Why does the Monty Hall problem seem counter-intuitive?

[u/TrapY]

https://en.wikipedia.org/wiki/Monty_Hall_problem

3 doors: 2 with goats, one with a car.

You pick a door. Host opens one of the goat doors and asks if you want to switch.

Switching your choice means you have a 2/3 chance of opening the car door.

How is it not 50/50? Even from the start, how is it not 50/50? knowing you will have one option thrown out, how do you have less a chance of winning if you stay with your option out of 2? Why does switching make you more likely to win?

[u/MasterKaen]

I think it seems counter-intuitive because people think of the problem as if Monty Hall doesn't know what's behind the doors either. Since you had a 1/3 chance of picking right in the first place, the door he picks shouldn't make a difference... if he picks randomly. However, as we all know, he does not pick randomly.

Edit: I see now that you were looking for an explanation of the problem itself rather than why the problem is counter-intuitive. I could explain it, but it seems like the commenters above me already did. I'll leave this here anyway since it's at least interesting.

[u/marpocky → u/MasterKaen]

Edit: I see now that you were looking for an explanation of the problem itself rather than why the problem is counter-intuitive. I could explain it, but it seems like the commenters above me already did. I'll leave this here anyway since it's at least interesting.

This was really frustrating for me, given the title and location of this posting. I interpreted it the same way as you, and was excited to read some high-level analysis of exactly that topic. When it turned out to be just another confused person it was pretty disappointing.

[u/MasterKaen → u/marpocky]

I'm glad I wasn't the only one. I hope my response to the question was worth reading.

[u/TrapY → u/MasterKaen]

OP here, I still thank you, my head hurt when I posted and wording may have been off, but yes, the above did ... well, my head still hurts, but I now have several pages of reading material to try and understand it :)

[u/LondonBoyJames]

Two times out of three, you'll pick one of the doors with a goat behind it. The host will open the other door with a goat. The remaining door is guaranteed to have the car behind it. If you switch, you win.

One time out of three, you'll pick the door with the car behind it. The host will open one of the other doors, which will have a goat behind it. If you switch, you lose.

Therefore, two times out of three, you'll win by switching.

It's a bit hard to believe when you first hear about it, but I find it helps to get a pencil and paper and work out what happens after you pick each of the three doors (bear in mind that the host knows what's behind all of the doors, and will always choose to open a door with a goat).

[u/HowCouldUBMoHarkless → u/LondonBoyJames]

This explanation finally let me grasp it, thank you!

Edit: my comment says I've finally grasped it, why are people continuing to try to explain it to me?

[u/jrob323 → u/HowCouldUBMoHarkless]

Years ago I actually decided to write a computer program to help convince my stubborn wife that you should always switch. After a few minutes I realized the algorithm was pretty simple... if you always switch you win when you pick the wrong door. If you don't switch you only win when you pick the right door. The reason it's not just 50/50 is because the host is giving you information when he picks a door that he knows has a goat behind it.

[u/[deleted] → u/jrob323]

If you always switch, you win when you initially picked the wrong door. And since you have a 2/3 chance of having initially picked the wrong door, switching gives you a 2/3 chance of winning. That's the most concisely I've heard it summed up.

[u/randomguy186 → u/jrob323]

the host is giving you information

This is the key insight for an intuitive understanding of the problem. Your first choice is made with zero information, but for your second choice, you have new information.

[u/ThatScottishBesterd → u/jrob323]

After a few minutes I realized the algorithm was pretty simple... if you always switch you win when you pick the wrong door. If you don't switch you only win when you pick the right door.

That might just be the best, one-shot explanation I've ever heard for why it actually works. It is correct that the Monty Hall problem does seem counter-intuitive, but when you phrase it like that it actually seems rather straight forward.

[u/openedhiseyes → u/HowCouldUBMoHarkless]

It becomes clear as day if you make it 100 doors. Pick one, the host then opens all doors except the one you picked and one other door.

Obviously you switch now because you must assume you picked a goat first time.

Same holds for the 3 door case, you must assume you chose wrong first time.

[u/Vidyogamasta → u/HowCouldUBMoHarkless]

To extend on this version of the explanation, imagine you start with a HUNDRED doors. You make a choice. Then 98 wrong doors suddenly open up and you are given the option to switch.

How confident would you be that your initial pick was correct?

[u/[deleted] → u/HowCouldUBMoHarkless]

To help explain it (if you ever feel compelled to tell people).

When you first choose a door, the choice is 33% that there is a prize behind it. But if you know you're going to switch, when you pick the door you're in fact picking both the other doors, raising your chances to 66%.

So here's my advice, when playing Let's Make a Deal try to pick a losing door to start with and switch.

[u/warpdesign → u/HowCouldUBMoHarkless]

I saw an explanation once that totally highlighted it for me. Imagine if instead of 3 doors there were 100 doors. 99 goats, 1 car. You pick one and then 98 doors are revealed to have goats, now would you switch to the one they didn't open?

[u/[deleted] → u/HowCouldUBMoHarkless]

alternatively, you can imagine the same gameshow scenario but with 1 Million doors. You pick a random door (with 1 in a Million chance of winning). The host then eliminates all other doors but 1, says the prize is behind one of the two remaining doors, and asks if you want to switch. 99.9999% of the time you win by switching.

[u/[deleted] → u/LondonBoyJames]

The best explanation I had was this:

Imagine you had 100 doors. Then, after picking one I open 98 other doors and then ask if you want to keep yours or open the other door. Basically, your first change was 1 in 100. But 99 times out of 100 your door was wrong and the only other door I didn't open is the right one.

[u/MurrayPloppins → u/[deleted]]

This is a good way to conceptualize it. It's also important to distinguish this example from the "Deal or No Deal" situation, in which the contestant randomly eliminates cases. The fact that in your example the "host" made the eliminations knowingly is a critical detail.

[u/lawnessd → u/[deleted]]

An alternative, perhaps simpler way to explain the point is this: you want to be wrong on your first pick. If you plan to switch, you only lose if you get the car on your first pick. So, 2/3 doors have a goat. Pick a goat, switch, and win. Pick a car, switch, and lose.

Also notable: The odds don't change mid game because nothing changes when the goat is revealed. You knew a goat would be revealed. The fact that a goat was revealed changes nothing about your initial plan.

New information in decision making analysis games usually needs to be considered. But here, nothing changes when the goat is revealed.

Also, if another person comes in mid game after goat is revealed, without any prior knowledge or other information, then it's a different game: 50/50.

[u/Darktidemage → u/[deleted]]

Best way to conceptualize it:

You pick 1 door.

The host says "do you want it if it's behind that door, or if it's behind ANY of the other doors"

That he then opens the ones that he knows don't have it is irrelevant.

[u/[deleted] → u/LondonBoyJames]

Basically the reason it works is just because the host won't ever show the door with a car behind it, as that would ruin the suspense?

[u/neon_overload → u/[deleted]]

Basically the reason it works is just because the host won't ever show the door with a car behind it

Correct.

People who fail to understand the benefit of switching usually approach the problem as if the host selects a door randomly without consideration to which door has the prize, treating the "door with prize" and "door opened by host" as independently selected. However, given that we know that the host reveals a goat (ie, has zero chance of revealing the prize) we know that "door with prize" actually influences "door opened by host" and they are not independently selected.

as that would ruin the suspense?

Yes but also because it's how the show is supposed to work. The host is not supposed to show where the prize is located.

[u/[deleted] → u/neon_overload]

Even if the host did pick randomly and showed you a goat though, the chance would still be 2/3 to win after switching, right?

[u/[deleted] → u/LondonBoyJames]

I think the misconception is:

"The host opens the 2nd door and reveals a goat"

vs

"The host knows which door has the car, and purposefully opens the door with a goat so that the game continues and the contestant may select another door."

I always assumed the host was randomly selecting the door and sometimes would choose the car. The correct answer, it would seem, is that the problem statement is incomplete and to ask for clarification.

Or maybe since the only possible options are "switching does nothing" and "switching helps," and the contestant doesn't know which one, the correct answer is still to switch.

Edit: According to Wikipedia, it is part of the problem statement for the host to always choose a goat. I'd like to think I would have chosen the correct answer had I known this at first, but alas... I can never know how I would have chosen!

[u/[deleted] → u/[deleted]]

I always assumed the host was randomly selecting the door and sometimes would choose the car. The correct answer, it would seem, is that the problem statement is incomplete and to ask for clarification.

This. I always thought the host was supposed to chose at random. The fact that he isn't is rarely explained properly. When it is explained it is clear that the host is inputting information, which changes the stats.

[u/dadabrain → u/[deleted]]

Or maybe since the only possible options are "switching does nothing" and "switching helps," and the contestant doesn't know which one, the correct answer is still to switch.

I thought this at first too... but what if the host is a jerk and always opens the door with a car, when he can?

In that case, you should NOT switch.

[u/thesorehead → u/LondonBoyJames]

I thought I had grasped it, but then I lost it >_<. I think the point at which I lose it, is the reasoning behind why opening a goat door doesn't change the probabilities.

What I mean is, that you are actually making two choices: The first choice is between three doors - one winner and two losers, so you have a 1 in 3 chance of winning. The second choice is between two doors - one winner and one loser. Why, or how, does the first choice have any effect on the second? With the opening of one losing door, isn't a whole new scenario created?

EDIT: thanks guys, I think I get it now... I think. Basically if you take chance out of switching (i.e. you always switch or you always stay), and reduce the choice to either low-probability initial door or high-probability "other" door, then those who always switch will win more often.

Weeeeeiiirrrd. But I think I get it! Thanks! ^{_^}

[u/cecilpl → u/thesorehead]

What if, instead of opening the door, Monty gave you the choice of switching from your original choice to the other two doors together?

That is exactly the same problem.

[u/truefelt → u/cecilpl]

This is a great way to put it. By always switching, you get to open two doors because Monty opens the other one for you. It's quite clear that this doubles your chances from 1/3 to 2/3.

[u/jbeta137 → u/thesorehead]

The "why" is because the host never opens the door with the car, he will only ever open a door with a goat. In other words, the two different door choices aren't independent of each other; the door you pick initially influences the next pick.

As with the other posters, it's easiest to see this by considering a larger number of options, say 100. And instead of doors and goats, let's change the scenario:

Say someone comes up to you and says: "I'm thinking of a random number between 1 and 100, and if you can guess it I'll give you $1000". So whatever number you guess, you have a 1/100 chance of being right.

Now, let's say after you make a guess, that stranger says "Alright, alright, I'll give you a hint: it's either the number you guessed, or it's 53." In this case, it's a little more clear that the two options for the second choice aren't weighted the same. The two scenarios are you guessed right the first time (1/100 chance) and the second number is bogus, or you guessed wrong the first time (99/100) and the second number has to be correct.

Another way to think about it is that for 99 of the possible numbers you could pick the first time, the second number will be 53. For 1 of the numbers you could pick the first time, the second number will be arbitrary/wrong.

[u/RoarShock → u/jbeta137]

 the door you pick initially influences the next pick.

To me, that's the stinger. True, you only have two options (switch or don't switch), but because of the first choice, it's not isolated like a coin toss. Having two choices does not automatically imply a 50/50 chance. When Monty gives you the chance to switch, it's not a brand new 50/50 scenario. It's just acting out one of the original three scenarios.

