What if there were TWO contestants and each of these two contestants were ardent mathematicians, both with an innate understanding of the principle behind the Hall paradox. After each choose a door, and the third unchosen door was opened to reveal a zonk, would each of the two players, if given the chance, leap at the opportunity to accept the other fellow's door? (And in this scenario the 2 scholars are rivals who would not agree to sell the car and split the proceeds) Hmm?
@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
I thought you were absolutely wrong until you got to the part about 'my hand' vs 'your hand'.. then it clicked and I understand now. Great explanation!
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.
@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.
@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?
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.
You had it right in the first part of the video then got it wrong afterwards. Allow me to show you why your second reasoning doesn't work because it actually would conclude that it is sometimes better to switch and sometimes better to stick: You initially choose door 2. Door 1 is opened and revealed to be empty. Well, the probability that the prize was in EITHER door 2 or door 1 was 2/3. So that should still be true. Door 2 should now have a 2/3 probability. I should stick this time. Huh!?
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.
It seams to me like the perfect gambling oppertunity especially knowing that most people make the wrong choice and stay with there first choice. To the 14 year old kid why not put a few bucks on picking the right door.
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!
I am only 14 years old, and I have two uncles (<50yrs) who studies mathematics at the university of Copenhagen. So they think they're pretty damn good at stuff like this.
About a year ago, I discovered the Monty Hall problem, and told them about, thinking they might find it interesting. A year later, I am still trying to prove to them how it works, and they refuse to believe it.
How do I show these old ignorant fools that I am right?
@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.
@AsbjornOlling Your uncles aren't fools.For someone who didn't have prior information(unlike the person who 'initially' had a choice between three doors until Monty revealed that one door contained a dud)the Mathematical odds would be 50/50.That is,if someone came along without knowing that Monty had dispensed with one door,the probability of that person picking the right door would be 50/50.Probability requires an agent whereas mathematics doesn't-I suspect that's where they're going wrong.
in round one, the percentage is a picking percentage, not winning, and in round 2 is a winning percentage, so how can you add both different types of percentages and say you have 66%? it doesnt make any sense, its like saying
@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.
I like your explanation, and I liked the deck cards example which you used to show us why it is like that. It's really good and it helped me understand everything. The problem for those who don't understand why is that there are only 3 doors, so it's hard to grasp. It's much easier to see it when you watch the cards example.
The Monty Hall problem has fooled many very intelligent people, but has not fooled the average person. The game is 2 games. The first one involves a 1/3 shot and the second a 2/1 shot. Assume I go on the show with my brother. He makes the first choice of door, say #1. Monty then opens door #2. I now appear and see 2 closed doors: #1 and #3. What are my chances of choosing the correct door? Yes, that’s right 2/1.
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
If you're still stuck on 50/50 as an answer, take a look at my video response "Proving The Monty Hall Problem." It'll give you a way to prove to *yourself* just what the correct answer is.
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.
This only works assuming that Monty Hall deliberately avoids turning up the car. If he chooses randomly, then there is a 1/3 chance he'll turn up the car, a 1/3 he'll leave the car face down, and a 1/3 chance you'll have the car face down. So if he didn't turn up the car, your chances would be even either way.
If he avoids turning up the car, then there is a 1/3 chance you'll have the car face down, and a 2/3 chance he'll leave the car face down.
It makes perfect sense. If you still don't get it, pay attention to the card game and think of the 3 doors as 3 cards. Have a friend set up the game with three cards and record the results of switching ten times and not smitching 10 times.
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.
You make the same mistake as Marilyn Vos Savant (a very rare thing for her). You assume the host will always give a second chance while stating that only in your answer and not in the problem. If you never switch, you have 1/3 chance of winning each game. If you switch at every opportunity and the host can choose not to give a second chance, then you win or lose at the host's sole discretion.
@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.
@InnocenceExperience Yes, it just means he forgot to state all the rules (or doesn't know them). But that rule is quite important. And your misunderstanding of the problem is the reason it is so important to make that rule clear. If the host wants you to lose and can choose when to let you switch, you do not get 2/3 probability by switching, you get 0. Unless you had won with your first choice, he would simply say "sorry you lose", and not allow you to switch.
@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.
@InnocenceExperience I'm a bit confused. you seem say the odds double by switching regardless of the rule or Monty's efforts (which is wrong, but Hollywood made the same mistake while making "21" so don't feel too bad). Then you say you are assuming he has to open and wants you to win. If the assumptions aren't necessary, why make them? If he has to open, then the odds double by switching. If he doesn't , but wants you to win, your probability by staying is 0 and goes to 1 if you switch.
@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.
If you stay, you win 1/3 of the time regardless the host's actions or intentions. If you switch, the host decides whether you win or lose. if he has to show a goat your probability is 2/3. If he picks another door randomly, it is 1/3 overall, and 1/2 when you get to switch. If he does as he pleases and he wants you to win, you always win, but if he wants you to lose, you always lose.
@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.
@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.
@InnocenceExperience You're welcome, and thanks for listening (reading). You should congratulate yourself on keeping an open mind and learning: that seems to be a rare quality.
