The Simple Question that Stumped Everyone Except Marilyn vos Savant
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- čas přidán 9. 02. 2022
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I have a question, if there are three doors, there is a 1/3 chance of winning, but if there is only two doors there is a 1/2 chance of winning. There is a 50-50 chance you will win, so what’s the point of changing. The first door you picked might be correct. Plus, isn’t it human psychology to give the person who did it correctly to pick something else so they go wrong?
The chart at 3:40 is wrong. Only getting a car is considered winning
I had to watch this video like ten times. I bet it could get into some mathematic gibberish. If you add them all up to 1, make a pie, and then take away 1/3, you are left with 2/3 of pie and 2 doors to give an equal slice to. You'd split them into 1/3 each, counting the total they came from, which would be half of what was left while using their logic. Help.
@@vaibhavk1342 does this make any sense? I had to watch this video like ten times. If you add them all up to 1, make a pie, and then take away 1/3, you are left with 2/3 of pie and 2 doors to give an equal slice to. You'd split them into 1/3 each, counting the total they came from, which would be half of what was left while using their logic. Help.
@@pheresy1367 That is why this question is sort of like fake news. no normal person accepts it. after the goats are shown its 50/50 of the last 2 doors. no normal person could think that in the final choice of the 2 doors THAT THEY SOMEHOW LOOK AT THE PAST.
I had the honor of having dinner with this lady while I was in college. Smart as hell, but very down-to-earth.
Wow, that's amazing.
She probably picked up the bill knowing there was a 100 % chance you would take it from her 😆
If youre the smartest person in the room? You're in the wrong room.
Lucky you. I'd love'd to speak to her for just few minutes.
You too?! The most remarkable thing about noshing with her is how she can pass things around the table telepathically. Oh, deary, I do hope she regaled your party with such feats. She even levitated all of us home after the dinner. Brilliant woman, that.
@@roberttyrrell2250 So what you're saying is, that she's always in wrong rooms?
I would have switched to door #2 as well but for a very different reason. I would have assume that the goats would need to be kept as far apart as possible so they would be less likely to incite each other into making noise and thus giving their relative positions away. Putting the car in between them would help keep them out of each other's sight. I might just have won the car because I knew more about goats than mathematics in that instant!
I had to give you a thumbs up, not because it was correct, but because it was damn good.
However, on a show that happens every week, the viewers would soon get wise if the car was always behind door 2.
lol!
@@mandolinic I think statistically, door #2 was more often the right choice. Everyone playing along at home would always yell DOOR NUMBER TWO!
So the trick then would be to guess door #3 and see if he shows you a goat behind door #1. If he does, you got the historical statistics and the live probability on your side.
They were schroedinger's goats
I am most impressed with the math professor who publicly admitted his mistake. It is so refreshing to see someone who will actually take responsibility for their errors, regardless of how embarrassing it may be. If only our politicians could show as much humility. Much respect.
True
Around here, when a vocal 50/50er finally figures it out, a typical response is to delete the thread so there is no evidence they were wrong. What's the opposite of humility?
@@Hank254hubris?
Or don't respond like a prick in the first place.
"I am most impressed with the math professor who publicly admitted his mistake."
Me to. How can you get a PhD in math and fail so hard at a simple probability problem?
People humbly and publicly admitting to be wrong, if only that existed today.
Too bad idiot Vos Savant failed to acknowledge her profound error. Too bad too those MIT profs are shown to be idiots twice over.
It's a new game. 1 in 2 chance, 1:1 odds.
Though Hall does not say it in these words, he asked this: "There is a car behind one of two doors. There is a goat behind the other. Which do you choose?"
It is irrelevant that Hall phrased it this way: "Do you wish to stay on Door 1 or switch to Door 2."
Vos Savant is an idiot as are those MIT nitwits.
@@DonLicuala It is almost a psy op her exist right down to her name "Savant".
What? My comment has nothing to do with her achievment, its about the humble gentleman who knew how to apologize.
I have no obligation what so ever to list anyones achievements, please look for an argument elsewhere.
Vos Savant is wrong still. She will be wrong even after she dies.
She has applied conditional probability math skills but from the wrong premise.
At the initial door opening to show one goat, the probability is 0%, odds, 0:0, chances 0 in 0.
The contestant is not trying to avoid two goats, but rather only one.
The probability of winning from the moment when an actual positive probability can be calculated, i.e., from the moment of two doors, is an equal probability of 50% to each door, as there are two options and no further information available.
And that is the only solution, the correct answer to the Let's Make a Deal Problem (LMADP). An alike problem "The Monty Hall Problem" (MHP) is a pseudo-realistic problem derived from the Let's Make a Deal Problem that illustrates an application of conditional probability assuming a contestant can win on Round 1 but does not and gets a second chance with updated info.
While the analysis of the MHP is self-referentially correct, it is inapt for the LMADP, which presents a contestant with a choice from two options, two and only two unopened doors.
The MHP would be appropriate if and only if there were three doors and a constest could win right away from picking the right door of three. Yet, in the LMADP, there is no deal when there are three doors.
The Rules to the LMADP are these, which are different from the MHP rules:
1. Say the name of the door. It does not matter because we're not revealing it. your odds of winning are 0:0., probability 0%, 0 in 0 chance.
2. Carol, from the doors not picked reveal a goat.
3. There could be a goat behind your door or a car. What door do you wish to name? You can name the same one as you did previously. You have a 1 in 2 chance, or 50% probability with the odds being 1:1 of getting it right. Once you pick we reveal the goat door first if you picked the car, the car door first if you picked the goat.
The question which wasn't answered here is that "why would the host open a door if they had the wrong option"?
Simpler explanation; assume you always switch:
If you initially picked a goat, you win. If you initially picked the prize you lose.
What's more likely?
This is the only thing I understood
@@thecoweggs i know right
I love goats. I can no longer afford to run a car... :-)
This video presents the presenter as being on your side, "helping you out." This is a bad assumption unless they're truly your friend, which is unlikely. Haven't you seen Slumdog Millionaire? :-)
So when they select something to lure you away from your initial choice of door, as an option for you, it can be a decoy, not the real deal. After all, if you chose that door first, it would have the same amount of probability from the standpoint where the presenter may not be on your side.
IQ is racist pseudo-science. Savant in name only.
The reason Mongoloids in Singapore and elsewhere have the highest IQs is because their youth was sacrificed for the god of money, and they did parroting memorization including of formulae which help in many IQ tests, at least to be used to patterns. This is why a Papuan tribal won't beat them: The tribal isn't dumber; just not used to those types of tests 'cause they DON'T NEED TO BE.
IQ tests use a one-size-fits-all appraoch, which is stupid, and proves IQ is stupid.
If I took an IQ test, it was probably disguised as some standardized test in Florida long ago.
I don't believe in that shit, so I refuse to participate. It's just like DNA testing: Different testing companies give different and contradicting results, so they're all scams.
If you are truly smart, you'll know better than to let others tell you how smart you are, when those others are self-entitled narcissist establishment people trying to dictate your mind.
@@scintillam_dei very very wrong
The best thing about this video is a reminder that when people publicly stated something incorrect, they used to express accountability and humility. That never happens anymore.
I know, and I hate it, we need better people in this world!
It doesn't matter what people think. it is what it is
in your imaginary world? sure
@@toxic_narcissist wdym?
@@toxic_narcissist without more context, your comment means nothing.
Now I get it. He will never pick your door, and 2 out of 3 times you have the goat, so in those 2 out of 3 times you have the goat, he reveals the other goat and only the car is left. You win. The only time you lose is when you have the car to begin with.
Yup.
Easy-peasy... yet to complicated for people thinking "two doors equals 50/50".
@@max5250 What I missed at first is the fact that he will never pick your door, so now you get lots of information about the remaining two doors.
no it's not like you understand it, if we go by your understanding the door you switch to would have a 100% chance of winning
@@oussama1811 No you lose if you already have the car. You have the car 1 our of 3 times.
Your assumption is mistaken. While it is true that 2 out of 3 times you have the goat before Door 3 is opened (or 4 chances out of 6), when he reveals the goat in Door 3 then there are only 4 possibilities left, not 6. So at the point that he reveals the goat you only pick the car 2 out of those 4 remaining times if you switch doors.
I've long understood the Monty Hall solution - but extrapolating the information scale to 100 doors - makes complete sense - knowing that the "one door" is hot - and that you have a 98% chance of being wrong on your first door choice.
it's 99% of being wrong and then the switch is 99% of being right. it's reversed.
revealing the doors does not give any hint on your first choice, so it's still 99% and not 98%
99%.
(Car is prize). Let’s say the chooser did not pick a door at all. The chooser does not know whether there’s a car or a goat behind ANY of the three doors. The host, however, knows what’s behind each of the three doors (ALL). They both stand in front of the three doors and the chooser does not pick anything at all. Now, the host reveals that there is a goat behind door #3. That eliminates door three as an option for the prize. Bye-bye! Now there are two doors left. That means there’s a 50% chance that the car is behind either of the two remaining doors. The high IQ vos Savant is declaring the idea that the chooser having originally simply “THOUGHT” (chosen) of what the winning door might be before the host opened door 3, CAUSES the remaining 50/50 ratio to change. To me that sounds like she is either not that smart OR OR OR, she is pulling ideas from things like quantum theoretical physics such as the “double slit experiment” and such. Like, the theory that when trying to guess if there’s a cat in the box, actually there is no cat in the box until you open the box, then it either does or does not manifest, because our reality is made up of potentialities rather than actualities, until they are “looked at” or measured -to be more precise (that’s how the theory goes, which is pretty much proven at this point). So, I think she needs to clarify what level of accepted reality or physics she’s pointing to in advance of her explanation, because the description of the solution by her and this video is implying that the chooser’s THOUGHTS (manifested as a supposed “choice” or predisclosed “guess” of where the car might be, affects the outcome. It’s similar to me arguing with you that nothing actually exists because atoms don’t reeeeally touch each other and sub-atomic particles are only energy in their smallest form. In a physics class that discussion or answer to a topic might be appropriate, but in an online forum about whether I have one or two cups on my table , it isn’t necessary appropriate. We obviously have different “levels” of reality we are talking about. The one to attend to needs to be disclosed in advance because she is LITERALLY saying that the difference between whether the chooser specifically made a choice (THOUGHT), or not, before the process of door openings began, AFFECTS the outcome. That’s not really actually fair according to the standard discourse. An online basic forum is not a quantum physics class. Thanks for reading.
PS. This might be going too far, but, it also makes me think she’s purposely trying to mess with people’s heads. lol. It worked ‘cause even the PhD’s were crying but funnily enough what they missed is she subtly didn’t disclose to them all the facts of the game. If I tell you we are living in a video game, am I lying? If I walk away and leave all the PhDs to fight about it, then they cry and appologise, am I smart?? Hell yeah! Am I right? Not if the rules of said game weren’t disclosed. lol. Sounds a bit narcissistic IMO. 🤭
@@SolutionsWithin You obviously did not understand the problem and are not smarter than Savant lol.
Not picking a door first is different than picking a door first, because then the host cannot pick this door and has to chose between the other 2 only. You picking a door eliminate this choice for the host.
That's why it becomes a 1/3 - 2/3 and not 1/2 - 1/2
This hurts my brain. But even high level mathematicians didn't understand it at first so I can't feel too bad for not getting it.
there's nothing to understand. She considered each door a floating variable. Each door is in absolute matter of fact, a constant. So there is no means to transfer probability from one door to another. Once the 3rd door is opened, there is no more question of what's behind door 1 or door 2. It is either a goat or a car. The feeling that you have a 33% higher chance when switching is based on the original probability you had to correctly guess between 3 doors. Once the 3rd door is eliminated you have a completely new expression. And the probability between two choices is always the same. Stay or change your mind, implies that the previous variable is still in play. It is not. You are not "switching", you are again choosing.
The difference between the two choices is 50%, probability is recalculated at every choice.
@@ZennExile incorrect.
@@vladimirdemirev4948 the reason why door are grouped is because they are different. One is the door you picked, the other two are the doors the host has to pick from. Your grouping doesn’t account for that.
The professor admitted they were wrong because they were wrong.
@@vladimirdemirev4948Say we take a bag with ten marbles in it, one orange and nine white. Whoever gets the orange wins. You pick a random marble from the bag and I then take the remainder, open the bag and select a marble, discarding 8 white ones. Who is more likely to be holding the orange marble.
Clearly the person who knows where the prize is will have a better chance of selecting it.
@@morbideddie well, I guess I am wrong. Seeing the 100 doors example changed my mind.
I will go with the excuse that binge-watching CZcams videos on different topics trains you to react first, without giving much thought ;)
First encountered this in high school. I tried to explain: "if you switch it's like picking 2 doors instead of 1", which convinced very few classmates. The teacher noted that I had good intuition and poor articulation. So true.
You explained it in simple to understand language. In fact, the best articulation!
But you are wrong.
I don’t understand how the extra odds don’t also add to the door you chose. They’re both still closed and options.
@@BasedGodGotenks Look at the 100 door example. The chance you chose the correct door from 100 doors is very low. The chance the winning door is amongst the other 99 doors is high. Now remove 98 of those 99 favourable doors and the remaining one has a very high probability of being the winning door.
