The Monty Hall Problem Explained

[@RapMastaZ]

Even though I understand the problem mathematically, your explanation of with the deck of cards is BY FAR the best intuitive explanation of the problem I've ever seen. My hat is off to you!

[@seanof30306]

The deck of cards example; more precisely, his reasoning behind that example, is utterly flawed. He says the person in control of the deck would only leave the winning card unexposed, when in fact, the person in control of the deck is simply leaving an unknown card unexposed. It could be anything. The person controlling the deck is always going to turn every card in the deck but one face up. All you know is, the card in front of you, or the single card left in the deck is the winner. One, or the other. 50/50. He's trying to apply game theory to the situation, but it's an utter fail. Of course, if one of the cards in the deck is the winner, that's the one he's going to leave face-down, but if he examines the deck and sees the winner is not in it, is it more likely that he will simply concede that you are the winner, or would he leave any random card face-down and hope you choose to switch to it? He's 50/50 there, too.

[@scotthumphreys4157]

Thanks Philip! My head was about to explode from watching another video and reading the comments. All it took was pointing out that the revealed door/cards weren't revealed at random. I just shook out a lot of tension from my upper back :)

[@henryporter101]

@Johnnyradionic.In original game,Monty can't show the car door-even if selected correctly.You have a choice,either you 'guessed' it,which has 1/3 probability or the door he's concealing has it.By switching,you go from 1/3 guess to an informed 'either/or' which makes it 2/3 probability.In your scenario,since 2 doors are selected,Monty is forced to tell the scholars that one of them 'guessed' correctly,so they may as well stick to their original guess which was 1/ 3 and became 1/2

[@KittenIgnition]

in the beginning it seems overcomplicated and not really mathematical, but about halfway through its explained pretty good. I still think they explained it best in the book "The Curious Incident of the Dog in the Night-Time", and in much less time (a page or so). also an awesome book.

[@newcastlemusic]

For the probabilities to pan out the way you have described Monty would need to inform the contestant that there is nothing behind door number one BEFORE HE OPENS IT. Or at least he must convey the fact that his choice of door is not random.

[@Birkirrey]

I've learned most of the stuff you explain in your videos before but you always manage to give me some new understanding of the subject. Excellent videos!

[@djbentendo3929]

The formula guarantees a constant with the goat being eliminated. The only way 2/3rds can be achieved as an answer is if you count goat A and goat B as separate choices (each only presenting one outcome for the second question of "would you like to change) and don't consider that choosing the car presents TWO options instead of a forced single option. You cannot view the goats as separate entities, and then say selecting the car initially can only count as one outcome.

[@ty2358]

The reasoning with the cards really made it click with me haha

[@InnocenceExperience]

@UTJhn that just means he forgot to state all the rules. Anyway, you only switch if you're given the opportunity. its 1/3 if you stick or change without Monty opening a door. And its 2/3 if you switch with Monty opening a door, which you can only do if he opens a door and gives you the second chance anyway.

[@PhilipBrocoum]

To sum up, if you pick the first door and he opens the second, you should stay, and that happens 1/6 of the time. If you pick the first door and he opens the third, you should also stay, and that also happens 1/6 of the time. So, regardless of how you count the number of possibilities, staying wins exactly 1/6 + 1/6 = 1/3 of the time.

[@newcastlemusic]

It's nice of you to clarify the rules of the game retrospectively "in the interests of precision", but you can't expect someone to accurately calculate the probability of a particular outcome unless the rules are explicitly stated a priori. A contestant on this game show, being faced with the problem as posed in the video, cannot assume that Monty's choice of door was not random. Such potentially erroneous assumtions could over-simplify the problem by limiting the list of possible outcomes.

[@InnocenceExperience]

@UTJhn the probability of the opened door goes to the unchosen one because it is selected (purposely) from twice as many options, so the 2/3 of the time when you don't make the right choice first time 9and choose the goat), the other goat is eliminated. To put it differently, you're (twice) more likely to choose a goat the first time, so if a goat is removed, its (twice) more likely to be car if you switch. That's it. ...Yes, its only 2/3 if the rule is followed. If its not followed...