@InnocenceExperience If the host picks another door randomly, then what is behind that door gives new information and new probabilities. Before he opens the door, the probability to win is 1/3 (whether you always switch or always stay). After he opens the door, if it was the prize, you lose (p to win=0) and you don't get to switch. If it's a goat, at this point the probability to win becomes 1/2 if you switch (and is also 1/2 if you stay).
@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.
@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...
@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?
@InnocenceExperience Ask yourself why the probability of the opened door goes to the unchosen door rather than the chosen one. When you correctly answer this, you should understand why it's only true if the rule is followed (actually, it is true as long as the host 's decision to open one of the other doors and show a goat is in no way influenced by the contestants first choice - but that is just adding more complication and possibly confusion).
@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...
@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...
@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.
Another way to explain why the probability switches to the unchosen door if the rule is followed is to look at the information gained. That information is only which of the other doors is opened which is not influenced by what is behind the chosen door and hence cannot change that probability (a result of Bayes' law). If the doors are chosen randomly, the goat is the new info, and that is equaly affected by both closed doors.
@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?
For people old enough to remember the show, Monty Hall did not ALWAYS offer the person a chance to switch. So anybody remembering the show rather than a made up ruleset gets fooled. This riddle really should be called something else.
@lejonzx to answer your queastion, you lose. but that will only happen 1/3 of the time. the other 2/3 of the times you win the car. don't forget we are talking about the long term probability over playing the game many times. not a single time. and since you don't know you lost the car until AFTER the game is over, you play the strategy that is guaranteed to win you a car 2/3 of the times instead of the strategy that wins you the car 1/3 of the times.
@lejonzx to answer your question, you lose. but that will only happen 1/3 of the time. the other 2/3 of the times you win the car. don't forget we are talking about the long term probability over playing the game many times. not a single time. and since you don't know you lost the car until AFTER the game is over, you play the strategy that is guaranteed to win you a car 2/3 of the times instead of the strategy that wins you the car 1/3 of the times.
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
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.
great job! You sure do know your stuff! I know how it works but I decided to watch it anyway, and now i know EXACTLY how to explain it if anyone asks me this problem. What's the probability that I subscribe to you? Without this vid, 50%. With this vid, 100%!
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.
@truthseekers666 That wouldn't make any sense. If he opens up your door and the prize is behind it, well, you've lost. If he opens up your door and the prize isn't behind it, switch, duh, and your chances increase from 0% to 50%.
I think I follow the logic but I wonder if anyone has tried this as an experiment. Has anyone actually tried this, let's say, a thousand time to prove or disprove that it works? I suppose I could give it a try IN MY SPARE TIME.
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.
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.
WOW!!!!!! thank u sooooo much!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! this helped me a lot and i greatly appreciate u taking your time 2 make this video!! this problem was referenced in a book (the curious incident of the dog in the night-time) which i had 2 read 4 school...and i was very confused but not anymore!!
This problem is flawed because it attempts to seamlessly social assumptions and mathematical ones, how is this not a problem that makes Monty Hall flawed from the get go?
I learned about the Monty Hall problem for years, but I never actually understood it on a meaningful level because I had always figured "What if you picked the right door, and Hall was just bluffing?" But after watching just half of this video, I feel like the answwer was so obvious, and that switching would be a matter of common sense. Thank you for explaining the problem in such a simple yet enlightening manner.
@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.
Or without bothering to add up all scenerios. In the first choice you have two thirds chance of being wrong. When Monty chooses he does not choose randomly. He will not choose your door and he will not choose the right door. Everytime that you are wrong he will leave the right door unchosen. You are wrong 2/3 of the time. So if you always switch you will win 2/3 of the time.
What is the probability of someone keeping or choosing is the real question here. Answer that then we can apply this. I'm probably the only rational person to oppose this scenerio but so be it.
you will have to divide by 2 to find an "unbiased" probability, independent of what you decide on as your set. Meaning if you have your set 1, and in set 2 one of the doors is guaranteed to be wrong, so for finding the "additional" probability you have to add, you instead define one of the set 2 doors as your new set 1, but once these probabilities are added up, it means you still have a 50% chance.
We'll say the doors are A, B, and C. A is the door with the prize. The scenarios are different only in which door you choose, so there's three of them:
You choose A. B is shown to be empty. Switching to C loses.
You choose B. C is shown to be empty. Switching to A wins.
You choose C. B is shown to be empty. Switching to A wins.
Alternatively, imagine the exact same scenario, but with more doors. Let's say 1000. Your odds of choosing right at first are 1/1000. 998 doors are opened and shown to be empty. You're left with one wrong and one right. The odds that your first choice was correct were 1/1000, and therefore the odds that the other choice is correct are 1-(1/1000), or 999/1000. Switching is almost guaranteed to win.
The number of doors is X. The odds of a switch winning are (X-1)/X.
However, you can define these sets regardless as you'd like (you can pick door 1, 2, or 3 and that will become your set 1).