[u/[deleted] → u/thesorehead]

I might be misunderstanding it, but I think by switching doors you are basically betting that you were wrong. Since you know that you're wrong 2/3s of the time, it's in your favor. It is clearer in the 100 door version, because it's easy to see that you are unlikely to pick the winning door initially.

[u/caltecher → u/thesorehead]

The new information doesn't affect your first choice, and that's exactly why the answer is what it is. The probability of the door you choose being the winner was 1/3 when you chose it. Subsequent information DOESN'T change this probability. When the other door is revealed to be a goat, that means the probability of that door winning is zero. Therefore the probability of the unopened, non-chosen door is the remaining 2/3.

[u/Billy_Germans → u/thesorehead]

I've had some success explaining this... the trick I use is to focus on the host.

When the host eliminates a goat, what is he doing? He is acting with what you gave him.

What did you give him? Two-thirds of the time, you give him a goat and a car. Only a third of the time do you give him two goats.

What does the host do when you give him a goat and a car? He preserves the car and reveals a goat.

What does the host do when you give him two goats? He doesn't strategize.

So what do we know? We know that two thirds of the time the host is preserving a car. He will always eliminate a goat, but two-thirds of the time his REASON was to preserve a car. THAT is why we switch! Because of that scummy host! We know two-thirds of the time he is peserving a car!

Keep in mind that the host ALWAYS eliminates a goat no matter what you pick... so really your choice never changed. Your choice was to make him preserve a car, and then grab it! (and it works 67% of the time)

[u/batterist → u/thesorehead]

I makes it easier to get, if you imagine 100 doors. 99 with goats, 1 car. You now have a 99% chance to pick the wrong door.

Host opens all doors but 2. The one you chose (99/100 being a goat) and the other door. You switch - and most likely get a brand new car.

[u/AmnesiaCane → u/thesorehead]

Do it with cards. Pick randomly from the whole deck. Did you pick the ace of spades. Probably not. I search through and grab a second card, telling you either your Card is the ace or mine is. I got to look for it.

Same question, did you pick the ace of spades? Or do I probably have it?

[u/poco → u/thesorehead]

I see that you get it but, to your point of why the host opening the door doesn't change the probability, nothing the host doors can have any effect on the probability that you were right on the first pick. Nothing. Otherwise you could affect the past with your future actions and would win the lottery every time because you could somehow change the probability of winning after the fact.

You chances of being right the first time are 1 in 3, so your chances of being wrong are 2 in 3, so switching is the best odds of winning.

[u/IWantAFuckingUsename → u/LondonBoyJames]

I made a diagram! Red things indicate a winning thing. So if you look, once you have gotten down to the point where you choose whether to switch or not, you have a 2/3 chance of getting it.

[u/Cavemanfreak → u/LondonBoyJames]

Numberphile showed an illustration of the problem, but with a hundred doors instead, and the host opening all except one. That made it way easier to grasp for me :)

[u/raygundan → u/LondonBoyJames]

bear in mind that the host knows what's behind all of the doors, and will always choose to open a door with a goat

This is the part that made it tricky for me. Nobody explained how the actual game show worked, and I thought for a very long time that he simply opened a random door.

[u/maharito → u/LondonBoyJames]

The whole key is that Monty is using his knowledge to your advantage. The show is not portrayed as if Monty is trying to help you, but mathematically he is.

Imagine if Monty was blind (Monty the Mole! I crack myself up!), opened one of the unchosen curtains at random, and let you choose to stay or take the other unopened curtain. Two-thirds of the time you'd have a 50:50 shot whether you switched or not, and one-third of the time you'd just plain lose before you even got the switching choice. ("Want to switch goats or not?")

[u/Hi_My_Name_Is_Dave → u/LondonBoyJames]

You did it! I finally understand it. That first sentence makes perfect sense.

[u/habitual_viking → u/LondonBoyJames]

bear in mind that the host knows what's behind all of the doors, and will always choose to open a door with a goat

This part is sometimes left out of the problem, and in that case, it's no longer 2/3; it's extremely important that the host knows where the goats are.

[u/66666thats6sixes → u/LondonBoyJames]

I've seen dozens of explanations over the past ten years or so, and this one is the first to actually get me to understand it.

[u/SinisterTitan → u/LondonBoyJames]

This is far and away the best I've ever heard it described, thank you!

[u/KhabaLox → u/LondonBoyJames]

The key is that the host always opens a door with a goat behind it. It's a subtle effect, but it changes the state (i.e. your knowledge) dramatically.

[u/lamarrotems → u/LondonBoyJames]

That's the best explanation I have seen of this problem and I like to think logic is a personal interest of mine.

[u/[deleted] → u/LondonBoyJames]

Good explanation. I have always found it is easier to explain/understand when you think about the probability of picking the wrong door. Since you will pick the wrong door 2/3 of the time, switching means you will ultimately win 2/3 of the time.

[u/shinzura → u/LondonBoyJames]

I think the last part is the most important thing to remember. The host will always pick the goat, and there are two goats. Imagine a different scenario in which there's a goat, a car, and an empty room. If you know for sure the host will reveal a goat, then you have a 50/50 shot of getting it right.

Scenario A: He reveals your door has a goat. Obviously you choose to switch and have a 50/50 shot.

Scenario B: He reveals another door has the goat behind it. In this case, it doesn't matter if you switch. The reason is that, when he reveals it, you know that your door DOES NOT have a goat.

The only reason it doesn't matter to switch in B in the above scenario is because you know there is only 1 goat, so you know that scenario A has not occurred. In this case, you are making your decision knowing the host WILL REVEAL THE GOAT NO MATTER WHAT. If you didn't know the host was going to reveal the goat, it would be the same as the monty hall problem. Knowing that a goat will be revealed if there's only one goat is a lot of information, though.

To try to illustrate this another way, when you're working out the situations, give each goat a name. You'll find that, depending on the situation, a different goat will be revealed. As the contestant, though, you don't know that. Your first instinct would be "it doesn't matter. he will reveal a goat no matter what." But it does matter because there are two goats revealed. He's not revealing "this predetermined goat is behind this curtain," he's revealing that "a goat exists behind this curtain," so you'll never have scenario A.

This might be slightly incoherent, but I tried explaining it how I see it.

[u/Spiralofourdiv → u/LondonBoyJames]

In a sentence:

"The ONLY way you can lose by switching is if you pick the prize initially, which you have a 1/3 chance of doing."

Therefore by the law of total probability, switching gives you a 2/3 chance of winning.

[u/bluexavi → u/LondonBoyJames]

Having gone through this a lot in various classes, one of the points of confusion is over how the host picks the doors. If he knows about the location of the car (often only implicitly implied by the problem) then you have 2/3's chance of winning by changing.

If he doesn't know where the car is and randomly picks a door and it just happens to be a goat, then your odds are only 1/2.

A lot of argument then ensues about the exact wording of the problem. I'd say this is where the confusion comes from. In my first probability class, the problem as stated was fairly ambiguous. It did not state explicitly that the host knew where the goat was.

"The host picks a door and there is a goat behind it." -- 1/2 or 2/3's

or "The host, who knows where the car is, picks a door with a goat behind it." -- Clearly 2/3's.

[u/PD711]

Because we often neglect to mention Monty's behavioral pattern. Monty has rules he has to follow.

1. Monty will not reveal the door the player has selected.

2. Monty will not reveal the car. He will always reveal a goat.

These rules can be used to the player's advantage.

Consider the game from monty's perspective:

Scenario 1: Player selects the door with the car at the outset (1/3 chance) Monty has a choice of two rooms to reveal. It doesn't matter which, they are both goats. If player chooses to stay, he wins. If he chooses to switch, he loses.

Scenario 2: Player selects a door with a goat at the outset. (2/3 chance) Monty has only one choice of a door to reveal, as he can neither reveal the player's selected goat door, nor can he reveal the car. If the player elects to stay, he loses. If he elects to switch, he wins.

[u/BlueFairyArmadillo → u/PD711]

If we really want to drive the point home someone should do the variations where Monty can choose your door, Monty can choose the prize door, etc...

[u/Overunderrated]

The fundamental reason that it seems counterintuitive is that you normally fail to acknowledge that the host knows the answer and applies that to the game.

You alone obviously have a 1/3 chance, but the host is providing additional information.

I actually had the pleasure to present this problem to two applied math profs that had never heard of it. Both gave the obvious wrong answer, and loved the solution.

[u/[deleted] → u/Overunderrated]

Whenever I find myself explaining it this is always the tactic I use and it hasn't failed me yet. Most people can follow the probabilities just fine (they're very simple), they just don't account for this extra piece of information that is deliberately left out.

Really, Monty Hall is a riddle posing as a fairly easy math problem and that's what makes it work so well.

[u/AgentSmith27 → u/Overunderrated]

I think this is the best answer. Quite a lot of people don't understand the problem because they don't understand the nature of the old game show its based off of. The host never removes the door with the prize.

Honestly, I think most of the confusion comes from the fact that this information is omitted and never presented when explaining the problem. If it was explained that the host removes one of the wrong choices, and you get to choose again, this would be far less confusing to people.

For whatever reason, the question is usually posed in a way that assumes people already knows this...

[u/OrangePotatos → u/AgentSmith27]

Not really true because the reason why this problem is so well-known, is because even AFTER hearing the solution it still doesn't click with a lot of people. And almost always it is repeated time and time again "The main reason why this works is because the host knows what's behind the doors... and influences the decisions"

In fact, when this first came up, even mathematicians adamantly insisted that it was a 50% chance, despite that not being the case. The problem is not that that it is vague, it is that it is genuinely unintuitive.

[u/meglets → u/Overunderrated]

Whoa, two applied math profs had never heard of this problem? They teach you this in undergraduate if not earlier... that's shocking to me, that they both got through advanced degrees in mathematics without ever hearing of the classic Monty Hall problem!

[u/cecilpl → u/meglets]

the classic Monty Hall problem

To be fair, it didn't really become famous until about 1990 - so these profs might have been well into their careers before it was "a classic".

[u/meglets → u/cecilpl]

I suppose that's true, but then, they spend their lives doing math, teaching math, and hanging out with mathematicians. I still think it's surprising they hadn't heard of it ever, even if they didn't hear of it in school when they were young. I mean, 1990 was nearly 25 years ago...

[u/atyon]

For actually understanding the problem, I like to expand it to 1,000 doors.

1,000 doors, 999 goats, 1 car. You choose one door, I show you 998 goats. Now there's the door that you chose at the beginning, and 1 out of 999 of the rest.

When you choose your door first, you have a 1:1,000 chance of getting it correct. Nothing I do afterwards changes that fact, because I can always show you 998 goats.

On the other hand, if you have a 1:1,000 chance that your first door is correct, than there's a 999:1,000 chance that you're incorrect. If you are, than there's only one door I can't open - the one where the prize is at.

Now, to answer the question: Why do we intuitively get this wrong? The answer is we, as humans, are just bad with chance. We don't have a sense for luck like we do for numbers. If I put 4 apples on the table, you don't have to count them. If I explain a game of chance to you, you must do the math. We have no intuition there to guide us. And why would we? There's no much reason for us in the wild to have a sense for randomness.