[@YesIamJames]

Awesome video, subbed. Apart from myself (I hope) you are the only person I've seen who has actually done a decent job of explaining this.

[@InnocenceExperience]

@InnocenceExperience I retract the sentence 'that could be a signal' because its irrelevant to the probability...if he only opens when you can win by switching then switching gives a 1 chance of winning. it doesn't matter whether you get the signal or not, although it would be handy if you did.

[@d4v3y01]

I still know people who say "thats rubbish...when monty opens 1 of the doors theres only 2 doors left so the probability MUST be 50/50". Its annoying that they just dont get it ,even with a good explanation

[@Bjowolf2]

Simply choosing between the two doors at random ( like you suggest - a 50 % chance of either door hiding the prize ) gives you a 1/2 * 1/3 + 1/2 * 2/3 = 1/2 probability of getting the prize. Therefore the strategy of always swapping has a (2/3) / (1/2) = 4/3 = ca. 1.33 advantage over the random choosing - or 33% better (on average! ). A considerable advantage that would pay off big time in the long run.

[@lavatrex6736]

plot twist: the car is all ruined, and the "goat" doors are actually goats riding in cars.

[@PhilipBrocoum]

When you split the initial choice into two, they now each have half the probability as the other choices. In other words, picking the first door (1/3) and having Monty Hall open the second door (1/2) has a probability of 1/6. However, picking the second door (1/3) forces Monty Hall to open the third door as his only option (1/1) and therefore has a probability of 1/3. So, while you could say that there are more possibilities, there are NOT more probabilities. They always add up to 100%.

[@smellen]

Thanks. That was the best explanation I've seen yet. In particular, I like how you showing that it comes down to choosing 1 door (1/3) versus choosing 2 doors (2/3). People want to ignore the preceding events, focusing only on the final event of choosing between two doors. Kinda like if you were flipping coins and the difference between "What are the odds of getting another 'tails'?" versus "What are the odds of getting 'tails'?" Oh god, I babble incoherently. I leave the teaching to you. :-)

[@PhilipBrocoum]

The analogy with picking a card is exact. If you switch, your odds of winning are 51/52. That is, the only way you can lose is if you *correctly picked* the ace of spades originally. The person showing you 50 cards **cannot change this fact**. If you don't switch, your odds of winning are 1/52. If you switch, the probability *must* be 51/52. see the description for credits.

[@InnocenceExperience]

@UTJhn thanks for your patience and help! I don't know what Bayes' law is as I've never studied probabilities...I just heard about this problem a couple of days ago and got interested in it. I think I'll leave it for now anyway...I'm satisfied with this. Have you studied this kind of thing much?

[@EliteFalco341]

The chart immediately made everything so much more clear.

[@InnocenceExperience]

@UTJhn Surely even if he opens a door randomly and its a goat, then its 2/3 to switch. That's because in that situation, the car is twice as likely to be in the switch option. Your first pick was twice as likely to be a goat than a car, as always. So if another goat is cancelled out, the switch is twice as likely to be the car. Alternatively put, the swtich option has been selected from twice as many options as stick and takes on the 2/3, whether Monty meant to eliminate the goat or not.

[@corujabuho8253]

I read two other explanations and watched two videos. Yours made it very clear! Thank you. This matters because I have to know this for my math course! You are a life saver!

[@SandBarAndYou]

@Stedwick Let's say the card scenario is changed slightly so that you pick a card initially but the other guy does not get to know which card you picked. Next he flips one card over and it's not the one and fortunately it's also not your card. The chance of your card being the one now becomes 1/51. It WOULD be at 1/52 had he known your initial pick and PROMISED not to flip it over(i.e original scenario). I know why this is true. Do you?

[@surawatthana]

Thanks for making the Monty Hall Problem clearer.

[@ZehManelCigano]

Ok, the thing is: -Suppose i have two aces and one king face down in my hand. -You pick one randomly. -The card you picked is probably (2/3) an ace. -Then i cross the other ace out. -At this point, there are only 2 cards left, but notice that the card you picked is still probably an ace. -Then,as this theory suggests, you switch your first pick to the other card i hold in my hand. ---That's the king.