Now if you arrived at the game late and missed the above process of elimination and just had 2 doors to choose from then your chances would be 50/50 because you do not have the information that points towards the more favourable door.
it doesnt make sense still fuck, how, every door has a 1 in 3 chance of being the car, revealing one door doesnt give the other door a higher chance as it was predetermined beforehand, its still 1 in 3, it cant just change cause you revealed the other door fuck like ik it makes sense to some people but it doesnt to me. EDIT: This is how i see it, in the start you have 1 in 3 chance, after goat reveal, you are left with 2 doors, 1 has a goat one has a car, so you are back to beginning, when deciding in that moment, you have a 50 50 chance you will get it
I look at it this way. There are only three possibilities:
1- you pick the car, so the host shows one of the 2 goats - then you should not switch door
2- you pick goat 1, so the host shows goat 2 - you must switch door
3 - you pick goat 2, so the host shows goat 1 - you must switch door
Therefore there is 2 out of 3 chance that switching your door choice will get you the car
@@user-ej9nl1ng9d lol - same principle
@@user-ej9nl1ng9d I know - I was just putting it simple
@@user-ej9nl1ng9dMost of the people who argue about it can't read a table. Many can't even follow John's simple logic. The biggest problem is getting them to realize 50/50 is wrong or they don't even listen to explanations of the right answer.
thank you - that is a good explanation
the key is that *the host always eliminates a losing choice*
the odds that the original choice was correct don't improve but the odds that the remaining non-eliminated door is correct do increase
it is easier to grasp for most people if the door count is say 10 or 20 or some other larger number
if all the choices are eliminated except the initial choice and one other there is a very high probability that the correct door is the one remaining door (the one not eliminated)
When you initially explained that the other door would have a 2/3rd probability of a car being behind it, i couldn't understand it one bit. But i loved the explaination including a 100 doors where 98 were removed. That explaination immediately clicked to me and now I get it! What an interesting question. I always love these kinds of probability questions cuz they make me use my brain in ways I don't get to use while studying 😅
3:27 - "This is contingent on the host always opening a door with a goat."
Yes, it is, which is why this restriction must be included within the problem as phrased, something the introduction to this video fails to do. That's actually a very common problem with this problem. It cannot be assumed that a goat had to be revealed simply because a goat was revealed unless the host's intention is incorporated into the problem. The host could have selected a door at random that simply happened to contain a goat. This legitimately changes the math to the Monty Fall/Blind Monty problem.
This failure to accurately phrase the problem is frustratingly common.
well If you watch the show is oblivious, you can also assume how the show would b er if you havnt watched it. But still the author of the video has a bad taste.
ohhhhhhhh ok this makes sense now
Oh this makes sense now. I was under the impression that the game host always open door 3 regardless. Which is why I was confused at 3:20, when the table showed scenarios "game 3" and "game 6" having a car behind door 3 which made no sense to me at first so I excluded those scenarios. But I understand now, thanks!
Came here to say this. In video it sounds like host might have just chosen a door at random, and it happened to have a goat. It should be stated that the host will NEVER open the door with the prize when he opens his own door after the contestant chooses.
Well if the host would pick the door that was chosen by the participant and there was a goat, there would be no question of switching the choice of doors.
And if it was one of the other doors and there was the car, the same thing applies.
So in my opinion it's self explanatory
Really impressed with the individuals that took accountability for ridiculing her and publicly apologized. That sort of character is in short supply
I always tell my kids four things:. 1) you panic, you die. 2) stupidity kills. 3) never do anything for which you will have to apologize for later. 4). And) (a biggie),.never take up a habit you're just going to have to break later on.
@@sheilalopez3983 Good word
Seemed to me "they" were sort of picking on her for not using the "math stats" as they did ?
Oh, of course; now let's hear it for the "Intelligence of Creative Thinking!?"
It seems THERE ARE different "kinds" and "types" of IQ " Tests." Experience and Education , two possibly different types ?
So we should be also asking WHICH IQ Test did Marilyn excel in, or on ?
MISSING FROM the video; Does this lady write and speak in both German and Italian ?I've always felt there is an extreme indication of high intellect in regard to peoples' abilities TO express themselves in foreign languages ?Seems there is a certain "type" or "kind" of logic it seems in learning to "relate " foreign language to one's own ability to speak and write in their native language ? It appears Marilyn 's " (by assumption?) that Marilyn had TWO "Native languages" yes ?
Her opinion of "public schooling" holds great merit. I remember a question required to be asked on a high school test , was "Who were the Phoenicians ?" The ABCD Answers included the answer " Venetian" . Most admitted later that they all misunderstood the word Phoenicians because they were all more "familiar" with Venetian Blinds, than historical terms of peoples and places! (Ha Welcome to American World History 101-we (all) need to repeat that one!?) Which btw lead to my last question (for you or Marilyn, ha ?) Is the inability to "spell" properly (in any language/especially ones native language ) indicate ignorance ?Duh...As a retired teacher, I submit I've become dependent on the Google Gargoyles ' offers for correction, which often just doesn't exist.
The robots tell me I've misspelled something, yet/while, offering no options with which TO correct it.
Good at Questions; Slow at the answers. Anyone ?
"I Ain't no Middleman"
Fred Gold & Lynda Faye
Copyrighted 2016 by LyndaFayeSmusic@gmail.com or Yahoo, if censored for using the word " God" too often?
I don't because they made her same mistake she did when they followed her idea.
@@gerardcote8391 ? Please explain! Thanks!
UPDATE: The 'Dunce of the Month' Award goes out to none other than joevarga. After four discussions he is still unable to read a simple list of possibilities, nor has he been able to calculate a single thing. Hope you feel as proud of your non achievements as the rest of us do.
At least they did realize that the math checks out in the end and that the initial doors do matter, not a 50-50.
"Yes, it makes sense to switch"
"As the number of doors that the game starts with declines, the odds of you winning by switching diminish"
Sadly once they realized that they turned to insults instead, but they do understand the idea behind the problem now.
@@Jurting His latest post says with 3 doors it is 50/50. He is just trolling.
Fool, the chart and trying to use a 100-door game to prove the 3-door game is bogus. In the 3-door case shown in the video, it makes NO difference if you switch in THIS case where door 3 is opened. I'm sure you're too immature to do as follows, and just with to hurl childish insults. When you grow up, do the following:
Make a drawing as described below. There are 3 outcomes for the game (car is behind door 1, 2, or 3), so:
Draw row 1 with 3 doors. Write "car" on door 1. Door 2 is blank. Draw a large circle on door 3 to show that it's been opened. This is case/game 1.
Draw row 2 with 3 doors. Write "car" on door 2. Door 1 is blank. Draw a large circle on door 3 to show that it's been opened. This is case/game 2.
Draw row 3 with 3 doors. Write "car" on door 3. Door 1 is blank. Draw a large circle on door 2 to show that it's been opened. This is case/game 3.
Now, according to the rules stated, the host will always open a door with a goat. So in case 1 (shown in row 1), if you switch you lose. In case 2 (row 2), if you switch you win. In case 3 (row 3), if you switch you win.
So, you win by switching in 2 out of 3 cases (case 2 and 3). Since it makes sense to switch in 2 out of 3 cases, you now think "It pays to switch". HOWEVER, in the case in the video, you selected door 1 and door THREE is opened, NOT door 2. Therefore the last row in the drawing does NOT apply since it shows door 2 being opened. Therefore, the case in the video is either game 1 or 2 (represented by rows 1 and 2).
There are only 2 possibilities for outcomes now. Either you switch and you lose shown in row 1, or switch and you win shown in row 2. So it makes no difference if you switch or not.
Now let's say that door 2 was opened instead of door 3. Strike row 2 then (instead of row 3), because it's either scenario 1 or 3 that applies. Again, in one case it pays to switch, and in the other it doesn't.
In the 100 door example your odds of winning increase greatly if you switch doors. It's pretty obvious. As the number of doors that the game begins with declines, the odds of you winning by switching diminish, and by the time it's down to 3 doors, the odds of winning by switching are down to 50/50, at which point it no longer matters if you switch or not.
@@joevarga1769 Nice wall of text. Sadly, 2 is still larger than 1 even if you wish it wasn't because you answered the question incorrectly initially because it's counterintuitive. Time to stop trolling and just admit that your intuition failed you in a solved probability problem that is famous *specifically* for being counter-intuitive, It's no big deal.
@@Jurting Copy and paste the part where I answered the question incorrectly initially.
Copy and paste the part where I admitted that I was wrong.
Think it this way: Once you've chosen a door, you are given an option to open either the one that you've chosen OR both of the other two. It's a cakewalk now, no?
Wouldn't it have been easier to understand if you just said switching is the same as picking two doors and staying?
@@klaus7443 Sure! Whatever works for you.
@@code-chaser
I already understand it, you don't. In the Monty Hall Problem the host must know where everything is, otherwise there is no advantage in switching if a goat is revealed. If the contestant is simply getting two doors instead of one then he couldn't care less about the host's knowledge.
A good way to think about this problem is: You first choose one door. You are then able to change your choice to BOTH the other doors. you get a car even if one of the doors have a goat behind it. This is the exact same thing as to show the goat beforehand.
This is an excellent way of looking at it
That is brilliant.
And the only way to lose the switch would be because you beat the 3 to1 odds against you when you first chose.
So it's always (3 to 1 against you) vs (2 to 3 for you by switching after the goat reveal).
(Ooops, I think my "further clarification" only served to complicate).
;-)
That is a great way to look at it. It takes a certain personality, I think, to be encouraged about odds of 2/3. Chance is not something you can predict -- it's chance. This is the difference between stochastic (your statistical analysis) and random (what actually occurs). There is a world of words and ideals and a world of things and occurrences. Some people love to argue over words and ideals, but we each may go home with a goat. Or a car.
Unless you did pick the car on your first choice.......
It's literally a 50/50 chance.
You either change your answer or you don't, and you either win or you lose
exactly. or pretend there's 100 doors. now you get to pick either (a) one specific door, or (b) the combined total of any 99 doors, where if there car were in any one of the 99 doors you'd win it. would you pick choice (a) or (b) ? okay now instead of 100 total to start with, let's do 3 total to start with.
When I first heard the "Monty Hall Experiment", I reacted the same way everyone else did. I am however a software engineer. So I decided to write a small program to prove them wrong. What I ended up doing is proving MYSELF wrong.
The program played 100 random games. The first 100 games stayed on the same door whereas the second 100 games switched doors. In the end, the program that stayed on the original pick won approximately a 33% of the time whereas the program that switched won approximately 67% of the time. I couldn't believe it. You do double your chances by switching doors.
Omg, I was just about to write the code after seeing this random puzzle. Unbelievable conclusion.
Yeah I was around ~15 when this happened and also wrote that program at the time, though I either had no idea what was correct or was pretty sure she was. Don't remember which!
I still don't understand how. You don't know what is behind the two closed doors . So the probability for you ( any subject) is 1/2 .
@@siddharthshekhar909 the chance of your original guess being correct is 1 in 3, no matter what door you guess, the host can open an empty door, thus your original odds of 1/3 are not changed by the host opening an open door. By switching you end up with 1/2
@@siddharthshekhar909 EXACTLY!!! This is what I said. I absolutely could not believe it! This is why I wrote the program! I wanted to prove them WRONG. There is no logic to their conclusion. But as it turns out I was wrong!!! I still can't explain it. But the computer doesn't lie. I am a software developer with 30 years of professional experience. The program took an hour to write. I had to run it like 10 times before I could believe it. I still can't explain why it works. Don't feel bad. There were MIT professors who felt the same way you and I did. And she is the smartest person alive so I wouldn't sweat it if you don't understand why. I don't either.
This is a very simple probability question. If your strategy is to not change, you simply pick 1 door out of three at random and so have a 1/3 change of being right. Full stop. If your strategy is to switch, there are 2 possibilities. a) your first guess is correct (1/3 probability) and you then switch and lose because you are switching away from the correct door. Total probability 1/3 of failure. b) your first guess is wrong. You are then shown an empty door so as your first pick was wrong and one door is empty by moving you have 100% change of being right. Total probability 2/3 x 100% = 2/3 of success. So if your strategy is to move you have double the chance compared to the "stick" strategy. Anyone who struggles with this has a very weak grasp of probability theory.
"Anyone who struggles with this has a very weak grasp of probability theory"
so, paul erdös had a weak grasp of probability theory?
i ask, because he as well struggled with the outcome.
Has anyone tested this? Like have 100 gameshow-like setups.
Nevermind, I realised how easy it is to test. Tried 16 ‘games’ and switch won 11 times while stay won 5. Pretty fun 😄
@@Dev1nci Try this one. Take a balanced coin. Toss the coin 10 times. Are 5 of them heads? Try tossing the coin 50 times. Did you get 25 heads? How many times did you need to toss he coin to get half of the outcomes "heads"?
Try this one and report back. How many times did you have to play the game to win 50% of the games? Same result as 16 games?
@@techhie1302 I understand that principle but that's a different thing. You need to wait for a large enough pool for the stats to be visible :)
It's actually quite simple, if you DID end up on the right one at first, which is 1/3, you lose switching. If you chose a goat door, which is 2/3, then switching WILL MAKE YOU WIN. ez
it's quite simple for all those, who aren't trapped in this "one out of two-->fifty fifty" illusion.
@KarlHeinzSpock it's 50/50.
@@EthanFaidley you're the next victim of the illusion.
take note of the fact, that this simple problem has been solved decades ago and the solution passed the review of several generations of mathematicians.
@@EthanFaidley
"it's 50/50."
It both doors would hold a car with 50% chance, that would also mean player can pick the winning door from 3 closed doors, with exactly the same success rate as a host, who get's twice as many doors, and knows where the car is.
@@EthanFaidley
Assume you stay with your first pick.
If your first pick is Goat A, you get Goat A.
If your first pick is Goat B, you get Goat B.
If your first pick is the car, you get the car.
You only win 1 out of 3 games if you stay with your first pick.
Switching means the opposite.
It's just basic math/logic kids understand.