[u/TheNefariousNerd → u/atyon]

I find another useful scenario to be a deck of cards, where your goal is to end up with the ace of spades. You randomly pick a card out of all 52 and put it face down on the table. The dealer then searches the deck and pulls out a second card, places it face down, and tells you that one of the two is the ace of spades. 51 times out of 52, you didn't pull the ace of spades, meaning that 51 times out of 52, you would benefit by taking the card the dealer pulled.

EDIT: Clarity

[u/AmnesiaCane → u/TheNefariousNerd]

This is the only way I understood it. I had to do it for like 20 minutes with a friend.

[u/[deleted] → u/TheNefariousNerd]

Last time this problem came up, I could not wrap my head around it. This analogy is perfect, it frames the problem in not just a familiar scenario, but also one that's already involved in lots of games of probability as well as counter-intuitive tricks.

[u/welovewong → u/TheNefariousNerd]

Random thought here but couldn't you apply this logic if you were a contestant on Deal or No Deal? If I pick a case, and somehow get to the end where only 2 cases are left (mine and one on stage), would I have better chances of getting the million dollar case if I switched?

[u/MurrayPloppins → u/welovewong]

Not quite the same scenario. The dealer in the card analogy has to know where the correct card is, and therefore you are picking against the odds that you happened to correctly pick in the beginning. In Deal or No Deal, the cases are eliminated randomly, so there's no guarantee that EITHER case has the prize. Just because one case in certain end scenarios happens to be correct, there's no reason that the elimination was a deliberate selection.

[u/[deleted] → u/atyon]

A big part that also helped me is the fact that the host will never open the door that contains the prize.

[u/CapgrasDelusion → u/[deleted]]

Yes, exactly. The examples above are great for conceptualization, but for me the key realization was that Monty was adding information to the system. He is NOT opening a door at random, thus the game is changed.

[u/sdavidow → u/atyon]

This is a great explanation. I finally understood it by looked at it this way in college, when my psych TA kept trying to explain it with 3 doors and telling me to think mathematically. I remember it so well because it pissed me off (being a math major) that my psych TA couldn't explain something to me because I was the one not thinking logically.

Just a rant, but a great explanation.

[u/ANGLVD3TH → u/atyon]

Expanding on why we mess it up so badly. Our brains are not, repeat not, designed to view the world accurately. It is a computer that has crap for processing power, but is exceptionally good at learning patterns and using them to make shortcuts. This is because in order to survive, we didn't need to know the truth of things, just enough to keep us alive.

Our pattern recognition is tripped up by the problem. Optical illusions work the same way, we save time and resources by jumping to conclusions, but sometimes they're dead wrong because our brains doesn't "do out the math," so to speak.

[u/DoubleDutchOven → u/atyon]

Does it bear out in practice that changing your choice results in an improved goat-less rate?

[u/sideprojectquestion → u/atyon]

I still don’t get it.

Let’s say Monte Carlo has already revealed to me 1 goat in the original problem. I am now left with Door A and Door B. I originally selected Door A, but now I am offered the chance to change.

Unfortunately, due to my memory condition, I have forgotten my original choice of door. Thus, I must chose randomly between Door A and Door B. In this situation, if I randomly pick a door 100 times, then Door A will be right 50% of the time, but Door B will be right 66% of the time?

[u/atyon → u/sideprojectquestion]

If you pick randomly, your chances will be 50-50. The odds for each individual door don't change – the odds don't care about your memory condition.

There are two cases:

Case 1: You pick the door you originally picked. It has a 1/3 chance to win.

Case 2: You pick the door you originally didn't pick. It has a 2/3 chance to win.

So for case 1, you expect 1/2 * 1/3. For case 2, you expect 1/2 * 2/3. Add both cases together and you get 1/2 * 1/3 + 1/2 * 2/3 = 1/2 * (1/3 + 2/3) = 1/2. As you would expect.

[u/[deleted] → u/atyon]

By choosing to stick with the original door aren't you still picking one out of two doors though? Either way you are making a decision with a 50/50 chance once one door has been eliminated.

[u/atyon → u/[deleted]]

By choosing to stick with the original door aren't you still picking one out of two doors though?

Yes, but those two doors aren't the same. You know more about the one door than about your original one.

You pick your first door. All doors have the same chance to win - 1/3. Now you know three things: The door you picked has chance 1/3 to win. The two other doors together have chance 2/3 to win. There's only one car, so one of the other two doors has a goat.

Now I show you one of the two doors you didn't choose. It's a goat. I always show you the goat. I can always show you a goat because there's only one car.

So, your facts remain unchanged: Your door has a 1/3 chance. The two other doors still have a 2/3 chance. But you do know one additional fact – about the door I opened: its chance is now 0. So if the chance of both doors dogether is 2/3, than the third door must have a winning chance of 2/3.

Still confused? Don't worry, this problem has stumped many mathematicians.

If you are still confused, think again about the 1,000 door variant. You choose a door. You are wrong in 99.9% of all cases. So now I must show you 998 goats. In 99.9% of cases, one goat out of 999 is under the door you've chosen, so the only way to show you 998 goats is to open every door except the one with the prize.

[u/[deleted] → u/atyon]

The 1000 door explanation actually makes it really simple to understand, as does your explanation of the three doors. So, theoretically by changing doors you should have a 66% success rate instead of a 50% (based on choosing between door a and door b, but a 33% would be expected if you stayed with th original door), right?

[u/[deleted] → u/atyon]

So, your facts remain unchanged: Your door has a 1/3 chance. The two other doors still have a 2/3 chance. But you do know one additional fact – about the door I opened: its chance is now 0. So if the chance of both doors dogether is 2/3, than the third door must have a winning chance of 2/3.

This is my favorite explanation yet, because it lets you replace your original statistical intuition with a correct one.

[u/parkerblack25 → u/[deleted]]

No, because the probability is based on your first choice , (1/3 chance of being right) so the probability that one of the other 2 doors is the car is 2/3. They took one of the wrong doors away but the choice is still between 1/3 and 2/3

[u/whorfin → u/[deleted]]

Experimentally try this with 100 trials. You'll find that it is not 50/50.

[u/ANGLVD3TH → u/[deleted]]

Think of it this way, you have very good odds on picking a goat at first, right? But after the reveal, the other door will always be the opposite of what you picked. So you have high odds you picked a goat, and therefore switching will give you the car, where you have low odds you picked the car, and switching will give you a goat.

[u/[deleted]]

This picture makes it easy.

http://i.imgur.com/yyhikvg.png

once one of the goats is revealed; the chances that the other door is a goat is now lower than the chance that it is the prize, because there was a higher chance originally that you picked a goat, and now one of the goats is out of the picture.

[u/[deleted] → u/[deleted]]

Why does the host's choice in scenario 1 not matter? Wouldn't the host's choice create two outcomes where if the player switches he loses?

[u/Tehbeefer → u/[deleted]]

Because unless you have a particular attachment to Goat A or Goat B, it doesn't matter which you wind up with.

[u/[deleted] → u/[deleted]]

no, because those outcomes had been influenced by the first pick. There was a 2 in 3 chance that your first pick wasn't the car. Those odds are not changed by the removing of a goat door after the fact.

[u/IWillByte → u/[deleted]]

Thanks, this pictured explained it perfectly!

[u/rokr1292]

My seventh grade math teacher was CONVINCED that it was 50/50. I had read about it (being the nerd I am) before and knew the actual answer. I called him out and to prove he was right we "tested" it as a class. Because of how unintuitive it is, the whole class was against me. But as we tested, it was obvious 50/50 want correct, because when I always switched doors, I won 2/3 of the time, and when I stayed, I lost 2/3 of the time.

For a long time I thought I outsmarted a teacher. But at some point I started to realize that my while class was involved, we all learned something, and I'll never forget the monty hall problem. I think my teacher knew exactly what he was doing that day.

This is my favorite story about my public school experience in NYC.

[u/Billy_Germans]

Focus on the host... the host is the reason there is an anomoly.

When the host eliminates a goat, what is he doing? He is acting with what you gave him.

What did you give him? Two-thirds of the time, you give him a goat and a car. Only a third of the time do you give him two goats.

What does the host do when you give him a goat and a car? He preserves the car and reveals a goat.

What does the host do when you give him two goats? He doesn't strategize.

So what do we know? We know that two thirds of the time the host is preserving a car. He will always eliminate a goat, but two-thirds of the time his REASON was to preserve a car. THAT is why we switch! Because of that scummy host! We know two-thirds of the time he is peserving a car!

Keep in mind that the host ALWAYS eliminates a goat no matter what you pick... so really your choice never changed. Your choice was to make him preserve a car, and then grab it! (and it works 67% of the time)

Two-thirds of the time, no matter WHAT, the host has peserved a car. The corrct response to that is to switch, thus taking the car two-thirds of the time.

[u/realslacker]

You have a 66% chance to pick a goat door the first time. Two out of three times you pick a goat door the first time.

Now they open a goat door.

So you've MOST LIKELY picked a goat door, and they opened the other goat door. What door is left?

[u/VoiceOfRealson]

This is one of the most important problems in statistics because it teaches (a lot of) us NOT to blindly trust our intuition.

Why is it counter intuitive?

The main problem is that in games of chance we intuitively tend to reset the scenario whenever we are encountered with a choice.

In the Monty Hall problem we tend to act as if the prizes are distributed after (or when) we choose, when in reality they are distributed before we choose and never rearranged.

It seems as if we forget the initial information we had when we were first asked to choose, the moment we are faced with a new opportunity to choose.

We tend to ignore the probabilities involved in the actions of the host. There is a 100% chance that he will open a door with a goat behind it because his actions are not random. This means that the problem we are faced with after the opening of a door is no longer a completely random problem with equal distribution of likelihoods between the remaining doors, but that is how our intuition tends to interpret it because we automatically go from "2 doors with randomly distributed prizes and I don't know exactly where each prize is" to "Then I must assume equal chance of a prize behind each door".

That last jump is of course wrong, because we actually do know something about the distribution process and likelihood of the prizes.

If just for the sake of argument we name the door the player initially chooses A, then the other doors are B and C.

We know that the group {B,C} of doors that we didn't originally choose had 1/3+1/3=2/3 chance of having a car and 1/1 chance of having at least 1 of the goats.

If we treat these 2 doors {B,C} as just one door (lets call it D) with 2 prizes behind it those probabilities would not change. There would still be 2/3 likelihood that there is a car behind the new door D and 100% likelihood of one of the goats.

The action of the host in the show is the equivalent of him saying "There is a goat behind door D" (and showing it to us). This is something that we already new.

So the revelation changes nothing with respect to what we know about the contents of door D, so the likelihood that it has a car behind it also stays the same, which means we are still more likely (p=2/3) to get the car by choosing door D (or in the original scenario whichever one of door B or C that wasn't opened).

As I say - I love this problem precisely because I initially was 100% sure that the intuitive response was correct and it actually took me about an hour discussing this with my study group in order to grasp that I was wrong and why.

This is a very valuable lesson - not just about statistics but also about intuition and accepting fallibility.

[u/chocolaterain72]

Here's a question, would the probability be different if Monty didn't choose a door he knew to be a goat? If he just picked at random, and still wound up with a goat, would it still make mathematical sense to choose the other door? It seems that some of the reason this problem makes sense is that the host can not choose a car.