[@InnocenceExperience]

@UTJhn if you have already made a wrong choice and he can stop you switching then you have 0 chance at that point but before both of your choices, you have 1/3 chance and for the first choice and if he opens a door with a goat then its 1/3 (stick) or 2/3 (change). Even with a host who wants you to lose, the worst he can do is stop you having the choice of switching and contain you to a 1/3 chance. I was assuming though, and I think you are supposed to, Monty has to open door n wants u to win.

[@Bjowolf2]

I was fooled too at first, even though I did take probability, bec. it's so very tempting to jump to the fifty-fitty assumption without thinking first. But after doing a few calculations I now see it clearly ;-) In the case (2 of 3), where there is a no-prize behind the orig. door, there HAS TO BE a car AND a no-prize behind the remaining 2 doors. So when the host reveals the empty dor, he is act. "telling" you where the car is ( in 2 of 3 cases!).

[@nickeaglesfield]

The fact that you get a choice (post elimination) means it is an independent event and what has happened previously is irrelevent. 50/50 chance. When this chappy was counting up his probabilities he took the probabilities from the beginning when in fact the probabilities should start from the 2nd choice. Its like getting the final 2 doors, shuffling them, then asking to choose. *independent event*

[@InnocenceExperience]

@UTJhn now, one other thing I can think of...what if he opens the second door every time but randomly? is it still 2/3 to switch? I'll have to think about it. I've always premised it on he wants me to win and thats how it makes sense to me.

[@AngryNerdBird]

Regardless of how circumstances change, the odds of success with the original door is 1/3. There is a 2/3 chance that the door with the prize is among the remaining two doors. REvealing one of the doors to be a wrong choice does not change the fact that the prize has a 2/3 chance of being behind the 2nd or 3rd door. Your chances of success are better by switching. It is NOT a simple 50/50 chance.

[@HumptyDumptyOakland]

The card example is exactly the same as the MHP just with more doors. If 50 cards that are not the A of Spades are intentionally removed from the residual pack of 51, then switching wins with a probability of 51/52 or about 98%. The behaviour doesn't have to be consistent just well-defined for the experiment being performed

[@InnocenceExperience]

@UTJhn yeah I think it stays 2/3 for switching even if he chose it randomly, as long as the door is already opened on the goat,right?

[@PhilipBrocoum]

I have another video about deal or no deal, but it's not explanatory. However, in the comments, I discuss in depth why deal or no deal works differently than the Monty Hall problem.

[@ShadowJ20]

@Gamebito Well if you don't change then the odds are 1/3. So it's not like it's impossible to get it right on the first pick. It is possible but less likely.

[@PhilipBrocoum]

If there's only three doors and two players, it is NOT always possible to open an empty door. Therefore, the game cannot be played. However, if the game is ONLY played if there is an empty door (i.e. only if one of the two contestants actually picks the door with the prize) and aborted otherwise, then the chance is 50-50 since it's symmetrical. Basically, you are never playing with three doors at all, just two.

[@GetMeThere1]

Good job. I ALWAYS need to find an intuitive grasp of such things, or else they mean nothing to me, and I can't really integrate them into my generalized thinking. When I first confronted this problem I mulled it over for quite awhile before coming up with the same idea as you present here (except I imagined a MILLION doors!). It's an interesting reminder that there are amazing truths all around us that aren't necessarily obvious.

[@SandBarAndYou]

Everything said before 4:10 was beautiful. You explained that the reason why you should gain no confidence in your initial pick is because the other guy is SUSPICIOUSLY leaving other cards face down. You are then correctly implying that there is nothing SUSPICIOUS about why he left YOURS face down. He HAS to leave yours alone simply to follow the game rules. Throughout the entire game you have no reason to EVER gain confidence in your pick. After 4:10 you go off the rails.

[@mgpo3]

No. Look at it this way. You pick door number 2. If the car is behind door number 2 Monty will open 1 or 3. If you switch your odds are still 2/3 that you are right, it just so happens that this is the 1/3 situation when you should have stuck with your original pick. If you pick 1, Monty can only 3 which provides the same circumstance, in which case if you switch to 2 you win. The only way to tilt the odds in his favour is if you pick 1, he opens 1 and then you have to guess 2 or 3.

