El problema o la paradoja de monty hall es un error: Se basa en un error de suma: Cada puerta tiene tiene 1/3 de probabilidad, si escojo una, las otra dos tienen 2/3 de probabilidad, eso es correcto, se ha sumado las probabilidades de las otras dos puertas cerradas, pero cuando una de las dos se abre, las suma no es válida, porque es como sumar peras con manzanas… es decir la probabilidad de la puerta abierta no se le suma a la puerta cerrada… Vámonos por otro camino, supongamos que las cabras son de oro, unos 50 kg de oro, y valen mas que el carro… asi quiero es la cabra y el abre una puerta con una cabra … La probabilidad de cada puerta es de 2/3, si escojo una, las otras dos valdrán 4/3 Una probabilidad de 4/3 es superior a la unidad, lo cual es infalible, pero no es real … la probabilidad real es 1/2, porque la probabilidad cambio a abrirse una puerta … se podría decir en este caso se redujo y en el anterior aumento, ambas quedaron en ½ se igualaron entre si …
This problem is very simple though it had me fooled initially! The key to understanding this problem is to realize that the only way to lose when applying the switching method is to *initially* pick the door with the car (which has a probability of 1/3). In that case, you'll switch away from the correct door. That's the only instance in which you'll lose. If, on the other hand, you initially pick a door with a goat (which has a 2/3 probability) Monty will show you the other door with the goat. In that case you'll switch to the with the car.
The reason it works is because you pick a door before the host, not after. If the host were to remove a goat door first, then it's 50/50, but in this case u selected a door with 3 possible rewards, 2 out of 3 being a goat. You force the host 2 out of 3 times to remove remaining goat door, so the other door the host couldnt open must have the car behind it.
Another way to think about it is tell the audience the full sorry with the original question. The host can't pick your door weather you are right or wrong so this naturally will give the other 2 doors the better odds.
Take it as Goat n1 and Goat n2. If u pick n1, he chooses n2-swich. If u pick n2, he chooses n1-switch. If u choose the car-he picks either n1 or n2. So 2/3 if u swap its the car.
@@lenahhhhh6568 oh my im happy for that, i dont remeber exactly, but i think this was exactly the case with me and then i tried to think about it alone and this is what i came up with
The Alaska Zoo got started because someone won a radio contest with the prize being a choice of cash or a baby elephant. The winner chose the elephant. And yes, there's a Simpsons episode based on that.
The host always opens a losing door in this scenario. Fucking obvious, Monty used to switch to winning doors. Just broads changing the rules to look smart
Here’s the quick version, if you switch, you’re basically saying, “I want to win a car if there’s a goat behind the door that I picked when all 3 doors were closed and I want to win a car if there was a goat behind the door that I picked when all three doors are closed. That means if we call our doors, “Doors: A, B & C” and the car is behind Door C, I can pick Door A and switch and I can pick Door B and switch and either way, I’ll win.
At best you have a 2/3 chance of winning, at worst you have a 1/3 chance of loosing. Options, A, B, C = 3/3 Initial pick, A = 1/3 so B, C = 2/3 Goat reveal, B = 0/3 so C = 2/3
Math has never been my strong suit, but I'm trying to make sense of the "always switch" logic all the same and it has me stumped. So as I understand it, the rules are that there are three doors. One with a car behind it, and two with a goat behind each of them. You want to win the car, and so your original choice comes down to a 1 in 3 chance guess at random. Wichever door you choose, the show host will then open one of the other two doors, revealing a goat, purposely ignoring the door with the car behind it if that is not the door you chose. This obviously does in fact take away some of the uncertainty: that door is now confirmed to have a goat behind it, which changes the odds. So why is the 33.333(etc)% chance from the revealed door offloaded entirely onto the other door you didn't choose instead of being evenly distributed between both of the remaining sealed doors?
The chances to win by switching is determined by the probability of you picking a goat and the host picking a goat. The chances for you is 2/3, for the host it's 1, so the chances of winning are 2/3x1=2/3. If the host didn't know where the car is then you would be correct that it's 50/50 if he revealed a goat because the chances to win by switching are not 2/3x1=2/3 but rather 2/3x1/2=1/3, same as in staying which is 1/3x1=1/3. In one case 1/3 pick cars , 2/3 pick goats and they all continue, and in the other case 1/3 pick cars and 2/3 pick goats but only half (host reveals car half of the time when he's left with a car door and a goat door) of those that pick goats can continue.
@Klaus 74: A very well-structured explanation, thank you! There is one thing I still do not understand however. You mention that the chances of winning when you switch doors are equal to 2/3x1=2/3. In other words, the chance of winning when you switch to the other door is equal to the chance that you picked a goat multiplied by the chance that the show host picked a goat. Why is this? Maybe this is an obvious one, but like I said mathematics aren't my strong suit. I *DO* understand how this same formula changes based on whether or not the host knows the position of the car though, and how this affects the odds of winning based on whether you switch or not if he does know where the car is, but leaves it at a 50/50 shot if he doesn't. (And also imposes a chance that he reveals the car behind the door after the first pick, ending the game.) That really helped me figure this out a little better, so thank you!
Glad it helped. Another way of looking at it is this....the chances of what you pick have to be the same as the chances of what's left over for the host. So when there's a 2/3 chance of picking a door with a goat then the host must have a 2/3 chance of having a car door and a goat door left over to him from which to reveal the goat and leave the car. So there's a 2/3 chance the revealed goat is the last one.
"You mention that the chances of winning when you switch doors are equal to 2/3x1=2/3. In other words, the chance of winning when you switch to the other door is equal to the chance that you picked a goat multiplied by the chance that the show host picked a goat. Why is this?" Two events must happen in order to win the car by switching. One, you pick a door with a goat, and two the host does as well. To calculate the probability of both those events happening you need to multiply their respective chances. For you to pick a door with a goat is a 2/3 chance and the host the chance is 1 because he 'must' reveal a goat according to the rules of the problem. So when you multiply the two events by their probabilities you have a 2/3x1=2/3 chance that both of you will pick a door with goats thus winning by switching. Let's take another example. You are in a chess tournament and you have to win the next two games in order to win the championship. One of the opponents you will face is is not as good as you are and you have usually defeated him in the past, perhaps three times out of four games. So the chances of defeating him again you can estimate as a 3/4 chance. The other opponent you will face is a stronger one and you estimate your chances of defeating him is 1/2. So before you begin play against either one of them the chances of beating both of them is 3/4x1/2=3/8.
@Klaus 74 and @ Georg Cantor: That cleared it up! I understand now. It's still really bizarre to me that the math works out this way, but I get it. Thank you two for your explanation, that was enlightening!
More specifically if your last selection is random, then your odds are in fact 50/50, however this does not change the odds of switching Vs not switching which is 2/3 and 1/3 respectively. Thus there are actually 3 scenarios, however if you count doing nothing at all, then your odds are 0% You always initially had a 1/3 chance of selecting the car and 2/3 chance of selecting goat. Thus by switching doors you're more likely to end up with the car, because the door you picked first up, was always more likely to be the goat and with one door and goat removed, the only remaining switch option is the car 2/3 of the time and the goat 1/3 of the time. Thus a random selection between the last two doors ( 50%), offers better odds than not switching (33.33%) at all, but not as good odds as switching (66.66%).
The easiest way to think about this is to just realize the whole "Monty opens a door" part is there to mislead you. The effective result of switch or stay is you either get to pick 1 door (original choice) or 2 doors (remaining choices). It doesn't matter that one of the two remaining choices is a door that Monty opened.