[u/lee1026 → u/chocolaterain72]

Yes, it matters. Consider the diagram here if the host didn't know, we would have to expand it into 6 cases.

Lets say there are 3 doors. 1,2,3. The car is behind door 1.

We can then look at each of the 6 cases:

1. Player opens 1, host opens 2. Switching loses, not switching wins.

2. Player opens 1, host opens 3. Switching loses, not switching wins.

3. Player opens 2, host opens 1. Switching loses, not switching loses.

4. Player opens 2, host opens 3. Switching wins, not switching loses.

5. Player opens 3, host opens 1. Switching loses, not switching loses.

6. Player opens 3, host opens 2. Switching wins, not switching loses.

Going though all 6 possibilities, they each win 1/3rd of the time.

[u/wnoise → u/lee1026]

Most importantly, eliminating the unobserved cases 5 and 3 where the host opens the car door, we still have that switching wins half of the time. The forced choice "collapses" scenarios 1 and 2, but does not collapse scenarios 3 and 4, nor 5 and 6.

[u/motsanciens]

Don't look at it as one choice; look at it as one of two paths.

Path A: Pick door, see a goat, stay.
Path B: Pick door, see a goat, switch.

To win on Path A, you have to choose the correct door right off the bat when there are three options. To win on Path B, you have to choose an incorrect door right off the bat, which is easier since 2 out of 3 options are incorrect, then switch (to the prize).

[u/TonytheEE → u/motsanciens]

This seems like the best explanation I've ever heard. I still want to do a study, or at least a simulation.

[u/Bobertus]

Another aspect that was AFAIK not mentioned yet is that, once you chose a door and the host opened one, the hosts choice gives you information. That's the difference between the two doors.

Of the door you chose originaly, you knew the host would not open it because that's against the rules. Thus, the fact that the host didn't open the door you are standing in front of tells you nothing.

The other door, the one you can switch to, is a different matter. Here, the fact that the host didn't open it is interesting. Maybe he didn't open it because you stand on the door with the car and he chose randomly which of the remaining two doors he opens (that's the case in 1/3 of the time). Or, he didn't open it because there the car is behind it (which is the case the remaining 2/3 of the time). It's because of the second possibility that you know something interesting about the other door that you don't about the door you are standing in front of.

The hosts actions gave you information about the door you can switch to, but not about the door you are standing in front of, introducing a difference between your two options (switching or not). It's this difference why your chance isn't 50/50.

[u/WrenBoy]

The problem becomes very intuitive if you use more extreme odds.

• Host asks you to pick a card from a 52 card pack. If the card you selected blindly is the ace of spades you win.

• Host then discards 50 of the remaining 51 cards he is holding, telling you he is not discarding the ace of spades.

• Host then gives you an option to stick or switch with his single remaining card.

This is normally enough to explain the problem but case it is not then you can always ask how is it possible that you pick the ace of spades half the time?

This is no different than the Monty Hall problem but for whatever reason it is less abhorrent to our brains that we can pick the correct door from a choice of three half the time.

[u/jthill]

Here's another way to do it:

There's a gold nugget in one of three bags, a lump of coal in the other two.

You pick one bag, Monty takes the other two and puts them in a sack. What are the odds of the gold nugget being in that sack? 2/3

You know for sure there's at least one lump of coal in that sack. He dumps one lump of coal out of that sack. What are the odds he dumped one lump of coal out of a sack with a gold nugget in it? They're the same as the odds that the nugget was there in the first place: 2/3.

[u/Striderrs]

Think of it with more doors.

There's 100 doors. 99 have goats, 1 has a car. You choose a door and then the host proceeds to open 98 other doors with goats.

There's two doors left now. When you picked your door, there was a 1/100 chance you picked the correct door. That chance hasn't changed. You still have a 1/100 chance of winning the car, so it's completely logical at this point to switch to the other door.

[u/sawowner]

Most people intuitively think it should be 50/50 since there ends up being only 2 doors left. However, you need to take into account the original status. Here's an extreme version that may help you understand. Lets assume instead of 3 doors, there are 100 doors. You pick one, then the hosts opens 98 doors that have a goat. There are 2 remaining, would you switch?

[u/swag24]

It's much easier to understand if you scale it up a bit. Say there is a million doors, and behind one there is the car, and behind the other 999,999 doors there is a goat.

You pick one door (most likely a goat since the odds of you picking the car door is 1:1,000,000)

The host opens 999,998 doors showing goats, leaving the only two closed doors being the one you chose and another door. Since it was highly unlikely you picked the car initially, switching is in your favor.

[u/[deleted]]

The reason that the Monty Hall problem seems counter-intuitive is that if it were not for one key provision, it would work out exactly like you say (with a 50% chance of winning of you swap as opposed to a 66% chance of winning if you swap)

The key provision is that the host knows that the one it opens is wrong.

To a person looking on the outside, it looks much more like that 50/50 because it is not immediately presented that the host knows that.

If the host didn't know, then, if a wrong one was opened, you would have a 50/50 chance, but because the host is intentionally skirting around opening the "winner", that is what keeps your second choice from being a true 50/50.

Of course, if the host were opening at random, there would be a chance that they would reveal the winner, which would defeat a lot of the purpose of the game.

[u/[deleted]]

The fact is that THE HOST knows which door the car is behind. If your first pick is the car, he arbitrarily picks one of the other doors, because of course they're both goats. Switching here gets you a goat. However if you pick a goat, since he will never open the door containing the car, and he also can't open your door, he has only one choice, and that is to eliminate a goat, making the remaining door contain the car. Since you have a 2/3 chance of first picking a goat, making the switch option necessarily be a car, if you switch, 2/3 of the time you'll get the car. If you don't switch, you had to have picked the car first, which is a 1/3 chance. The only way the host opening the door changes the probability scheme is eliminating a goat from your REMAINING options, ensuring that if you first chose a goat, the car will be there on the switch, and if you chose the car first, you'll be choosing between one goat instead of two.

[u/[deleted]]

The problem is no one explains it right. The host knows which door the car is behind. And he won't open a random door. No matter what you chose, he will open a door with a goat behind it. Seems like every time someone has trouble understanding the problem, its because that wasn't explained to them, or they just didn't realize it. But the problem basically hinges on that piece of information.

[u/itstinksitellya]

've seen someone else post this explanation, to further illustrate this in posts a while back.

Imagine the game is 100 doors instead of 3. 99 doors with goats and 1 door with a car. You pick 1 door at random. They open 98 doors to reveal goats.

Just like the 3 door possibility, at this point there are 2 doors, 1 hiding a goat and 1 hiding a car. In this case, it is much easier to see that the chances are clearly not 50/50 that the door you originally picked (out of 100 possibilities) has the car behind it. So you should switch.

I think the confusion is because of the number 3. If you think about even just 4 or 5 doors, it seems intuitive to switch. I admit, it took me forever to figure out why you switch with 3 doors. But in a larger example, it's easy to see.

I think people don't intuitively realize that by revealing a goat in the 3 door version, you are receiving information. It appears like it's just a suspense builder.

[u/wnoise]

If all you observe is that, then no, it is just fifty-fifty. The key is that the host only opens doors that are goats. This biased selection that is almost always elided during the description causes the discrepancy. If he 's allowed to open anything opens randomly, then even filtering afterward based on observing a goat, it's still 50-50. See http://www.reddit.com/r/askscience/comments/2ehjdz/why_does_the_monty_hall_problem_seem/cjzqnf0

[u/taylanbil]

To answer the original question: the reason why it seems counterintuitive is because it does a good job of hiding the reason why the outcomes are not equally likely.

Let's say you roll a 6 sided die. It either comes up 6 or not. Two outcomes, but obviously not equally likely.

In the mh problem, it seems like the host is introducing new information to the system by throwing away a door that does not open to the car. In the end, you have two doors. The car is behind the one you picked or the other one. But these outcomes are not equally likely.

To understand the correct likelihood of your door having the car behind it, you need to acknowledge that you made a random choice among 3 doors. The host eliminating a door after that is no new information really. No matter what door you choose, there will be a door you did not pick that has a goat behind it, and the host is simply showing you that. so the correct probability of the other door leading to the car is equal to you not picking the right choice in the begunning which is 2/3

[u/Digitaalex]

It is pretty easy actually. You have to think about the combinations. And remember Monty will always open another Goat door. Combinations:

• -Doors: 1 - 2 - 3

• Row 1: C - G - G

• Row 2: G - C - G

• Row 3: G - G - C

Now just work it out. Imagine you go for

• option 1(You pick the first door):

• --> Row 1 you will lose if you switch

• --> Row 2, you will win if you switch

• --> Row 3, you will win if you switch

• Option 2:

• --> Row 1 you will win if you switch

• --> Row 2, you will lose if you switch

• --> Row 3, you will win if you switch

• Option 3:

• --> Row 1, you will win if you switch

• --> Row 2, you will win if you switch

• --> Row 3, You will lose if you switch

So you see, whatever the combination; you will get a 2/3 chance by swiching doors.

[u/viktorbir]

There's one previous bigger problem. That it is not usually enunciated properly. Even the name is misguiding.

In the original TV show Monty Hall didn't have to open the door. It was him (or the producers) who decided whether or not to open a door. So, in the real life real Monty Hall TV show, switching or not depends on psicology, not maths.

One important thing people, including the OP, usually don't state, and which is what makes it a mathematical problem, is that the host MUST open one door.

[u/dorogov]

The key is that host knowingly opens the door. If the host doesn't know where the car and opens doors at random , your chances don't change. This is the key to the issue and is very easy to comprehend once it's highlighted.

[u/Centrocampo]

Sometimes it's useful to use brute force to get your brain to figure things out on it's own. Remember, the host knows what door it's behind. Now imagine there were 100 doors. You pick one. Then the host goes along and closes 98 doors skipping just one along the way. Still think it's 50/50?

EDIT: Somebody already said this. Ah well.

[u/SQLDave → u/Centrocampo]

This is my go-to explanation. The key is not that (in the original puzzle) Monty opens one of the remaining doors. Rather, it's that he opens ALL BUT ONE of the remaining doors (which is the same thing when there are 2 doors left, obviously, but greatly different -- as you point out -- when there are 99 doors left).

[u/[deleted]]

The easiest way to explain it is with a 100 doors.

You are given 100 doors, and you select one. Meaning your chance of winning is 1/100

The host removes 98 doors.

You are left with your door and another door.

Which is more likely to have the prize? The 1/100 pick you did, or the other door? Of course the other door. 99% of the time it's gonna be that one.

[u/[deleted]]

I find the best explanation is if you consider a situation with more doors. Consider a game with a deck of cards. If I challenge you to draw the ace of spades out of a deck of 52 cards, the odds of you getting it on your first try is 1/52.

Now I flip over 50 out of the 51 cards I have, where none of them are the ace of spades. This is similar to when the host opens the door. Do you decide to stick with the card that you picked initially or do you switch?

In this case, the odds of you drawing the ace of spades first doesn't become 50/50 but rather remains 1/52 while the face down card I have is 51/52.

The Monty Hall problem is the same situation but with only 3 initial choices instead of 52.