[@PhilipBrocoum]

Yes, you could do it that way, but then not all possibilities have the same probabilities. You can ALWAYS split things up to create more possibilities. For example, I could say there are two possibilities when rolling a die, you can get an odd number or an even number. You could then contradict me and say that there are FOUR possibilities: you could roll a 1, 3, 5, or an even number. While true, it's misleading because some rolls have a probability of 1/6 and others have a probability of 1/2.

[@InnocenceExperience]

@UTJhn okay, I get it now. with rule: 1/3 you choose car, 2/3 you choose goat and he removes and goat leaving car. Random game: 1/3 you choose car, 1/3 you choose goat and he removes car, 1/3 you choose goat and he removes goat, leaving car in switch. In the game with rule, he is weighting it towards switch 2/3 of the time, whenever you don't choose car. Phew! I had to think hard to get that fully. I am relating to you as a teacher now.

[@PhilipBrocoum]

@AsbjornOlling I'm not sure, this is a pretty standard problem in mathematics circles, so anyone with a degree in mathematics should be very well familiar with this problem. The reason it's famous is not because mathematicians had trouble with it, but because the general public finds it so fascinating. Maybe your uncles never took probability? Maybe 50 years ago the puzzle wasn't famous so they never learned it? I don't know.

[@PhilipBrocoum]

@henryporter101 He opens a door and it's empty. What more do you want?

[@Gamebito]

What happens if the Car was behind door 2, because they knew you would choose door 3 for changing?

[@arnonuhm6922]

Amazingly well explained, the best explanation I have heard so far. Kudos to you!

[@1BMPP]

@THUMPBUMPDUDE there is technically no "2nd time", these two instances are not mutually exclusive and directly change the other

[@newcastlemusic]

At 8:05 in your video you state "there is a zero percent chance of the prize being behind the door that Monty Hall opens". Yet when you pose the problem at the beginning of the video, you say "I'm going to open up door number one, and show you that there is ... NOTHING BEHIND IT!". Note that you said what was behind the door AFTER you opened it. It is not clear at that point whether the choice of door was random, or whether you choose door number one knowing there would be nothing behind it

[@henryporter101]

I know one is twice as likely to win if one switches if they're in the game from the beginning.That's because the dispensing of one card gives information that hitherto wasn't available to the person in the game.Initially,the person who was making the selection had no information,so the chances of him/her getting it right were one in three.Similarly,if someone has no prior information concerning Monty's revelation of the dud card,to them,the probability is 50/50.It's a matter of perspective.

[@mocrg]

I'm starting to get it now, after 18 years of thinking about it.

[@SandBarAndYou]

@Johnnyradionic No, at that point, the probability of one contestant's door and the others is the same.

[@lejonzx]

My question is though, while statistically it is correct the prize is in the next door. What if you correctly chose the prize on the first try.

[@Lenoxuss]

@prateekcertain I'm curious: Is "luck" equivalent to a 50% probability, or what? Or… say someone played this game 3 million times and won all of them, would you be remotely surprised? If it's "all luck", then you shouldn't be; all outcomes should seem equally likely to you. Instead, in the real world, probability is what tells us what to expect in the long run.

[@SandBarAndYou]

The first choice you make (as in the one you may change your mind on later) DOES matter. That's because THAT will be the door that Monty Hall is locked out of. He can't choose that door. That is what "locks in" the probability of that door to an unchanging 1/3. That is the true explanation of this problem which is seldom successfully pointed out.

[@InnocenceExperience]

@UTJhn even if he opens a goat door by accident or from random selection, you are in the same situation, probability wise, as if he did it on purpose, right?

[@somedaythewave]

i feel like there is suppose to be 8 possibilities instead of 6, because when you choose the car on the initial choice there are 2 possible choices for a door to be taken away as oppose to initially choosing nothing and having only one possible door taken away. is this not correct? i imagine a goat and a rock instead of two nothings, to distinguish

[@SandBarAndYou]

Door 1's chance stays the same not because the chance of that door was 1/3 to begin with (probabilities potentially change with new knowledge). And door 1's chance stays the same not because you know he was avoiding your door and not because you know he was avoiding the prize but because of a combination of the two. The distribution 1/3 2/3 0 is created because the host was avoiding your pick AND the prize. Without that COMBO, we get 1/2 1/2 0.