If I choose door 1, and door 2 is eliminated, then set 2 contains door 2 + 3. But isn't this randomly biased to whatever you decide as your set 1, and shouldn't therefore it's probability with respect to set 2 be considered? Meaning you add the probabilities of the two sets.
Okay, if someone can explain to me why my method of thinking about this is wrong, I'd be happy. While I agree the maths in this is correct, I still feel it's only 50%. The reason for that being that you make the choice to split this into two sets.
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
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.
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%.
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.
You're amazing. I am a smart cookie who has never been able to grasp this problem, because of how it was explained and how I was trying to understand it. You could not have done a better job covering all the bases of potential false assumptions and confusions. Thank you!!!
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. :-)
In response to naysayers, yes, Monty Hall knows what's behind the doors. I don't think there is anyone in the world who assumes that Monty Hall is opening doors at random; that seems like a really silly way to do the riddle.
Still, in the interests of precision, let it be known that Monty Hall is cheating.
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.
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.
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.
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.
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.
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
Frankly, I think anyone who doesn't get it after watching this video just isn't trying. This is my far the best explanation of the Monty Hall Problem I've ever seen. Great Job.
@scimike22 - I'm still under the impression that Montey Hall ALWAYS opens a goat door after you make your first choice, and that you ALWAYS get to switch if you like. If this is true, and you as the player ALWAYS know this going in, then there are never 3 doors.
If you know that Montey will ALWAYS negate one door no matter what, then the odds are 50 / 50, as there are ALWAYS two doors to choose from, never three doors.
This comment has received too many negative votesshow
You're trying to carry over the 1/3rd probability to a situation with only 2 CHOICES!
Your card-game analogy was disingenuous at best. The only way the card game could work in the case of having 2 doors to choose from was if you only flipped over 25 cards, leaving 26 still face down.
It's a logic problem you're trying to figure out with pure math.
There never ever were 3 doors as one would always be eliminated.
If we played with 1/2 a deck and the ace WAS! still in there, you have1/26NOT1/52
Two choices are not necessarily equivalent merely because there are two of them. Two choices:
a) You are the first guy to disprove an interesting mathematical quirk that actual mathematicians don't consider as up for debate
b) You are just another guy who doesn't understand something simple and interesting.
Are these two choices equally likely? Of course not. One is about 99% and the other is 1%. I won't say which is which, because I'd hate to hurt your feelings.
I've watched this twice now, but I still don't see how the probability hasn't become 50/50. You're factoring in all three doors, when the probability hasn't changed. 3 doors, 2 are wrong. When Hall removed one door you have 2 and can switch. The fact you chose a door before he removed one is IRRELEVANT to the situation as you can now pick between one of two. One door was NEVER GOING TO WIN since Hall would always remove it. You're simply looking at the problem wrong. There are only 2viable drs
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*
Shuffling them would introduce another variable, and you'd be right. However, the problem as stated stipulates that Money Hall does not open a door randomly- he will never accidentally open the door with the prize.
If you think this video is inaccurate, I invite you to test it yourself empirically. It's been done before, and switching ends up with a 66% success rate.
What most people forgot is to take into account the information given by the preceding events. Take this example : "Here are two doors. Prize is behind door number 2. Which door will you choose ?" Are you going to say "Oh it's just 50/50 chance" ? Of course not, because you have more information that merely a random throw. Here it is exactly the same, the information is just not as plainly obvious. The host can give you this bit of information, because *he* knows where the prize is.
This comment has received too many negative votesshow
You're making two assumptions though. First that the host is purposefully leading or misleading you, and second that he is not constricted by the rules of the game. If you choose correctly or incorrectly he is always going to remove one door. Once you realize the first choice and being shown three doors is simply a farce to make you over-think the situation it becomes blindingly simple.
It IS 50/50 because you ARE only given two choices with your second guess.
Watch the video again. Maybe a few times. I know it's counter-intuitive at first, but at some point you should realize it's true. It wouldn't hurt to read an entry-level mathematics text on how to compute the probability of event A knowing the outcome of event B.
What if there were TWO contestants and each of these two contestants were ardent mathematicians, both with an innate understanding of the principle behind the Hall paradox. After each choose a door, and the third unchosen door was opened to reveal a zonk, would each of the two players, if given the chance, leap at the opportunity to accept the other fellow's door? (And in this scenario the 2 scholars are rivals who would not agree to sell the car and split the proceeds) Hmm?
~ Johnny Radionic ™
Johnnyradionic 2 weeks ago
@Johnnyradionic No, at that point, the probability of one contestant's door and the others is the same.
SandBarAndYou 2 weeks ago
@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
henryporter101 4 days ago
I thought you were absolutely wrong until you got to the part about 'my hand' vs 'your hand'.. then it clicked and I understand now. Great explanation!
CapumOfOuttaSpace 3 weeks ago
12 people need to learn math
REDTEAM22003 4 weeks ago
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.
SandBarAndYou 1 month ago
@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.
Stedwick 1 month ago
@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?
SandBarAndYou 1 month ago
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299KPS 2 weeks ago
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.
