This video is nearly ten years old, and about every fifth comment or so is about wanting the goat instead of the car. You're all so very clever and original.
RyoMiianPlayz No shit, Sherlock. You left the same comment a million other people did *because* just like them, you typed up the first obvious thought that popped into your head and sent it out into the universe. People like you clog up the internet with repetitive, unoriginal thoughts, and the worst part is that you can't be bothered to read what anyone else has to say and then you expect people to read your bullshit.
Ross Sadler I literally don't understand how anyone can be like you. I said I know that other people left the same comment. Just cause other people have the same thought as you doesn't mean it's an "unoriginal thought". I just literally can't rap my head around the fact that this is so important to you that I know that I'm "unoriginal"
I'm not aware of probability being despised by many that study mathematics. And in my experience, probability and statistics were by far the easiest higher level math courses I took (I majored in Mechanical Engineering, minored in math.) I find the process of taking something that appears straightforward, and uncovering a surprising underlying principle to be a thing of beauty. And in many cases, I feel like a hero battling villains when I see data that has been presented in a misleading way and I get to identify it and call it out.
@@whitey6317Making sense and being counterintuitive are not mutually exclusive nor is making sense and being intuitive the same thing. When some thing "makes sense" means that knowing the "solution" or "the way a X works", you can reason it out. Being counterintuitive means that just via intuition you will have a hard time arriving at the answer. Intuition usually does not examine all the possibilities but reasoning, i.e. the part of your brain that can make use of logic to make sense of things does (if done properly). Intuition on the other hand makes decisions based on the amount of data that you receive at a glance (you can train it of course to start inferring the answers you'd arrive at with reasoning, but that requires building experience). The perfect example is why the regular person (which includes me) will think (or rather intuit) its a 50/50 chance. People don't consider that the chance of winning the car is bigger because you can make the choice. They assume that having one or two choices does not make an impact on the probability and even if they notice the fact that the host will always go for a door with the goat in the first reveal they only think they understand the implications. Regular person thinks like this: "Its one in three, but because the a goat will be revealed on the first reveal its really 50/50, so because I want the car therefore I want to hit it on first try and keep the door on the 2nd change thus producing less anxiety." Upon thinking of it a little more, its easy to observe that the anxiety level will not change (it might even rise if the person doubts their initial choice - which is a likely occurrence) and its easy to observe that its wishful thinking to hit the lower probability. A person who can flip their intuition may think: "Well, since the first thing to be revealed will be a goat, and definitely not the door I chose, I should be aiming to hit a goat so the host is forced to reveal the other goat, and really hitting the goat is the more likely reality, then I should just embrace the goat and then release it." This shows a surprising level of acceptance towards the possibility failure which stems from understanding and acceptance of the limitations of the way this choice works (under the assumption its not rigged) as opposed to the wishful thinking of the regular person's reaction to this. I hope the above makes sense.
The Monty Hall problem only gets the answer it does if you add another hidden rule: that Monty always has to make the offer of another door. If the likelihood of his making the offer differs depending on the subject's first choice in an unknown way then you cannot determine the correct choice. If Monty for instance, likes screwing with you and only offers the possibility of another door when the person makes a good first choice, then you reduce your chance of winning by switching.
A surprisingly astute comment - many gainsayers wrongly assume the revealed door is eliminated from the probability equation when in fact, as you illustrate, it remains and has 3/3 chance of goat and 0/3 chance of car. While it may be eliminated from the set of 'doors a contestant who wishes to win would choose' or even from the set of 'doors offered to the contestant to switch to', it remains as the door whose conditional probability is 0/3 car following the reveal, the other doors now having 1/3 and 2/3.
Well, I wanted to leave a unique brain fart, just point something out that I assume nobody or hardly anyone has pointed out so far. Thanks for congratulating me on that. :-)
Senjiu Kanuba Your comment merits being taken seriously. Many people simply counter the advice to switch with 'But I want a goat', in a limited attempt to subvert the expectation that the car is desirable without respecting the core purpose of the MHP as a math problem. However the theoretical chance of trying to win the goat is an effective counter to misguided solutions such as the Ablestmage Error, where the doors are reduced to two, and Buxtonian Logic, where the remaining two doors' chances are not divisible in a different way to before the reveal.