In fact, Monty opening the door does not mislead but tells you that the 2/3 chance is all on the final door... which it is every time. A, B, C=3/3 A=1/3 ... B, C=2/3 B=0/3 ... C=2/3
It's in the first act when Jude Law is in Kevin Spacey's math class. Kevin Spacey literally just asks his class this problem, Jude Law gets the solution right, and then I think he leaves.
I remember that scene, but they did a poor job explaining it to the casual viewer. I was convinced that the odds remained 50/50. They should have spent an extra minute explaining WHY the odds increased, like they did in this video. :)
But Marilyn's IQ turned out to he incorrect. Its actually around 132 i believe. Which means its not that high. Her 228 score was tested at a young age and for a didferent test, it was a ratio IQ, so while she may be intelligent her score cannot be above 170, as said by Alan S. Kaufman, a psychology professor and author of IQ tests, writes in IQ Testing 101 that "Miss Savant was given an old version of the Stanford-Binet (Terman & Merrill 1937), which did, indeed, use the antiquated formula of MA/CA × 100. But in the test manual's norms, the Binet does not permit IQs to rise above 170 at any age, child or adult. And the authors of the old Binet stated: 'Beyond fifteen the mental ages are entirely artificial and are to be thought of as simply numerical scores.' (Terman & Merrill 1937). ...the psychologist who came up with an IQ of 228 committed an extrapolation of a misconception, thereby violating almost every rule imaginable concerning the meaning of IQs
@@ifisawyourreplyiwillanswerback Of course it's based on chance; no one disputes that. The question is not whether it's based on chance but does your chance improve by switching. The answer is yes, if you stay with your original choice of door, your chance remains at 1/3 but if you switch your chance is doubled to 2/3.
I don't think three would be accurate in any capacity though since I feel like no matter what, the game show would *not* want you to win that car. Just knowing gameshows and their mechanics, they only let people win the not-so-great-but-still-worth-something prizes ("cheap" stuff that isn't cheap to the average person) to keep people coming onto the show. Any expensive prizes will always be kept for the company (they're most likely props) so that they don't lose money and go bankrupt since there'd be no more show (look at how often Jeapordy "gave out" cars and boats and cruise trips and shit. You know it's a scam. Just like those million dollar prizes were most likely paid in installments and after taxes you probably got less than half of it anyway). So yes, her solution would work just fine because there's no way in hell the company would give away a free car that's probably worth more than the host's annual salary. I can't believe these studies went through all this work and completely missed the economics of the gameshow itself haha.
I say you pick the door closest to him because if the prize is in yours he will open the closest to him assuming he's lazy but if he skips the 2nd door it's probably in the second door
but thats another assumption of his behavior, not to mention the fact that depending on the setup, he might not have to walk over to a door, the doors might be controlled by someone in a booth.
But iocane comes from Australia, as everyone knows, and Australia is entirely peopled with criminals, and criminals are used to having people not trust them, as you are not trusted by me, so I can clearly not choose the door in front of you.
We could actually test this out by sampling the real data - all of the shows are there and for each game, there would be the initial Door choice, the Door opened by Monty, whether the switch was made, snd where the good prize was. I smell a dissertation!
Is this still controversial ? Its pretty easy to understand. You just have to take into consideration the fact that the door Monty chooses to open is dependant on the first door you choose. That is, if you choose a door with a goat (66% chance) then Monty is forced to open the other door with a goat. Therefore by switching you transfer that 66% chance to the door with the car
Think about it this way. If the theory is right (and it is) and you switch, you have a 2/3 chance of winning the car. If the theory is wrong (and it isn’t) and you switch, you still have a 50-50 chance of winning the car.
If Monty revealed the car, why would he offer you a chance to switch between two zonks? So either he's supernaturally lucky to always reveal a zonk and thus not deflate the drama of the choice OR (and more likely) he will always reveal a zonk and thus Marilyn's solution is correct.
That's a scenario that doesn't happen. Of course Monty knows but all that's important is that when the goat is revealed the contest should switch. Options, A, B, C = 3/3 Initial pick, A = 1/3 so B, C = 2/3 Goat reveal, B = 0/3 so C = 2/3
“Would stir up so much controversy in academic and intellectual circles” There’s 0 controversy about this in academic and intellectual circles. Everyone with the most remote understanding of statistics has no issue at all with this. It’s people who don’t know statistics and are stubborn to trust their instincts over logic that are experiencing the controversy, nobody else.
Naturally weird Nope, who do you think would waste time preparing statistics on that that I could send you lmao. My job is as a statistician, I know what people who deal with statistics do and do not struggle with in statistics, and at least unless American statisticians are significantly less educated than European ones I would be very surprised to hear that any of them struggle with this particular issue. If your argument is that without backing statistics I can’t make any argument, then here’s a dosey for you: since the statistics don’t exist, infographics show pulled that “fact” out of their asses too. At least I base mine on interacting with hundreds of statisticians over the years and being aware with the statistical curriculum and hence that anyone with a more than basic understanding of stats should have 0 issues with this.
The problem is too simple for statistical analysis. The simplest proof is this: 1. I f you switch you always get the "opposite" of your original pick. 2. 2/3 of the time your original pick will be a goat. 3. Therefore 2/3 of the time a switch will give you the car. (1/3 of the time you will lose by switching)
Realistically, He’s always going to pick a door with a goat to show you, and he’s always going to ask if you want to switch to the unpick/unopened door, so it doesn’t matter whether you switch or not, odds = 50%
That isn't realistic because the chances of what you pick are the same as the chances of what's left over for the host. Since there is a 2/3 chance of you picking a goat then must be also a 2/3 chance the host is left with a goat door and a car door from which to reveal the goat and leave the car.
Seth Webster, If the host *_must_* reveal only a goat, and never the car... then if your first choice is one of the two goats (which is a 66% likelihood), then him revealing the second goat means that the only door left is the car. If you pick a goat, and the host reveals a goat, the only door left is the car. If instead, you pick a car, and the host reveals a goat, the only door left is a goat. Thus, if you switch, the result is that you *_always_* end up with the opposite of what your first choice was. So if you know your first choice is 2/3 chance of being a goat... then switching will mean a 2/3 chance of being a car, because switching always results in the opposite.
Simpler version. Your first choice will be wrong 2/3 times, then the host must reveal the other wrong option, so you switch and win. Therefore you guarantee a win 2/3 times by switching and that's all you need to know.
I've used this trick to make sense of this paradox in the past: the odds you've chosen the correct door are 1/3. So you should definitely switch because you probably did not make the correct choice in the first place.
@@lucyrhodes7010 the host always reveals a losing door. If your first choice was a losing door and you switch you automatically win. The chance of choosing a losing door first time is 2/3. Therefore 2/3 of the times you should switch
I looked at this problem half a year ago so now I'm revisiting it with the same opinion as before. Here the way I see it. You have a 1/3 chance of choosing the right door the first time, so, you are more likely to choose a goat at first. With only two doors left, and the assumption that you have already chosen one of the goats, the remaining door that he didn't choose to open would have the car. This would obviously not work all the time, but if you played the game multiple times you would win more often if you switched.
Yes if the host knows where the car is and always must reveal a goat. If he didn't know where the car is and randomly opened a remaining door (or if the wind blew it open) and it happens to be a goat then it's 50/50.
In the real world you are usually only given the choice of two doors to start with. The third door is only shown as a possibility afterward. Then a fourth, then a millionth. But if given the choice of three doors at the start, and on the assumption that monty will never reveal the car (which he never has) the original premise is correct from pure mathematics.
I recall watching a documentary way back when showing how those old game shows were often times rigged to make contestants win or lose based on how much the audience liked them. If you were well liked, you were more likely to win because it would please their audience and increase their views.
simple solution for winning by switching you would have to take door with goat in your first choice which has probability 2/3 but for winning by staying you would have to pick door with car which has probability 1/3... correct me if i am wrong.