[@djbentendo3929]

After one goat is removed, which is constant, you only have 4 different outcomes, 2 wins, 2 losses. You base the second question's probability off of the DIFFERENT outcomes of the first. So if G1 = x, G2 = y, C = z... X chosen: YZ ZY Y chosen: XZ ZX Z chosen: XZ YZ The second question asks at this point the probability of getting X/Y or Z. Only each unique outcome may be counted. That leaves XZ YZ ZX ZY The probability of swapping if the first variable was your choice is 2/4, or 50%.

[@noobsdumb]

ok i get what you're saying but if this was a game show and they knock out one door, by logic, if they were trying to get you to lose the game, that way they would make more money, they wouldn't offer the chance unless you have the right door right? By the problem, that right and i get it but it real life i think that by logic it's probably better to stay? I dont know..... i just am kinda confused

[@SoDarkTheCon]

When the host opens that door, its odds do go to zero, but it is only an illusion that the other door would then become a 66.66% chance winner, this is the human error in the equation. Realistically, the removal of any one door effects the odds of both other doors at the same time, and in this scenario, it is guaranteed that the removed door will be a wrong choice. Your choice now is not between the two other cards at all, but the 2 left. 50/50, wright or wrong.

[@Meades17]

It's all about percentages, the car could in fact be behind door #2 but it goes like this. three doors, 33% each 33% 33% 33% you chose door #2 with 33% and door number one is open, door number two still has a 33% chance but door number 3 now has a 66% chance, therefore you change doors.

[@SandBarAndYou]

If someone doesn't know which door the contestant initially chose but witnessed door 3 being "exposed," then to them the chances are correctly 1/2 1/2 0 (the chances of doors 2 and 3 together CHANGED!). Anyway, the best explanation of the 1/3 2/3 0 is this: The chances of door 1 remain 1/3 because there is no reason to "gain ANY confidence" in a door that was avoided by the host not because the prize may be there but simply because the host is not allowed to open it per the rules of the game.

[@IlGreven]

I've always thought this could be moved over to Deal Or No Deal, but does the fact that no one knows where the million is in that game change the parameters?

[@DanPLC]

This is an excellent explanation of the Monty Hall problem.

[@owenjones1746]

Excellent!! Now I really do understand! Thank you.

[@truthseekers666]

do the rules work in that the first door you choose cannot be a door Monty Hall chooses to open to show you. If so then this makes sense. If he could choose to open your first chosen door as his peek door then you have not improved your chances surely.

[@ZehManelCigano]

Correction: The king is the thing you want to choose, not the ace.

[@henryporter101]

Good video,however,when you say"Since he's telling you which door is empty,he might as well be giving you both doors,since you wont mistakenly walk away with the empty one ".No,Monty can only tell you that 'a' door is empty.If he gave information that revealed that two doors are empty,then one would be able to get the car every time,instead of 2 every 3 times.Great video,but I don't think the analogy holds.

[@djbentendo3929]

Also, the card question is ridiculous. The chance your card was an ace of spades on the first pull is 1/52. Once you eliminate 50 cards from the deck, regardless of reason, and then present a single card and say "NOW what's the chance that if you swap with the one I had, it's the Ace" means you're asking a completely new question on probability. You can't just factor behavior into math unless it's a consistent behavior.

[@IlGreven]

...but if he doesn't know where the car is, he could show you the car before you get the chance to switch. Thus, his choice of door is independent of where the car is. If he knows, and is forced to show you a door with nothing, his choice depends on where the car is. This changes the game significantly.

[@R2MintOptions]

I explained this in two other similar videos, but I failed to mention that in a real life game show, even knowing the probability, I'd stay with my 1st choice simply b/c 1) it's a game show and I'm not likely to be on it again. 2) 66.6% is still not 100%, i.e. if I switched and lose I'd be more disappointed then if I stayed and lose. But if it's over multiple trials, definitely switch b/c 66.6% is always better than 33.3% chance of winning.

