+João Vítor M It's SUPER-critical (and often overlooked in the introduction to the problem) that Monty KNOWS where the prize is AND will intentionally open the goat-door initially. Once you understand that, the solution becomes nearly simple.

I have a better explanation: you will ALWAYS win if you initially choose a goat and then swap. ALWAYS. What are the odds of choosing a goat initially? 66%. So 66% of the time you choose a goat, he shows the other goat, then you switch to the car and win. Thats it.

+João Vítor M It becomes even more clear why you should always switch, if there were 10 or 1 million doors at the beginning. Then Monty practically gives you the car every time.

@@mjlv3862 ooooooooohhhhhhhhh

Jea me to

That makes so much more sense ty

@@pranavsanthakumar106 Agreed. I was lost, until this explanation.

The concentrated probability explanation is equally valid but more purely mathematical, and that is why a lot of people struggle with it.

You said that much more succinctly than I did.😀

The explanation I posted before reading yours is identical...but you said it way more concisely. I do think it's worth adding that in the two cases where your initial guess is wrong, Monte has only ONE possible choice, which "gives away" that you MUST switch to win.

Or it could've just been explained normally through maths....

Yeah but some people need to understand the consepts or else they won't accept the math to be anything more than "theory", but some do get it that way.

To me, it was pretty simple. You know your first choice has only a 1 in 3 chance of being correct, and it's more likely to be one of the other two doors. It's like Monty is pointing you to where it probably is (2/3 chance) by showing you where it isn't.

Thx that helped even more

i still feel like there is something missing that makes your logic not valid

Please don't be mad, it's just that I have a fundamental disagreement with one crucial point of the argument and I'd like to know if someone can enlighten me on my error, if there is one. If I choose the first door, then the moment the second door is revealed to be a goat, that means scenario 2 is not in play anymore. You can't still count it as a possible scenario. Scenario 2 has a car behind the second door, but that conflicts with the goat that has been revealed to be behind door two. Therefore, after the first reveal, there are only two scenarios left, scenario 1 and scenario 3. I don't see why the scenario that was just eliminated has its probability merged into the remaining scenarios. Or, to show an example like Rundvelt has done: Scenario 1: Car / Goat / Goat Scenario 2: Goat / Car / Goat Scenario 3: Goat / Goat / Car Let's say you pick the door on the *left*. The *middle* door is revealed to have a *goat*. You do _not_ switch. Scenario 1: Win -Scenario 2- (Eliminated, due to goat behind middle door) Scenario 3: Lose Let's say you pick the door on the *left*. The *middle* door is revealed to have a *goat*. You _do_ switch. Scenario 1: Lose -Scenario 2- (Eliminated, due to goat behind middle door) Scenario 3: Win. So after a goat has been revealed to be behind the middle door, there are only two possible scenarios left (#1 & #3), not three scenarios. Therefore I'm puzzled why this seems to be incorrect. Where is the crucial point I seem to be mistaken about?

Dharma Jannyter You are mistaking Scenario for Door. They are scenarios. I'm showing all the possible combinations for a single door. The reason is because all other doors will be the same. But let's do them all just to understand the issue. I've put (Scenario #) beside the trials that I represented in my previous post. Let's stick with the same door. You Pick Door #1 -> Prize is behind Door #1 -> WIN (Scenario 1) You Pick Door #1 -> Prize is behind Door #2 -> LOSE You Pick Door #1 -> Prize is behind Door #3 -> LOSE You Pick Door #2 -> Prize is behind Door #1 -> LOSE (Scenario 2) You Pick Door #2 -> Prize is behind Door #2 -> WIN You Pick Door #2 -> Prize is behind Door #3 -> LOSE You Pick Door # 3 -> Prize is behind Door #1 -> LOSE (Scenario 3) You Pick Door # 3 -> Prize is behind Door #2 -> LOSE You Pick Door # 3-> Prize is behind Door #3 -> WIN So, as you can see, there are 9 possible combinations, and out of those 9 combinations, 3 result in wins. Now, let's examine what happens if your strategy is to switch. You Pick Door #1 -> Prize is behind Door #1 -> LOSE (Scenario 1) You Pick Door #1 -> Prize is behind Door #2 -> WIN You Pick Door #1 -> Prize is behind Door #3 -> WIN You Pick Door #2 -> Prize is behind Door #1 -> WIN (Scenario 2) You Pick Door #2 -> Prize is behind Door #2 -> LOSE You Pick Door #2 -> Prize is behind Door #3 -> WIN You Pick Door # 3 -> Prize is behind Door #1 -> WIN (Scenario 3) You Pick Door # 3 -> Prize is behind Door #2 -> WIN You Pick Door # 3-> Prize is behind Door #3 -> LOSE So, as you can see, there are 9 possible combinations here too. And out of those 9 combinations, 6 result in wins. Above, I was doing scenarios to show what happens for one door (as all other doors will show the same pattern). Hopefully this is a clearer explanation.