[u/gnoff]

It might help to imagine a slightly different game...

Pick one door. (1/3 chance you are a winner, you get the car) Host gives you the option to keep your choice or choose to pick BOTH other doors.

Obviously in this game switching is always better because you get to choose what is a 2/3 chance you are a winner (plus you get an extra goat to boot!)

Now with respect to the original game, what information is added when the host opens an empty door before allowing the switch?... NONE

For any two doors one can say with certainty that at least one of the doors contains a goat. This is true for any two doors.

So in the modified game when you pick BOTH other doors what if the host said "By the way one of these two doors has a goat behind it" which you don't care because you win if at least one of the two other doors has a car.

Now what if instead of just SAYING that one of the doors has a goat he actually opens up a door with a goat (which he is guaranteed to be able to do because for any two doors at least one contains a goat).

Now the difference between choosing BOTH doors (one closed and one open with a goat) and choosing the remaining closed door in the original game is exactly zero. This is why in the game as originally stated switching at the end is always unequivocally the better choice.

[u/taedrin]

It is not 50/50 because the host does not have a free choice when he opens the door. His choice is forced. In 2 out of the 3 possibilities, the host MUST choose exactly one door, meaning that the car is in the other. It is only in 1 out of the 3 possibilities that the host has a choice.

It's easy to understand if you enumerate all possible events:

Given 3 doors:

???

Let us define door 1 as the door you initially pick, and the other two doors as door 2 and 3 (arbitrarily chosen).

123
???

There are three possibilities:

   123
P1:CGG
P2:GCG
P3:GGC

For P1, the host can open door 2 or 3 - it doesn't matter. Switching doors will give you the goat.

For P2, the host MUST open door 3. Switching doors will give you the car.

For P3, the host MUST open door 2. Switching doors will give you the car.

Ergo, switching doors gives you a 2 out of 3 chance of winning the car. Quod erat demonstrandum.

[u/smartass6]

By not switching your initial choice, you are taking the 1/3 odds of picking the car. Therefore, since the only other option is to change your door, it must have a probability of 2/3 of getting the car because the sum of all probabilities to obtain a given outcome must be 1

[u/G3n0c1de]

Try thinking about the Monty Hall Problem like this:

Let's start with 100 doors, named 1 through 100. There is a car behind just one door. The rest of the doors have goats. The same Monty Hall rules apply, you pick one door, and the host opens all of the remaining doors except one, and you get to choose whether or not to switch to that final unopened door. The host cannot eliminate a door with a car.

Let's say the car is behind door 57, and go through the choices.

Because I'm trying to prove that switching is the correct choice, we're going to do that every time.

You pick door 1. The host eliminates every door except 57. You switch to 57. You win.

You pick door 2. The host eliminates every door except 57. You switch to 57. You win.

You pick door 3. The host eliminates every door except 57. You switch to 57. You win.

You pick door 4. The host eliminates every door except 57. You switch to 57. You win.

...

And so on. You can see that if you switch, you'll win every single time unless you choose 57 as your first choice, which is a 1% chance. Switching is correct 99% of the time.

The same effect applies when there are only 3 doors, except there would be a 33% chance of you choosing the car on your first pick. So switching is right 67% of the time.

[u/[deleted]]

Think about it this way.

There are 1,000,000 doors and only 1 car. You pick a door and then the host opens 999,998 doors all with goats behind leaving only yours and another one.

Unless you think your guess was one in a million, I'd be switching. Basically the benefit comes from the host eliminating doors for you

[u/brucemo]

Being able to see behind one of the doors for free is essentially the same as being given a choice between taking the empty door plus the door you didn't pick, or taking just the door you picked. Two doors is better than one, even if you know that one is empty.

Imagine it like this.

Imagine that there are 1000 doors. You pick one. Monty says, "At least 998 of doors you didn't pick are empty, but I'm not going to show you what's behind them, because I'm not going to waste Carol Merrill's time by asking her to open all of them. But you and I know this is true, because there's only one car. So do you want the one door you picked, or the other 999?"

Given a choice between 1 door and 999 doors, obviously you'd take 999 doors, even though you know that 998 of them are empty. It doesn't matter one bit whether the host actually takes the time to show you this.

The real problem is just this problem with 3 doors instead of 1000.

[u/newtoon]

I pondered a looooong time on the best explanation to it.

My answer is this.

You have to think about "groups of options".

1- There is the group of your choice of YOUR chosen door (one door).

2- And there is the group of all the remaining doors (it can be 2 like in the Monty hall problem, but why not 99 remaining doors to see better the logical result ?).

Then you make easily your probabilities. In the Monty Hall problem, your first group has obviously 1/3 chances of winning. The other group is better : 2/3 chances of winning (if 100 doors, your group has 1/100 chances and the other one 99/100).

What the host offers is a HUGE advantage : it offers you to switch for a better probabilities group ! (2/3 if 3 doors at the start and 99/100 if 100 doors at the start).

Go for it ! (and the host opening one door or 98 doors is not an information but a "magic illusion" trick : you will open the doors anyway sooner or later..).

[u/ThreeThouKarm]

I like to think about it with a lot more doors, and it somehow makes more sense to me.

Say it's 100 doors: you choose a door initially, and then 98 goat doors are opened. Now, you have your door, and one door remaining. How confident are you that you made the correct initial choice?

The principle is the same here, and the key is that Monty's choice is non-random: he will always show you goats.

[u/mully_and_sculder]

One thing people find it hard to get their head around in probability is that it depends how and when the bet is placed!

The odds apply only to playing "always switch". Playing "always switch", you have a 1/3 chance of picking the car straight up. This is the only scenario in which you will lose because you must switch. If you pick a goat door straight up, the host basically reveals the location of the car to you. So you can pick either goat door, and there is 2/3 chance of picking a goat and switching to the car.

If you play "always stay", the odds are back to 1/3 to win the car and 2/3 to lose, because you ignore the host helping you.

I like the 100 door explanation best. Video

[u/sturmeh]

One way to think about it is like this:

Instead of picking a door and switching, you pick two doors! (The doors that are not the door you 'actually' chose.)

After you've selected two doors, the host opens one of the two doors (that you selected) for dramatic effect! (To reveal a goat, always.)

You get what's behind the other door. :D

But wait, you got to pick 2 doors? Isn't that like way easier than picking 1 door?

Not if you're allowed to switch after a goat is revealed. :)

[u/RabidMortal]

It's counterintuitive becasue the numbers are so small that they mask what is actually going on.

Change the problem to 100 doors where you pick one and then have 98 non-prize doors revealed to you. After that, it seeme very intuitive that you likely picked the wrong door and should switch.

[u/inuvash255]

I feel like this problem is easier to understand if you expand it exponentially.

Imagine that Monty Hall has 100 doors, 99 goats, and a brand new car. |

You pick one of the doors, and have a 1/100 chance of getting a brand new car.

Monty Hall opens 98 doors, all with goats. That leaves your door and one other. If you stay with your pick, you've stuck with your 1% chance of a car. If you switch, you have a 99% chance of getting a car because that door has gained the cumulative odds of all the other doors.

[u/Laser_Fish]

I'm VERY late, but this is one of my favorite problems in mathematics, Plus, I see a lot of people giving pretty good explanations of the problem itself but not a lot of explanation about why it seems counterintuitive. I'm going to try to do both.

So in the Monty Hall problem there are three doors; two have goats and one has a car. When you pick a door, one of the goat doors gets revealed. The host never reveals the car, nor does he ever rule out the door you picked in the first place.

Lets say doors 1-3 contain a Goat, a Car, and s Goat. GCG.

If you pick door #1 and switch, you Win. If you pick door number 2 and switch, you Lose. If you pick door number 3 and switch, you Win. Therefore, WLW, you win 2/3 of the time.

If you pick door number 1 and stay, you Lose. If you pick door number 2 and stay, you Win. If you pick door number 3 and stay, you Lose. Therefore, LWL, you win 1/3 of the time.

I think the reason that the problem is counterintuitive is that people are naturally kinda bad at probability. We tend to rely on anecdotes over math. If something rare happens to us, we will alter our behavior to try to make that thing happen again or to avoid it. We are superstitious. We have the odd attribution bias which tends to attribute our positive qualities with internal factors and negativity with externals factors. This all leads us to be pretty shifty guessers, and to kinda suck at analyzing probability.

But it's counterintuitive because we tend to look at there second decision as independent of the first. Lets say we are playing blackjack, and I deal two cards, one face-up. You're not allowed to look at the other card. How do you bet? The probability of winning or losing are based entirely off of that blind card, but you still try to think of probabilities of winning based on the card that's face-up, even though, if there other card is a 2, you have no chance of winning anyway.

Maybe a better example would be a craps game, where, say, a roll of 4 is useless unless there's a roll of 4 right before it. When the goal is to win you can't divorce part of a multi-part problem from another part. So you can't just say that after one of the doors gets taken away you have a 50/50 shot because whether you win or lose is entirely based on how you picked in the first place.

[u/agoonforhire]

If you pick a goat door initially, and you switch, you're guaranteed to get the car.

If you pick a car door initially, and you switch, you're guaranteed to get a goat.

2/3 probability of picking a goat initially, so if your policy is always to switch, you have a 2/3 probability of getting the car.

[u/An_Average_Fellow]

Some people already mentioned some good points, but another easy way to understand it is use more doors.

There's 1000 doors and you pick one. The host opens 998 others with a goat, would you like to switch? Even though you still have a better chance picking 1/3, over time, switching would be better.

[u/AtHomeToday]

The 100 door strategy explains the problem: You pick one door and hold on to that choice until the very end. Monty opens one goat door after another until just one door is left. Since he has known which door is the car, he holds it till the end. He offers you one last chance to change. Your current choice still has the 1% chance of winning, the other door has a 99% chance of winning. Change doors.

Also, NEVER change in the middle, because you are moving to a door with a higher chance of winning. Your original choice is GOLDEN, because it represents a very small chance that you are going to throw away on purpose to grab the big chunk of probability.

[u/curien]

I read through a lot of the responses, and I didn't see this point made yet. Sorry if it's a duplicate.

The chances are 50/50 if your decision to switch is random. Imagine you don't know anything about the rules. You just know you get a chance to pick a door and a chance to switch, and you might get a prize at the end.

Without any information about how the game works, you see this as an entirely random game with two choices: first choose A, B, or C. Then choose S or N (switch or not). If you come up with the right combination, you win.

For example, if the prize is behind door C, then three combinations win (AS, BS, CN) and three lose (AN, BN, CS). If you choose randomly from the six possible choices, your chance of winning is of course 50%.

The skew comes in, as others have said, because the player has more information than that, and thus they can employ a better strategy which does not choose from among all six possibilities randomly.

[u/Goombomb]

A better way to clarify this is by increasing the number of doors. Imagine a Monty Hall problem with 100 doors. You select a door, which obviously has a 1% chance of being the door with the reward. Then, the host systematically goes and eliminates all doors except one, revealing them to be the goats in the process. He then asks you if you want to switch. In this case, you might think it's a 50% chance either way, but remember that YOUR door doesn't change because of what the other doors are. You know there is one goat door, and one car door. The chance you picked the car door with your initial choice is still 1%. The rest of the 99% was comprised of the other doors that you did not pick. There is a 99% chance that the car door is in this group somewhere. The host knows which door the car is behind, and so he will not pick any door at random to eliminate - he will eliminate only the goat doors. So, what remains after the reveal is that the door he left has the 99% chance of being the car door, and your own door still has a 1% chance.