[@Bjowolf2]

Because the goat ( or no-prize ) will not appear equally often (on average!) behind the original door and the remaining 3rd door. Try it with a deck of cards - or use a die to simulate it ( 1 - 4 means a goat / no-prize behind the original door ( p(goat) = 4/6 = 2/3 - 5 & 6 means a car ( p(car) = 2/6 = 1/3 ). Always doing the swap is - ON AVERAGE! - twice as good as always staying with your original choice ( (2/3) / (1/3) = 2 ! - It is NOT 1 to 1 (or 50 /50 )!.

[@PhilipBrocoum]

@SandBarAndYou Well, the odds do change… with more information, it becomes apparent that switching is better, which wasn't apparent before. Please see my extended description on this video by clicking "show more" above for more details.

[@PhilipBrocoum]

If you don't switch, the only way you can *win* is if the car is behind your door. If you switch, the only way you can *lose* is if the car is behind your door. There is a 1/3 probability that the car is behind any particular door, so switching *doubles* your odds of winning. This is a plain fact.

[@ABlackbirdOnline]

another way of making it clear is: let's imagine there are 100 doors, with only one prize. you choose one, and then i open 98 doors to reveal there's nothing behind them. even though the logic is the same, it seems much more obvious in this example that the car is more likely to be behind the other door. This doesn't exactly explain how it works, but it's useful for proving the theory to someone who disagrees with it.

[@InnocenceExperience]

@UTJhn amazing! you'll only win by switching 1/3 of x amount of games, if he chooses at random. But, couldn't you also say that 2/3 of the games where he chooses at random AND its a goat, you'll win by switching? That's like a subset of the x amount above (1/2 of x), which is identical to him choosing the goat on purpose. The part (1/2) of the 2/3 (1/3) is restored to you if you only count those games. So when you are confronted with an open goat door, its 2/3 to switch in that in instance...

[@RunSushiRun]

So, let me see if I get this right. There's no guarantee you'll get the prize, it's only statistically where the prize should be?

[@henryporter101]

@Stedwick An explanation of Bayesian probability would be good.

[@harshithb]

Beautiful! I FINALLY managed to understand the Monty Hall Problem!

[@Kaufmoe]

First video on this that made me understand why switching is the right choice :-)

[@superyahoobrothers]

Well its not

[@jadenkhor]

But what if the right door was number two despite the whole 2/3 chance of door 1 and 3?

[@IlGreven]

So, if you get to the last two cases of "Deal or No Deal", $1,000,000 v. a penny, you should always switch cases? Knowing where the car is is a BIG deal with this problem. If Monty can show you a car at any time, then switching returns to a 50/50 proposition. Of course, half of Monty's schtick is to fool you into thinking he doesn't know where the car is.

[@henryporter101]

What is the 'it' that you're referring to?If someone has two doors to choose between,and they have no information regarding what's behind either door,then the probability of them choosing the right door-the one with the prize,is 50/50.I wont say that you fail at Bayesian probability-I'll assert that you are unfamiliar with it.

[@PhilipBrocoum]

In my exact words from the beginning of the video where I explain the rules, "The host of the show, Monty Hall, then opens up one of the other two doors to show you that it's empty." It doesn't get much more clear than that. Maybe I could have been more emphatic about it, but I also stated multiple times throughout the video that Monty Hall was cheating. I do not believe that anybody watching my video will come away confused on this issue.

[@superyahoobrothers]

Wrong

[@siowliang101]

Let's put it into 3 conditions: 1) the 1st box is car and you choose it, after host reveal the box 3 and you swap with box 2, then you get no car. 2) the 1st box is not a car and you choose it, after host reveal the box 3 and you swap with box 2, then you get the car. 3) the 1st box is not a car and you choose it, after host reveal the box 2 and you swap with box 3, then you get the car. Hence, out of 3 conditions you have 2 chances to get a car, so better to swap =)

[@SandBarAndYou]

Probability distributions change with knowledge. For example, if door 3 is revealed to be empty, then its probability drops from 1/3 to 0 and the probability of door 2 increases from 1/3 to 2/3. Yes, this means the chances of doors 2 and 3 COLLECTIVELY stay at 2/3. But look at door 2 alone. Its probability CHANGED. Hence my point. It is dumb to say "the chances were THIS therefore they should still be THIS" without fully explaining why. Their math is right but their explanation is missing.

