6y ago
Indeed, this is what made it click for me. Switching is equivalent to saying "I didn't get it right the first time", which is a good bet it has a probability of 2/3. Keeping the same door means "I bet I got it right the first time", which is only 1/3 probable!

emmelaich 6y ago

I think one way people get it wrong is by arguing from symmetry; if I choose 1 and 3 is shown to be a goat, then that's the same as choosing 2 and being shown 3 is a goat. So what's the diff? Which is wrong but seductive.

martincmartin 6y ago

Has anyone gone through old episodes of Let's Make A Deal, and looked at the success rate of those who change their guess, vs those who don't?

RickJWagner 6y ago

I'm just stunned anyone remembers Monte Hall. Kudos to the author.

nullc 6y ago

> Since you don’t care about goats, this makes no difference. Says you. I choose the goat!

anonymousiam 6y ago

Does (Did) Monty always make the offer, or did he only make the offer to switch sometimes? Perhaps he could have only made the offer if the contestant had chosen the winning door. This would change the odds considerably.

justusthane 6y ago
I don’t know about the game show, but the premise of the classic problem is that he always makes the offer.

m12k 6y ago

When you pick a door, there's three possible scenarios: A) You picked the car door B) You picked a goat door C) You picked the other goat door. All three are equally likely. But in B and C, the host only has one goat door available to open (because you already picked the other goat door) so since he has to show you a goat, he is effectively forced to tell you where the car is, since that must be the third door that has neither been picked nor opened yet - and switching doors is how you act on that information to get the car. It's only in scenario A that he gets a choice of which of the two doors to open. So there's a 2/3rds chance that he's telling you exactly where the car is, so you're better off acting on that info by switching.

jtsuken 6y ago

How about this? import numpy as np def play_monty_hall(rounds=100): car = np.random.randint(low=1,high=4,size=rounds) first_door = np.random.randint(low=1,high=4,size=rounds) switching_wins = car != first_door staying_wins = np.logical_not(switching_wins) print(sum(switching_wins)) print(sum(staying_wins))

jtsuken 6y ago
Result from 1,000,000 games: Switching wins: 667,299 Staying wins: 332,701

bumbledraven 6y ago

The kind of answer to this problem that I see most frequently uses logical reasoning. I don't think that's the best approach to start with, because it is abstract and disconnected from reality. It has a high chance of error. I think people would do better to write a computer simulation of the problem, or, for non-programmers, to design a game with dice that simulates the problem. When you execute the simulation, you pretty quickly realize what's going on. Then you can use logic or formal reasoning to put your intuition into words. A couple people here have said something similar (ctrl-f "simulation" on this page). According to https://www.mwsug.org/proceedings/2010/stats/MWSUG-2010-87.pdf https://www.mwsug.org/proceedings/2010/stats/MWSUG-2010-87.p... , even the great mathematician Paul Erdős wasn't convinced that switching was better until he saw a simulation: > Vazsonyi ran the program 100,000 times. Erdős watched the results of the simulation. The simulation results indicated that by switching, the odds of winning are indeed two out of three. Finally, he was grudgingly convinced that switching was better. He did not like it but seeing was believing. He could not argue with the results.

kgwgk 6y ago

For the “The host knowledge and deliberate choice are irrelevant, if he opened a random door nothing would change.” crowd: If he’s going to open a random door he may just leave it to you (because you don’t know anything). The problem becomes: 0) you’re presented three doors, there is a car behind one of them 1) you pick one door 2) you choose one of the other doors and open it: there is a goat Now, you have the choice between keeping your initial door and switching to the other unopened door.