SandBarAndYou 1 month ago
You had it right in the first part of the video then got it wrong afterwards. Allow me to show you why your second reasoning doesn't work because it actually would conclude that it is sometimes better to switch and sometimes better to stick: You initially choose door 2. Door 1 is opened and revealed to be empty. Well, the probability that the prize was in EITHER door 2 or door 1 was 2/3. So that should still be true. Door 2 should now have a 2/3 probability. I should stick this time. Huh!?
SandBarAndYou 1 month ago
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henryporter101 1 month ago
Best explanation I've seen so far; could have saved me a few hours of speculation back when I didn't get it :)
stenblank 1 month ago
Draw a tree diagram. One branch says might get car. another branch says might not get car. that ways it shows why u should switch. lol.
Desmonddd2002 2 months ago
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.
R2Mintus 2 months ago
It seams to me like the perfect gambling oppertunity especially knowing that most people make the wrong choice and stay with there first choice. To the 14 year old kid why not put a few bucks on picking the right door.
Chrisbaker525 2 months ago
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!
Birkirrey 2 months ago
I am only 14 years old, and I have two uncles (<50yrs) who studies mathematics at the university of Copenhagen. So they think they're pretty damn good at stuff like this.
About a year ago, I discovered the Monty Hall problem, and told them about, thinking they might find it interesting. A year later, I am still trying to prove to them how it works, and they refuse to believe it.
How do I show these old ignorant fools that I am right?
AsbjornOlling 2 months ago
@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.
Stedwick 2 months ago
@AsbjornOlling Your uncles aren't fools.For someone who didn't have prior information(unlike the person who 'initially' had a choice between three doors until Monty revealed that one door contained a dud)the Mathematical odds would be 50/50.That is,if someone came along without knowing that Monty had dispensed with one door,the probability of that person picking the right door would be 50/50.Probability requires an agent whereas mathematics doesn't-I suspect that's where they're going wrong.
henryporter101 1 month ago
Beautiful! I FINALLY managed to understand the Monty Hall Problem!
harshithb 2 months ago
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in round one, the percentage is a picking percentage, not winning, and in round 2 is a winning percentage, so how can you add both different types of percentages and say you have 66%? it doesnt make any sense, its like saying
x + y = 2x, where x is 33%, it doesnt make sense.
ImageGround 3 months ago
WHERE IS MY WON CAR!?
osheetidied 4 months ago
this is the perfect explanation, and this was solved in my class, but probabilty doesnt win you cars, luck does.
prateekcertain 4 months ago
@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.
Lenoxuss 4 months ago
I like your explanation, and I liked the deck cards example which you used to show us why it is like that. It's really good and it helped me understand everything. The problem for those who don't understand why is that there are only 3 doors, so it's hard to grasp. It's much easier to see it when you watch the cards example.
mediantiba 4 months ago
The Monty Hall problem has fooled many very intelligent people, but has not fooled the average person. The game is 2 games. The first one involves a 1/3 shot and the second a 2/1 shot. Assume I go on the show with my brother. He makes the first choice of door, say #1. Monty then opens door #2. I now appear and see 2 closed doors: #1 and #3. What are my chances of choosing the correct door? Yes, that’s right 2/1.
Bathfire1 4 months ago
don't understand this, no matter what....the second time around you are still chosing between 2 items....so a 50/50 chance
THUMPBUMPDUDE 4 months ago
@THUMPBUMPDUDE there is technically no "2nd time", these two instances are not mutually exclusive and directly change the other
1BMPP 3 months ago
Sorry, your explenatin is WRONG!!!
sami0elias 5 months ago
I'm a philosophy student and we do this in a class called Theory of the Rational Choice. :)
milicalex 5 months ago
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milicalex 5 months ago
I HAD 3 HAHAHAHAHA AWESOME
ThatAwesomeUsername 5 months ago
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
d4v3y01 6 months ago
saw this shit on 21
charbroiledmonk1033 7 months ago
isn´t it 50/51 chance in the example? because one card was already taken away
DerInsahne 7 months ago
DAYUM this was a great explanation. great monotone voice but easy to understand thnx
XxMaplerzxX 7 months ago 8
This has been flagged as spam show
If you're still stuck on 50/50 as an answer, take a look at my video response "Proving The Monty Hall Problem." It'll give you a way to prove to *yourself* just what the correct answer is.
kirkbocek 7 months ago
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lets try this way. you walk in and theres 3 doors but one is already open with a goat and monty says pick one of the 3. now what are the odds?
exactly.
jubileeshine 8 months ago
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.
henryporter101 8 months ago
@henryporter101 He opens a door and it's empty. What more do you want?
Stedwick 8 months ago
@Stedwick An explanation of Bayesian probability would be good.
henryporter101 8 months ago
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This only works assuming that Monty Hall deliberately avoids turning up the car. If he chooses randomly, then there is a 1/3 chance he'll turn up the car, a 1/3 he'll leave the car face down, and a 1/3 chance you'll have the car face down. So if he didn't turn up the car, your chances would be even either way.