Rundvelt I can understand those possibilities you've enumerated, but I think the moment we gain information (the goat reveal) the probabilities adjust as we now have more knowledge about what could not have been possible from the start. I've prepared all the possibilities as I see them, but in order not to create more confusion than necessary I'm holding it back for now and instead try to explain the main problem I'm having in understanding this: Imagine we're at the stage where we've chosen a door and a goat was just revealed. Imagine further that our memory gets wiped right after that. Now we see 2 closed doors & a goat. That means we have the choice between two doors and the car could be behind either one of them. Choosing the door we had chosen before the memory-wipe would be like sticking to the door. Choosing the door we haven't chosen before the memory-wipe would be like switching the door. So, would you say that before the memory-wipe the door I would've _switched_ to had a 2/3 win-chance, but _after_ the memory wipe that same door had only a 1/2 win-chance? If you do, then how is the pre-memory-wipe situation any different in terms of making a decision to switch than the post-memory-wipe situation?

Yes! I kinda got it from the 'concentration of probability' idea, but by flipping the problem, i.e. it's not "What's the chance I've picked the car" it's "what's the chance I've picked a goat" I finally get it.

tanukigalpa OOOOH ITS ALL ABOUT THE INITIAL PROBABILITY THAT YOU CHOSE A GOAT OR CAR

+Double Shrekt And that Monty knows whats behind each door, and so indirectly leads you to higher odds of winning, if you switch.

yes that's also how i understand it, i think it's easier that way.

EXACTLY, its better to say you have a 2/3 chance that you get a goat rather than saying there is a 2/3 chance that you will get a car when you switch

+strayster2 No you where not wrong. We where not given enough information to fully understand the concept. This man explained that the host knew where the items where. I for instance was under the impression that everything was random

+David Arencibia Ah. That really is the rub, isn't it?

+David Arencibia of course the host knows where the car is... Do you really think he would reveak the car and say " oops lets retake this"...

+Sundqvist And even if the host revealed the car you then should CHANGE your door to the "car door", right?

I faced the exact same situation. I thought I 'solved' the paradox and found out that the probability is, in fact, 50%.

You are wrong... if he hads 0% of being right then why you add 33% to the other door insted of adding 16.5% to each door?

trululasoxd scotti Because of the order that the doors are picked, When you pick there are 3 doors. Each has 33% at that point (pure random). Your door will never change from 33% until the final reveal. But by adding information to the system, however, you CAN change the odds of the remaining doors because the remaining picks are non-random. Suppose there are 100 doors. You pick one at random (1% chance). Monty then opens 98 bad doors because he has information of which are bad. Now there are only two doors left (yours and one other). You think your door has 50% now? I say your door still has 1%, and Monty's door has 99%.

raven lord I think your explanation is even better than the video's.

even after watching the vid, i still didn't understand. But reading ur comment, I did :D

Honestly for me when I think of this problem the way I remember it is that Monty isn't being random. That's why the probabilities get skewed. Noticing non-random actions is the most useful technique for dealing with this class of problem, which most other solutions do not help with.

Great explanation, Monty knows, that's the key. Now suppose there are 10 doors, after you pick, Monty shows you 8 doors with goats.... you better switch to the one door Monty didn't show you.

falcon tinker that’s different as you added in more doors love

Ugh! Then PLEASE explain it to me! I can’t get beyond 50/50 chance.

yes ok . my brain can understand this , thank you ,

+nekogod This one line you wrote is literally the simplest and best way to describe the problem.

+UsuriousCactus8 The problem is that people incorrectly assume that once a goat is revealed, the probability of our originally selected door containing a goat changes instead of remaining fixed. Unfortunately, people often don't understand that the deliberate, non-random nature of the goat elimination is the key, and there doesn't exist a very simple one sentence explanation for that.