the_af 6y ago
Even in this scenario, you should switch too. Failing to switch is equivalent to saying "I chose correctly the first time", which has 1/3 probability of being right.
- kgwgk 6y ago
  No, in this sceneario it has a 1/2 probability of being right. P(I chose correctly the first time | all doors are closed) = 1/3 but P(I chose correctly the first time | I opened one of the other doors at random and there was a goat) = 1/2 Without loss of generality, we can say that I picked door A and opened door B findind a goat. P(car@A | goat@B) = P(car@A and goat@B) / P(goat@B) = P(goat@B | car@A) P(car@A) / P(goat@B) = 1 x 1/3 / 2/3 = 1/2 In case it's not clear where the formulas above come from: P(car@A and goat@B) = P(car@A) P(goat@B | car@A) is equal to P(goat@B and car@A) = P(goat@B) P(car@A | goat@B)
  - the_af 6y ago
    Hmmm. Makes sense. I iterated a simulation of 10000 cases, discarding every case where the second random door reveals a car (because this breaks the game; at this point there is nothing to guess anymore) and it gives a probability of winning of only 0.42 if you switch. So it's neither 1/2 nor 2/3. Curious!

rahimnathwani 6y ago

Assuming you understand the rule (that Monty will always open a door you didn't choose, and that door always has a donkey behind it), then one way to understand the game is: You can choose to have: - The most valuable prize that's behind door A, or - The most valuable prize that's behind doors B or C When you look at it this way, it's obvious that you would rather have the most valuable prize from the 'other' doors, and the way to do that is to switch. Then you don't need to think about whether the probabilities change once Monty opens the door with the donkey. If you need convincing, here's a simple python script based on the above: https://gist.github.com/rahimnathwani/2b6ca328a74b37b952c75d14b6b34a93 https://gist.github.com/rahimnathwani/2b6ca328a74b37b952c75d... If you're still thinking about whether the probabilities change, consider whether 'Monty opens a door with a donkey behind it' is new information. It's not because he always does that. And the two doors you picked are fungible/identical except for physical position, as both are in the set of doors you didn't pick. So which one he opens is irrelevant.

nobaelazum 6y ago

Some explanations of this problem are far too complicated. This is how I explain it. 1/3 of the time, you guess correctly on your initial guess. If you switch, you'd be wrong. 2/3 of the time, you guess incorrectly on your initial guess. If you switch, you're right. So when switching, the expecting outcome 1/3 of the time is 0, but the expected outcome 2/3 of the time 1. (1/3)0 + (2/3)1 = 2/3

kgwgk 6y ago
You have solved a different problem. You didn’t include in your a analysis the “he opens one of the other doors and there is a goat” bit. The original problem is more complicated.

Dryondristica 6y ago

This problem is good because it illustrates a good way to think through any counter-intuitive problem - increase the numbers to amplify the effect. I remember being taught a similar thing in physics class, to make the weight stupidly heavy or the charge stupidly large, and then see what happens. This kind of thinking can be applied in a lot of different areas.

theelous3 6y ago
An even easier way to achieve this is to just say there are 100 doors, you pick one, he opens 98 goat doors and asks you to if you want to switch. I have yet to meet someone who didn't get it from that. No need to have this complex evolving ruleset and so on.

trashtester 6y ago

3 Traits can cause a person to get the Monty Hall problem wrong: 1) Stubborness: "I've made my pick, and I'm sticking to it, options will just make me double down!" 2) Suspiciousness: "I'm faced with a man in a suit, looking like a salesperson, and he's trying to make me change my mind. I think he is up to no good!" 3) Stupidity (relative to those that get it right): "I overestimate my ability at statistics, and I think there is an even chance of a car between each door after one goat has been revealed!" And for those who get it right, not from luck, but from understanding the statistics, there is always the Two Envelopes problem that is very similar, yet so much harder.

tpoacher 6y ago

I had given a very similar answer here: https://stats.stackexchange.com/questions/41208/the-sleeping-beauty-paradox/410602#410602 https://stats.stackexchange.com/questions/41208/the-sleeping... in an attempt to draw a parallel to the Sleeping Beauty paradox.