If he avoids turning up the car, then there is a 1/3 chance you'll have the car face down, and a 2/3 chance he'll leave the car face down.
MindinViolet 8 months ago
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MindinViolet 8 months ago
well game shows must of totally wanted to sue his ass
shealdinho 8 months ago
i finally get it
iceymanjack 8 months ago 2
It makes perfect sense. If you still don't get it, pay attention to the card game and think of the 3 doors as 3 cards. Have a friend set up the game with three cards and record the results of switching ten times and not smitching 10 times.
john17972 9 months ago 2
I think you just wrote my maths assignment
garymofoak 9 months ago 2
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.
Optimus6128 11 months ago
so you raise the possibility of being right by about 17%. not a big number, but better odds none the less.
tomitstube 11 months ago
Yessssssss!!!!!!!!!!!! Next time somebody asks me to switch doors I'm going to laugh in his face and run off into the wilderness.
newsense87 11 months ago
Great explanation!! :)
TheToeBrand 11 months ago
You make the same mistake as Marilyn Vos Savant (a very rare thing for her). You assume the host will always give a second chance while stating that only in your answer and not in the problem. If you never switch, you have 1/3 chance of winning each game. If you switch at every opportunity and the host can choose not to give a second chance, then you win or lose at the host's sole discretion.
UTJhn 1 year ago
@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.
InnocenceExperience 9 months ago
@InnocenceExperience Yes, it just means he forgot to state all the rules (or doesn't know them). But that rule is quite important. And your misunderstanding of the problem is the reason it is so important to make that rule clear. If the host wants you to lose and can choose when to let you switch, you do not get 2/3 probability by switching, you get 0. Unless you had won with your first choice, he would simply say "sorry you lose", and not allow you to switch.
UTJhn 9 months ago
@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.
InnocenceExperience 9 months ago
@InnocenceExperience I'm a bit confused. you seem say the odds double by switching regardless of the rule or Monty's efforts (which is wrong, but Hollywood made the same mistake while making "21" so don't feel too bad). Then you say you are assuming he has to open and wants you to win. If the assumptions aren't necessary, why make them? If he has to open, then the odds double by switching. If he doesn't , but wants you to win, your probability by staying is 0 and goes to 1 if you switch.
UTJhn 9 months ago
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InnocenceExperience 9 months ago
@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.
InnocenceExperience 9 months ago
@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?
InnocenceExperience 9 months ago
@InnocenceExperience Sorry, but no.
If you stay, you win 1/3 of the time regardless the host's actions or intentions. If you switch, the host decides whether you win or lose. if he has to show a goat your probability is 2/3. If he picks another door randomly, it is 1/3 overall, and 1/2 when you get to switch. If he does as he pleases and he wants you to win, you always win, but if he wants you to lose, you always lose.
UTJhn 9 months ago
@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.
InnocenceExperience 9 months ago
@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.
InnocenceExperience 9 months ago
@InnocenceExperience You're welcome, and thanks for listening (reading). You should congratulate yourself on keeping an open mind and learning: that seems to be a rare quality.
UTJhn 9 months ago
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@UTJhn I don't get this sentence: 'If he picks another door randomly, it is 1/3 overall, and 1/2 when you get to switch.' Can you explain please?
InnocenceExperience 9 months ago
@InnocenceExperience If the host picks another door randomly, then what is behind that door gives new information and new probabilities. Before he opens the door, the probability to win is 1/3 (whether you always switch or always stay). After he opens the door, if it was the prize, you lose (p to win=0) and you don't get to switch. If it's a goat, at this point the probability to win becomes 1/2 if you switch (and is also 1/2 if you stay).
UTJhn 9 months ago
@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.
InnocenceExperience 9 months ago
@InnocenceExperience Sorry again. There are 3 possibilities with probability 1/3 for each.
p=1/3 your first choice is the prize (you lose by switching).
p=1/3 the host opens the door with the prize (you lose without getting the chance to switch).
That leaves only p=1/3 that you will win by switching (which is the same as if you stay).
You forget that part of your 2/3 is used up when the prize door is opened.
UTJhn 9 months ago
@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...
InnocenceExperience 9 months ago
@UTJhn ....even though you'll only win that way 1/3 of the time overall (because 1/2 of the time you'll never get that far).
InnocenceExperience 9 months ago
@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?
InnocenceExperience 9 months ago
@InnocenceExperience Ask yourself why the probability of the opened door goes to the unchosen door rather than the chosen one. When you correctly answer this, you should understand why it's only true if the rule is followed (actually, it is true as long as the host 's decision to open one of the other doors and show a goat is in no way influenced by the contestants first choice - but that is just adding more complication and possibly confusion).
UTJhn 9 months ago
@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...
InnocenceExperience 9 months ago
@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...
InnocenceExperience 9 months ago
@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.
InnocenceExperience 9 months ago
@InnocenceExperience Congrats! it seems you truely have it this time.
Another way to explain why the probability switches to the unchosen door if the rule is followed is to look at the information gained. That information is only which of the other doors is opened which is not influenced by what is behind the chosen door and hence cannot change that probability (a result of Bayes' law). If the doors are chosen randomly, the goat is the new info, and that is equaly affected by both closed doors.