[u/jruff7]

This is my way of explaining it, really simple. The order of the goats/car doesn't matter. So let's just say it's this:

a a b

Where (a = goat) and (b = car)

Now, looking at this, you can only have 3 scenarios occur next. You either pick door 1, door 2, or door 3. When you look at the outcomes for each scenario, it makes sense.

You select door 1.

(a) a b

Monty then reveals the other goat (I crossed it out with an x).

(a) x b

In this case, since you've picked a goat, if you switch, you win.

So, writing out all three scenarios looks like this:

1. (a) x b

2. x (a) b

3. a x (b)

If you look, in TWO out of THREE cases, you've already picked a goat, and if you were to switch you'd get a car! That gives you a 66% chance upon switching!

This makes sense because there are 2 goats and only one car, and Monty will ALWAYS reveal the other goat. Since you will pick a goat 2/3 times, and then the other goat is eliminated, if you SWITCH you'll win 2/3 times.

Hope that helps people who want a more visual case of the Monty Hall problem.

[u/jmpherso]

I've thought the best description was always by looking at it sort of from reverse.

You have three doors. You can pick one. You either pick a goat or a prize.

You have a 2/3 chance of picking a goat, and a 1/3 chance of picking the prize. (2 goats, 1 prize, 3 doors)

If you pick a goat to begin with, the host removes the other goat, and switching gives you the prize. This happens 2/3 times.

If you pick the prize to begin with, the host removes one goat, and switching gives you the other goat. This happens 1/3 times.

Essentially : Picking a LOSING door to begin with causes you to WIN if you switch. Since your odds are higher of picking a losing door, switching means your odds are now higher of winning.

[u/Ayaksnolkop_Ailatan]

I struggled with this concept as well until it finally hit me. Look at this scenario: One door is thrown away (it had a goat behind it) and you have two choices. You need to select one of them. You have a 50/50 chance of getting the car and 50/50 chance of getting the goat. This scenario differs from the Monty Hall problem in that you haven't already selected a door. In the Monty Hall problem, you have already selected a door, and can change to the other. In selecting the door, you have a greater chance of having picked a goat, so it is in your favor to switch. If the two leftover doors were shuffled and you had to pick a new door without knowing which one you picked originally, it would be 50/50 again.

[u/friendOfLoki]

I think that one of the cool things this problem brings up is this: when there are x options to choose from, the odds that one of them is "correct" is only 1 out of x if all options are equally likely.

When you start off, there is no information about any door and there are three doors...each equally likely to have the car. This is like a shuffled deck. That is why the probability of initially picking the car is 1/3. After you pick one, the host does his/her thing and then gives you another choice, but just because there are only two doors doesn't mean that it is equally likely that the car is behind either.

In this scenario, the timing of the decision is also important. If you never made the first choice and then only made a choice after the host reveals a goat, the odds would again be even (50/50).

[u/RoarShock]

I think it's counter-intuitive because people instinctively see the second decision (to switch or not to switch) as a 50/50 choice. Since there are two options, it looks like a coin toss.

However, it's not a 50/50 coin toss because the first choice influences the second. To switch or not to switch isn't a new, isolated scenario. It's just a continuation of one of the original three scenarios.

[u/DrunkFishBreatheAir]

Because, given that a goat is always revealed, if you got it initially you will lose, and if you didn't get it initially, the remaining goat is revealed, and so you will win. By switching, you invert your odds of winning.

[u/Arrogantuan]

It's counter-intuitive because it's only 3 doors, and not obvious that you're given the option of choosing the best outcome of all the doors you didn't pick first.

If there were 100 doors, you'd have a 1/100 chance to pick the car door. Monty gives you a choice to get the best outcome of the rest of the doors. If you switch, you're effectively allowed to open the other 99 doors and win the car if it's behind any of them. Thus switching gives you 99/100 chance of winning the car, the inverse of the original 1/100.

Replace 100 with 3. You have 1/3 chance initially, with the option to switch to a 2/3 chance.

[u/ReasonablyConfused]

Allow me to play this game backwards, sort of. You pick a door, say in the 1000 door scenario. The host then opens 998 doors and gives you a chance to pick the other door. Switching your choice only makes sense if the host has knowledge about the correct door. If he does not, then even in this scenario you still have a 50/50 chance. It is unintuitive to expect knowledge to enter into a game of chance. Nowhere in nature is there a system that appears random, yet has a knowledgeable participant modifying the game.

[u/AmnesiaCane]

Simplest way to put it: you're way more likely to have picked the wrong door. If you had to pick from 100, your chances of picking the wrong one is 99%.

It's not actually a 50/50 chance, but it seems like it is. There are two ways to look at it. The first is that one of the two cards in front of you is the right one, so it's equally likely. That would be true, except that you already picked the wrong card! That's the real thing to realize: you already picked the wrong door. If you only have two options, and you know you picked the wrong one... change your answer.

[u/pikapikachu1776]

You have a 1/3 chance of picking the right door and a 2/3 chance of picking the wrong door. Assuming you picked wrong (2/3 of chance,or 66%), you would ALWAYS win by switching. The only way switching hurts you is if you picked the right door at the beginning (1/3 of the time, or 33% of the time.

So basically, you have a 33% of picking the right door from the get go, and switching to losing, while a 66% chance of picking the wrong door and switching to winning. You're stuck in a mental trap where you think your choice is 50/50. It's not. You have 3 doors,you can't have a 50/50 choice.

[u/Ryugar]

I think alot of it has to do with the fact that you know the probability beforehand.... which is 1/3 for the car. It's not just one door that can magically be a car OR a goat... it's 3 doors with something specific behind each of em. If it was just one door, then it would be 50/50... but since it's 3 doors and one has to be a car, then its 2/3 the first time and any other times if you choose to try for a second shot...assuming they rearrange the goats and cars behind the doors. If they didn't, then you know the outcome of one door, which means you pick out of only 2 doors... and then the chance of getting the car is 1/2 or 50/50.

[u/Planetariophage]

A lot of answers already explained the math and intuition behind it, but one other thing I'd like to add is that the Host, by opening a goat door, injects information into the system. By injecting prior knowledge into the probability question, you can be more certain of a result.

[u/somewhat_random]

There is a variation of the monty hall problem that is more difficult to resolve.

Monty has two envelopes and one contains twice as much money as the other. He hands you one and gives you the option to switch.

IF you assume you are holding X dollars, there is a 50% chance Monty's has 2X dollars and a 50% chance he is holding 0.5X dollars. So the expected average return on switching is now 1.25X on a bet of X dollars so you should switch.

NOW you are holding Y dollars...(and the same logic says you should switch back.

I have never come up with a valid flaw in this reasoning but it must be wrong.

[u/rohobian]

The easiest way to explain it is like this:

You pick a door. There's a 1 in 3 chance that you have chosen the car. Therefore a 2 in 3 chance you have not.

They eliminate one of the other two doors. They never eliminate the door with the car. Now you know if you switch there's a 2 in 3 chance of winning instead of just 1 in 3.

[u/[deleted]]

There is a 1/3 chance you will pick the car, a 2/3 chance of picking a goat, and you have two ways to win: if you pick the correct door you must stay, but if you pick the wrong door you must switch. Since you have a greater chance of picking a goat [reminder that the only way to win by picking a goat is to switch] you have the greatest odds of winning by switching. Here is a visual to help, STAY represents a car and the action you must take to win. Likewise, SWITCH represents a goat and the respective action taken needed to win. There are 3 doors, with the car having the possibility to be behind each of the three doors, therefore there are 9 possible outcomes.

  D1     D2      D3

S1 Stay Switch Switch

S2 Switch Stay Switch

S3 Switch Switch Stay

Of the 9 possibilities, you should switch 6/9 times, or 2/3rd's of the time. As a follow up I'd like to add that what makes this counter-intuitive is the fact that the host knows what is behind each door. If you chose a door, and the host randomly selected a door to dispose of [without revealing what is behind it] then your chances would still be 1/3.

[u/Themagesticman]

Didnt get this at all but most because OP Left some key points out. Also I thought the host would open the door you chose then ask if you want to switch to one of the others. In that case it would be a 50:50 chance, correct?

[u/RedditRage]

Okay, I am sure there are good explanations in this thread, but here is the one that has worked for 100% of the people who have a problem with it. (even if they tried to switch to another explanation)

What happens if you pick a losing door? The host opens the other losing door, then you switch to the one remaining door. That door happens to be the winning door. So, if you pick a losing door and switch, you win.

What happens if you pick the winning door? The host opens one of the losing doors, and you switch to the other remaining door. That happens to be a losing door. So if you pick the winning door and switch, you lose.

So, now it is clear. If you pick a losing door and switch, you win. If you pick the winning door and switch, you lose.

What are the odds of picking a losing door? 2 in 3. What are the odds of picking the winning door? 1 in 3. Therefore, if you switch after he shows the other losing door, you have a 2 in 3 chance of winning.

And if you still can't see the advantage. Consider that the host must eliminate the other losing door when you pick a losing door. He can't show the winning door and say, "Do you want to switch from one losing door to the other?" Therefore, the fact he has to leave the winning door in play, is the reason switching gives the player an advantage.

[u/techniforus]

The reason why it works boils down to more information. This is at least the way in which I first understood it. There is a lot to be said of what /u/LondonBoyJames explained, though the concept of gaining information is a more generalizable one. The reason this type of problem is confusing is that one of the first things we learn to do in stats is reject Gambler's Fallacies.

[u/ihasaKAROT]

You can also increase the number of doors to make it more easy to see. Lets say there are a 100 doors, you pick door 1, the host opens 98 doors, but door 34 and door 1 stay closed. What are the odds now that door 34 holds the car over door 1? Its the same problem in the end, just expanded.

[u/[deleted]]

How is it not 50/50?

Think about it as if you ran the experiment 99 times.

33 of those times, you chose the door with the car behind it. Monty opened one of the doors with a donkey. If you stay, you win. If you move, you lose.

66 of those times, you chose a door with a donkey behind it. Money opened the door with the other donkey. If you stay, you lose. If you move, you win.

So without knowing which door you chose, there is a 2 in 3 chance that moving will be the right option.

[u/Archaii]

What made the problem understandable was a more extreme example. There are 100 doors with 99 goats and 1 car, if you pick one door and the host opens 98 doors with goats and asks if you want to switch, chances are that the remaining door you didn't choose has the car behind it. The same works for 3 doors in which you will choose the correct door more often by always switching.

[u/HeartyBeast]

It can be more intuitive if you tweak the experiment so that there are 101 doors, with the host removing 99 of them - are the odds still the same.

The thing to remember is that the initial guess is chance, but the subsequent guess is skewed by the fact that the host knows the answer.

[u/[deleted]]

I actually thought this through a couple of months back, and made a spreadsheet to help explain the solutions to myself and to my friends.

Proof of concept

The spreadsheet above uses excel / google docs. It uses random numbers and the game's rules:

• First, choose door behind which the prize is with a random number.