steerablesafe 6y ago

Monty hall is a conditional probability problem. Conditional probability problems are typically not intuitive and require very precise definition of the condition. Rule of thumb: don't meddle with the condition. An algebraic, Bayes-theorem solution to the problem: The relevant events: A_x = "contestant first picks door x" B_y = "Monty opens door y, revealing a goat" C_z = "price is behind door z" We want to calculate P(C_z|A_x & B_y) for certain combinations of x,y and z. I assume x=1, y=2 for the following calculations (A = A_1, B = B_2). Assumptions: P(C_z) = 1/3, the price can be behind any door with equal probabilities A and C are independent, the contestant has no prior knowledge of the placement of the price P(B|A&C_3)=1, that is Monty opens door 2 with probability 1 if the contestant first opened door 1 and the price is behind door 3, Monty deliberately picks the door with the goat, very important! P(B|A&C_2)=0, Monty never opens the door with the price. P(B|A&C_1)=1/2, Monty equally randomly picks between two doors when he can. Now substitute into all the probabilities: P(C_3|A & B) //probability for winning when switching = P(A & B|C_3)*P(C_3) / P(A & B) = P(B|A & C_3)*P(A|C_3)*P(C_3) /(P(B|A)*P(A)) // P(A|C_3) = P(A) due to independence = P(B|A & C_3)*P(C_3) /P(B|A) //expand denominator = P(B|A & C_3)*P(C_3) /( P(B|A & C_1)*P(C_1) + P(B|A & C_2)*P(C_2) + P(B|A & C_3)*P(C_3) ) // use P(C_1) = P(C_2) = P(C_3) = 1/3 = P(B|A & C_3) /( P(B|A & C_1) + P(B|A & C_2) + P(B|A & C_3) ) // substitute all our assumptions above = 1 / (1 + 0 + 1/2) = 2/3 We can see that the condition B is not trivial and requires precise knowledge of Monty's strategy. Meddling with this condition results in different outcomes.

317070 6y ago

There is another important assumption: what if Monty knows which door has a car in advance and: 1) if you pick the car, Monty opens up other doors as in the regular problem 2) if you do not pick a car, Monty just opens your door and says "unfortunate pick, contestant". If Monty is an adversarial agent (which is not an unfair assumption in television land), then your strategy changes. So an important assumption to make for the Monty Hall problem to work, is that Monty announced what he will do after you pick your door _before_ you pick your door. Only in that case do you have your counterfactuals correct and in that case it is indeed better to swap doors. It is a hidden assumption in most probability theory type of answers.

andrewla 6y ago
This is at the heart of why so many people find this unintuitive. People who know the right answer are assuming that this is a repeated game, where every time Monty Hall does the same thing. Every instance given in the linked article includes the assumption that you will always have a choice. If you encounter this situation, how can you possibly know whether you would be offered the choice had you picked the wrong one?

klausjensen 6y ago

...I am not a very smart man, and after reading this - it still did not click why the correct choice is to switch. So I went on youtube.. Watching the first 3 mins of this 5 min video made it click: https://www.youtube.com/watch?v=4Lb-6rxZxx0 https://www.youtube.com/watch?v=4Lb-6rxZxx0

tracerbulletx 6y ago

Just using more doors (both for the total, and the number he opens) is probably enough to make it clear. Like 1000 doors. If I choose one at random I am pretty unlikely to be right on that first choice. But then he opens 998 of the unchosen doors, knowing he won't open the door with the car, I would think oh if it was originally in any of the 999 unchosen doors it is in the one he didn't open.

vanusa 6y ago

But that was the beauty of the original MH problem: It's so everyday and normal seeming -- and while too cryptic or opaque -- just opaque and non-obvious enough to make it, even for many smart and well educated people -- an exquisitely slippery trap to fall in.

arh68 6y ago

The only "surprising" concept in Monty Hall is that a game show host would do anything at all to help a contestant. I don't know if this is just an Americanism, but the default assumption is this guy's trying to trick you out of your money. To explain Monty is helping you is, at all times, a stretch of the imagination.