It baffles me that many people still don't understand this simple math problem. This is like the math equivalent of a toddler sticking shapes in a toy box.
At first I couldn't understand it, but then I thought that when picking the first door you have 1\3 chance to win, so because of that it's better to switch door since there is a 2/3 chance that the other two doors were winning, but it is reduced to only one door when one is opened, while you still have 1/3 chance on your door if you don't change Does that thought process make sense?..
In the MHP yes because the host knows where the car is and must reveal a goat. If he didn't know where the car is and randomly opens a door not picked by you and it has a goat then it's 50/50.
Klaus 74 Only the host has 1/2 chance of picking a goat (assuming he doesn't know where the car is), if he picks a goat, randomly or not, the chances of you getting the car if you switch remain the same in every case
What are you talking about??? If the host doesn't know where the car is then 1/3 of the time it is behind your door, 1/3 of the time it's behind the door you can switch to, and 1/3 of the time it's behind the host's door. The chances to win by switching is 2/3x1/2=1/3, same as in staying which is 1/3x1=1/3. So if a goat is randomly revealed then it doesn't matter if you stay or switch.
If you take 99 games and the host always reveals a goat then 33 pick cars, 66 pick goats and they all continue. If the host does not know where the car is 33 will still have picked cars, the host picked 33 cars and there are 33 cars (not 66) behind the door that you can switch to. All those that picked cars can continue, and half of those that picked goats can continue. 50/50.
"That means there's a 2/3 chance the car is behind one of the other two doors. Those odds don't change because Monty Hall shows a goat behind one of the doors -- whether he knows what's behind the door or not." You are wrong when it comes to the host not knowing where the car is. What's so good about leaving twice the cars for the host when half of time he leaves it for you to switch with and half of the time he reveals it?
Why? There aren´t just 3 but 4 options. (In the example I will assume that you always pick the first garage.) AZZ and he shows you the first Z AZZ and he shows you the second Z ZAZ and he shows you the second Z (because you pick the first) ZZA and he shows you the second Z (because you pick the first) There are 4 options and in two of them you should change. (2/4 or 1/2)
Since you are 1/3 likely to pick the car door ar first, the two first cases should fall into that 1/3, so each will have 1/3*1/2 = 1/6. You cannot compare those cases that have 1/6 with the others that have 1/3.
I think the first two cases are something different. Then there are 4 options and if you pick one of the first two you shouldn´t change and if you pick one of the last two you should. It is two different cases if he shows you goat one or two.
There are 4 options but two are half as likely as the other two. In order to get probabilities by counting cases you have to make sure that everything you count as a case is equally likely. If we count cases as you are doing, we are assigning two for having hit the car, and 1 for each goat, so if we played a lot of times, like 100, we should have selected the car about in 50, the goat 1 about in 25, and the goat 2 about in 25, which is absurd. There is no reason to think that we are going to hit the car door firstly more often than the other two. Imagine you played 900 times. Let's put the numbers, assuming that the sample proportions exactly match the probability, that is, you should pick each content 1/3 of the time. 1) In 300 games your door has the car. 1.1) In 150 of them the host reveals the goat1. 1.2) In 150 of them the host reveals the goat2. 2) In 300 games your door has the goat1. In all those 300 the host reveals the goat2. 3) In 300 games your door has the goat2. In all those 300 the host reveals the goat1. - If goat 1 is showed, you can only be in case 1.1) or in case 3), so you are under a subset of 450 games. You win by staying in 150 of those 450 (case 1.1) and by switching in 300 of those 450 (case 3). • 150 represents 1/3 of 450. • 300 represents 2/3 of 450. - If goat 2 is showed, you can only be in case 1.2) or in case 2), so you are under a subset of 450 games. You win by staying in 150 of those 450 (case 1.2) and by switching in 300 of those 450 (case 2). • 150 represents 1/3 of 450. • 300 represents 2/3 of 450.
Think like this, - There are 1000 athletes, who have been ranked among themselves. You are asked to guess the athlete who is ranked 1. You choose one of them. Now divide 1000 athletes in to 2 groups. The person you chose is in group A, and rest 999 men are in the other group B. Now Monty Hall eliminates 998 of the athletes . Now u are asked to choose whether you want to stick to ur first choice or switch to the other athlete. Now see, the last man standing in group B has defeated 998 men, and till now u have no information about the ability of the person u first chose. Which athlete you want to choose? The person who u know has defeated 998 men , or the person about whom u know nothing?
I mean this is all assuming that he wants you to lose. Assuming he doesn't care about the outcome and just wants to build tension and suspense, than he would show one of the losing doors every time, making any logical meaning behind showing one losing door irrelevant. It's for the views not the wins and loses.
People who didn't get it. There are 3 doors..A,B,C. Probability is first equally divided..i.e 33% in A, 33% in B, 33% in C. You selected A(doesn't matter now) So, right now, probability of car in your selection(A) is 33%, while in B+C combined(others) is 66% Now, the guy showed you one door out of B and C, and there was a goat behind it.(Say C) That means your initial choice was 33%. Other was 66%..now, out of others C got eliminated and now B has 66%...while your initial choice is still 33% because B& C's combined was 66% and C got eliminated.
That is absolutely wrong B+c ≠ 66% Cause we know for sure that one of b or c has a goat So when separating the 3 doors into 1 single door vs 2 doors combined It becomes 50/50 because 1 of door b or c = 0% We just dont know which one so the host opens that door Now we have literally 50/50 shot at the prize
It's just some condition probability, thats most people with no mathematical background find hard to understand. It's not that complex, you don't need to be an expert to understand it but I guess most people don't really understand non intuitive maths and are just arrogant.
Random Internet User I believe most people understand this theory, after all, as you also said, it's not a complex one. People are just getting it wrong the first time, because they don't really think about it and just go with their gut.
they just word this terribly, all they have to say is, " you are more likely to pick the wrong door first, then after he removes the other wrong door, the right one will remain 2/3 of the time
Ethan Barnes I think the wording isn't that bad, sure it can be a bit confusing, but IMO i think the explanation is what confuses the regular viewer more so. I think it makes more sense by making a table with all the possibilities and then counting them to find the probability at the start and then when a goat is revealed.
Random Internet User: The question really should state whether the host chooses randomly or always chooses a door with a goat. If that's not specified, it's more natural to assume the host chooses randomly, which makes the 50/50 solution correct.
Surely the easiest way to look at which is the better option would be to say, you pick a door, you can either keep that door or swap it for the other two? That is what is actually being offered isn't it?
You are doubling the chances of winning in both cases but for different reasons. In one instance the host must know where the car is and must reveal a goat in order to double the chances (otherwise it's 50/50) and in the other case swapping for the two other doors does not require of the host to know where the car is at all. One is a conditional probability problem with a subset of doors having 0 and 2/3 chances of having the car while the other is an unconditional probability problem with a subset of doors having a 1/3 and 1/3 chance of having the car.
@@klaus7443 l was implying the chance changed in anyway Klaus, more how to explain how the ratio is how it is. Imagine Monty didn't open the door until after the contestants had chosen whether to keep their original or swap. Ie Monty gives the option to keep original or opt for the other two doors and contines by saying "one of the other two is definitely a goat allow me to take it away" By doing this the odd on whether their original pick is correct haven't changed in the slightest. I thought this may have been an easier explanation of the 1/3 or 2/3 option.
@UCPsKVRwX525ZRoaISJl3Ijg Hi Klaus, I'm bot disagreeing with you in anyway. Other than, this video is about the ratio and the controversy surrounding its eventual conclusion.
I won't discourage you for it's usage then. I'm just speaking from experience in dealing with many here that know switching doubles the chances of winning but they also say it doesn't matter as to whether or not the host knows where the car is or not. If they know the answer but not the reason why then good luck in helping those that think it's 50/50 with that explanation.