To be honest, I found the problem too simple to take it seriously but I can see people having trouble with it. It's the same problem with "You have 3 similar looking keys, one of them opens the door, you remember which ones you tried already. Which one is most likely to be the key, the first, the second or the third one you try?", people don't try to figure out the likelyhood of each event, they just answer on gut level (or with system 1 as opposed to using system 2). That's why you learn this stuff in school. Or a more popular problem because it might be actually important to someone: "You doctor has a diagnosis tool that returns positive in 99% of the cases when someone has the illness and negative in 1% in that case. It returns positive in 10% of the cases when you do not have the illness and negative in 90% of the cases then. One in ten thousand has the illness. You are diagnosed positive. Is it more likely that you do have the illness or not?" What's your first guess? And what do you think is the actual answer? Take a moment and think about it. The solution is that in a million people 100 have the illness, of those 99 are diagnosed positive and 1 is negative. Of the 999900 people that don't have the illness 899910 are diagnosed negative and 99990 are diagnosed positive. So if you are diagnosed positive you are either one of the 99990 who don't have the illness or one of the 99 who do have it. So your chance of actually having it is 99/100089 which is about 0.1%. That means, in this setup, if you are diagnosed positive you have about 99.9% of not having the illness and 0.1% of having it. If you are diagnosed negative you have a one in 899911 chance of having the illness and about a 99.9999% chance of not having it. So it's a relatively reliable test if it returns negative but if it's positive you should probably take another test, unless you can live with a one in thousand chance of having that illness. So I agree with you in that many people just don't think enough. :)
I disagree with your simplification. The entire logic of the problem relies on the fact that they are revealing a certain door to you. With your simplification, you don't know what is behind the doors on your first pick, so you have a 2/3 chance of being wrong. Then you say that you are probably wrong so you should switch. Well, you still don't know if you are wrong, and you don't know what is behind any of the other doors, so you are effectively eliminating your own choice making it a 50/50.
Here are all possible outcomes of swapping vs. all possible outcomes of not swapping. I've numbered the goats to make the illustration a bit easier to understand. Swapping: player picks goat 1, host reveals goat 2, player swaps to car - player wins player picks goat 2, host reveals goat 1, player swaps to car - player wins player picks car, host reveals goat 1 or 2, player swaps to goat - player loses Swapping odds - 2/3 for player Staying: Player picks goat 1, host reveals goat 2, player stays - player loses Player picks goat 2, host reveals goat 1, player stays - player loses Player picks car, host reveals goat 1 or 2, player stays - player wins Staying odds - 1/3 for player
Say there are 1000 doors, and only 1 car still. Once you choose a door, the host opens 998 other doors with goats behind them, leaving a single door unopened, plus your initial door. What are the chances if you switch, vs if you don't? Would that not make for a very, very, very easy game in which you are almost assured to win?
+David G Yes, it would. As you might expect, swapping will win 99.9% of the time while staying only wins the other 0.1% of the time. Examples using more doors are often used to help people who don't "get it" because it becomes more clear that swapping is simply betting on our first choice being wrong when we start with a near-impossible chance of selecting the car off the bat. Counterintuitively, the more doors we start with the easier it is to win the game!
Drieks It shouldn't, but it does (to some people). It's a lot more obvious when you're only 0.1% likely to pick the grand prize from the start that swapping is almost certainly the only way to win. When the chances are 1/3 against 2/3, lots of people give themselves permission to regard the chances as "close enough" to 50/50 as to not make a difference. Using a thousand doors is a way to force certain people not to ignore the diminished probability of selecting the grand prize off the bat.
+TedManney Thanks, Ted. I couldn't have offered a better way to convey what I was actually trying to say. And thanks Drieks, I see your point ~ the 1000 door explanation does not in any way help a person who is attempting to solve this problem mathematically.
Think of it this way.....When you PICK initially, you have a 33% chance of guessing correctly, and a 66% chance of guessing incorrectly. Now, if you SWITCH doors, after asked by Monty, you ONLY LOSE if you picked the CORRECT door initially! If you DON'T switch however, the door that Monty opens (with the goat, after your selection) is just a INDEPENDENT event, and has NO impact on your initial selection! This is a GREAT problem....I have taken stats classes, advanced mathematics, and have even studied this specific problem in particular, yet every time it continues to amaze me!
@Steve It's just the contrary. The contestant chooses a goat door in 2 out of 3 times on average. On the other hand, the host knows the positions and cannot reveal the contestant’s selection and neither the door that hides the car, which means that everytime the contestant has chosen a goat door (2 out of 3 games), the other door the host leaves closed is which has the car. So, we will always end with two doors, but the switching one will have the prize in 2 out of 3 times, not in 1 out of 2.
Steve, you can't have a 1/2 chance of having the car behind any of the three doors you pick unless you change the total probability from 1 to 1.5 in your explanation. That makes it wrong no matter how you word it.
The total probability in the original question is 1 because there is one car among three doors. The total probability in your explanation must also equal 1...but it doesn't. In order to have all three doors with a 1/2 chance of having the prize if selected would require a total probability of 3x(1/2)=1.5 which is 1/3 more than there are cars.Two of the three doors can have a 50/50 chance if, and only if, the other door that was opened had a 1/3 chance of having the prize but didn't. Therefore your explanation must be wrong.