UTJhn 9 months ago
@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?
InnocenceExperience 9 months ago
@InnocenceExperience You're welcome.
Depends what you mean by "studied". Formal training in science (phys/chem, ended ~ 20yrs ago), and I like and work various types of puzzles.
You might like to google "DDWFTTW" and check out Coolaun's YouTube videos for another brain twister.
UTJhn 9 months ago
this guy's voice is soooo gay
theazn07 1 year ago
For people old enough to remember the show, Monty Hall did not ALWAYS offer the person a chance to switch. So anybody remembering the show rather than a made up ruleset gets fooled. This riddle really should be called something else.
Sebadiah23 1 year ago
It pisses me off when people still think its 50%.
chad301894 1 year ago 16
@chad301894
bleach4711 5 months ago
This has been flagged as spam show
ur videos suck
0099132646 1 year ago
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.
YesIamJames 1 year ago
Where the fuck is my Porshe?
oogrooq 1 year ago
VOICE CRACKS EVERYWHERE!!!!!! AHHHH!!!!
PSDGaming 1 year ago 2
BS
alexmigda 1 year ago
21!!!!!! movie
555darkman 1 year ago
thx for the vid, never got it before
kirador218 1 year ago
Sheldon get out of youtube go back to big bang theory ó.õ
TheJuancarlosbraxD 1 year ago
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.
lejonzx 1 year ago
@lejonzx to answer your queastion, you lose. but that will only happen 1/3 of the time. the other 2/3 of the times you win the car. don't forget we are talking about the long term probability over playing the game many times. not a single time. and since you don't know you lost the car until AFTER the game is over, you play the strategy that is guaranteed to win you a car 2/3 of the times instead of the strategy that wins you the car 1/3 of the times.
webmail111 1 year ago
@lejonzx to answer your question, you lose. but that will only happen 1/3 of the time. the other 2/3 of the times you win the car. don't forget we are talking about the long term probability over playing the game many times. not a single time. and since you don't know you lost the car until AFTER the game is over, you play the strategy that is guaranteed to win you a car 2/3 of the times instead of the strategy that wins you the car 1/3 of the times.
webmail111 1 year ago
@webmail111 Yea, i get you.
lejonzx 1 year ago
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
noobsdumb 1 year ago
i'll switch, but i'll still lose >.<
hippymobster 1 year ago
1:08 to 1:15 two door voice cracks HAHAHA sexy
bubblesofhell 1 year ago
i dunno how i got to this video but it was really interesting i actually watched the whole thing
thegrapebeyond 1 year ago
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.
ABlackbirdOnline 1 year ago
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?
RunSushiRun 1 year ago 3
great job! You sure do know your stuff! I know how it works but I decided to watch it anyway, and now i know EXACTLY how to explain it if anyone asks me this problem. What's the probability that I subscribe to you? Without this vid, 50%. With this vid, 100%!
wpinrui 1 year ago
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.
truthseekers666 1 year ago
@truthseekers666 That wouldn't make any sense. If he opens up your door and the prize is behind it, well, you've lost. If he opens up your door and the prize isn't behind it, switch, duh, and your chances increase from 0% to 50%.
Stedwick 1 year ago 4
I think I follow the logic but I wonder if anyone has tried this as an experiment. Has anyone actually tried this, let's say, a thousand time to prove or disprove that it works? I suppose I could give it a try IN MY SPARE TIME.
jondeeca 1 year ago
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.
GetMeThere1 1 year ago
When one door is revealed, that door is no longer part of the equation.
conkerxxx1 1 year ago
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.
KittenIgnition 1 year ago
Remeber a key assumption that makes your decision RATIONAL is that your chances are NEVER 100% guaranteed.
you will not ALWAYS win the game, numb3rs fucked this explanation up PROPER
people get lost in the idea that it now becomes a matter of certainty.
DarkKnightBob1o1 1 year ago
WOW!!!!!! thank u sooooo much!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! this helped me a lot and i greatly appreciate u taking your time 2 make this video!! this problem was referenced in a book (the curious incident of the dog in the night-time) which i had 2 read 4 school...and i was very confused but not anymore!!
once again THANKS SOOOO MUCH!!!
lovebieber23 1 year ago
Very well explained :)
Frag1ty 1 year ago
This problem is flawed because it attempts to seamlessly social assumptions and mathematical ones, how is this not a problem that makes Monty Hall flawed from the get go?
dmcgraye 1 year ago
@dmcgraye whoops *attempts to seamlessly blend social assumptions
dmcgraye 1 year ago
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dmcgraye 1 year ago
I learned about the Monty Hall problem for years, but I never actually understood it on a meaningful level because I had always figured "What if you picked the right door, and Hall was just bluffing?" But after watching just half of this video, I feel like the answwer was so obvious, and that switching would be a matter of common sense. Thank you for explaining the problem in such a simple yet enlightening manner.
MrMeowmix9994 1 year ago 2
Off topic: I really really hate the new YouTube comments format. I can't tell what the hell anybody is saying in response to anybody else.