Sadly, it's far too hard for idiots.
The Monty Hall problem and people's approach to understanding it is very interesting. Another way to think about the problem not covered explicitly in the video is the fact that only one independent choice is being made in the game. That choice is the players initial guess when there are 3 doors. The host isn't making a meaningful independent choice since they have to reveal a non-prize door only from the doors not guessed initially, and the results of the decision whether to stay or switch are entirely dependent on the initial guess(when there were 3 doors). If the player initially guessed the prize door(a 1/3 chance) and they switch they lose. If the player initially guessed a non-prize door(a 2/3 chance) and they switch they win.
"If the player initially guessed the prize door(a 1/3 chance) and they switch they lose. If the player initially guessed a non-prize door(a 2/3 chance) and they switch they win."
I first heard this riddle in the 2008 movie "21", and until today I never understood why switching doors after the reveal of a goat would increase the chance of winning, but the way you phrased it made it click for me, so thank you, i finally get it :)
that is probably the only explanation i've ever heard for this that makes sense. because no other explanation, even when presenting empirical evidence, actually draws that connection.
For me the biggest mystery is why your explanation is not immediately obvious to everybody. Lots of people deny it, even after someone carefully explains it to them....
@@tonybrowneyed8277 yea I think they're missing the difference between the host revealing a random no prize door, and the actual Monty hall rules. That would give 50/50 odds and the Monty hall game looks the same on any individual round. But the host not being able to reveal the same door as the players first guess completely changes the odds and the nature of the game.
" If the player initially guessed the prize door(a 1/3 chance) and they switch they lose. If the player initially guessed a non-prize door(a 2/3 chance) and they switch they win." omg i finally get it tyy
Almost every smart or educated person says that school is not the best way to learn and still nobody tries to change it.
It's the only affordable way for most.
The 1948 and 1988 encyclopedia sets of the "Book of Knowledge" were my favorite source of learning. I LOVED them! I always checked the Public library a lot. My fiction was mostly from Silver Age comics that helped me much with my art. I loved studying electronics. I didn't need school for any of that.
Now days, Professionals seem to get their credentials out of a "Quaker Jack Box."
I prefer self-teaching.
@@karangupta1825 Yes. Expand your passions and have creative hobbies.
"The book of knowledge" was very pictorial and was suited for all ages. The simple facts of the topic were clearly explained and the information became gradually more complex for when you are older. It was laid out like the internet with many links in the index's. There were plenty of do it yourself projects. I build my first radio from the 1948 book at 11 and it was fantastic!
I'll admit I didn't see the logic behind it until thinking through each case scenario. I fell for the "now there's two doors, so it's 50/50" trap.
I think another way to explain it is that the constant is that you have a 1/3 chance of choosing the correct door.
There are only 3 possible scenarios--the winner is Door 1, Door 2, or Door 3. You can only choose one door, so you can only be correct in one of the 3 scenarios.
Which means, regardless of the other steps such as crossing one door off the list, you ARE correct in one scenario and you ARE incorrect in two.
Therefore, there is always a 2/3 chance your original choice of door was wrong, which is equivalent to saying you have a 2/3 chance of winning if you change your choice. Same as saying your original choice always has a 1/3 chance of being right, and so you always have a 1/3 chance of ultimately being right by not changing that choice.
For anyone struggling to get their head around this, the simplest, easiest to understand explanation is this: you have a 1/3 (approximately 33%) chance from the start of picking right...that means the other 2 combined have a 2/3 (approximately 66%) chance of having the prize. He isn't going to open your door first regardless. The moment he picks one of the other two, it is more information, because he can only open the one that doesn't have the prize if one of the two unchosen doors happen to have the prize. Odds were 66% that one of the other two had the prize, but he removed one that didn't and since he couldn't remove one if it did, it is a tell...the odds were originally still 66% for both the doors, and he eliminated one of them but the odds are still 66% from where you first choose. Hence, it makes sense to switch. The reason it isn't a 50/50 proposition is because "Monty" has information that he is acting on, and by not being able to manipulate your chosen door, doing so is inputting that information to the decision by revealing which of the remaining two had the "prize" if you happen to be wrong on your 1 in 3 initial guess, and odds are you were indeed wrong. Switching will be correct about 66% of the time I would imagine.
good explanation. Perhaps it's easier to just focus on your prize?
It's easier to pick wrong, and if you stay, you would end up with the wrong prize because the prize behind your door never changes.
Couldn't it also be interpreted as him trying to sway you though? This has too many variables. For example, what if I already picked the right door and he could have just chosen either 2 or 3 and it wouldn't matter. What if he's trying to discourage my current answer by making me second guess myself, which often results in switching doors? At the end of the day the idea that switching increases your odds is reliant on the idea that you didn't pick the right door first. And that he could have just picked either door because neither and the prize and he didn't have to avoid it.
Not to mention. If I've already picked the wrong door. Then what does he stand to gain by giving me the chance to switch? He has absolutely no motive to give me a second chance and not only that. But narrow down my choices and increase my odds of winning.
It's just not a smart move to make and only a fool would do this. His smartest move would be to try and sway me away from my current decision by making me second guess myself.
ALWAYS choose door #3
The way I look at it is he initially had a 1 in 3 chance of being right with door #1. After door #3 is revealed he now has a 1 in 2 chance regardless of whether he chooses door#1 or #2. Revealing door #3 gives no specific information about either of the other 2 doors. So I agree with you. In my mind they have neglected to revise the chances of door #1 being correct after door #3 is revealed.@@trapidtrap2612
Imagine how intelligent the person who created this problem would be
probably in the 125 - 150 range.
@@arandomguy46 wtf
@@epicmorphism2240 The creator probably didn't know. It was just a game. Maybe after decades of hosting the game they ended up with a "gut feeling" that is better to switch but they probably thought it was 50% too
@@andressoto739 i was commenting ln CRB‘s ridiculous comment
noobs create problem pros solve it
I think that some people find hard to wrap their mind around this concept because they fail to understand the very nature of probability. It’s not about being 100% right, it’s about being more likely
i just don't get it. if there are still only two doors wouldn't the odds be the same.
@@chestnut1279 Two doors left proves that you can either switch from your car to a goat, or from a goat to the car. The host could have simply asked you that same question when you first picked a door without doing anything else.
@@chestnut1279 I'm thinking the same thing. My understanding of this brain teaser is that changing your answer after the host reveals one of the fail doors is most correct but still not a guarantee as picking the first door was 1/3 vs changing the answer now making the odds 2/3. Its less likely you got it right on your first choice but it is possible. I think the logic of "two doors means equal odds" only works if the host revealed the door you picked to be the wrong door and you still had two options. But nothing makes me feel low IQ more intensely than brain teasers. I suppose the real question is do you think its more likely you got it right on your first try when you had three options?
@@chestnut1279 Yes. Sheep being lead.
If the contestant was given the opportunity to swap his 1 door for the hosts 2 he would almost certainly accept the offer leaving themselves with 67% chance.Don’t forget the host knows where the car is and has to open one of his which has to be a loser leaving the remaining 1 of his 2 which still has a 67% chance
True! If you initially pick the car you will lose upon switch. If you initially pick the goat you will win upon switch. However picking the car is way less likely... compared to initially picking a goat. So switching has highest chances of winning.
"Way less" is pretty vague. It would be better to say "half as likely" which is accurate.
I've always thought that's the simplest, clearest way to look at it. Don't try to figure out the probabilities. Just list the possible outcomes. It's a small enough data set that it's not a big deal to make a comprehensive list. There's only three places the car can be. There's only three initial choices you can make. And there's only two second choices you can make. That's 3 x 3 x 2 = 18 possible scenarios (9 scenarios for staying and 9 for switching). If you stay, there are 3 ways to win and 6 ways to lose. If you switch, there are 6 ways to win and 3 ways to lose.
The way I see it is that you have a 2/3 chance of getting a goat behind your door. There are also two facts to establish:
-The host will not choose the door you have chosen
-The host will choose a door with a goat behind it
Since you have a 2/3 chance of getting a goat, and the host will always choose the other goat, that means 2/3 of the time the car will be in the other door. Essentially every time YOU get a goat behind the door you choose, the car has to be in the second door, since the host will choose the other goat. And since you get a goat behind your door 2 out of every 3 scenarios, that means the car will have to be behind the second door 2 out of every 3 scenarios as well.
Well worded. Regards!
If you like goats and cars, you can't really lose.
I guess I still don't get it. "The host always chooses the other goat, that means 2/3 of the time the car will be in the other door." I ask why? Why wouldn't the probability shift to say the car will be in the door you originally picked 2/3 of the time and therefore don't switch your choice of door.
@@waakca The host must always reveal a goat, but it must be from the doors that you did not pick. That's the rule of the game.
It means that whenever you start selecting a goat, the revealed goat will necessarily be the second one, so the car will necessarily be in the switching door. And you had 2/3 chance to start selecting a goat.
Only in the 1/3 cases when you start selecting the car, the other door that the host leaves closed will have a goat.
@@RonaldABGwtf are you trying to say?
The way I approach this is by thinking that the initial choice can be one of three alternatives: Goat 1, Goat 2, or Car. This then plays out into three (and only three) possible scenarios:
1) If you chose Goat 1, and Goat 2 is revealed, then switching would have given you Car.
2) If you chose Goat 2, and Goat 1 is revealed, then switching would again have given you Car.
3) If you chose Car, and either Goat 1 or Goat 2 is revealed, then switching would have given you Goat 1 or Goat 2.
Thus in 1) and 2), switching works, and in 3), it doesn't. Thus switching is right 2/3 of the time.
Yup, it really is that simple but people still get locked in to 50/50 for some reason and hold on to it like a pit bull.
You are not correct, and I will explain to you why, very simply.
You correctly differentiated between Goat 1 and Goat 2 in alternatives 1) and 2).
In alternative 3), you do NOT differentiate between Goat 1 and Goat 2 by stating , "either Goat 1 or Goat 2"
Alternative 3) should be, " If you chose car and Goat 1 is revealed, then switching would have given you Goat 1"
Then alternative 4) which needs adding, "If you chose car and goat 2 is revealed, then switching would have given you Goat 2"
Thus, in 1) and 2) switching works, but in 3) and 4) it doesn't.
The result is 2-2.
Speaking of a pit bull...
@@Hank254 LOL!!!
@@georgebliss964 How can you NOT make a probability tree for this problem? Contestant pick car, host leaves goat. It has only ONE branch!!!
So , I've known the Monty Hall problem since 2 decades, watched 100s of youtube videos on it too. But nobody cared to tell me that not many believed Marilyn vos Savant when she gave the correct answer. (Off course Steve Selvin gave the real proofs and solution with 3 prisoners problem).
*EDIT* : I had no idea that *so many dudes* would be *triggered* by this comment. I'm a dude and I'm aware that a man's ego is just *too fragile* . But imagine being triggered by a harmless comment, almost all of the replies are by butthurt young boys and men. LOL
we already knew the ans if we watched other videos about that
I picked the 2nd door also. They said the car was behind #1 at the beginning. I listen very well
And this helped move mankind forward...how?
@@anti-apathy9715 1) Nobody said it did. It didn't have to.
2) Discouraging, patronizing women in the field of science, mathematics , etc. is not a trivial matter.
(Although i do believe this could happen to anyone, but being a woman made it worse for her)
Point is, such incidents prove to be a hindrance for little girls and young women who are already brainwashed by the society to think that they are not as good as men.
@@anti-apathy9715 ego trip
3:33 You will be surprised how many people beleive that even in the 100-door game the player has a 50/50 odds between the remaing pair of doors.
Of course it is only 50/50.
@@slind3517 No, by switching it is a 98% possibility, when the original pick was 1% Perhaps you should actually study exactly what the 100 door scenario is before commenting
This 50/50 thing plays large in the history of the MHP, even among university profs.
They are constantly trying to say that if rules said that the host would open that door at random, the game reverts to 50/50. It obviously does not.
I'll give a simple and a more detailed explanation. Read the detailed one if you still do not understand, after the simpler one. This question assumes that the host will NEVER open your chosen door, and will eliminate all doors containing goats save for one door (which may or may not contain a goat).
Simple:
In the 3-door question, your initial choice has a 1/3 chance of being correct. If you are incorrect, (2/3 times), and the host opens/eliminates another incorrect door, then other door is correct (there can only be 2 goat doors, so the remaining one must be a car). If your initial guess was correct, (1/3 times), then the remaining door after the host eliminates another one will be incorrect. Notice how this only happens 1/3 (~33.3%) of the times, while the other door will be correct 2/3 (66.6%) of the times.
Detailed:
Let's start with the hundred door question, as it's easier to understand. For the hundred door question, you first choose a random door. Your probability of getting the prize is 1/100, 1%. The host then eliminates 98 doors and leaves another one for you to choose between. According to the intuitive logic, there's 2 doors, so the chances are 50/50. That is incorrect. Your first choice was a 1% chance of getting it right. The host didn't eliminate it because it was YOUR choice, regardless of whether or not there was a goat behind that door. You had originally a 99% chance of getting it wrong, and him eliminating all the other doors except one does not change that fact, as 99% of the time he will eliminate all remaining goat doors, leaving the door with the car. Only upon the 1% chance that you originally chose the car door, will he ever leave you with an option with a goat.
The same logic applies to the 3 door question. You had a 1/3 chance of getting your first choice correct. If your first choice was wrong (which will happen 66.6% of the time), and he reveals an incorrect (goat) door, the safest bet would be to choose the other door.
I saw another guy in the comments run both simulations (100 or 3 doors) in code, and the results are pretty much synonymous with what Ms. Savant predicted. There have been many other simulations summarizing the same thing. And, you could always read Wikipedia's explanation on this.