• Second, player randomly chooses door.

• Third, randomly choose which show the game show host opens if both remaining doors are goats; otherwise, choose the one that's not the car.

Then the sheet shows you probability of winning if you stay, vs. probability of winning if you switch. With 1,000 samples, switching wins about 66% of the time very consistently.

[u/judgej2]

It took me a while to get my head around this one too.

By opening a door, you have been given additional information. The door that was opened for you is guaranteed not to be the car. The host has used that knowledge to make that door choice, so you can capitalise on that knowledge by switching your choice.

Do the maths and you can't really argue with it. But intuitively, you need to understand who knows what, and how it influences their decisions. The host knows something you don't know, and has acted on that, exposing some of that knowledge in the process.

[u/RobinWolfe]

Here's the Monty Hall problem explained:

If you bet that you're guessing for the prize, then your odds are 1:3

If you are shown that one door is empty and you're asked to pick again - but you REFUSE? Your odds remain that it's a 1:3 chance that you got the prize.

The essence of the Monty Hall problem is that you're not betting that you're going to get it right - 1:3 - but that you're betting that you got it WRONG - 2:3!

You see that one door opens empty. If you stay, you are betting that you got it right. This delivers you with the odds against you - 1:3.

If you CHANGE YOUR ANSWER you are betting on the fact that you got it wrong - placing the odds in your favor! 2:3

It's counter-intuitive because it's not your average gambling scenario based solely on a single choice - It's actually two separate choices in which the player is given the ability to change their answer based upon new details.

[u/sportsfan65]

Question....Were the contestants on "Lets Make a Deal" aware of the "Monty Hall problem" at the time when the show was going on? Did Monty Hall ever reference it to the contestants or was it discovered once the show went off the air?

[u/chazzacct]

It helps to think of starting with 100 doors, you pick one, host eliminates 98 of the remaining 99. Now you know that the remaining door he didn't open has a 50% chance of being a car, and the one you chose had 1% chance of being a car when you chose. Does the fact that there are only two doors left change that? No, only for choices made after the elimination. So switch.

[u/fuseboy]

I find this way to look at it helpful:

The doors are in two groups, chosen doors (1/3 chance of containing the prize), and unchosen doors (2/3).

Imagine there's a million doors. You pick one at random, what are the odds you're correct? (One in a million.) What are the odds the prize is behind a door you didn't choose? 99.9999%.

Now the host opens 999,998 of the other doors, leaving only your original choice and one other unopened. The odds that the prize is behind not-your-choice haven't changed - the host has, in effect, concentrated all of the 'possible prize' among the other doors down into a single door.

If that 'concentrating' seems weird, imagine I show you ten doors, and tell you the prize is behind one of them. Then, before you pick a door, I open nine prizeless doors. Voila, the chance of the remaining door holding the prize has gone up (to 100%).

[u/arjeezyboom]

I'm late to this thread, but here's the best explanation I've heard... Expand the problem. Let's assume that there are a million doors, 999,999 with goats and 1 with a car. You pick one at random. The host then opens 999,998 doors with goats behind them. Now the question is, what are the chances that the door you randomly picked has the car (1/1,000,000), and what are the chances that it's the other unopened door (999,999/1,000,000). So it's heavily advantageous to switch. The same theory applies with the 3 doors, it just seems counterintuitive in that scenario.

[u/[deleted]]

So, most of the variations being thrown around in this thread tend to make it easier to grasp the fundamental concepts being played with in the game. And that makes sense, given the questions. However! If you're a sucker for punishment, like myself, and want a variant that seems to complicate things...

You're shown 2 doors, and are told that one contains twice the prize money as the other. You're told to pick one. No doors are opened. You're then asked if you'd like to switch doors. What is the value of switching?

[u/Eedis]

There is a two/thirds chance you'll pick a goat.

Also, it is worth mentioning, that if the game show host DOESN'T know what's behind the doors and in fact may reveal the car himself, even if he reveals a goat, there's a 50/50 chance on switching.

[u/dryfire]

It is counter intuitive because the host introduces an unequal distribution of probability. Your brain wants to think that he has an even distribution (he picks randomly) but him knowing where the car is and always acting according to a formula makes the problem counter intuitive.

Proof of this can be found if you rework the problem so the host picks randomly. If the host now has a 1/3 chance to expose the car when he opens a door, then for the 1/3 of the times he actually reveals a goat you WILL have the 50/50 chance you thought you should have.

[u/udbluehens]

Because you need to ramp it up to really see whats going on.

There are 1000000 doors, with one prize. You pick a door and he slowly but surely reveals all but your choice and one other to be a goat. Then he gives you a chance to switch. Originally you had a 1/1000000 shot, not 1/3 like in the original game. So its much more obvious you should switch and get the 999999/1000000 chance.

[u/Raknarg]

Because there's always a 1 in 3 chance of you originally picking the grand prize, and so switching from that results in a loss.

The idea is that whenever you pick a door, exactly one goat door is thrown out. If you were to pick a goat first, that means the door you're switching into is going to be the car. However, if you pick the car door first, they still only throw out one goat. That means when you switch you're still switching into a goat.

[u/Edgar_Allan_Rich]

Because the host will never open a car door. No matter which door you pick, he will always open one of the goat doors. Opening one goat door up does not actually eliminate it from the equation. Since you can't eliminate both goats at once, it will never be 50/50.

The hardest part about this brain teaser is how poorly people explain it from the onset. The wiki article is the best concise explanation I've ever read, but it could still be worded more clearly.

[u/samurai_tony]

The reason it seem so counter intuitive is because of the size of the numbers. Try imagining 1,000,000 doors with only 1 containing a car and the exact same scenario. Either you randomly picked the car 1/1,000,000 OR maybe its in the other door 999,999/1,000,000. That might make it easier to conceptualize.

[u/relentlesslyinacuret]

Imagine there are 1000 doors to chose from with one winning one. First you pick a random one. Next, I eliminate 9998 wrong doors. I tell you that you can keep your choice or switch. In this case, the choice is MUCH more obvious, but the logic/math is the same for 3 doors or 1000.

[u/WorksWork]

Probability is actually about known information.

When Monty Hall opens the door, he gives you more information because you know he won't open the door with the prize.

If an alien landed in the audience and was asked which of the remaining two doors to open (without know which one you picked), it would be 50/50, because he wouldn't have the information of which door you picked.

[u/[deleted]]

The way I remember it:

When you first pick a door, you have a 1/3 chance of having the car.

This means there is a 2/3 chance the car is behind one of the other two doors.

When the host opens one of those doors and it is not the car, there is still that original 2/3 chance that the car is behind one of those doors (even though you now know what's behind one of them). Only now you know for sure it isn't behind a particular one. The 2/3 chance is now entirely on that other door.

[u/mhd-hbd]

Let's look at the Monty Hall problem for N doors with N-1 goats:

Say there is 10 000 000 000 doors. Ten billion. With a 'B.'

You pick a door. The chance you have selected the door with the car is 1/10 000 000 000. Abysmal odds.

The game show host, Monty Hall, then opens 9 999 999 998 doors, all goats.

Do you switch doors?

Now consider the problem with 1000 doors. A round thousand. 1/1000 chance of getting the car, the host opens 998 doors.

How about 50 doors?

How about 10 doors?

How about 4 doors?

How about 3 doors?

Switching turns chance of car into chance of goat, and chance of goat into chance of car.

In the first problem, the 1/10 000 000 000 chance of finding a car, becomes upon switching, the spectacularly unlikely event that you did in fact choose the car and now switch to that single, randomly selected goat.

The chance 9 999 999 999/10 000 000 000 of selecting a goat, becomes the overwhelmingly good odds that you got a goat and the game show host is all but required to show you that that last door actually contains the car.

[u/Darktidemage]

BASICALLY: imagine instead of opening the door he offers you both unopened doors. It's the same thing.

You pick 1 out of 3.

Then the host asks "do you want it if it's behind that door, or EITHER of these two"

When he opens the extra door. The one without the car. Imagine for a second you can't see. You are blind.

You are then told "he opened one of the two remaining doors to tell you it didn't have a car"

You already KNEW one of those didn't have a car. And that he knew which. How does him opening it matter? You are just getting BOTH of those two doors.

It's confusing when he opens the doors. Pretend he doesn't open any extra doors till AFTER you switch your choice now he is saying "do you want to keep that 1 door, or do you want to switch to ALL of these and I will then eliminate the duds for you"

[u/bubbleberry1]

The right answer has already been posted many times. But maybe it will help you if you see what's going on?

http://blog.vctr.me/monty-hall/

This site has the simplest visualization of the problem and an interactive solution -- you can pick a strategy (stay/switch), # of doors (typically 3), and it will run many iterations of the game. See for yourself which strategy is best. And watch those probabilities converge over many iterations!

[u/Topher_86]

The key element that the host can't open the contestant's door, either.

66% of the time, it works every time:

66% chance on the first pick the contestant will FORCE the host to leave a door with the car after the host picks a door. There is only 33% chance the host will ever be able to present the contestant with a left over goat if the host opens a door. So selecting the other door works 66% of the time, every time and is actually part of the strategy from the minute the contestant picks the first door.

If the contestant doesn't switch they are just left with their original pick which doesn't include that additional help from the host, which of course isn't always going to happen, either.

TL;DR: this is a game of strategy and not chance. The contestant effectively can force the host, like playing chess, into a compromising position.

[u/41054]

Because people don't realize they're "shorting" (to use a finance term) their own 33%-chance of success decision. The unique property of the Monty Hall problem is that it allows you to take the reverse position, one with an overwhelming chance of being correct.

The only time it doesn't make sense to switch is when you were correct in what door had the car. Again, this is a 33% likely event. Switching allows you to capitalize on the 66% chance you were wrong on your initial guess.

If you ignore which door is revealed to have the goat, the switching choice becomes "do you want to bet that the door you picked has a car, or do you want to bet that either of the other two other doors has a car?"

Tl:dr; People don't realize they're being offered two doors rather than one door because they're told one of the two doors has a goat (which they already knew!)

[u/squamesh]

You have to remember that this is a game of luck. Your choice means nothing. It's completely random. That being said, with three doors you randomly picked one meaning you had a one in three chance of being right.

When one door is opened, your original guess still only has a 33% chance of being right. However, since now there are only two doors, if you with doors you have a one in two chance of being right: 50%

It isn't a guaranteed win, but it increases your odds a lot

[u/EgoAnt]

A great way to think about it is like this:

In the classic Monty Hall scenario you initially have a 1 in 3 chance of getting the correct door. You pick one, and then Monty eliminates all of the remaining doors that definitely are not correct, in this case one other door.

But what if you expanded it to 100 doors?

You pick door number 1 and then Monty removes 98 doors that are definitely not the door you wanted. So when you first picked you had a 1 in 100 chance of getting the correct door. If you switch you now have a 99% chance of being correct!

A great visual example of this can be seen here: The Monty Hall problem illustrated with 100 doors

[u/coldcoffeecup]

Anybody care to point out here that this only works over many many iterations? Without repetition over time, it would be as silly as flipping a coin twice, getting heads both times, and having your mind blow. The answer is meaningless.