+Pine Fe beginner programmer here: may I ask why you would use a times variable (especially as a double) when it is essentially the same number as the count variable?

maxid87 You're right, he could have either initialized count to 1 and then printed (success/count), which would have made for 100,000,000 iterations instead of the 100,000,001 iterations he's getting with the way he set it up, or he could have printed (success/(count+1)) and stopped at < 100000000 instead of

+TedManney yeah I was just wondering why one would do this in some example code that is usually supposed to use as little code as possible. Also it seemed inefficient to initialize a double when it only stores integer values.

I dont feel it was imprecise. Now that I understand it both feel valid. But definitely for me, reframing the concept as the odds your first pick contains the car cemented the idea--allowing me to accept the other mental model of probability "concentration" which has different intuition benefits. Kind of like how both analog and digital clocks accurately represent time, but analog quickly gives a rough perception of duration left in an hour or day and digital more quickly gives a precise answer for the exact number of minutes remaining. Both tools accurately represent time, but answer different questions better.

Just because there's only 2 doors remaining does not mean there is a 50/50 chance.

Eetarsaurus That only works when the moderator knows where the prize is. If he opens random doors with the chance of revealing the prize early, the chances split up evenly for the doors.

The original video made the example with 100 doors and people still didn't understand it, even though it was clearly obvious. Going higher should make it even more obvious, but I think some will still not understand it xD

Eetarsaurus Shouls have watched the video first. He said just that. Sorry mate. :D

Eetarsaurus Troll post is Troll :D

Yeah, your going through each step makes it more clear to me, but I've found by reading comments that the solution seems to click into place for people in slightly different ways, which is interesting. Most other comments make no sense to me.

I can disagree. The 2 in three is correct, but the "if the host didn't know is NOT correct in the second half of the video. Let's grant thatbthe host doesn't know, and 1/3 times opens the door with the car...that changes the odds to 0....bit if he opens a goat door the odds to switch are 2 in 3, we BOTH now know it was a goat door, and the prior knowledge of the host is irrelevant...he got that bit wrong

Thanks for helping me understand

The third door doesn't play a part because he is always going to open a door with a goat no matter what anyone says.

I just say "With the always-switch strategy, you only lose when you guessed right, and that only happens one out of three times. The rest of the time you win".

OK, so then don't switch, because there's a 2/3 chance you have a goat.

You can sell the car and buy more goats. You have greater +EV of goats by switching.

😂, smart man, goats are a great investment.

@@euming got me there

@@driziiD dude, that's great.

you're welcome

Shunya No, it does affect the decision, it merely does so in a trivial way.

Neat explanation Dean. Please stick around - there are harder nuts to crack, some persevering with their own 50/50 blogs for over six years and not worked it out!

For me the visualizing the 100-door version did the magic. 99 goats, 1 car, the chance of you picking a goat is 99%. If you do pick a goat then the other door MUST have the car because Monty has to reveal 98 goats no matter what. In other words, the only case you lose after switching is when you have initially picked the car. Which has the probability of 1% (or 33% with 3 doors). This is somehow much easier for me to grasp than the original 3-door problem

your'e wrong, but the answer is still switch. say you always pick door A, you would think the odds would be 1. W, L, L 2. L, W, L 3. L, L, W but option 2 and 3 are the exact same, it makes no difference what order the doors are in.

Morgan Bartholomew Well spotted. I'm sure the OP meant: "Draw three doors three times. Put the car behind #1 in the first row of doors, behind #2 in the second row and behind #3 in the third. First row if you pick #1 Monty will pick #2 or #3; if you switch you lose. Next row if you pick #1 ... etc". That's what I saw (by not reading it properly).

I've never actually seen someone on the internet admit they were wrong. Bravo

mickavellian It is not a "50/50 choice". You have failed to understand the argument he gave here based on the principles of probability. Always switch, like he says, that doubles your chance of winning.

mickavellian You are the incoherent one here. Your other i-word might apply too;) You have missed the point of the whole Monty Hall probllem. Once the contestant sees one opened door with no prize behind it, it is no longer 50/50. Sure, that is counterintutiive. But real math and real science are full of results that are counterintuitive to those who have not learned the concepts. Here, the main concept that is so importans is "conditional probability". It is notoriously difficult to learn.

mickavellian There are many things you got wrong in your long screed, but the key one is this: "With two doors you are going for heads or tails (50/50 .. correct?)" No, not correct. Once the game show host opens a door revealing no prize behind it, he has leaked information about where the prize is, so it is no longer 50/50; you now know (if you understand conditional probabilities) that your odds of winning are 2/3 if you switch, 1/3 if you stay.

mickavellian You still haven't figured out your objection is without merit. What happens if the door is taken off and another contestant brought on it entirely irrelevant. You have changed the conditions, so the conditional probabilities are different.

mickavellian No, he is wrong and so are you. You simply do not understand probability at all -- just like the many people Monty Hall took advantage of on the game show.