Stedwick 1 year ago 23
@Stedwick Thanks, It really explained a lot. My math teacher tried to explain but your was much better! :D
danne696 1 year ago
What happens if the Car was behind door 2, because they knew you would choose door 3 for changing?
Gamebito 1 year ago
@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.
ShadowJ20 1 year ago
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
sammysamster2 1 year ago
ok - you are correct - for the theroetical problem
but in real life, has anyone ever run a game show like this?
In practice, the host will only offer a swap sometimes.
This makes the problem very different. You will now want to judge, e.g. from the hosts's body language, is he trying to trick you?
richardmullins44 1 year ago
to tindfors et al, you are right. If the host always offers a swap, it is best to swap.
However, in real life, often the host will only sometimes offer a swap. This makes the game very different.
It seems unlikely that anyone has ever run a real quiz show, where the player is always offered a swap.
In other words, the answer (always swap) is correct but "may have little application to real life".
richardmullins44 1 year ago
but there is still a 1/3 chance of u gettting it wrong right?
lv70rouge 1 year ago
Or without bothering to add up all scenerios. In the first choice you have two thirds chance of being wrong. When Monty chooses he does not choose randomly. He will not choose your door and he will not choose the right door. Everytime that you are wrong he will leave the right door unchosen. You are wrong 2/3 of the time. So if you always switch you will win 2/3 of the time.
cunnidvd 1 year ago
The chart immediately made everything so much more clear.
EliteFalco341 1 year ago
Wow thanks heaps mate. I Subscribed by the way. =)
WhackaJacka 1 year ago
What is the probability of someone keeping or choosing is the real question here. Answer that then we can apply this. I'm probably the only rational person to oppose this scenerio but so be it.
MasterAdam100 1 year ago
you will have to divide by 2 to find an "unbiased" probability, independent of what you decide on as your set. Meaning if you have your set 1, and in set 2 one of the doors is guaranteed to be wrong, so for finding the "additional" probability you have to add, you instead define one of the set 2 doors as your new set 1, but once these probabilities are added up, it means you still have a 50% chance.
powerfury 1 year ago
It's not quite that complicated.
We'll say the doors are A, B, and C. A is the door with the prize. The scenarios are different only in which door you choose, so there's three of them:
You choose A. B is shown to be empty. Switching to C loses.
You choose B. C is shown to be empty. Switching to A wins.
You choose C. B is shown to be empty. Switching to A wins.
So, in 2 out of 3 scenarios, switching wins.
tlndfors 1 year ago
Alternatively, imagine the exact same scenario, but with more doors. Let's say 1000. Your odds of choosing right at first are 1/1000. 998 doors are opened and shown to be empty. You're left with one wrong and one right. The odds that your first choice was correct were 1/1000, and therefore the odds that the other choice is correct are 1-(1/1000), or 999/1000. Switching is almost guaranteed to win.
The number of doors is X. The odds of a switch winning are (X-1)/X.
tlndfors 1 year ago
Comment removed
chrisl1202 1 year ago
However, you can define these sets regardless as you'd like (you can pick door 1, 2, or 3 and that will become your set 1).
If I choose door 1, and door 2 is eliminated, then set 2 contains door 2 + 3. But isn't this randomly biased to whatever you decide as your set 1, and shouldn't therefore it's probability with respect to set 2 be considered? Meaning you add the probabilities of the two sets.
powerfury 1 year ago
Okay, if someone can explain to me why my method of thinking about this is wrong, I'd be happy. While I agree the maths in this is correct, I still feel it's only 50%. The reason for that being that you make the choice to split this into two sets.
Set 1: Your 1/3rd probability door.
Set 2: Your 2/3rds doors.
powerfury 1 year ago
Excellent explanation.
vDubrivation 1 year ago
sick!
kasraPISHYAR 1 year ago
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
somedaythewave 2 years ago
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.
Stedwick 2 years ago
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%.
Stedwick 2 years ago
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.
Stedwick 2 years ago
k i get it
somedaythewave 2 years ago
@Stedwick Nooooo!!!! Let's look at three scenarios and see if it was best to choose staying or swapping. Behind 2 doors are goats, the other 1 a car.
YOU CHOOSE:
car > goat removed > stay: car, swap: goat
goat > other goat removed > stay: goat, swap: car
goat > other goat removed > stay: goat, swap: car
Since the gameshow host knows ONLY to pick a goat, and never a car, your chances increase. sorry lol.
ItsGordie 1 year ago
Nice way to explain it with the cards!
OneShot187 2 years ago
I'm starting to get it now, after 18 years of thinking about it.
mocrg 2 years ago
amazing honestly the cards explanation made me laugh a bit, I am telling a lot of people about this thanks again 5 STARS
-CCV334
CCV334 2 years ago 3
Top Notch!!!
sweetcornnugget 2 years ago 2
Very well explained, I was curious how it worked. The deck of cards analogy was very helpful.
punkman911 2 years ago 3
You're amazing. I am a smart cookie who has never been able to grasp this problem, because of how it was explained and how I was trying to understand it. You could not have done a better job covering all the bases of potential false assumptions and confusions. Thank you!!!