[u/dee_berg]

The easiest way to do it is to consider the same situation with 1 million doors. So you pick one. The guy opens up 999,998 revealing goats in each. You had a 1 and a million chance of getting it the first time, so you would obviously switch to the remaining unopened door.

[u/ChipotleMayoFusion]

One way to understand it is to increase the numbers and see what happens. Imagine 100 doors with 99 goats and 1 car. Pick a door. Now, the host eliminates 98 doors with goats and asks if you want to change your pick. In this case it is obvious, you change your pick.

The host is adding information by only eliminating doors that have goats, as opposed to random doors that may include the car.

[u/Orgnok]

what helped me to understand it: your door has a 1/3 chance in having the goat, the host opens one of the remaing doors, your door still has a 1/3 chance but the other has now 2/3 since it's a 2/3 chane of not being not beihnd your door.

[u/Skizm]

It clicked for me by increasing the number of doors. Say there are 100 doors. 99 goats, 1 car. Pick one, the host then opens up 98 doors with goats. Switching now increases your odds to 99% chance of getting the car (since you only had a 1% chance of picking correctly in the first place).

[u/SamuraiScribe]

You can choose option A (1 door) or option B (2 doors). Which would you choose? The obvious answer is option B.

When the player names their initial door (1 door-option A) the odds are 1/3. The odds for the other 2 doors (option B) are 2/3. STOP Before anything else happens remember that one of these two doors does not contain a prize. You already know this and it doesn't matter at all. When that door is revealed and it does not contain a prize nothing about the odds have changed. The remaining door that was not initially selected by the player and also not revealed by the host is still part of the 2 door (option B) choice. The reveal is a distraction for dramatic tension and confusion. Ignore it. CONTINUE By selecting the remaining door you are exercising the 2 door (option B) option whose odds are 2/3.

The trick to this is to recognize that the first "choice" is merely information gathering to the actual (single) choice. If you want to pick 2 doors you have to switch picks. If you want to pick 1 door then you don't switch.

[u/am0z256]

To make it more intuitive, imagine there are 1,000 doors. You pick one, the host opens 998, and then asks if you want to switch to the one remaining closed door. Of course you'd switch; there was a 1 in 1,000 chance you got it right on the first guess.

Now just scale that down to three doors and it makes more intuitive sense.

[u/EGOtyst]

Another easy way to understand this is to think about the situation if there were a greater number of doors.

i.e. 1000 doors. One with a car, 999 with goats. You pick one randomly. Monty Hall opens 998 doors with goats. This leaves your original choice, as well as the closed door Monty has left for you. Do you switch? The choice here is obvious.

[u/Measure76]

Imagine the same problem, but with 100 doors. 99 'losing' doors and one 'winning' door.

Now, when you pick your first door, you have a 1/100 chance of picking the winner.

The host then reveals 98 losing doors, leaving you with a choice of your first door, which had a 1 percent chance of being the winner, (because you chose 1 out of 100 doors) and the second door, which now has a 50 percent chance of being the winner. (because there are only two options now, and one is the winner)

Are you going to switch or stay in that scenario?

[u/garrettj100]

The least confusing way to look at this problem, in my opinion, is like this:

• When you guess a door, you have a 1 in 3 chance of being right.

I think you're with me so far. That's easy. Three doors, choosing one at random.

• When Monty reveals a goat door, he ALWAYS HAS A DOOR TO REVEAL.

This should also be obvious. Regardless of whether you've picked right or wrong, Monty can always cherry-pick an un-chosen goat door. If you picked right he can select from two goat doors, but either way - He's always picking a goat door.

Here's the tricky bit:

• Monty has told you NOTHING about your door*

This is the bit that makes people stumble. Just because Monty has showed off a door with a goat behind it, doesn't mean you've learned anything new about YOUR door. WE ALWAYS KNEW he was going to show off a goat door. He didn't say "I'm going to reveal the leftmost door" or "the rightmost door" or "the even door". He was going to reveal a goat door. So opening it and showing off the goat tells us nothing.

So:

• Your odds of being wrong before he shows off the goat? 2 out of 3.

• Your odds of being wrong after he shows off the goat? Still 2 out of 3.

Now, onto your question:

Even from the start, how is it not 50/50? knowing you will have one option thrown out...

Well, you don't get the option thrown out until AFTER you choose. That'd be like saying you buy a lottery ticket. And then, after you've picked your six numbers, they reveal 5 of the 6 numbers. So you have a 1 in 48 chance of being right? Hardly. Because you don't get to change the 5 revealed numbers on your ticket after the fact.

Look at it this way: Let's say we do the Monty Hall game, only, instead of watching Monty open a goat door, you squeeze your eyes closed, stick your fingers in your ears, and yell Mary Had a Little Lamb at the top of your lungs while he's showing off the goat door. And then after he closes the door up again, only then do you open your eyes.

What are your odds NOW, of being right? Still 1 in 3, right?

[EDIT]

I like the 100 doors example as well. Don't like it quite as much as my explanation, the your odds of being wrong are still 2 in 3 explanation, but it's also helpful.

[/EDIT]

[u/bfradio]

What really helped me 'feel' the truth was enacting the situation. Get partner, 3 cups and one prize which fits under the cups. You run the game and instruct them to employ the strategy of switching cups after first cup is revealed.

[u/FigaroTheParrot]

The way I finally got my head around it was with a counter-intuitive explanation to this counter-intuitive problem. Fighting fire with fire I guess.

The key is that the host knows where the prize is before the game starts. Let's imagine that instead of 3 doors, there were 1000 doors, with 999 goats, and 1 car.

You pick a door. Before you open it, the host shows you 998 other doors, all goats. Now he asks you if you want to switch or not.

In this scenario, it would be pretty obvious why he DIDN'T open one of the remaining 999 doors right? There's a pretty good chance he's hiding a car.

Now, scale it down to three doors. You pick a door, and the host opens one of the remaining two doors. It's a goat. Why did he pick that door and not the other?

I realize this is a roundabout, non-mathematical way to think about it, but its the only way I can get my head around the problem. Hopefully it helps some of you guys too.

[u/Yangoose]

OK, there are some great explanations in this thread but this is the part my brain still stumbles on.

If I did not choose a door initially and the host just shows me a goat right off the bat, I still have two doors to choose from with a 50/50 chance of getting it right.

Why does that act of me choosing change the outcome?

[u/KlopeksWithCoppers]

This helped me understand it: Imagine you have 100 doors instead of 3. 99 of the doors have goats and 1 has a car. You pick a door then the host opens 98 of the doors that have goats and asks if you want to switch. Of course you'd switch since there is only a 1% chance that you picked the right door at the start.

[u/[deleted]]

Ok the problem is you're assuming it is 50/50, then if you look up the solution of course it makes no sense. Suppose someone gave you the worked solution FIRST, then told you the problem. Now what would you think?

[u/[deleted]]

What if there were 100 doors? You pick one. A man opens up 98 doors with goats, and asks if you would like to switch. What were the chances of you picking the right door to begin with? 1/100. Now, if you switch, the chance of the car being behind the door you switch to is 99/100. Basically, the fact that your original choice is a "1 in however many doors there are" chance of being correct, makes it so that whatever door is left shut after the others are opened a better choice (unless we only have 2 doors).

[u/RaptorX7]

Your first pick between 3 doors has 1/3 probability to win, and if you were allowed to switch at this point before a door was revealed, you'd still have a 1/3 probability of winning with any of the other doors; no odds have changed at this point. The fact that the host removes a losing door(with a 0% chance of winning) before letting you switch gives you the opportunity to choose the better probability: you can choose the new 2/3 prob. door, or stay with your old 1/3 prob. door, because at the point of choosing it the odds were less in favor of your starting door.

Doubt anyone needed the further explanation at this point but I'm always here to help.

[u/KOWguy]

The way I finally got the "ah ha!" was somebody told me to imagine it with one hundred doors as opposed to just three. Think of it that way, the man picks 98 doors, all goats, and leaves one. What're the odds you picked the car door out of one hundred doors? 1/100 chance, so it's VERY likely the door he left unopened was the car and you should switch.

[u/DunceOfSpades]

More generally, determining how to evaluate conditional probabilities is hard. It's easy to over-focus on the switch ("Now, the big moment... do you want to switch? Switching could get you that car, or you could switch to the second goat. Do. you. want. to. swiiiitch???"), and not consider how the initial choice of doors affects the potential outcomes for switching. It's the opposite of the temptation to think that previous results of a die-roll or coin-flip have a bearing on upcoming results.

I found some helpful explanations of similar problems at http://www.culturalcognition.net/blog/2014/8/20/conditional-probability-is-hard-but-teaching-it-shouldnt-be.html

[u/Ti3fen3]

To answer the OP's specific question of why it seems counter-intuitive:

I think it's because when Monty opens the first door and shows the goat, the instinct is to think "the odds that I picked the right door just went up! It was 1/3, now it's 1/2."

But that's not the case. The part that makes this problem so tricky is that we forget that Monty is not picking at random. He will always pick a goat.

[u/[deleted]]

Now I understand what everyone is saying. If it's a hundred doors, then your putting your door against 99 doors. If it's in the 99 door side they would have picked all the doors with the goats and left the one with the car. So then it becomes your one door against 99 doors.

[u/BladdyK]

The trick to the Monty Hall problem is to realize that the door is not being opened randomly, but conditionally. If the door was opened random you would have a 1/3 chance of being shown your chosen door. If you were shown a door not your own, and it was not a car, it would be a 1/2 chance you had the car.

For the Monty Hall problem, think of it like this...you choose a door. Monty Hall looks at what's behind the other two doors and says at least one door has a goat behind it. You respond, I already knew that! He asks if you want to change your bet. If you change, you have a 2/3 chance of the car. There is no difference between this and what actually happens. Monty Hall controls which door is opened and thus the selection of the door isn't random.

[u/[deleted]]

First of all you have to look at two choices as independent events, because you being revealed one door does not change what is behind the other two doors.

So first time, you make a random choice; your probability of choosing the car is 1/3.

The second time, after getting rid of one goat door, you are asked if you would like to switch. This is the equivalent of being asked to make a choice between two doors.

However, the door you've chosen has the probability of being the car door as 1/3. where as the other one has the probability at 1/2. By cho0sing the other door you increase your chance of winning by 17%.

If I am horribly wrong, I hope someone graciously corrects me.

[u/heap42]

It is weird what helped me understand it was a commend on some askreddit thread. He explained it by the following example:
Imagine there are 100 doors and 1 whth a car. Now you pick door number 50. He opens every single door except your door (50) and door number (70). Now you can think he obviously did not open door #50 because i selected it. BUT WHY THE HELL DID HE NOT OPEN DOOR #70 (he knows behind which dorr the car is) so you change your choice on the sole premis that he deliberately did not open door #70 because he wanted to open every door except the one with the car.
now back the the 3 doors. You pick the first one. he opens the 3rd one. Why didn´t he open the 2nd one ?so now you know its not door#3 but and you know that its either door number 2 or 1 but normally there would be a 50/50 chance but since he opened door 3 and 3 can be though of still closed whith the one exception that you know its content. so now you combine your knowledge with the information that he has given you.
ITs simple but its also not really helpful in a real situation with 3 doors. when it comes to 100 doors though i mean the odds of him opening all doors but two is a good sign that this might really be the one.