Yes! That is the whole essence of the problem. And so, so many people can't see it. If contestant picks door #1, and Monty offers to let contestant swap his unopened initial choice for BOTH of the other unopened doors, it would be a very stupid contestant who wouldn't take Monty up on his offer. In regular play, that's essentially what happens.
Yes it makes sense... It works because the host know where the prize is, so for the sake of continuing the game, the host is always gonna keep picking up the goats until all the choices are exhausted. You can visualize it better if you have many more doors. So if that's the case, you'd ask why did the host open that goat door instead of the one in the middle? Because opening the middle door exposes the prize hence the end of the game. So it makes more sense to switch... Or course it's probability. Can still lose
Exactly. The game has to always have the premise that you can win based on your choice. The situation can never be that you lose the game stick or switch. Monty must know since he would never reveal the car, always a zonk. Knowlege you have effects the odds.
it's not 50/50 solely because when you make your first pick, you're picking between 3 doors not 2 THEREFORE you're more like to have chosen wrong BECAUSE there are more wrong choices than right ones.... it would be 50/50 if say you got to pick after the host opened one wrong door, but you're picking BEFORE that when there's 2 wrong doors.
imagine there are a thousand doors instead of on 3. You choose one, and then the host reveals 998 of the doors leaving just a door and the door you picked. which door do you think it is? It's better to switch because there is less chance you got it right and more chance that the other door is right.
An easier way to explain the problem: 3 doors, 2 doors with goats inside and 1 door with car inside. Pick a door, your chance to get the goat door is 2/3. This means you are more likely to get the goat instead of the car in the first place. So switching gives you better chance to get the car.
Lmao I thought it was common sense to choose to switch to the door that the host _wouldn't_ open at first. I mean you could either go for it or have a pet goat so..
It's not difficult IF you are able to break down and isolate every possible solution. In this case, the car can be behind each of three possible doors and you don't switch, or the car can be behind each of three doors and you DO switch. That is a total of six possibilities to analyze, which is manageable.
The assumption must be that Monty has no freedom of the procedure of the show. Otherwise he could open the chosen door with a goat in it right in the beginning. The fact that he wouldnt open the door would tell me that i have chosen the right door.
Isnt Savant's solution a little over the top? I thought the answer is so much easier: If you pick a goat in the beginning , the host will reveal the other goat, so switching means that you'll get the car. Howvwer, if you pick a car in the beginning, the hose will only reveal one of the 2 goats, so switching means you get a goat. (Both are reversed if you don't switch) There are 2 goats and 1 car though, so the chance of originally getting a goat is 2 thirds.
I have a theory as to the wording of Ms. Savant's solution...but I could be wrong. It was rather lengthy but it could have been due to the wording of the question that she was sent by a reader. In that question it was not clear as to whether or not the host knew where the car was and had to reveal a goat. So she answered back in a way that addressed that uncertainty.
It's easier to understand if it is assumed that what you choose BY YOURSELF is "always going to be most likely wrong". If Monty doesn't help you, you only have 1 out of 3 chances to get the car and 2 out of 3 chances to get a goat. If you play the game multiple times, which prize are you most likely to get more? The car 🚗or a goat?🐐🐐 A goat because there are 2 goats and only 1 car. Now let's say you chose a door, and like we talked about it's "always going to be most likely wrong" because you are more likely to choose a goat BUT this time Monty helps you out. Of course Monty won't show you the car or open the door you chose, so he ALWAYS opens the door with a goat. Remember we assumed the door you choose initially is "always going to be most likely wrong" and Monty ALWAYS opens a door with a goat and also not the door you chose, so that means the door that is left is "always going to be most likely correct". With that logic, it always makes sense to switch whatever door is left IF you want to go by the most likely probability to win. If you think your spidey senses are more reliable then go with that. 😂
Hey Infographic show . Have you ever done BRICS vs European Union?if no.then do this it'll be pretty interesting and your video could hit some couple of million view. Thanks
I don’t get why no channel ever just says, you’re probably gonna get it wrong, so when the other wrong one is opened you switch to the 2/3 chance that that is the right one is the one you didn’t pick and that the one that the show host didn’t open.
Did you get it? 🤔
Nope
yes, thanks
El problema o la paradoja de monty hall es un error:
Se basa en un error de suma:
Cada puerta tiene tiene 1/3 de probabilidad, si escojo una, las otra dos tienen 2/3 de probabilidad, eso es correcto, se ha sumado las probabilidades de las otras dos puertas cerradas, pero cuando una de las dos se abre, las suma no es válida, porque es como sumar peras con manzanas… es decir la probabilidad de la puerta abierta no se le suma a la puerta cerrada…
Vámonos por otro camino, supongamos que las cabras son de oro, unos 50 kg de oro, y valen mas que el carro… asi quiero es la cabra y el abre una puerta con una cabra …
La probabilidad de cada puerta es de 2/3, si escojo una, las otras dos valdrán 4/3
Una probabilidad de 4/3 es superior a la unidad, lo cual es infalible, pero no es real … la probabilidad real es 1/2, porque la probabilidad cambio a abrirse una puerta … se podría decir en este caso se redujo y en el anterior aumento, ambas quedaron en ½ se igualaron entre si …
Honestly NO!!
Yeah.
Just listen to the door to see if there is a goat
do a fake roar. XD
General Grievous haha i thought about it too
General Grievous
IQ of 1000 right here that is actually genius
The real solution is in the comments
IQ OVER 9000
The infographics show: 50/50 is the „commoners solution“
Holt: you‘re fired
Come on Captain, the night shift is keeping you and Kevin apart. You two just need to bone.
@@meowthtroplaysvideogames BONE !
How dare you detective diaz I am your SUPERIOR OFFICER
BOOOOOOONNNNNNNEE
What i do in my bedroom detective is none of your business
I literally saw this puzzle in Brooklyn 99 yesterday
BOOOOOOOONE?!
and in the end it worked lmao i died
NINE NINE!!!
NINE NINE!
I literally saw this puzzle in the Ask Marilyn column back in the 90s.
This problem is very simple though it had me fooled initially! The key to understanding this problem is to realize that the only way to lose when applying the switching method is to *initially* pick the door with the car (which has a probability of 1/3). In that case, you'll switch away from the correct door. That's the only instance in which you'll lose.
If, on the other hand, you initially pick a door with a goat (which has a 2/3 probability) Monty will show you the other door with the goat. In that case you'll switch to the with the car.
this somehow makes sense
The reason it works is because you pick a door before the host, not after. If the host were to remove a goat door first, then it's 50/50, but in this case u selected a door with 3 possible rewards, 2 out of 3 being a goat. You force the host 2 out of 3 times to remove remaining goat door, so the other door the host couldnt open must have the car behind it.
Another way to think about it is tell the audience the full sorry with the original question. The host can't pick your door weather you are right or wrong so this naturally will give the other 2 doors the better odds.
@@jdbaker82 The full SORRY?
Best
Take it as Goat n1 and Goat n2. If u pick n1, he chooses n2-swich. If u pick n2, he chooses n1-switch. If u choose the car-he picks either n1 or n2. So 2/3 if u swap its the car.
thank you!!!!
Watched an entire 7 min video and didn't get it, you and your 2 line comment made it way clearer. Thanks !
@@lenahhhhh6568 oh my im happy for that, i dont remeber exactly, but i think this was exactly the case with me and then i tried to think about it alone and this is what i came up with
@@zxkredo yeah I had to made a schema of the situation myself ahah, real brain teaser
idk having a goat would be cool
yeah it has a lot of uses
Panzerker milk, cheese, meat, companionship
Sara Q. As a goat owner, I can confirm it is awesome👌👌
@@dachshunddoggo2764 companionship after meat?