Still totally wrong.. I mean I dont get it.. no Im kidding :D This is even better an explanation then the other ones, ohnestly. But they were good too :)

Those who don't get it with this explanation ... Please do yourself a favour and give it up to try :-)

This absolutely kills you, doesn't it? An explainable logic (well, statistics) problem that some people refuses to believe is true. While well maintained, I could feel the absolute rage in your gestures and complete disbelief people couldn't or wouldn't understand.

Thank you brady! this is the key piece of information many people seem to miss, and many describers seem to leave out, im very happy you clearly stated how one of the key factors is the hosts obligation and knowledge to open a zonk, and showing the odds before switching I don't think this problem could be explained any better than you have here, cheers!

***** I tried with code, using random function in Java after 1 million experiments I get about 66% (like 66.6331% , 66.7482% and etc. ) chance of winning. Or someone could try rolling dice like few hundred times if they do not trust computers :)

Do you mean betrayed or portrayed?

The fact that you have a 2/3 of selecting the goat initially, and knowing Monty will always show you a goat, you have a better chance of winning the car by switching. Not always, though

​@@kuperlilu5340im sure it was portrayed

Yup, that is exactly what it is

***** This. Going through every possible state is a pretty irrefutable way of coming to an answer.

Monty wouldn't be able to open the door with a goat every single time, if he didn't have information about contents of the doors.

It makes so much more sense when there’s more doors

Okay I was upset that I still didn’t understand but you putting it this way made it click instantly and I feel silly for not understanding it from the video. Thanks!

You are either asking if switching is better BEFORE the third door is opened, or after. If the third door is a variable then yes it favors the player no doubt, but once that door is revealed and your answer is dependent upon this information, then the third door being NOT opened or being the CAR are no longer variables so yeah.

lmao

Thanks, it's more clear now!

Yeah I prefer the explanation in the video but this is valid too.

And there is also the implied rule where the host cannot open the door the contestant has chosen, or there is no game

@@jmak9376: "there is also the implied rule where the host cannot open the door the contestant has chosen" It can't be implied. It must be spelled out. That was the 3rd of my list of 4 important things to emphasise: - the host must then open one of the two doors you have not selected.

You are either asking if switching is better BEFORE the third door is opened, or after. If the third door is a variable then yes it favors the player no doubt, but once that door is revealed and your answer is dependent upon this information, then the third door being NOT opened or being the CAR are no longer variables so yeah.

maybe he will (or already did) no guarantee. What is the probability he will?

Ro Jay His birthday was 2 days ago (and he's still alive) :P So 100%...

How about now?

Just to keep you guys in check about ten months later. Monty Hall is still kickin'.

RIP Monty Hall, September 30, 2017

Now try explaining to someone that swapping does *not* confer an advantage in Deal or No Deal because the eliminations of the cases are blind and random. It can be literally impossible to get some people to accept this, they insist that DoND is a special application of the Monty Hall problem.

Jeez... the video is wrong!!! All of this quite 'frighteningly' shows how people are so easily influenced to believe something, which is clearly incorrect. No wonder that propaganda and fake news are having such a dire consequence on our society. 'Think' for yourselves people... don't let those in authority or with slick videos do that for you!

@@sebastianwrites You need to do some thinking yourself. The host doesn't have a door that has a 1/2 chance of anything to leave for you to switch with.

thanks for watching

It's good, isn't it? But then you get annoyed at those who don't get it.

It really is that simple. Funny how some people twist themselves into pretzels trying to over-complicate this.

its flawed

how does montys knowledge change this as long as the one he chooses is the goat even if it was by accident in that scenario the chances are 2/3 if you change

Well there is actually a trick to it. It's that Monty knows and his hand was forced. Had you arrived at the same scenario where you pick a door, the a goat is revealed, however Monty didn't know where the car was. Your chances are actually 50/50. This happens because rather then converging two option into one, we completely eliminate what was a possibility, that he could have eliminated the car. If Monty knew, odds are 33/66, if he didn't they are 50/50.

reread please: "There's no MORE trick to it."

Your swap will not ALWAYS be a car. 1/3 your swap will be a goat. 2/3 it will be a car.

@@jpg7616 I did say "If you do" in my first statement. Perhaps I should have written it in all caps to clarify?