The one-hundred door example makes it very clear. No one should assume they guessed the correct door in that scenario. The pagentry of a Game Show might make people falsely presume the host is hoping to trick the contestant into switching their pick, but unless the host is actively trying to persuade you to select one door over another there's simply no trick to be had. This line of thinking is precisely why I got this wrong upon initially hearing of it years ago. It's still shocking to see how venomous some of her detractors were.
I thought the host was wanting the contestant to lose the car, thus influencing my answer.
I dont understand this, it makes no sense. if you choose a door and the other is eliminated then the probability should distribute evenly between all of the remaining doors, not all of them except for your choice. Can someone explain why it does this?
@@versenova5531 door 2 and door 3 combined add to a 2/3 probability but you know door 3 isnt it so that 2/3 probability has to be for door 2 thats how i understand it
Exactly, saying that the host will open a door with a goat, before you make your pick, is must for it not to influence your answer in this scenario.
@@versenova5531 you have 33.33...% chance to guess it right, which means you have 66.66...% to choose WRONG. Switching will automatically give you the price 66.66...% of the times. You're more likely to choose wrong, so switching is a better choice 2/3 of the times.
That's why it's even more clear with 100 doors, you have 1% chance initially, but after revealing a goat behind 98 doors, it's your best option to switch 99 out of 100 times.
It makes total sense when mapped out. I guess the difficulty comes in understanding why "switching" doors increases your odds at all. I got hung up on the "switching" part having any impact, instead of realizing that it's making a new selection with better odds. We used to do simple stuff like this in grade school, it's kinda crazy how a little bit of language can subvert your logical faculties.
"instead of realizing that it's making a new selection with better odds"
It is not making a new selection with better odds, but swapping your lower odds for better odds.
@@max5250 Now I get it, thanks max ; )
@@anthonydenn4345 Welcome back dude.
I still don’t understand why switching will increase your odds of winning. If you take away your first selection, meaning you never made a choice, are you still going to choose the one out of the 2 doors that didn’t get eliminated?
@@jacobcutrer
Switching increased odds of winning because host get to pick from two doors therefore, he gets a door with a cat twice as often than player does.
When he opens his door with a goat, we know which door holds a car twice as often as the door initially picked by the player.
There is a very simple way to understand this problem/solution. Her genius is that she saw it instanstly. When you make your first guess, you have a 1/3 chance of winning. So, the other two doors, together have 2/3 chance of winning. So, if after choosing, say door A, I give you the choice of choosing both doors B and C - you win if EITHER door has the car. Of course you would switch - you have a 2/3 chance of winning.
Now, suppose, after choosing door A, I show you door B and it's a goat and give you the choice of switching from door A to door C. Of course, you switch - why? Because you are really choosing door B and C - and you win if either door has the car - AND I just showed you that the car is not behind door B. I could have waited - just offered you to switch to doors B and C, and then showed you that the car isn't behind door B - but I didn't, I just showed it to you first.
And that's why she is a genius - the rest of us have to get the answer and then figure out why it's right.
I'm not a genius, but when I was presented this problem, I saw it right away because I only thought about switching to the winning side. I realized that removing a goat does not effect the chance of having the car.
Why would you change your choice to BOTH other doors if one of the doors has been already disclosed? The choice can be either to keep your FIRST choice or opt for the SECOND door. You are left with 2 doors and you don’t know behind which door is the car? The odds are the same. Easy peasy.
@@mr.jollybear5180Switching to the other side is, in a way, getting both doors because if the car WAS on the host's side when he had 2 doors, then it's guaranteed to be on his side when he has only 1 door left.
In other words, it's easier for the host to land the car on his side, and once it's landed on his side, he cannot get rid of it. Therefore you should switch to the side that has more room for the car to land.
@@mr.jollybear5180 think of this differently. Imaging that at the beginning you are given the choice: door A or both door B and C - if the prize is behind door A and you choose door A, you win; on the other hand, if you choose both door B and C, if the prize is behind either, you win. What would you choose? Of course, door B and C. Why? Obviously, you win 2/3 times.
Now imagine. You choose door B and C and then the emcee shows you what's behind door B and it's a goat. Should you switch to door A or keep door C?
The thing to bear in mind is that disclosing information doesn't change the probabilities. For example, if the emcee shows you that behind door B is a goat, that didn't change the probabilities associated with choosing both door B and C - switching means going from a 2/3 chance of winning to a 1/3 chance.
I don't understand why people have difficulty with this problem. It has always been intuitive and trivial to me. (Maybe I'm a genius I guess.)
I look at it this way: On the first selection, I probably selected a goat. (Basic probability: 3 doors, 2 goats. You picked a goat!)
When the host shows you the other goat, and gives you the option to switch doors, they're basically handing you the prize.
Why would anyone think it was 50/50 on the second choice when they obviously selected the first goat on their first pick? 🤔
I can list a thousand statistical problems tougher than this one. If people can't do this one, don't do any career involving stats lol.
"I don't understand why people have difficulty with this problem."
Because they see two doors and instinctively associate 50% chance to each door.
They also don't realize once you pick your door, you can't "pick again" with only two doors, since one door holds your previous selection, while other door holds the opposite of your door.
But that's my point. If it was intuitive for them to see 50/50 in their second selection, why wasn't it intuitive for them to see 33/33/33 in their first selection? It's just strange. It's as though most people have no memory of their first selection. The field of statistics involves a lot of conditional probability problems. This problem is an example of that.
Example: If I have a bag of 10 marbles, with 5 red and 5 blue, my odds of drawing a red marble change on every selection from the bag and they change depending on what I pulled out, assuming the selected marble stays out of the bag. @@max5250
@@taekwondotimeThe vast majority of 50/50s think probability is limited to 1/# of choices. They almost always understand the 33/33/33 but they can't accept the idea that two doors could have different probabilities.
I remember reading this from her as a kid. Loved working through the logic; it's just counting. The thing people don't get is that the 'host' (originally Monty Hall) doesn't behave randomly; he always shows you a goat. Just count out the cases, which is what she did, and the answer is obvious.
If he doesn't behave randomly, the it ain't a probability question. It's more along the lines of 'form' like at the horse races . We then talk more about odds . No strict rigourous standard calculations for odds . It something settled on by experts . However the probability calculations are ½ for a choosing from two given options for which absolute randomised conditions apply.
@@philip5940 you're confused. It is a probability question, but the difference is it's about posterior versus prior probabilities. just write down all possible options; pick one at random, and consider all of the different ways that the prize might be distributed. Count what happens when you switch or stay. you have 1/3 chances of getting it right by guessing out of the box. He then shows you one of the places the prize ISN'T (that's the part that's not random). Now you have 2/3 chances of winning if you switch. Why? Because there was 2/3 of a chance at the beginning that it was in one of the spots you DIDN'T pick, and he showed you one of the places where it isn't, so those odds still apply; you have 2/3 chances of getting the prize if you switch. It's a classic problem that is now commonly taught in probability, known as the Monte Hall problem (he was the host of the game show). If you don't believe me, write down 3 doors, put the prize behind one of the (doesn't matter which), let's say it's behind door 1. Now say you pick door #1, and it's behind door #1. He shows you either 2 or 3 (where there's a goat), if you switch you lose. Say you originally pick door 2. He has to show you door 3, where there's a goat, if you switch, (to door 1) you win. Let's say you pick door #3; he has to show you the goat behind door 2, you switch, you win. Count it up; 2/3 chance of winning if you switch.
It's also ironic you're arguing about his given that the entire point of this video is to explain to you how your answer is wrong.
@@bartonanderson1106 seems that the entire point of the video is to show the power of spin and to highlight mass gullibility . ⅓ and ⅔ probability when given three choices are now a phantom that have ceased to exist The new state is two choices for which the probability is 50/50 .
@@philip5940 I just explained the entire logic to you; you're obviously not capable of understanding this, but here's another chance: en.wikipedia.org/wiki/Monty_Hall_problem
@@philip5940 Since the host always removes a goat that is not which the player picked and neither which contains the car, then that means that the other he leaves closed is the winner one as long as the player starts failing, and that occurs 2 out of 3 times on average, not 1 out of 2.
This is better seen in the long run: imagine you played 900 times. In the first selection you are equally likely to select the option that has each content, so in about 300 times you would get which has goat1, in 300 which has goat2, and in 300 which has the car. In total, 600 times a goat and 300 times the car.
As the host will reveal a goat from the two doors that you did not pick in all the 900 games, in the 600 that you already have a goat, the revealed goat must be the second one, and so the car is in the switching door. Only in the 300 games that you had the car behind yours, the switching door will have a goat.
So, always two doors left, but yours results to be correct 300 times, and the other that the host leaves, 600 times.
Just reverse engineer it.
If you always switch, you will only lose if you had first selected the car.
There is a 33% chance to lose when you do select the car on the first try. Leaving you with a 66% chance to win if you always switch.
Way too logical. And a good way to think about it
But having a chance and actually proving you are correct is two different things. The car could have been behind door #1 and then everyone using their math ability to choose #2 would lose. Sometimes you need luck too. That is how nature works. You cannot predict anything definately. You can only guess.
@@edwardspencer9397 Sure this game depends on luck. But the chances to win are higher.
@@edwardspencer9397 that's why it's called probability. Out of three doors you will win one in three times if you stay with your original door. Out of 100 Doors you will win one in 100 times if you stick with your original door. It would be ridiculous to not switch unless you didn't want to win
@@edwardspencer9397 yes you need 'luck', but by always switching you will be 'lucky' 2 out of 3 times. Whereas, if you do not switch, you will be 'lucky' 1/3 times. Therefore, by switching you increase your 'luckiness'. Considering this, you may be less fortunate if you merely guessed at what to do.
You select a door with 1/3 chance of getting the car. This means remaining 2 doors has 2/3 chance of having the car combined. Host always opens one of them, so that 2/3 chance is now accumulated in the last door. That's why you always switch. Mathematical professionals at the time really could not figure it out first glance?
"Host always opens one of them, so that 2/3 chance is now accumulated in the last door."
Probability is always "accumulated" behind one door, since a car can be behind only one door, not two of them, and a host, who can see content of the doors with his own eyes, see which door holds a car.
I get it now : let's say you play 10000 times, in 1/3 of the cases the car is behind your door, and in 2/3 it's a goat. And in those 2/3 of cases the host must eliminate the other goat so the car remains in game. So in 100% of those 2/3 of cases, the car is behind the door he chose.
Yup.
But you are not playing 10000times!!!! You are playing only once.
@@jolel2566 Since when number of tries affect probability?!
@@jolel2566 It helped me figure it out, but it is the same with one tine :
1. you chose Door 1
2. you have 1/3 of chance of the car being behind
3. you have 2/3 of chance of a goat being behind.
4. in that more common case (2/3) the host selects the right remaining door for you...
For people who still don't understand - Think about it with 10 trillion doors. First, you pick any random number out of 10 trillion. Then the host eliminates all the doors (numbers), except yours and one other number (remember, one of the two remaining numbers is correct). Would you switch? Obviously, because the chances you randomly picked correctly out of 10 trillion is basically zero.
Some people still couldn't understand this with 100 doors, and I think it becomes even easier to understand when you increase the total number pool.
Oh that makes sense
Thank you, I saw this explained on another video, I don’t understand why mathematics people are disagreeing with this simple fact of probability, it’s not that hard to understand
I still don't agree with this no matter if its correct because 1 out of 3 is not 1 out of trillion or 100 or 10 its still out of 3 so it's completely different.
@Krichnu It is the same; that's the point. It doesn't matter if you agree, because the math checks out.
Find the comment on this video where the guy wrote a program and ran this simulation 10,000 times with 3 doors (or whatever the number was).
Can you imagine how you would feel if you switched and were wrong? @@MrMcSnuffyFluffy
One way of wrapping your mind around this question is by thinking of it this way: Originally there are 3 doors, so your odds of picking the correct door are 1/3 and your odds of picking the wrong door are 2/3. The host is then going to reveal a goat, which seems like new information, but you already knew that at least one of the doors you didn't choose had a goat, so it's actually not telling you anything you didn't already know. Asking you to switch is really asking you if you think that your first choice was wrong, and uses the same odds as when you made that choice, so the odds that you chose wrong the first time are 2/3.
So switching comes down to thinking you made the wrong choice the first time. The odds change with the revelation of one goat. This is new information, not just "seemingly." The 1/3 or 2/3 scenario is now irrelevant.
@@diannemccarthy8685 No, the odds don't actually change, even though intuitively it may seem like they should. The way that I explained it above is the explanation that I personally find the easiest to understand, but if you don't find it convincing, just search for the "Monty Hall problem" in the search engine of your choice, and you should be able to find plenty of other explanations. When I first came across this problem a few years ago, I did a lot of reading about it myself, and I remember that there were even websites that allow you to play this game over and over, and they keep track of your win/lose record for when you stay and when you switch, and you'll see that over time the win percentage for switching averages to 66.6% while the win percentage for staying averages to 33.3%.
No, given the mind set of the Host, your chances of winning are BETTER than the problem suggests! Monty WANTS you to win! More fun, more excitement, better ratings, more sales of every item offered as a "prize" on the commercial, (ahem, "Show") The clue he gives you is that your odds got better after your first choice. Monty wants to make your chances of winning as near certain as he can, without being obvious.
@@lindsaymitchell9300 The show producers are always telling Monty what his next move should be. Anything that makes mo money.