@@DenkyManner Well, yeah. Can't get any closer than being part of someone else.
The real question is how many people wanted the goats and would the show let the contestants have the goat. I'd sue for the goat.
The Alaska Zoo got started because someone won a radio contest with the prize being a choice of cash or a baby elephant. The winner chose the elephant. And yes, there's a Simpsons episode based on that.
I think the contract doesn't let you keep the zonks.
you know what, i’ll just pick a door and get it over with
2complicated4me
Its easy, open all doors. You win
The video over complicated a very simple problem.
The host always opens a losing door in this scenario. Fucking obvious, Monty used to switch to winning doors. Just broads changing the rules to look smart
Here’s the quick version, if you switch, you’re basically saying, “I want to win a car if there’s a goat behind the door that I picked when all 3 doors were closed and I want to win a car if there was a goat behind the door that I picked when all three doors are closed. That means if we call our doors, “Doors: A, B & C” and the car is behind Door C, I can pick Door A and switch and I can pick Door B and switch and either way, I’ll win.
At best you have a 2/3 chance of winning, at worst you have a 1/3 chance of loosing.
Options, A, B, C = 3/3
Initial pick, A = 1/3 so B, C = 2/3
Goat reveal, B = 0/3 so C = 2/3
“Always account for variable change”
I would want a goat because I could keep it and sell it for a lot of money.
Craftee Madee u can bring more such goats by selling a car lol
Or you can buy another goat and breed a lot more
Math has never been my strong suit, but I'm trying to make sense of the "always switch" logic all the same and it has me stumped.
So as I understand it, the rules are that there are three doors. One with a car behind it, and two with a goat behind each of them. You want to win the car, and so your original choice comes down to a 1 in 3 chance guess at random. Wichever door you choose, the show host will then open one of the other two doors, revealing a goat, purposely ignoring the door with the car behind it if that is not the door you chose. This obviously does in fact take away some of the uncertainty: that door is now confirmed to have a goat behind it, which changes the odds. So why is the 33.333(etc)% chance from the revealed door offloaded entirely onto the other door you didn't choose instead of being evenly distributed between both of the remaining sealed doors?
The chances to win by switching is determined by the probability of you picking a goat and the host picking a goat. The chances for you is 2/3, for the host it's 1, so the chances of winning are 2/3x1=2/3. If the host didn't know where the car is then you would be correct that it's 50/50 if he revealed a goat because the chances to win by switching are not 2/3x1=2/3 but rather 2/3x1/2=1/3, same as in staying which is 1/3x1=1/3. In one case 1/3 pick cars , 2/3 pick goats and they all continue, and in the other case 1/3 pick cars and 2/3 pick goats but only half (host reveals car half of the time when he's left with a car door and a goat door) of those that pick goats can continue.
@Klaus 74:
A very well-structured explanation, thank you! There is one thing I still do not understand however.
You mention that the chances of winning when you switch doors are equal to 2/3x1=2/3. In other words, the chance of winning when you switch to the other door is equal to the chance that you picked a goat multiplied by the chance that the show host picked a goat. Why is this? Maybe this is an obvious one, but like I said mathematics aren't my strong suit.
I *DO* understand how this same formula changes based on whether or not the host knows the position of the car though, and how this affects the odds of winning based on whether you switch or not if he does know where the car is, but leaves it at a 50/50 shot if he doesn't. (And also imposes a chance that he reveals the car behind the door after the first pick, ending the game.) That really helped me figure this out a little better, so thank you!
Glad it helped. Another way of looking at it is this....the chances of what you pick have to be the same as the chances of what's left over for the host. So when there's a 2/3 chance of picking a door with a goat then the host must have a 2/3 chance of having a car door and a goat door left over to him from which to reveal the goat and leave the car. So there's a 2/3 chance the revealed goat is the last one.
"You mention that the chances of winning when you switch doors are equal to 2/3x1=2/3. In other words, the chance of winning when you switch to the other door is equal to the chance that you picked a goat multiplied by the chance that the show host picked a goat. Why is this?"
Two events must happen in order to win the car by switching. One, you pick a door with a goat, and two the host does as well. To calculate the probability of both those events happening you need to multiply their respective chances. For you to pick a door with a goat is a 2/3 chance and the host the chance is 1 because he 'must' reveal a goat according to the rules of the problem. So when you multiply the two events by their probabilities you have a 2/3x1=2/3 chance that both of you will pick a door with goats thus winning by switching.
Let's take another example. You are in a chess tournament and you have to win the next two games in order to win the championship. One of the opponents you will face is is not as good as you are and you have usually defeated him in the past, perhaps three times out of four games. So the chances of defeating him again you can estimate as a 3/4 chance. The other opponent you will face is a stronger one and you estimate your chances of defeating him is 1/2. So before you begin play against either one of them the chances of beating both of them is 3/4x1/2=3/8.
@Klaus 74 and @ Georg Cantor:
That cleared it up! I understand now. It's still really bizarre to me that the math works out this way, but I get it. Thank you two for your explanation, that was enlightening!
More specifically if your last selection is random, then your odds are in fact 50/50, however this does not change the odds of switching Vs not switching which is 2/3 and 1/3 respectively. Thus there are actually 3 scenarios, however if you count doing nothing at all, then your odds are 0% You always initially had a 1/3 chance of selecting the car and 2/3 chance of selecting goat. Thus by switching doors you're more likely to end up with the car, because the door you picked first up, was always more likely to be the goat and with one door and goat removed, the only remaining switch option is the car 2/3 of the time and the goat 1/3 of the time. Thus a random selection between the last two doors ( 50%), offers better odds than not switching (33.33%) at all, but not as good odds as switching (66.66%).
The easiest way to think about this is to just realize the whole "Monty opens a door" part is there to mislead you.
The effective result of switch or stay is you either get to pick 1 door (original choice) or 2 doors (remaining choices). It doesn't matter that one of the two remaining choices is a door that Monty opened.
In fact, Monty opening the door does not mislead but tells you that the 2/3 chance is all on the final door... which it is every time.
A, B, C=3/3
A=1/3 ... B, C=2/3
B=0/3 ... C=2/3
How many of you saw this in the movie 21
i did..it was Kevin Spacey, right ?
Angan Bhattacharyya yup
Hi, can you explain, the situation in the movie? I don't remember it. Would be nice. Maybe i will see the movie again. :-)
It's in the first act when Jude Law is in Kevin Spacey's math class. Kevin Spacey literally just asks his class this problem, Jude Law gets the solution right, and then I think he leaves.
I remember that scene, but they did a poor job explaining it to the casual viewer. I was convinced that the odds remained 50/50. They should have spent an extra minute explaining WHY the odds increased, like they did in this video. :)
But Marilyn's IQ turned out to he incorrect. Its actually around 132 i believe. Which means its not that high. Her 228 score was tested at a young age and for a didferent test, it was a ratio IQ, so while she may be intelligent her score cannot be above 170, as said by Alan S. Kaufman, a psychology professor and author of IQ tests, writes in IQ Testing 101 that "Miss Savant was given an old version of the Stanford-Binet (Terman & Merrill 1937), which did, indeed, use the antiquated formula of MA/CA × 100. But in the test manual's norms, the Binet does not permit IQs to rise above 170 at any age, child or adult. And the authors of the old Binet stated: 'Beyond fifteen the mental ages are entirely artificial and are to be thought of as simply numerical scores.' (Terman & Merrill 1937). ...the psychologist who came up with an IQ of 228 committed an extrapolation of a misconception, thereby violating almost every rule imaginable concerning the meaning of IQs
Average is 90 to 110. 138 is 'very superior'. Doesn't matter anyway, her IQ doesn't affect the mathematics of probability.