[@InnocenceExperience]

@UTJhn haha I deleted my last post because I understand what you've been saying about the importance of the rules! it does matter whether he had to open the door or not and whether he wants you to win/only opens if you will win, because he could be only opening doors when switching is the winning thing to do. That could be a signal. You are very clever. Thanks for improving my understanding of it.

[@starfox300]

This doesn't really convince me somehow. Have people tested this?

[@willoughbykrenzteinburg]

The probabilities do not change really. The easiest way to think about it is that you have a 1/3 chance that you picked a car to begin with and a 2/3 chance you did NOT pick a car to begin with. Everything else is irrelevant. Those odds do not change. When Monty reveals a goat (which he will always do), nothing changed. You still have a 1/3 chance of having a car, and a 2/3 chance of NOT having a car. Monty just reduces the number of doors that represent that 2/3 chance.

[@SandBarAndYou]

The host must pick a door that... RULE 1) ...is empty RULE 2) ...is not the contestant's first pick Door 1: When you notice the host doesn't pick your first pick, don't think ANYTHING of it. Don't gain or lose confidence in door 1. The host is simply following RULE 2. Chance: 1/3 Door 3: The host reveals it is empty. Chance: 0 Door 2: By elimination, the chance is 2/3. Your confidence in this door has increased. And why not? After all, the host avoided this door and it ain't because of RULE 2.

[@nactan]

@JustinDejong Uh... no. Monty Hall would never open the door concealing the car. Remember, he knows where it is, and opening that door will ruin the game.

[@sammysamster2]

hi, im doing my math hw, and i had to look up lots of different ways of solving the prob, and i just wanted to say, that i thought urs was the best gd job

[@JeeJeez1]

Whichever door you pick, Monty will choose one he knows isn't a car and open it. So the first choice doesn't really matter at all. The 2nd choice is between two possibilities, two doors. So 50/50 chance to get the right one.

[@Rentphotographer]

It means that Monty Hall works 2/3 of the time. But since 2/3 beats 1/3, always go for it.

[@AngryNerdBird]

Incorrect. See, your initial choice has only a 1/3 chance of success, and there is a 2/3 chance that the door with the prize remains behind one of the other two. However, thanks to the host, the "remaining doors" is now just a single door. However, there is guaranteed to be at least ONE wrong door among those two, so there is still a 2/3 chance that one of those two doors included the prize, and therefore if you switch to the remaining door you have a 2/3 chance at success.

[@Optimus6128]

It's funny how people are still arguing about this problem. For me, what makes the difference is that you make a first selection before letting the host choose another door from the two remaining to switch. Now, if this didn't happened, but the host removed on door with a goat and let you choose from the remaining two then it would be 50%. The first thing that persuated me about 66% is a computer programm. It really did around 66%. This is science. What remains then is to find why this happens.

[@MrMZaccone]

Why wouldn't it be 50 out of 52 times leaving the 1 out of 52 chance still in your hand?

[@InnocenceExperience]

@UTJhn …1/3 of the time, you lose as soon as the random choice is the car…if its not the car, then the car could be behind either door equally, giving 50/50. That 1/3 where you lose goes over to the unchosen door with the rule because the host is purposely avoiding the 1/3 where he puts the car out of the game. He’s sort of swinging it in your favour. cont...

[@SandBarAndYou]

The whole point of conditional probability is the understanding that knowledge of an event changes the probability distribution of the event. Therefore to use an argument that says "the probability that THESE doors had the prize at the start was X and so it should STILL be X" is not explaining anything. This argument is refusing to allow the distribution to CHANGE with knowledge. And it CAN change with knowledge.

[@bendswny]

You knew one of the other doors was empty but you didn't know which one