@@zackreagin8384 The odds of the second door are increased after seeing the third door open, IF the game (which regulation I ignore) had the rule that BOTH a door must always been open AND the switch must always been offered. If showing an open door and/or the option of switching are not the rule of the game but may be decided by the host unpredictably, the host of the game might use the door open or the switch option as a way of deceiving the player, at times. That would not guarantee that switching to the other door carries more chances to win, in my opinion...
Wow I never heard of this brilliant woman but I love her and especially enjoyed humbling so many "experts", the likes of whom we usually place too much faith in.
The solution proposed in this video can be tested experimentally with observation.
[1] Create a table with 100 little air tubes coming up from the bottom, all arranged in a grid.
[2] The table has sensors around each tube that can detect very slight pressures, such as a pea or pressure on the cup from a human hand.
[3] A computer is connected to an air pressure source, and controls a valve on each tube.
[4] Setup 1: Put inverted cups over all air tubes.
[5] Setup 2: Put a pea under 1 cup chosen by a random number generator on the computer.
[6] Bring in the tester and an observer, neither of whom know where the pea is.
[7] The observer tells the tester there is a pea under one of the cups, and to select one cup at random.
[8] Tester selects and holds down cup #54.
[9] The computer sends a strong puff of air under all cups except, let's say, #76, which was chosen for one of two reasons:
[9a] #76 has the pea under it, OR
[9b] #56 has the pea under it, so #76 was chosen at random so there would be two cups left on the table.
[10] The observer now tells the tester to release cup #54 and turn over cup #76.
Result of experiment: If the solution presented in this video is the correct one, and the tester transfers his/her selection to whatever cup was not originally selected, the tester will select the cup with the pea 99 times out of a hundred. Would it really work that way (allowing for small variations in probability)?
Explanation: Cup #54 was indeed a choice of separating 1 from the remaining 99, giving it a 1% chance of being correct. However, cup #76 was among the 99 not selected originally. So, the non-selected group of 99 had a 99% chance of including the cup with the pea. So, by changing the selection to cup #76 you are going to the remaining cup out of that original group where the chances were 99% of including the pea.
I think I get it . . . but it feels wrong for some reason. 🤨
You ever seen "Good Will Hunting?" Marilyn is basically that in real life. In the film there is a scene where Will is sitting in his professor's office and another professor walks in with a math problem he had been working on for a long time. Will solves it in a few minutes, leaving the professor highly embarrassed. This is what Marilyn did here.
@@JohnSmith-ys4nl We dont at all know that she did that.
We dont know how she first reacted during the first time she encountered this problem.
Wow, air tubes! COMPUTERS!!!
You sure did pick an unnecessarily elaborate method to test the MHP. You could just get a friend, sit down at a table, and test it 30 times, switching, and then 30 times sticking
Recording results.
Not satisfied, do it another 30 times.
But in fact, the answer will become glaringly obvious long before the first 30 are done.
I found the easiest way to explain the problem is:
You had a 2 out of 3 chance of picking goats and only 1 out of 3 chance of picking the car.
Once the host reveals a goat, there are only 3 outcomes of a switch:
Scenario A) You picked the car at the start. If you switch, you lose.
Scenario B) You picked Goat #1 at the start. If you switch, you win, since Goat #2 is gone.
Scenario C) You picked Goat #2 at the start. If you switch, you win, since Goat #1 is gone.
You can only lose if Scenario A happens, and you win if Scenario B OR C happens. Therefore, you should switch since you have 2 scenarios where you win, versus only 1 where you lose.
Thank you so much,your explanation made it make sense to me!
I find it amusing how people can be so naive. If the host reveals one of the door had a goat
Then your Scenario B and Scenario C is EXACTLY THE SAME AS IT HAS 1 door with a goat and 1 door with a CAR!!!
@@shafin141 I guess you didn't fully understand the explanation. You have to think about the problem like I explained it, as in, including the odds of you picking correctly in the first place. Yes, after one door is removed, there's a 50/50 chance that the car is behind one of the remaining doors. HOWEVER. You have to take into account the fact that your INITIAL choice only had a 33% chance of being correct. Those odds have not changed.
Try extrapolating it to 100 doors, and 99 goats. You have a 1% chance of picking correctly at the start. Now the host removes 98 incorrect doors, and you're left with 2 doors. It's still technically a 50/50 chance of there being a car or a goat, but you're not going to switch still? You have new information now. The host has essentially removed the majority of the wrong possibilities from the equation, so the odds of you now picking correctly if you switch have become 99/100, versus your intial 1/100 odds.
@@dchang8619 Sorry you are confusing yourself by believing that the initial 33%, 1/100 and 99/100 remain. They are gone now and dont carry over. Its now 50:50 and your original choice is just as likely as switching.
@@kenandlynboyle9300 it’s really not though. Try it yourself in practice, not in your head. Try switching 10 times, and not switching 10 times. See how many times you win by switching and how many times you win when you don’t switch. I promise you that you will win more times by switching. This has been mathematically proven, so there’s no real use trying to argue against it. Just try and understand why it’s the case.
I can't find the comment again to give credit, but someone in the comments here helped me understand it intuitively. It gets much easier if you think about the odds of *losing* rather than the odds of winning:
- If you *always* switch then you *only* lose if you picked the car to begin with.
- What are the odds that you picked the car to begin with? 1/3.
- So the odds that you lose if you switch are 1/3.
Learned about this in middle school. It was a really fun concept.
To get the answear you just have to brute force it.
1.If I don't switch and the correct door is Number 2 (for example)
A. I choose door 1, door 3 is eliminated, I don't switch, I lose.
B. I choose door 2, door 1 or 3 is eliminated, I don't switch, I win.
C. I choose door 3, door 1 is eliminated, I don't switch, I lose.
So in the instance that you don't switch you have 1/3 chance of winning
2. I switch doors Number 2 is still the correct answear.
A. I choose door 1, door 3 is eliminated, I switch, I win.
B. I choose door 2, door 1 or 3 is eliminated, I switch, I lose.
C. I choose door 3, door 1 is eliminated, I switch, I win.
We have a 2/3 chance of winning if we switch.
"So in the instance that you don't switch you have 1/3 chance of winning"
Yup. A short cut for that scenario is to simply pick one of the three doors and open it since what's behind the door won't change no matter what the host does. Doing that it is easy to see why the first door has a 1/3 chance. The people who insist the doors have an equal 1/2 probability can never seem to pick one door out of the three and win the car half the times.
I get that but these fuckers in the show can lie :D
@@alenngkIt's not a real show... it is just a math question.
I couldn't wrap my "intuition" around this and was pretty convinced that Marilyn vos Savant might be wrong. I spent a few hours writing up a Java application to simulate this question. I’ve run this a number of times, and the results are always pretty similar. My application randomly chose doors, randomly chose whether to stay or change, and the host randomly decided which door to open (for those times when the simulator contestant guessed correctly and either remaining door could be opened). In fact, the lady with a 228 IQ clearly put me to shame….like she did all those Mathematicians.
After the simulation proved to me that I was wrong, I figured out how my intuition had failed me. I was thinking that opening a door offered no new information as you could open up a “goat” door 100% of the time. The new information is NOT that there is a goat behind one of those two doors, it’s the fact that IF your 1/3 pick was wrong, the other door has been upgraded to being a guaranteed winner. This was explained in the video, but I had to see it for myself.
Here is one of the runs from my simulation when using 3 doors:
------------------------------------
Num Games Played: 1000000
Num Games Stayed Door: 499875
Num Games Changed Door: 500125
Num Games Won When Stayed Door: 166345
Num Games Won When Changed Door: 333990
Percentage Wins when Staying: 0.33277318
Percentage Wins when Changed: 0.66781306
Here is one of the runs from my simulation when using 100 doors:
-------------------------------------
Num Games Played: 1000000
Num Games Stayed Door: 499478
Num Games Changed Door: 500522
Num Games Won When Stayed Door: 5101
Num Games Won When Changed Door: 495542
Percentage Wins when Staying: 0.010212662
Percentage Wins when Changed: 0.9900504
Just do it with a deck of cards.
Pick one
How often is it the ace of diamonds? (or whatever you say is the winning card?)
The rest of the time the other door (card) wins
id say that the explanation this video said is a little bit weird, and doesnt really teach you why it works like that. my explanation would be that if you pick a door, you have a 2/3 chance for getting a goat. using that, when the door is revealed as goat, you know the one your on is 2/3 chance of goat, meaning the other one is 1/3 chance of goat, so 2/3 chance of car.
I missed be stupid , as when you make a choice , then change your mind , how on earth can that increase your chances ?
Great work by the way .
@@MrKrueger88 because you are effectively getting two doors instead of one if you switch.
Think about it with 100 doors. Then the host gets rid of 98 losing doors.
Is it more likely you picked the right door to begin with?
Or is it more likely the prize is behind one of the other 99?
You get the other 99 by switching.
I still don't get it. You pick a door, that has not been opened and you have no information of what is behind it. Another door is opened with a goat leaving you two doors with no information of what is behind either of them, you now have a 50/50 shot at having the right door. What makes door number two more attractive? You have zero information on either unopened door. Why would door number one not be equally attractive? You have no information to go on with it either. How can the odds of the first door be diminished if there is no info to diminish it? Now, this is not a three door quiz, it is a two door quiz, informing you of what one door has just eliminates it from being pertinent, the quiz becomes a heads or tails proposition.
The confusion people have, is that the door being opened by the host is NOT RANDOM. The host opening the door KNOWS it's NOT the winning door. If the host opened the door randomly then the odds wouldn't change.
Right, and the random opened door might accidentally reveal the car.
Not really. Because if the car was behind door 1 and you chose door 1, then the host could have opened either of the other doors at random. In that case, switching would have caused you to lose. However, that doesn't change the answer to the question. The question simply stipulated if your chances would be better for you to change doors, not if changing doors would guarantee you the win. And the answer is still yes, that technically, logically, and mathematically, switching, even in that specific circumstance, would have increased your chances of winning. For example, if you look at the example the narrator gives with 100 doors. If you just so happen to pick the door with the car as your first choice, and they eliminated 98 of the remaining doors, that other door still would have a higher chance of being the correct one, except for the fact that you had the dumb luck to happen to pick the one door that wouldn't have helped you.
@@TalkingHands308 The host does nothing randomly, he knows the winning door.
@@TalkingHands308 The host doesn't randomly reveal the car if you picked a goat. That's what they meant. This makes your odds after switching 2/3 instead of 1/2 or 1/3. You will still lose if you switch after originally picking the car.
@@CarMoves you've missed the point of my comment...
I think there's a simpler way of explaining this puzzle. If you choose the correct door and switch you will be wrong. If you choose the incorrect door and switch you will be right ( because the other wrong door gets eliminated), and you choose the wrong door 2/3 times
I'm surprised no-one mentioned the fact that 1990 is in the twentieth century - unless I have missed something.
Yeah, I caught that too
Imagine being born in the past millenium. How old you must be...
very common in the sciences.
historians are assholes for not starting with a zeroth century.
Put it this way: Everyone would switch in the 3-door case when the host does NOT open one of the doors but offers you to take the two doors instead of the one you took.
What I didn't realize was that the host obviously only opens a door different from the one I picked. So the odds remain 2/3 for goat.
Wow! That is the clearest way of describing the situation I've seen.
"Stay with YOUR original 1:3 chance or take OUR 2:3 chance". The revealing of the goat was just a trick to make it seem as if it's become 50/50 to switch or stay.
What a great explanation, well said! It's obvious explained this way. I also like how Savant's analysis of all outcomes is so simple and really proves it, and yet the top mathematicians who criticized her didn't think to do such a simple analysis.
The host knows what is behind every door. The offer to change my choice is only given after I picked a door already. Obviously the host is trying to fool me.
If I was given 2 choices from the start: pick a door, or pick 2 doors, the answer is obvious.
@@Tombecho If you think the host is benevolent: Switch. If you think he doesn't know where the car is but just got told where one goat is: Switch. If you think the host is mean: Stick with your choice.
4:10 NOW I get it. Two out of every three times you make the wrong choice so switching will lead to the right choice two out of three times.
Weirdly that does make sense.
Exactly. The thing people don't seem to understand is that the host must open one of the two wrong doors. He cannot open the door with the prize in it (this defeats the point of the game) or your door (which would render the decision obvious). If the host were randomly opening doors (and could open your door or the one with the prize in it), then this wouldn't work.
@@user-wu4bo1hz3p what if not randomly opening door? Just not opening door that is yours and.prize ?
@@yuquoint6633 Huh? That's what he's doing, which is why it works.
@@user-wu4bo1hz3p Ah finally i get it, thanks for explanation.
I used to be amazed on the quality of these videos, but seeing how much they've improved in just a few year, blows my mind.
Appreciate that! Are you noticing a quality in recent videos from 2024 compared to older ones?
@@Newsthink absolutely, from the audio to the graphics. However, your story telling has always been unbeatable, so no difference in that to be honest.
If you switch door you will lose if the goat is behind it.
The probability for this is 1/3. So the probability to win is 2/3.
I find it incredible that professional mathematicians spoke up and made asses of themselves before actually working the problem out
Kind of like people who post similar rants on CZcams comment sections
:)
(not talking about you, btw)
indeed, it appears so logic to me that it's better switcch
At this point in my life I’m not lol. So many experts come with a strong sense of truth mixed with pride.
This is the issue she brought up.. educated doesn't necessarily equal smart.. and most often not.
@@Arthur.H.Studio what do you mean?
The correct answer is quite obvious if you start with say 10 doors, pick one and then the game show host opens 8, to reveal goats. That leaves your first choice, which was most likely wrong, and another which is not.
Yup.