MalkaRose 2 years ago 26
@MalkaRose This is not a difficult problem. You are not as smart a cookie as you think.
cunnidvd 1 year ago
good job. can't make more sense.
bif2212 2 years ago
that is brilliant.
this problem has been driving me nuts, thanks for explaining it easily and simply
CakeMace 2 years ago
show, show! Not "how you showing". Maybe I should proofread my comments first (ah, but it's YouTube).
ellenellenwatermelon 2 years ago
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. :-)
ellenellenwatermelon 2 years ago 2
In response to naysayers, yes, Monty Hall knows what's behind the doors. I don't think there is anyone in the world who assumes that Monty Hall is opening doors at random; that seems like a really silly way to do the riddle.
Still, in the interests of precision, let it be known that Monty Hall is cheating.
Stedwick 2 years ago
Comment removed
newcastlemusic 2 years ago
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.
newcastlemusic 2 years ago
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.
Stedwick 2 years ago
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.
Stedwick 2 years ago
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.
Stedwick 2 years ago
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.
newcastlemusic 2 years ago
This has been flagged as spam show
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
newcastlemusic 2 years ago
Frankly, I think anyone who doesn't get it after watching this video just isn't trying. This is my far the best explanation of the Monty Hall Problem I've ever seen. Great Job.
scimike22 2 years ago 47
@scimike22 - I'm still under the impression that Montey Hall ALWAYS opens a goat door after you make your first choice, and that you ALWAYS get to switch if you like. If this is true, and you as the player ALWAYS know this going in, then there are never 3 doors.
If you know that Montey will ALWAYS negate one door no matter what, then the odds are 50 / 50, as there are ALWAYS two doors to choose from, never three doors.
daltydodo 11 months ago
this problem has been around for years: see Bertrand's box paradox from the 1800s.
The 'disbelievers' should test it out for themselves... switching will win 6 times out of 9.
speedbunny 2 years ago 3
If you could not change your choice than your math would work, but since you can it's a logical fallacy to see this as a three door problem.
JustinDejong 2 years ago
This comment has received too many negative votes show
You're trying to carry over the 1/3rd probability to a situation with only 2 CHOICES!
Your card-game analogy was disingenuous at best. The only way the card game could work in the case of having 2 doors to choose from was if you only flipped over 25 cards, leaving 26 still face down.
It's a logic problem you're trying to figure out with pure math.
There never ever were 3 doors as one would always be eliminated.
If we played with 1/2 a deck and the ace WAS! still in there, you have1/26NOT1/52
JustinDejong 2 years ago
You are an idiot.
Two choices are not necessarily equivalent merely because there are two of them. Two choices:
a) You are the first guy to disprove an interesting mathematical quirk that actual mathematicians don't consider as up for debate
b) You are just another guy who doesn't understand something simple and interesting.
Are these two choices equally likely? Of course not. One is about 99% and the other is 1%. I won't say which is which, because I'd hate to hurt your feelings.
baggaroo 2 years ago 4
Baggaroo, you are my less intelligent brother. Shame.
caveman1144 2 years ago
I've watched this twice now, but I still don't see how the probability hasn't become 50/50. You're factoring in all three doors, when the probability hasn't changed. 3 doors, 2 are wrong. When Hall removed one door you have 2 and can switch. The fact you chose a door before he removed one is IRRELEVANT to the situation as you can now pick between one of two. One door was NEVER GOING TO WIN since Hall would always remove it. You're simply looking at the problem wrong. There are only 2viable drs
JustinDejong 2 years ago
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*
caveman1144 2 years ago
Shuffling them would introduce another variable, and you'd be right. However, the problem as stated stipulates that Money Hall does not open a door randomly- he will never accidentally open the door with the prize.
If you think this video is inaccurate, I invite you to test it yourself empirically. It's been done before, and switching ends up with a 66% success rate.
Gned11 2 years ago 3
What most people forgot is to take into account the information given by the preceding events. Take this example : "Here are two doors. Prize is behind door number 2. Which door will you choose ?" Are you going to say "Oh it's just 50/50 chance" ? Of course not, because you have more information that merely a random throw. Here it is exactly the same, the information is just not as plainly obvious. The host can give you this bit of information, because *he* knows where the prize is.
twilight1138 2 years ago 2
This comment has received too many negative votes show
You're making two assumptions though. First that the host is purposefully leading or misleading you, and second that he is not constricted by the rules of the game. If you choose correctly or incorrectly he is always going to remove one door. Once you realize the first choice and being shown three doors is simply a farce to make you over-think the situation it becomes blindingly simple.
It IS 50/50 because you ARE only given two choices with your second guess.
JustinDejong 2 years ago
Watch the video again. Maybe a few times. I know it's counter-intuitive at first, but at some point you should realize it's true. It wouldn't hurt to read an entry-level mathematics text on how to compute the probability of event A knowing the outcome of event B.
twilight1138 2 years ago 5