@braden7180 no
They say that you can win 2/3rd of the time but what about that “1/3rd”? This game is purely chance based.
@@ifisawyourreplyiwillanswerback Of course it's based on chance; no one disputes that. The question is not whether it's based on chance but does your chance improve by switching. The answer is yes, if you stay with your original choice of door, your chance remains at 1/3 but if you switch your chance is doubled to 2/3.
Just pumping out videos! I love it! Thanks!
Yep!.....Very Beautifully explained....
*I'm impressed* 😄
Reply
Thank you!
I don't think three would be accurate in any capacity though since I feel like no matter what, the game show would *not* want you to win that car. Just knowing gameshows and their mechanics, they only let people win the not-so-great-but-still-worth-something prizes ("cheap" stuff that isn't cheap to the average person) to keep people coming onto the show. Any expensive prizes will always be kept for the company (they're most likely props) so that they don't lose money and go bankrupt since there'd be no more show (look at how often Jeapordy "gave out" cars and boats and cruise trips and shit. You know it's a scam. Just like those million dollar prizes were most likely paid in installments and after taxes you probably got less than half of it anyway). So yes, her solution would work just fine because there's no way in hell the company would give away a free car that's probably worth more than the host's annual salary.
I can't believe these studies went through all this work and completely missed the economics of the gameshow itself haha.
I say you pick the door closest to him because if the prize is in yours he will open the closest to him assuming he's lazy but if he skips the 2nd door it's probably in the second door
but thats another assumption of his behavior, not to mention the fact that depending on the setup, he might not have to walk over to a door, the doors might be controlled by someone in a booth.
But iocane comes from Australia, as everyone knows, and Australia is entirely peopled with criminals, and criminals are used to having people not trust them, as you are not trusted by me, so I can clearly not choose the door in front of you.
@@tomleonard830 😂😂😂
I think all the doors have goats behind them😂😂😂.......
Who actually used Skillshare?
Mr. Moji Gaming me
Me
I tried it, it's not bad
I tried it it sucks hard and you need a credit card aka its ASS
Rob Spagrenetti a 2 month free trial isn't that bad (especially if you have summer vacation)
this is so confusing alexa play despacito 3 ft dj khaled
3 foot D.J Khaled.
We could actually test this out by sampling the real data - all of the shows are there and for each game, there would be the initial Door choice, the Door opened by Monty, whether the switch was made, snd where the good prize was. I smell a dissertation!
Car salesman:
*Slaps the infographics show*
This bad boy can fit so many fucking amazing well-animated videos in it
Is this still controversial ? Its pretty easy to understand. You just have to take into consideration the fact that the door Monty chooses to open is dependant on the first door you choose. That is, if you choose a door with a goat (66% chance) then Monty is forced to open the other door with a goat. Therefore by switching you transfer that 66% chance to the door with the car
kevin a but what if you chose the one with the car?
Usually this channel distracts from studies. Rn it is helping me.
Think about it this way. If the theory is right (and it is) and you switch, you have a 2/3 chance of winning the car. If the theory is wrong (and it isn’t) and you switch, you still have a 50-50 chance of winning the car.
Infographics show your a really intestine person I've watched almost all of your videos you an amazing youtuber
If Monty revealed the car, why would he offer you a chance to switch between two zonks? So either he's supernaturally lucky to always reveal a zonk and thus not deflate the drama of the choice OR (and more likely) he will always reveal a zonk and thus Marilyn's solution is correct.
That's a scenario that doesn't happen. Of course Monty knows but all that's important is that when the goat is revealed the contest should switch.
Options, A, B, C = 3/3
Initial pick, A = 1/3 so B, C = 2/3
Goat reveal, B = 0/3 so C = 2/3
@@lrvogt1257 Yes. That is Marylin's solution.
I hope you start to cover more paradoxes and dilemmas
“Would stir up so much controversy in academic and intellectual circles”
There’s 0 controversy about this in academic and intellectual circles. Everyone with the most remote understanding of statistics has no issue at all with this. It’s people who don’t know statistics and are stubborn to trust their instincts over logic that are experiencing the controversy, nobody else.
Sebastian Anderson can you back that up ?? With some of the stats just curious
Naturally weird Nope, who do you think would waste time preparing statistics on that that I could send you lmao. My job is as a statistician, I know what people who deal with statistics do and do not struggle with in statistics, and at least unless American statisticians are significantly less educated than European ones I would be very surprised to hear that any of them struggle with this particular issue.
If your argument is that without backing statistics I can’t make any argument, then here’s a dosey for you: since the statistics don’t exist, infographics show pulled that “fact” out of their asses too. At least I base mine on interacting with hundreds of statisticians over the years and being aware with the statistical curriculum and hence that anyone with a more than basic understanding of stats should have 0 issues with this.
The problem is too simple for statistical analysis. The simplest proof is this:
1. I f you switch you always get the "opposite" of your original pick.
2. 2/3 of the time your original pick will be a goat.
3. Therefore 2/3 of the time a switch will give you the car.
(1/3 of the time you will lose by switching)
cool . i am just a high school kid and i really like probabilities so i thought maybe i could get some new insights into it.
Roger Bodey What you’ve just done counts as statistical analysis :)
Love ur channel
Realistically, He’s always going to pick a door with a goat to show you, and he’s always going to ask if you want to switch to the unpick/unopened door, so it doesn’t matter whether you switch or not, odds = 50%
That isn't realistic because the chances of what you pick are the same as the chances of what's left over for the host. Since there is a 2/3 chance of you picking a goat then must be also a 2/3 chance the host is left with a goat door and a car door from which to reveal the goat and leave the car.
Seth Webster,
If the host *_must_* reveal only a goat, and never the car... then if your first choice is one of the two goats (which is a 66% likelihood), then him revealing the second goat means that the only door left is the car.
If you pick a goat, and the host reveals a goat, the only door left is the car.
If instead, you pick a car, and the host reveals a goat, the only door left is a goat. Thus, if you switch, the result is that you *_always_* end up with the opposite of what your first choice was. So if you know your first choice is 2/3 chance of being a goat... then switching will mean a 2/3 chance of being a car, because switching always results in the opposite.
Simpler version. Your first choice will be wrong 2/3 times, then the host must reveal the other wrong option, so you switch and win. Therefore you guarantee a win 2/3 times by switching and that's all you need to know.
I've used this trick to make sense of this paradox in the past: the odds you've chosen the correct door are 1/3. So you should definitely switch because you probably did not make the correct choice in the first place.
Yes but after one has been revealed you have a 50-50 change, what makes switching the door better?
@@lucyrhodes7010 the host always reveals a losing door. If your first choice was a losing door and you switch you automatically win. The chance of choosing a losing door first time is 2/3. Therefore 2/3 of the times you should switch
How dare you, Detective DIAZ. I. Am YOUR. Superior. OFFICER!!!! BoooooooooooOOOOOOOOOOOOONnnnnnnneeeeeeeeee????
I looked at this problem half a year ago so now I'm revisiting it with the same opinion as before. Here the way I see it. You have a 1/3 chance of choosing the right door the first time, so, you are more likely to choose a goat at first. With only two doors left, and the assumption that you have already chosen one of the goats, the remaining door that he didn't choose to open would have the car. This would obviously not work all the time, but if you played the game multiple times you would win more often if you switched.
Yes if the host knows where the car is and always must reveal a goat. If he didn't know where the car is and randomly opened a remaining door (or if the wind blew it open) and it happens to be a goat then it's 50/50.
Watching your every videos for a long time and I like each video
And I commented on many
Hope you remember me
*But will SkillShare cure crippling depression?*
In the real world you are usually only given the choice of two doors to start with. The third door is only shown as a possibility afterward. Then a fourth, then a millionth.