Most of the people do understand this with bigger number of doors, but some still think with 3 doors there is no use in switching, although even in that case you are switching probability of 1 door for probability of 2 other doors, which is twice better.
That's nonsense. No matter what door you pick, 8 of the 9 remaining doors WILL have goats, and if the hosts removes 8 doors he HAS to pick the 8 with goats. (He can't pick 1 with the car, if it's in the 9.) Him being forced to pick the 8 doors with goats does not mean your first choice was most likely wrong!!!
@@blucat4
Are you seriously trying to imply a player, having 9 doors with a goat behind, and only 1 door with a car behind, has equal chances of picking either the door with a goat, or door with a car?!
Of course the fact host has to remove 8 doors with goats doesn't affect that, nor does anyone claim something as ridiculous like that, since we know future events cannot affect past events (player picks his door BEFORE host removes 8 doors with goat).
@@blucat4 There was a 1 in 10 chance you were right in your intial choice and so a 9 in 10 chance you were wrong. So the probabiity you were wrong was 90%.
@@NickLawson-ht6fe You all keep making the same mistake, which is to move the odds that existed in the initial conditions to the end conditions. Also, I said it wrong. Yes, my initial choice was most likely wrong, at the beginning. It was 10%, same as all the other doors, 10%. After the host removes 8 doors, each which HAD a 10% chance, their chance has now dropped to zero. 8 doors with 0% chance. Can't you see, that AFTER the host has removed them their odds do not affect the new conditions. The host HAD to removed doors with goats, so now the information we have is that there are two doors left, each with 50% chance.
I think the thing that confuses people is they think there is a chance the host will pick the car. This is wrong and were I fell foul of the odds.
If you pause at 3:28 and then look at just the first 3 games, ignoring the result column, and you pick Door 1, they table shows all of the possible actual positions of the items in the first 3 rows.
Now the host is not going to show the car, this changes the odds.
Most of us initially are thinking, if I choose door 1 every time, and he choose door 3 every time, then I have a 1 in 3 chance, but he also has a 1 in 3 chance to pick the car. BUT.... he is going to alter his decision, because he knows what is behind the doors and he isn't allowed to pick the car.
So instead of him choosing door 3, if he sees the car, he will switch to door 2.
That is were the odds change.
He has altered his choice, knowing that he has to pick a goat.
Game 1, if you switch you lose.
Game 2, he chooses door 3, because he can't pick 2.
Game 3, he chooses door 2, because he can't pick 3.
The fact that he is bound by the rules not to pick the car is what throws the balance, he cannot make a random choice.
Again, it is the assumption that both are choosing random doors, but his choice is 2/3 of the time made for him.
Majority of people who are claiming: Two doors equals 50-50 chances, are the ones who do understand that host haven't picked a car, but they still don't understand elementary probability.
Actually, if there were blatantly expressed rules that Monty chooses from those 2 doors at random, it would make no difference whatsoever as to the proper player choice.
And if it is unclear to the player whether or not Monty chooses at random it still makes absolutely no difference as to the players choice
Under these 3 possible scenarios, if the player gets to switch, it gives her better chances of winning the car if she switches. In any of the 3 possible scenarios, whenever a goat door is displayed, the player has a 67% chance to get the car by switching.
Any time Monte opens a goat door, (and notice, he always does that!), it is to the players benefit to switch.
The fact that (if Montys rules were that he opens that door at random), the player chances become equal, ( 33% stick , 33% switch) is meaningless as to what the player should do.
(Actually there is no game where Monty opens a car door, he always opens a goat door. It matters not at all how or why that door happened to be opened)
The question the puzzler asks is "does the player have a better chance by switching" Yes, she does, regardless of any ambiguousness of the stating of the MHP So you dont get to claim that you didnt get the answer correct because of some supposed "ambiguousness."
@@archimedesmaid3602
Wrong.
It is of utmost importance that host knows where the car is, in order for a player to have twice better odds by switching.
The mere fact that the door with goat is opened is not enough since it is important how was that door opened.
By opening the door with a goat randomly, host is destroying half of better odds by switching since he will open the door with a car in half of the situations where a car is behind two doors player haven’t picked.
@@max5250 What the host "knows" means nothing as to what is presented as the MHP. In every MHP there is a goat door opened. Any time there happens to be a goat door opened, the player chances increase to 67%, by switching. It matters not one bit, what process the decision was made to open that goat door. IN EVERY MHP THERE IS A GOAT DOOR OPENED
Marilyn did not make a mistake, or leave any ambiguousness that matters which will change the proper decision of the player, or your proper answer to this puzzler.
The only thing asked here is "can the player increase her chance by switching". The answer is "yes she can" and she can in EVERY MHP ever presented.
Yes, if the rules blatantly stated that Monte chooses by random, obviously the overall player chances are 33% by sticking, 33% by switching, and Monte keeps the car in 33% of the games because presumably, if by random, he chooses that car the player doesnt get it. But nothing like that is said in the MHP. It just says "a door is opened revealing a goat". In each and every instance like that the player is going to double her chance of winning the car, to 67%. And again, that happens in EVERY MHP ever presented.
And also note, many people state that the chances become 50/50. 33/33 is by no means 50/50
You are not thinking clearly, and you are WAY overthinking the problem
@@archimedesmaid3602
You are contradicting yourself.
In your first reply here you said:
"Actually, if there were blatantly expressed rules that Monty chooses from those 2 doors at random, it would make no difference whatsoever as to the proper player choice."
But now you say:
"Yes, if the rules blatantly stated that Monte chooses by random, obviously the overall player chances are 33% by sticking, 33% by switching, and Monte keeps the car in 33% of the games because presumably, if by random, he chooses that car the player doesnt get it."
So, which one is correct, the first or the second statement, since both cannot be correct at the same time?!
"Marilyn did not make a mistake..."
I agree on that. It is quite clear from the question, that host is not opening the door with a goat randomly, but by knowing where goat is.
What I wanted to emphasize is the fact that it is not enough to say that "the door with goat is opened" but it is also important to know how it is opened, in order to calculate probability correctly.
Ah, I get it now. You pick a door:
Outcome 1: The prize is in the door you picked. So switching would lose the game.
Outcome 2: The prize is in the 2nd door you did not pick. And since monty hall only removes the goat door, switching would win the game.
Outcome 3: The prize is in the 3rd door you did not pick. And since monty hall only removes the goat door, switching would win the game.
Switching has 1/3 chance of losing and 2/3 chance of winning!
this actually does it for me, at first i thought monty hall problem is only logical to switch when you have more than 3 doors
@@tactic2569 well if the total num of doors is 3 + x, with x being more doors added. If monty hall removes 1+x amount of goat doors,
Then the chance of winning is 2+x/3+x and losing is 1/3+x.
Meaning it would be even more logical to switch if more doors are added and monty removes more goat doors.
It still makes absolutely no sense to me. There are two doors after you remove the 1.. q of the two doors has the prize.. I don't get it lol
I got it a different way-- in the 100 door example, you pick a door randomly, then 98 wrong doors are removed. Your door can't be touched, but the right door also can't be touched. Your door was originally chosen at random, so switching doors would be better, since you have more information to make the choice with.
Although this method probably wouldn't work for any other problem and I just didn't know any other way to rationalize it..
Here's an extra thing:
I also thought, "hey, the probabilities just merged into one door! That's cool." Without really understanding why it happened.. :/
@@joeyeddy9344 even if probability of something is 99%, it's still possible that 1% is the final answer.
Choose a car first and switch, you get a goat.
Choose a goat first and switch, you get a car.
There is a higher chance of you choosing a goat first, so you should always switch.
I think the reason people get this wrong is that they don't think about the difference between the host showing you a goat after you picked a door, and the host showing you a goat before you picked a door. That latter would make it 50/50.
How would the former not be a 50/50?
@GrArtista it's easier for the host to land the car on his side, and once it's landed on his side, he cannot get rid of it. (he can only get rid of the goat) Therefore you should switch to the side that has more room for the car to land.
@@GrArtista2 out of 3 times, you pick a goat. This means that 2 out of 3 times, the host is forced to reveal the other goat and the door you didn't initially pick has the car. The difference is because if the host reveals a goat first, he has a choice of two every time, but you pick first, he only has a choice 1 out of 3 times.
There's an easier explanation to the original Monty Hall question. Since the host KNEW where is the prize, why didn't he choose door number 2 instead of 3?
Huh?
@@HollyGraham-wl6ye huh?
Because the host is not playing the game, he is presenting the game. He is not trying to keep the car.
Huh?
Me: I'm thinking of a number 1 to 1,000.
You: 629
Me: The answer is either 629 or 401. Do you want to change your answer?
You: Yes.
Great analogy. Thanks for sharing that.
Doing that with a number between 1 and 3 is an exact analogy, and also useful
It really is that simple. Sadly, only some people realize this when the numbers are expanded to 52, 100, 1000, or 1,000,000 as you stated.
@@normvargas2799 When you have a much larger number of choices, the probability of using this method goes way up. When you have only 3 choices, using this technique your chances of being wrong is still 1/3. In the opposite extreme, it would be like saying if A has a 50.1% of being correct and B has a 49.9% of being correct, obviously one should choose A because it has a greater likelihood, but with these probabilities they are so close that it doesn't matter.
It is important to note that the 1/3 to 2/3 probability is right only if the host who opens the door knows what is behind the doors. If he doesn't, then the probability is 50/50.
The problem is not so buffling after all.
Its also important to know if you also know this information prior to picking the door.. Because that also changes the probability.
@@bodybuilder6350 i do not think so. Could you explain?
Now i understand what you mean. You are right.
It is important to note that if the host reveals the car, the probability of losing goes up to 100% because the car has been removed from your choices.
What this tells us, is that it's a waste of time to think about what the host knows or doesn't know, or what the host's intentions are.
@@alkinooskontopodias5919 if you know or don't know that the host knows or doesn't know. So thats like 2x2= 4 different probabilities depending on what information is known. I am sure you bring in even more play of words and rules that would increase the variables. From the original question we are limited in our knowledge. We didn't even know if he would open the right or wrong door (initially) to expose a car.
That goat question always perplexed me, I thought it was a joke until I realized that people actually think it's a hard question.... ?! The answer is so obvious, I can't imagine there is even anything to discuss
The people who believe it is 50/50 also think it is a joke.
@@Hank254yeah true
I thought everyone knows about this. its a 1 in 3 chance of winning. once they open a door it is still a 1 in 3 chance of winning if you do not switch the doors but if you do switch it is 2 in 3. If you make a selection when the odds are 1 in 3 it cant change the odds by revealing one of the doors. It can only change the oods if you make a new selection. The simple reason why that makes sense is because if you chose the correct one off the bat they wouldnt open a door to give you the option of switching so they only open a door if you didnt choose it to begin with which changes the possible combination outcomes.
I find that people have a heck of a time realizing that (if one picks and sticks), actions after that decision make no difference whatsoever as to the outcome
If monte does nothing, she will win 33% of the games
If MONTE opens a goat door, she will win 33% of the games.
And if Monte would even open a car door (when one is there) she would win 33% of the games.
Also, I think some of these do not realize the simple concept that if we know "she will win 33% of the games", that means that in any one game the chances of winning by sticking is "33%"
Love her well deserved roast at the entire community at the end 😂😅
actually it wasn't an insult it was more her a opinion which i believe is correct, people always refer to intelligence when one is successful or is a failure
Shame on the comunity for letting themselves being tricked with an statistics trick. It is a simple sophistry. But no, changing your first option will not give you more probability to get the prize. The presenter can not open the cars door nor the contestan first election, that means that the posible 4-2 change-nochange will end in a simply 2-2.
People learn much better when they're allowed to follow their interests. If the subject is something that bores you, you'll only retain the information for a required time (test date, usually), but if you are interested, you'll track down information and fill-out the subject more thoroughly. This is the way I educated my youngest son - he chose the subjects and the timing. He graduated top-of-class in Navy Submarine School and is now stationed on a nuclear sub based in Hawaii (his chosen profession)
ADHD compounds this problem big time. Someone with ADHD is borderline incapable of learning things that don't interest them at all, yet I believe it becomes an advantage when we really are interested in the subject because we can devote hyperfocus to it.
@@jasondashney no kidding, I'm just a teenager, and yet I can understand, and I know about, many things that adults, even in my field of interest often don't know.
I was let to go my own direction, and I'm great at it!
I believe it. That's great you understand that about yourself. Keep that in mind when you decide what to do with your life.@@Metal_Master_YT
@@jasondashney thank you! :D
I also have ADHD, but I feel like I can "tame" it and use it to my advantage, kind of like you said.
This is interesting to hear. The school-system bothers me alot. For me, voluntariness is essential for sustainable learning.
which property of the human brain makes so many people fail in understanding the solution of this problem?
well, i decided to call it the "1 out of n thinking". we tend to assume, that a choice 1 out of 2, 1 out of 3 and so on, has always equally spread probabilities 1/2 and 1/3 and so on.
since so many simple problems can be solved by this "1 out of n thinking", we tend very strongly to apply it, even if it doesn't fit.
when exactly is the "1 out of n thinking" useful?
1)if there is a mechanism, which spreads the probability exactly equally over the n given choices
or
2)if we don't have any further information, how exactly the probability is spread.
what dos this mean, regarding the monthy hall problem?
it means, we tend strongly to think "1 out of 2"-->fifty/fifty, neglecting that we already know something different about the probabilities.
Agreed. I would say the 1/n thinking is the limit of most peoples' understanding of probability. They learn that shortcut, which usually works, and that's all that sticks. The bigger question for me is why do so many refuse to even consider alternative explanations? Why do they start from such insulting, closed-minded positions?