But if given the choice of three doors at the start, and on the assumption that monty will never reveal the car (which he never has) the original premise is correct from pure mathematics.
50 views, 14 likes, RUclips is finally in Order
Moon Pie Eater 50 views 78 likes sorry buddy
51 views 93 likes RUclips is back
I recall watching a documentary way back when showing how those old game shows were often times rigged to make contestants win or lose based on how much the audience liked them. If you were well liked, you were more likely to win because it would please their audience and increase their views.
Then you must watch the movie 'Quiz Show' starring Ralph Fiennes, based on a true story that matches your comment perfectly.....cheers!
@@klaus7443 one question tho... Why didn't you give a like to the comment?
simple solution
for winning by switching you would have to take door with goat in your first choice which has probability 2/3
but for winning by staying you would have to pick door with car which has probability 1/3...
correct me if i am wrong.
If you pick a door with a goat behind it and choose to stay, do you get to keep the goat?
Because that prize would be the greatest of all time.
Better than school
It baffles me that many people still don't understand this simple math problem. This is like the math equivalent of a toddler sticking shapes in a toy box.
At first I couldn't understand it, but then I thought that when picking the first door you have 1\3 chance to win, so because of that it's better to switch door since there is a 2/3 chance that the other two doors were winning, but it is reduced to only one door when one is opened, while you still have 1/3 chance on your door if you don't change
Does that thought process make sense?..
In the MHP yes because the host knows where the car is and must reveal a goat. If he didn't know where the car is and randomly opens a door not picked by you and it has a goat then it's 50/50.
Klaus 74 Only the host has 1/2 chance of picking a goat (assuming he doesn't know where the car is), if he picks a goat, randomly or not, the chances of you getting the car if you switch remain the same in every case
What are you talking about??? If the host doesn't know where the car is then 1/3 of the time it is behind your door, 1/3 of the time it's behind the door you can switch to, and 1/3 of the time it's behind the host's door. The chances to win by switching is 2/3x1/2=1/3, same as in staying which is 1/3x1=1/3. So if a goat is randomly revealed then it doesn't matter if you stay or switch.
If you take 99 games and the host always reveals a goat then 33 pick cars, 66 pick goats and they all continue. If the host does not know where the car is 33 will still have picked cars, the host picked 33 cars and there are 33 cars (not 66) behind the door that you can switch to. All those that picked cars can continue, and half of those that picked goats can continue. 50/50.
"That means there's a 2/3 chance the car is behind one of the other two doors. Those odds don't change because Monty Hall shows a goat behind one of the doors -- whether he knows what's behind the door or not."
You are wrong when it comes to the host not knowing where the car is. What's so good about leaving twice the cars for the host when half of time he leaves it for you to switch with and half of the time he reveals it?
Monty hall problem is annoying
Why? There aren´t just 3 but 4 options. (In the example I will assume that you always pick the first garage.)
AZZ and he shows you the first Z
AZZ and he shows you the second Z
ZAZ and he shows you the second Z (because you pick the first)
ZZA and he shows you the second Z (because you pick the first)
There are 4 options and in two of them you should change. (2/4 or 1/2)
Since you are 1/3 likely to pick the car door ar first, the two first cases should fall into that 1/3, so each will have 1/3*1/2 = 1/6. You cannot compare those cases that have 1/6 with the others that have 1/3.
I think the first two cases are something different. Then there are 4 options and if you pick one of the first two you shouldn´t change and if you pick one of the last two you should. It is two different cases if he shows you goat one or two.
There are 4 options but two are half as likely as the other two. In order to get probabilities by counting cases you have to make sure that everything you count as a case is equally likely. If we count cases as you are doing, we are assigning two for having hit the car, and 1 for each goat, so if we played a lot of times, like 100, we should have selected the car about in 50, the goat 1 about in 25, and the goat 2 about in 25, which is absurd. There is no reason to think that we are going to hit the car door firstly more often than the other two.
Imagine you played 900 times. Let's put the numbers, assuming that the sample proportions exactly match the probability, that is, you should pick each content 1/3 of the time.
1) In 300 games your door has the car.
1.1) In 150 of them the host reveals the goat1.
1.2) In 150 of them the host reveals the goat2.
2) In 300 games your door has the goat1. In all those 300 the host reveals the goat2.
3) In 300 games your door has the goat2. In all those 300 the host reveals the goat1.
- If goat 1 is showed, you can only be in case 1.1) or in case 3), so you are under a subset of 450 games.
You win by staying in 150 of those 450 (case 1.1) and by switching in 300 of those 450 (case 3).
• 150 represents 1/3 of 450.
• 300 represents 2/3 of 450.
- If goat 2 is showed, you can only be in case 1.2) or in case 2), so you are under a subset of 450 games.
You win by staying in 150 of those 450 (case 1.2) and by switching in 300 of those 450 (case 2).
• 150 represents 1/3 of 450.
• 300 represents 2/3 of 450.
Ronald I see your point...
the first two probability’s do not have the same probability’s as the rest
I love this channel so much...
This comment will be lost forever
Bob Elogeg your forever lost comment has been commented from France.
roucoupse And now from Reunion Island,Indian ocean
Orange Juice are you really there it is beautiful
Bob Elogeg Thanks ! And yeah,I live there and I was born there :)
Ok
Think like this, -
There are 1000 athletes, who have been ranked among themselves.
You are asked to guess the athlete who is ranked 1.
You choose one of them.
Now divide 1000 athletes in to 2 groups. The person you chose is in group A, and rest 999 men are in the other group B.
Now Monty Hall eliminates 998 of the athletes .
Now u are asked to choose whether you want to stick to ur first choice or switch to the other athlete.
Now see, the last man standing in group B has defeated 998 men, and till now u have no information about the ability of the person u first chose. Which athlete you want to choose? The person who u know has defeated 998 men , or the person about whom u know nothing?
I would ask the car keys and press the unlock😂😂😂😂
Gaurav Shanbhag
this cracked me up ≥3≤
love your videos
Am i the only one who didnt understand the explanation?
Losqualo Telli I’ve been scrolling through the comments and yours is the one that helped me understand, thanks
I've seen this problem before and somehow this is the first explanation that makes sense to me
I mean this is all assuming that he wants you to lose. Assuming he doesn't care about the outcome and just wants to build tension and suspense, than he would show one of the losing doors every time, making any logical meaning behind showing one losing door irrelevant. It's for the views not the wins and loses.
Yay i got free skillshare
People who didn't get it.
There are 3 doors..A,B,C.
Probability is first equally divided..i.e 33% in A, 33% in B, 33% in C.
You selected A(doesn't matter now)
So, right now, probability of car in your selection(A) is 33%, while in B+C combined(others) is 66%
Now, the guy showed you one door out of B and C, and there was a goat behind it.(Say C)
That means your initial choice was 33%.
Other was 66%..now, out of others C got eliminated and now B has 66%...while your initial choice is still 33% because B& C's combined was 66% and C got eliminated.
this is the explanation
That is absolutely wrong
B+c ≠ 66%
Cause we know for sure that one of b or c has a goat
So when separating the 3 doors into 1 single door vs 2 doors combined
It becomes 50/50 because 1 of door b or c = 0%
We just dont know which one so the host opens that door
Now we have literally 50/50 shot at the prize
*Common Sense*
Seems fair enough
Very good explained,, Thank you👍
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MegaChibi ......
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But it's a win/win situation. Either you get a goat that brings company in your live or a nice car.
It's just some condition probability, thats most people with no mathematical background find hard to understand. It's not that complex, you don't need to be an expert to understand it but I guess most people don't really understand non intuitive maths and are just arrogant.