@@Hank254
"The bigger question for me is why do so many refuse to even consider alternative explanations? Why do they start from such insulting, closed-minded positions?"
i'm not shure about it, but i believe, it is a mixture of things:
1)you're pretty shure, you're right and someone else contradicts repeatedly. this is annoying, even if you're not right.
2)no one is happy with being wrong and less happy with being told to be wrong.
3)if a person doesn't want to feel a little shame, it is 'better' to feel anger instead.
4)to many people, not being right is equal to loose 'dignity'. they prefer to claim they're right, even if they know, they're not.
in my life, i ran into many people, whose intelligence was very different from mine.
there is a tendency, that quite stupid people claim stupid things, loudly and very confident.
sometimes, i found myself as well on the side of the louder, confident, more stupid ones.
it was for example, when i thought that one of my university math teacher was wrong.
i learned, that such persons usually are not wrong and you better think five times before you contradict them......
...on the other hand: the more intelligent someone is, the more he/she tends to doubt him- or herself.......
@@HollyGraham-wl6ye wrong.
@@HollyGraham-wl6ye
ITS 2/3 TO 1/3 YALL
THE 3RD CHOICE WASN'T ELIMINATED> U DUBBLE UR TSCHAANSSESS
MYSTERY DATING GAME KID LOL THE DREAM DATE BEHIND THE DOOR
YO SUCK
@@HollyGraham-wl6ye SPAM SE SPAMMER MAN YO KNOW ITS DUBBLE CHANCE IF YO CHANGE
I remember a Q&A column she had entitled, Ask Marilyn.
I was skeptical until I wrote this: There are 3 options:
I select a car, then the host reveals a goat and the remaining door must contain a goat.
I select a goat, then the host reveals another goat and the remaining door must contain a car.
I select a goat, then the host reveals another goat and the remaining door must contain a car.
In 2 of 3 cases the remaining door contains a car... so I should switch to it.
Correct.
She cheated. Door 3 cannot be the car, so games 3 and 6 cannot be indicative of increased probability by switching.
@@polygontower Not after it is shown to be not have a car.
Now do the same, but assume you had selected door 2 on the beginning...
@@kelseysmart9472 the point of the game is that the host is eliminating one of the doors from the equation. From there you are choosing to switch your answer to the remaining door or not. Wether that is door 2 or 3 I do not believe matters.
The most simple way to understand the problem is that, only when you choose the door having the car, switching gets you a goat. Now, you only have 1/3rd chance of that happening, that is rather to your advantage. So, in 2/3rd of cases you won't choose the door having the car, hence switching fetches you the car.
When you pick the first time you have a 1 in 3 chance of winning. But what really happened is that 1 door got removed and you was ask to pick a second time your odds no matter the door and with no other info became 1 in 2 chance's of winning no matter if you picked the same door or changed doors.
@@RipMinner no, door with a goat got removed.
The door you chose still have 1/3 chance of having the car same as before.
Yea, basically, you want to choose a goat first , then you're guaranteed to win by switching. Since THERE'S a 2/3 chance of picking goat, you got a 2/3 chance of winning. The only way to lose is to pick the car first, only 1/3 chance.
@@hmm1778 Because 2=3?
@@hmm1778 no...your chances are 1/2 now. Consider the 3rd door was never there. What now?
Without a doubt the best explanation of the Monty Hall problem I have ever seen. I finally understand it! Thank you!
Exclamation?
No you didn't
😂
I think this one is better. (Monty Hall Problem (best explanation) - Numberphile) because it emphasizes the importance of Monty knowing the location of the goats. I think lots (if not most) people get lost because they think it's a completely random situation.
Think of trying to first pick a door with a goat. Try to do that. You have a 2/3rds chance of picking a goat door. When he shows you the other goat door, switch. This method gives you a 2/3rds chance of winning. More information always helps.
only when the information is useful...this is still a 1 in 3 chance for each door. it only helps if the host wants you to win and you trust them to be leading you correctly...otherwise the "additional linformation" is useless.
It really makes no sense…
Why does the extra 33.3% (of the 3 options) go to any specific option? Where is the logic for that?
If you have for example a coin with 3 sides. But if it lands on the extra side, you reroll (ignore it)… You still have 50% chance of being right.
Where does the idea that the extra % goes to the other option than the one you chose at the start come from?
@@heihvyegs It makes total sense. probability can be calculated by enumerating all possible scenarios. Write them down for the three doors and see what you get.
@@tjmozdzen I’ll write them down later.
But… You can eliminate everything before the final option “Do you change from A to B?” Because you will always end up at that same point…?
@@heihvyegsThe logic cant be found easily in a 3 door example. There probably you will have to do the experiment 100 times to get an advantage. The real benefit is shown in the 100 doors example. Imagine you run the test of 100 doors 100 times and you stick to your original choice each time. You will win once. But if you change all the time, you will win more than once. Cause more doors means more information. In 3 doors the information gain of revealing a door is the minimum and difficult to perceive. You just have to magnify the problem to get "faster" wins. It is the same with the stock market. Big companies have billions to invest and manpower to do it multiple times ino order to get that 0,0000001% profit which is mathematically guaranteed and real big if your assets are in 100s of billions. You will never bother to spend 6 months of your life to win 1 dollar out of 10K, though, just by yourself. The latter has generated a perception that its impossible to get a "vector" in randomness. But normal distribution is the first proof that randomness indeed can have vector to show you direction.
The best way to understand this is to imagine doing it many times. If you employ the strategy of always staying with your original decision, you'll only win a third of the time.
They did it on Mythbusters. Switching works.
@@archwaldo Anyone who can write macros is a mythbuster here xD
It's not hard to understand the pyramids !! By vibrating within the earth's "B" field they could communicate as well as travel intergalactically. My goal is to educate others. Just for clarification. The center of a magnet is by far the most powerful. AC current Radio frequency signals could not travel without the "B" field. The oscilloscope never shows the "B" field but it is there. otherwise it would show a short circuit as the two polarities collide. The "B" field pushes everything forward. Things you can do to see the "B" field:1> vibrate the "B" field of a magnet near a trickle of water and watch the water repel the same.2> Vibrate the "B" field on plastic or glass and watch the item lose weight!! You must understand, don't try this on iron or metal; it will distort the magnetic field . 3> Vibrate on Granite rock and it will become weak. and You can cut and shape ONLY with copper tools. AGAIN pay attention to the magnetic field you can't use iron tools for this action. . 4> I have not proven this part yet, but you should be able at the correct Two frequencies to levitate non metallic objects. FYI I think that's why most UFO's don't show on radar they are not metallic.
I have come to believe we are all operating on the magnetic "B" field. The entire universe is 100% magnetic, we are an algorithm of the same.
Think about this : The pyramids were communication and transportation device's of yesteryear: When you install such a huge mass within the "B" field of earth they intercommunicate via the "B" field. FYI I have teaching credentials and much much more. Want some Proof, Check out Coral Castle in Florida on CZcams, all his generators were set up in the repel mode to maximize the "B" field into the earth FACT!! The stones at Stonehenge and KT were part of the "B" field system, The stones would vibrate at frequencies that would stimulate the crops and many other things. Those were the true power plants of yesteryear. WB6HUN/1958
I remember when she presented this puzzle in the newspaper. I was one of those who was skeptical, so I set up a quicky computer program to simulate it over many trials to see what would happen. I was surprised to find that she was absolutely right! lol
I cannot believe it.
@@cvn6555 Why can't you believe it?
What about intention in this equation. The people at Let's Make A Deal know where the winning door is, so their intention could be to help or to harm. Is that not a variable.
@@joewhite4170 Yes, that is why they only open a door that does NOT have a car behind it when they open one of the doors. ;-)
I just laid it out like at 3:24. It took 2 minutes, and made it apparent why it's better to switch in a concrete and intuitive way.
A classic case of conditional probability that can be solved using Bayes' Theorem. Rather telling that supposed experts in probability theory didn't try that first before sounding off. People posting "PhD" and "Professor" after their name is attempting an argument from authority, which really shouldn't cut it in a mathematical discipline...
The best explanation ever of the game and the solution, Thanks
@@johnp.johnson1541
Assume you stay with your first pick.
If your first pick is Goat A, you get Goat A.
If your first pick is Goat B, you get Goat B.
If your first pick is the car, you get the car.
You only win 1 out of 3 games if you stay with your first pick.
Switching means the opposite.
It's just basic math/logic kids understand.
Sadly, it's far too hard for johnp.johnson1541, the idiot among idiots, and his parents, the ultimate idiots.
I love that the professors were humble enough to publicly retract their first responses
Whether their retractions were praiseworthy depends on why they did so.
They probably wouldn't have written them to a man in the first place
Nah! They were more worried about how they would be received if they did not. The humble would not rebuke someone in public, as in a letter to the editor!
@@Elle-ht3km There was an extreme amount of sexism back then, and there still is a shocking amount now, but I don't think all of them were because she was a woman. At least a couple acknowledged that she still had higher IQ than them, and thought she simply made an obvious mistake. Even the cleverest people can do incredibly dumb things sometimes. I'm sure even Marilyn oversaw something really dumb in her private life at some point; and they probably thought it was one of those, but in public. They did grew heated to make sure she "corrects" the mistake and thought it was so obvious that it would severely undermine the population's understanding of probability. It was the exact opposite, but I can definitely see why they'd think it was a mistake.
At least one of those letters was definitely sexist though, and knowing those times, probably a lot more. There's probably a lot of enchancement on it too, after having been told she's more intelligent than you are and then given actual proof of that statement. Thinking someone make a dumb mistake when you're both sexist and jealous would probably compound and get you extremely heated up. And then after she proves you wrong, and you get personal proof that you're dumber, you'd either get more embarassed and angry or you'd be embarassed and start respecting her.
#MeTOO
She literally did a move whose only logic is used to the fullest in Minesweeper.
And that is the true genius behind it.
We make these mathematically rational decisions when playing minesweeper and don't even realize we are doing it.
She points out how that basic algorithm can apply to a gameshow with only a 1 x 3 row of potentially mined tiles, and everyone looses their minds.
The first tile clicked in minesweeper is the 2nd safest tile on the board, and is never mined.
Making the chosen door a door that reveals other doors' contents.
I remember playing minesweeper. It came down to luck at the last move while the rest were calculated. Brilliant reference.
I was amazing at Minesweeper and yes, I remember using similar logic to determine the best odds of not choosing a mine. I was wrong some of the time using what I thought was 'stellar logic', but most of the time I was correct.
It's good. Typically with Minesweeper I stuck to 100% probabilities. I don't remember many games where guessing was really important.
The first two doors went from having a 1-in-3 chance to having a 1-in-2 chance, so there's no advantage to switching to door #2. The showing of a goat behind door #3 had an equal effect on door #1 and door #2. Ya'll think the door #2 went from a 1-in-3 chance to 1-in-2 chance, which it did, but then you discount that the probability for door #1 happened to change the same way. You incorrectly stuck to its probability before door #3 was opened. Its probability reality changed when door #3 was opened. Marilyn showed her intelligence, not in her answer, but in her ability to persuade others to agree with her answer.
Assume you stay with your first pick.
If your first pick is Goat A, you get Goat A.
If your first pick is Goat B, you get Goat B.
If your first pick is the car, you get the car.
You only win 1 out of 3 games if you stay with your first pick.
Switching means the opposite.
It's just basic math/logic kids understand.
Sadly, it's far too hard for idiots.
Your math skills are poor. There are three closed doors....
A) The door the contestant picked Probability of 1/3
B) The host's door that must have a goat Probability of 0
C) The host's other door Probability of 2/3
If B is opened then A and C remain unchanged.
Wow. Nobody has ever thought that.
If only there was a video that explained why it's not 50 50
Geez, just GEEZ!
Is it so FREAKING difficult to realize that the initial pick has a 33% chance , and that monte opening other doors does not alter that simple fact?
And is it SO hard to realize that those other 2 doors have a 67% chance of the car being behind one of them?
And that if you switch, you get the care every time it happens to be behind one of those other 2 doors?
But yet, reportedly about 90% get it wrong, even maths professors. This is indeed a great puzzler.
for all those, who claim "one door eliminated--> game reset".
imagine, you choose 1 out of three, where 1 door hides a win.
then, you add 97 doors, each with goats behind.
now, what would you say?
"100 doors-->new game, chosen door has 1/100 odds"???? 🤔🤔🤔
You are comparing eliminating 1 door to eliminating 98 doors, it's a strawman.
@@EthanFaidley
Assume you stay with your first pick.
If your first pick is Goat A, you get Goat A.
If your first pick is Goat B, you get Goat B.
If your first pick is the car, you get the car.
You only win 1 out of 3 games if you stay with your first pick.
Switching means the opposite.
It's just basic math/logic kids understand.
Sadly, it's far too hard for Ethan, the idiot among idiots.
@@EthanFaidley
"strawman."
the only thing you prove by calling it strawman is, that you don't understand properly what i wrote, or, in the other case, you simply want to deny that i'm right.
The way my luck runs I could choose all 3 doors and I would still lose.
Nah. I would choose 4 doors (in a gameshow with 3 doors) and still lose.
Please don't be self-deprecating
🤓@@Davewutsup
You’d get a goat. How can you consider that losing?
This is so unnecessarily complex!!! It's easy! If you chose one of the goats, then the two doors left are goat, and car. The host shows you the goat, so because the car is the one left, you would win the car *if you switched*. If instead, you had originally chosen the car, the two doors left are goat and goat. The host shows you one of the goats, so in this case, you would win the car *if you do not switch*. Because the probability from the beginning of you having chosen a goat is 2/3, that means you have a 2/3 chance that switching the door will get you the car!!!
yes, that's all.
What are you all smoking? The elimination of the goat and the 3rd door does not make switching now 2/3. lol, it just makes your original choice now 50% right or 50% wrong... instead of an original chance of 33.3333333333% 😂
@@JCMW-hw9jl
Why don't you test your clueless claim?!
Ask one of your friend to hide a ball behind 1 of 3 identical boxes, and you select your box.
When you do that, as your friend to show you one of the other two boxes, which doesn't hold a ball behind.
Then inspect where the ball is:
- if it is behind the box you picked, add 1 to the score of "staying"
- if it is behind the second box, add 1 to the score of "switching"
Repeat at least 30 times or more.
If you are right, and both boxes had a ball behind equally likely (50% chance) scores will be approximately equal (couple of points difference).
But, if you are wrong, "switching" score will be approximately double the "staying" score.
Now, I wonder what will happen....
LOL
@@JCMW-hw9jl We have tested it, you are wrong. The question is, why don't _you_ test it.
@@JCMW-hw9jl No, because the elimination was not randomly made. The host already knew the locations and was carefully avoiding to open the player's choice and also which contains the car. Those two doors are not the same 2/3 of the time, so revealing the third one will not make them stop being different in those 2/3.
Some people understand it by thinking that the host is showing you where the car is everytime you start failing -> in the other door that he leaves closed besides yours. And you start failing most of the time, not 1/2, so better to trust on him.
It took me awhile to wrap my head around this, but it's really quite simple now that I get it. There's a higher probability that the car is behind one of the two doors you didn't pick.
"There's a higher probability that the car is behind one of the two doors you didn't pick."
Exactly.
And also, the fact that host opens the door with a goat, doesn't change better odds of a host in any way.
@@max5250but how is there a higher chance it’s behind a door I didn’t pick
@@GoyMaster
You didn't picked 2 doors, not just one.
Two doors are twice as likely of having a car behind than a single door.
@@GoyMaster Put it simply, there are more wrong doors when you first pick so its more likely you got the wrong door.
@@HollyGraham-wl6ye
"ODDS OF THE HOST ? THE HOST- DOES NOT WIN ANYTHING"
If player haven't picked the door with a car, can you guess where that door is?
If player decide to switch, whose door he will get?
"THERE ISNT -- ITS BEHIND THE DOR OR NOT"
Yup, it is either behind the door or not, now, if you could only calculate probability.....
ROTFL
Ok…here is an explanation.
Let's imagine the car is at Gate 1:
We choose Gate 1: presenter can only show Gate 2 or Gate 3
If we change, we lose
We don't change, we win
We choose Door 2: Presenter can ONLY show Door 3
If we change, we win
We don't change, we lose
We choose Door 3, Presenter can ONLY show Door 2
If we change, we win
We don't change, we lose
If we change, in 2 (of the 3 scenarios) we win
The key is that, in 2 of the 3 scenarios, which are the cases in which we choose the door where the car is not, the presenter can ONLY show us the other door where it is not... therefore, in those 2 cases ( of 3) if we change we win.
The same analysis can be done for the assumption that the car is behind Gates 2 and 3.
Only someone brilliant could figure it out.
I think a more intuitive way of understanding this problem is recognising how the question has changed from the first choice when offered the second choice.
The chance of being correct (selecting the car) on the second choice is dependent on being incorrect on the first choice (having selected a either goat). Therefore, the chance of selecting the car in the second choice is the chance of NOT selecting the car in the first choice, which is the chance of selecting a goat on the first choice. This is 2/3.
A very logical explanation. Thank you.
No, you have the choice to keep your door, or pick the other one = 50%.
@@R391s And the odds of your door having the car are 1/3 and the odds of host's door having the car are 2/3.
@@R391s Let’s run through the different scenarios/outcomes (allowing for where the car is and whether you choose to stay or change your answer):
Car behind door 1 and you keep your 1st answer
- Choose Door 1. Shows Door 2 or 3. You stay at 1. Therefore: WIN
- Choose Door 2. Shows Door 3. You stay at 2. Therefore: LOSE
- Choose Door 3. Shows Door 2. You stay at 3. Therefore: LOSE
Car behind door 1 and you change your 1st answer
- Choose Door 1. Shows Door 2 or 3. You change to 3 or 2. Therefore: LOSE
- Choose Door 2. Shows Door 3. You move to 1. Therefore: WIN
- Choose Door 3. Shows Door 2. You move to 1. Therefore: WIN.
The same outcomes will occur if the Car was behind door 2 or door 3.
You may notice that if you stay with you answer you LOSE twice and WIN once. Therefore, your probability remains at 33,33% (as before the door was revealed).
However, if you change your answer you WIN twice and LOSE once. Therefore, your probability of choosing the car goes up to 66,67% when you use the strategy of changing your answer.
Showing the above by force is not ideal, but it illustrates it visually very well (imo). However, I would implore you to consider looking at the Bayesian Statistics proof or running a simulation (with Python for example) if you don’t trust the above at face value.
the chance is rather illusion, if someone else chooses the other door then what happens
I knew what this was... I remember the series Numb3rs where charlie explained this exact thing with 3 cards, 2 goats and a car winning price and I was (as always when and how he explains things) shocked and fascinated. But in the show he explained it pretty much the same way and the conclusion was "always switch! your odds are always higher because you picked twice so you had a 2/3 chance instead of a 1/3" something like that. And I think that show/episode was like 20years ago. crazy, but man I miss this show
You missed the whole point, which is that the quiz master KNOWS which door the car is behind. There is a 1/3 chance you got it right the first time. IF you got it wrong, probability 2/3, then the car is 100% certain to be behind the door the quiz master left closed.
in 2 out of 3 cases where they show 1 of the goats (not your goat or car) you win by switching. (There are only 3 outcomes in combination).
I believe one of the circles of hell in Dante's inferno is trying to explain the Monty Hall problem to people who can't understand it and who smugly insult you for saying the correct answer
A fool takes no pleasure in understanding, but only in expressing his opinion. Proverbs 18:2
Have the person trying to understand the problem play the role of Monty Hall using playing cards or something with someone who will always switch. They should soon realize that Monty Hall will have the car 2/3 of the time.
Host is not allowed to interact with door that was initialy selected that is why its chance for having a car do not change.
So you ending with tree doors one with 1/3 chance(the one selected when no information about one of the doors was known), one with zero chance as the host is allowed to open only the door whithout car and the third door takes the whole propability of the group of doors that host is allowed to interact with => 2/3.
I actually read the Parade article. I had 5 classes of math in high school and one chapter was on probability. One thing that stuck in my mind was 'All RANDOM choices must add to 1.' So if I had a 1 in 3 chance the first round, and Monty's choices are not random, then my second choice is actually a 2 in 3 chance.
Yes but you can choose to choose the same door
@@MrSanford65 No, you already chose that door and it had a 1 in 3 chance, not choosing the other door means you still only have a 1 in 3 chance. It only changes to 2 in 3 if you switch doors.
The table is flawed logic because of the labeling of the doors. And that's why it becomes so confusing. Only the first two entries of the table are valid. The rest of the entries are actually just permutations of the first two.
Imagine the doors had no numbers or arrangements and you didn't even know which one you picked, only that you picked one (of three). Now you were told, you thought you picked one of three, but, it's been reduced to one of two. Would you like to change doors?
There is literally no sense in changing.
I'm saying that I look at it this way: Instead of picking door 1 of three...
I pick one door out of three.
This is where the table becomes redundant. The first two entries are the only useful ones.
After studying it for about an hour, I see that I am wrong. I have to say that is very counter intuitive. Thank you for posting... 👍
After the first door is opened, the problem is changed. The remaining doors have a 50% chance of having the car.
But when you chose the first door, it had a 1/3 chance of having the car. So you get to 50% when you choose the other door.
This clicked for me in that it’s contingent on the “host” actively showing a “goat”. Which, in retrospect feels dense of me to not assume that originally. Thank you.
But only knew this from 21 and annoying friends pretending they can explain it. I can’t recall if 21 presents it this way
Simplest way to explain this, you have a 2/3 chance to choose the goat, and 1/3 chance of choosing the car.
Its always better to switch since your most likely to choose the goat hence giving you a 2/3 chance of winning.
I understood the logic like this:
1. If you switch, you will always get the opposite of your first pick.
2. You are more likely to pick a goat the first time.
Yep.
Yet, this sound like differential calculus to some people on CZcams.
Another way to try to grap it with intuition is to think about it the other way around.
Initially the Chance to pick a goat is 2 out of 3 and to pick the car is 1 out of 3
So the chance to pick the wrong door is 2:3. At this time the host has only ONE door left to open, as YOU already picked the one with a goat.
Therefore by switching the door you CONTAIN this 2:3 probability. In 2 out of 3 cases you picked the wrong door and the only one left is the one with the car as the host ALWAYS opens one with a goat.
Or out of three shots you wouldn't put the goat side by side, also understanding how the conditioned brain works.
Unfortunately it's not clear how it's not 50%. Even if I picked the right door the first time (say door 2), the host would have picked the wrong door (say door 3) and offered me door 1 or 2. The same applies if I'd picked door 1. So to me I don't see how the hosts actions increases the probability of the unpicked door having the car.
@@nitinullas it's not 50% because there is a much deeper intelligence at work, you have the social norm mathematical equation that would give door one the better chance. 😉 and then you have the social norm that only sees what's right in front of them. Think deeper or overthinking makes for a much wider probability of intelligence.
Hint: shallow thinkers of black and white only see 50% chance.
@@nitinullas It is quite clear.
1/3 times you pick the car. This means your odds of winning by staying are 1/3. By staying you only win when you initially chose the car.
2/3 times you pick the goat. This means your odds of winning by switching are 2/3. By switching you win by initially choosing a goat and switching to a car.
@@nitinullas Well, the probability does not change. That is part of the reason. And the host itself is not free to do what he wants - he is restricted in his actions as he is only allowed to open a door with a goat. If he would open doors freely and the doors would contain like envelopes with a key or a "you lost" note the probability would not change.
Think about it this way. Your chance to pick the correct door is one out of three. In this case the chance that the car is behind one of the other two doors is two out of three.
This does not change when the host opens a door. The chance you chose the correct door is still one out of three and the chance that the car is behind one of the other two doors is still two out of three BUT the host opened a door with a goat as he is not allowed to open the door with the car. Therefore the chance that the door you did not chose contains the car is two out of three.
The point is. You chose door A. Car can be behind A or B or C.
If it is behind A the host can either open door B or C. If you switch, you lose.
If the car is behind door B the host can ONLY open door C. If you switch to B you win.
If the car is behind door C the host can ONLY open door B. If you switch to C you win.
Even in this small example you can already see that you win in two of the three cases if you switch.
It's funny how so many of the replies below think they are giving us a simple "look at it this way" explanation but they are not making it any clearer, usually because of bad grammar, typos or because they've worded it in a way that makes more sense to them than it does to a reader. I'll go home and have a puff or two and figure this out.
it's funny, how so many people stick to the simple and one minded
"1 out of n-->chance must be 1/n" theory.
one could believe, that a problem, that was solved in 1975 works the way, the solution proposes rather than ones own first thoughts about it.
@@KarlHeinzSpock 🤣🤣🤣
@@GuyGabriel-eu7hb
"🤣🤣🤣"
very impressive "argument".
@@KarlHeinzSpock Was I making an argument? Nope. But you're original reply is hilarious, giving us more bad grammar and typos. 😂😂😂
@@GuyGabriel-eu7hb so you don't even know the possible meanings of the word "argument"?
i knew, you were not an intellectual highlight, but i didn't consider your brain being that poor.
That's mind blowing!
Technically out of 2 doors you have a 50 50 shot . But because we started with 3 doors switching doors at the end will always produce a better result. Her math is perfect.
It's easy af, by switching you pick 2 doors, the wrong one and the unknown one, that's 2/3 or 66%. If you don't switch you pick only 1 unknown door, that's 1/3 or 33%.
Some people believe it's always 50% but you have to start counting your propability from the very beginning when there were 3 doors, not after one was openend. Because the first choice also matters.
disagree. You flip a coin, neglecting the edge probability, you have either tails or heads. When you flip you always have 50% chance probability of landing either heads or tails. However, flipping 50 heads in a row is extremely unlikely and thus getting a 51st head in a row is, too, extremely unlikely, despite having only 2 outcomes.
Lavant's high IQ is garbage and false or she cheated on her test. IQ test measures several things: verbal, reasoning, memory, processing speed. Yes, those 4 combine into 1 IQ score, still, I would love to see her true published IQ test. Until then I call bullshit.
@@vkaa3k190 Chance for any of the doors to have a car is 1/3. Then the host is opening one door. But host is not allowed to open your door -> your door is a separate group of propabilities -> the door that is revealed transfer its probability to the other door in its separate group...
Or
You make decision to stick with your initail choice(when all 3 doors were unknown) = you are ignoring that extra information provided to you by the host so your chances do not change...
@@vkaa3k190Mate, this problem has been resolved by mathematicians, not just Vos Savant, since the 90s... you're making a right fool of yourself here.
@@fleurdemaiko lots of mathematicians have called out her bs, so you are the one making a n**ger out of yourself. And you don't have to have a PhD in math to figure out the probabilities.
@@vkaa3k190 The fact that it is astronomically unlikely that you might flip 50 heads on a row is irrelevant to the MHP.
But if you did that , the 51st flip has a 1 in 2 chance, The past, in this instance has exactly nothing to do with the future
If you stick to the first choice, the chances are 33%, opening other doors cannot change that fact. If you switch, you get the car if it is behind either of the 2 other doors, which happens 67% of the games.