Random Internet User I believe most people understand this theory, after all, as you also said, it's not a complex one. People are just getting it wrong the first time, because they don't really think about it and just go with their gut.
r/iamverysmart
they just word this terribly, all they have to say is, " you are more likely to pick the wrong door first, then after he removes the other wrong door, the right one will remain 2/3 of the time
Ethan Barnes I think the wording isn't that bad, sure it can be a bit confusing, but IMO i think the explanation is what confuses the regular viewer more so.
I think it makes more sense by making a table with all the possibilities and then counting them to find the probability at the start and then when a goat is revealed.
Random Internet User: The question really should state whether the host chooses randomly or always chooses a door with a goat. If that's not specified, it's more natural to assume the host chooses randomly, which makes the 50/50 solution correct.
Surely the easiest way to look at which is the better option would be to say, you pick a door, you can either keep that door or swap it for the other two? That is what is actually being offered isn't it?
You are doubling the chances of winning in both cases but for different reasons. In one instance the host must know where the car is and must reveal a goat in order to double the chances (otherwise it's 50/50) and in the other case swapping for the two other doors does not require of the host to know where the car is at all. One is a conditional probability problem with a subset of doors having 0 and 2/3 chances of having the car while the other is an unconditional probability problem with a subset of doors having a 1/3 and 1/3 chance of having the car.
@@klaus7443 l was implying the chance changed in anyway Klaus, more how to explain how the ratio is how it is.
Imagine Monty didn't open the door until after the contestants had chosen whether to keep their original or swap. Ie Monty gives the option to keep original or opt for the other two doors and contines by saying "one of the other two is definitely a goat allow me to take it away" By doing this the odd on whether their original pick is correct haven't changed in the slightest. I thought this may have been an easier explanation of the 1/3 or 2/3 option.
@UCPsKVRwX525ZRoaISJl3Ijg Hi Klaus, I'm bot disagreeing with you in anyway. Other than, this video is about the ratio and the controversy surrounding its eventual conclusion.
I won't discourage you for it's usage then. I'm just speaking from experience in dealing with many here that know switching doubles the chances of winning but they also say it doesn't matter as to whether or not the host knows where the car is or not. If they know the answer but not the reason why then good luck in helping those that think it's 50/50 with that explanation.
Yes! That is the whole essence of the problem. And so, so many people can't see it. If contestant picks door #1, and Monty offers to let contestant swap his unopened initial choice for BOTH of the other unopened doors, it would be a very stupid contestant who wouldn't take Monty up on his offer. In regular play, that's essentially what happens.
Here's the simplest solution: If you don't know what door to choose, always account for variable change. But still make the switch.
Yes it makes sense... It works because the host know where the prize is, so for the sake of continuing the game, the host is always gonna keep picking up the goats until all the choices are exhausted. You can visualize it better if you have many more doors. So if that's the case, you'd ask why did the host open that goat door instead of the one in the middle? Because opening the middle door exposes the prize hence the end of the game. So it makes more sense to switch... Or course it's probability. Can still lose
Exactly. The game has to always have the premise that you can win based on your choice. The situation can never be that you lose the game stick or switch. Monty must know since he would never reveal the car, always a zonk. Knowlege you have effects the odds.
BOOOOOONE
it's not 50/50 solely because when you make your first pick, you're picking between 3 doors not 2 THEREFORE you're more like to have chosen wrong BECAUSE there are more wrong choices than right ones.... it would be 50/50 if say you got to pick after the host opened one wrong door, but you're picking BEFORE that when there's 2 wrong doors.
But lets say that u have picked right, there are still more chances that you have picked wrong and then u would switch and lose
Can you do "what If the earth was flat" + What bad circumstance we would get of it.
imagine there are a thousand doors instead of on 3. You choose one, and then the host reveals 998 of the doors leaving just a door and the door you picked. which door do you think it is? It's better to switch because there is less chance you got it right and more chance that the other door is right.
easy as that
This was my fifth grade science fair project
Winner winner
Goat dinner !!
Just knock on the door loud to scare the goat
Keep Going
Our logic professor presented this game-problem in class. We all lost miserably.
That's Bayes theorem i have studied in class 6
An easier way to explain the problem:
3 doors, 2 doors with goats inside and 1 door with car inside.
Pick a door, your chance to get the goat door is 2/3.
This means you are more likely to get the goat instead of the car in the first place. So switching gives you better chance to get the car.
And none of them considered goat bleating from inside the room probability...😢
How dare you detective Diaz I am YOUR SUPERIOR OFFICER
Professor Scott Steiner will solve this problem easily: put Monty Hall in the Steiner Recliner until he confess.
Lmao I thought it was common sense to choose to switch to the door that the host _wouldn't_ open at first.
I mean you could either go for it or have a pet goat so..
Damn who wouldn’t want a goat lmao 😂
It's not difficult IF you are able to break down and isolate every possible solution. In this case, the car can be behind each of three possible doors and you don't switch, or the car can be behind each of three doors and you DO switch. That is a total of six possibilities to analyze, which is manageable.
The goat would scare me!
Captain Holt brought me here.
BONE!!!!
my brain just exploded
The assumption must be that Monty has no freedom of the procedure of the show. Otherwise he could open the chosen door with a goat in it right in the beginning. The fact that he wouldnt open the door would tell me that i have chosen the right door.
4:05 best example
Isnt Savant's solution a little over the top? I thought the answer is so much easier:
If you pick a goat in the beginning , the host will reveal the other goat, so switching means that you'll get the car.
Howvwer, if you pick a car in the beginning, the hose will only reveal one of the 2 goats, so switching means you get a goat. (Both are reversed if you don't switch)
There are 2 goats and 1 car though, so the chance of originally getting a goat is 2 thirds.
I have a theory as to the wording of Ms. Savant's solution...but I could be wrong. It was rather lengthy but it could have been due to the wording of the question that she was sent by a reader. In that question it was not clear as to whether or not the host knew where the car was and had to reveal a goat. So she answered back in a way that addressed that uncertainty.
What if I'd rather have a goat than a car?
Sell the goat for money to get a cheaper car lol
I clicked on this one when I saw "Monty Hall" . I immediately knew who he was & loved that show as Kid. Wayne Brady is the Host nowadays I think?
Stopping at 3 minutes... just....OWWWWW!!!!!! ^_^
It's easier to understand if it is assumed that what you choose BY YOURSELF is "always going to be most likely wrong". If Monty doesn't help you, you only have 1 out of 3 chances to get the car and 2 out of 3 chances to get a goat. If you play the game multiple times, which prize are you most likely to get more? The car 🚗or a goat?🐐🐐
A goat because there are 2 goats and only 1 car.
Now let's say you chose a door, and like we talked about it's "always going to be most likely wrong" because you are more likely to choose a goat BUT this time Monty helps you out. Of course Monty won't show you the car or open the door you chose, so he ALWAYS opens the door with a goat. Remember we assumed the door you choose initially is "always going to be most likely wrong" and Monty ALWAYS opens a door with a goat and also not the door you chose, so that means the door that is left is "always going to be most likely correct".
With that logic, it always makes sense to switch whatever door is left IF you want to go by the most likely probability to win. If you think your spidey senses are more reliable then go with that. 😂
Hey Infographic show .
Have you ever done BRICS vs European Union?if no.then do this it'll be pretty interesting and your video could hit some couple of million view. Thanks
My head is still spinning..
I don’t get why no channel ever just says, you’re probably gonna get it wrong, so when the other wrong one is opened you switch to the 2/3 chance that that is the right one is the one you didn’t pick and that the one that the show host didn’t open.
I wouldn’t mind a goat. You’d get free milk
Assuming of course, it's a female goat.
You really don't want to try to milk a male goat.