@Steve "except your chances change to 1/2 when they open a door containing nothing. there are only two possible solutions left. your selection now has a 50% chance to contain the prize." That logic must then apply to this as well....except your chances change from 1/3 to 1/4 when they add a door containing nothing. There are four possible solutions left. Your selection now has a 25% chance to contain the prize. Lol!!!!!!
One way to see this problem intuitively is to tweak the conditions of the game: Instead of three doors, imagine a million doors, with all goats and one car. After the contestant chooses a door, the host opens all remaining doors except one. Now it is obvious that the odds are overwhelming that the car must be behind the door he did not open.
I always was confused about why they switched even though it was explained to me. Now I understand. You are basically guaranteed to pick a goat, and the host reveals everything else you didn't pick except for one door. The host obviously isn't going to reveal the car, so you should switch to the one he didn't reveal.
@@TheSpacePlaceYT The car still might be behind your door (which he also didn't reveal) It's always two probabilities and the first one doesn't matter. It's the Monte Carlo fallacy in disguise.
@@johnrobertson93 yeah no. This is like saying the lottery is 50/50 because you win or lose. At first you have 1/,1000,000 chance of picking a car. Or 999,999/1,000,000 of picking a goat. So when he opens all other doors(which doesn’t change what your choice was) it’s more likely you picked the goat and swapping will leave you with a goat one time every million. It’s not a debate lol it’s literally solved
This is fun that ppl are still arguing it. Here's probably the simplest explanation (simplest I can think of anyhow). A random number between 1 and 3 (inclusive) is chosen, this is where the car is placed. Again another random number from 1 to 3 is chosen, this is the contestant's choice. Let's say the contestant will never switch, so the door the host reveals is immaterial, he's going to stick with his original choice. There are 9 total cominations, 11,12,13,21,22,23,31,32,33 and as you can see, only 3 of them the contestant picked the car and will win. 3/9 = 1/3. So by not switching, 1/3 chance of winning (2/3 of losing) and by the converse, 2/3 chance of winning (1/3 of losing) if switching.
First of, you listed permutations, not combinations, just pointing that out there. Second, you post out in one set of permutations (11, 12, 13) implying that there will always be three doors remaining at the second choice, which is not true since only two doors are left. It would be like this: 11, 12, OR 11 13 since the host does not remove their door choice. For the first set of permutations, the car remains in either door 1 or door 2. So if it was in door 1, and you stay you win. If it was in door 2, and you stay you lose. Staying has a 50% chance of winning. Consequentially, changing has a 50% chance of winning. For the second set of permutations, the car remains in either door 1 or door 3. So if it was in door 1, and you stay you win. If it was in door 3, and you stay you lose. Staying has a 50% of winning. Consequentially, changing has a 50% chance of winning.
+TheGreatslyfer You said: "For the first set of permutations, the car remains in either door 1 or door 2. So if it was in door 1, and you stay you win. If it was in door 2, and you stay you lose. Staying has a 50% chance of winning. Consequentially, changing has a 50% chance of winning." You concluded that staying had 50% because you were only focusing in the number of options, forgetting you have more information about one door than about the other, and this occurs because one door was selected randomly but the other on purpose with knowledge. (Remember the host knows the positions and must reveal a goat after your selection from the other non-selected two.) Let's see: 1) The number 1 is still an option because it was your selection and the host couldn't reveal it regardless of its content. 2) The number 2 could have been removed if it had a goat. It was not a guarantee for it to be in the second round. For example, if number 3 had the car and number two had a goat, the host would have been forced to reveal the 2. But he didn't, number 2 survived a possible elimination, which increases the likelyhood of it having the car. The other door the host leaves closed will be the correct in all cases you failed at first, so it wins 2/3 of the time.
I think the easiest way to understand this problem is to just imagine the doors open the whole time. Then you can conceptualize why changing doors results in a win more often than not. Say the truck is in the left door every time. You pick the left door. The host opens the middle. You switch to the right. You lose. The truck is again in the left door. You pick the middle door. The host MUST show you the right door. You switch. You win. The truck is in the left door again. You pick the right door. The host MUST show you the middle door. You switch. You win again. And that’s why you win 2 out of 3 times. Because there’s a 66% chance that the host is going to be forced to eliminate the other goat door.
Yes, imagine the doors being open all the time. These are the only possible options: The only 4 examples possible if the car is in door 1: 🚗🐐🐐 Example 1. Choose door 1, host reveals goat in door 2. Stay = you win, change = you loose = 50/50 % chance. Example 2. Choose door 1, host reveals goat in door 3. Stay = you win, change = you loose = 50/50 % chance. Example 3. Choose door 2, host reveals goat in door 3. Stay = you loose, change = you win = 50/50 % chance. Example 4. Choose door 3, host reveals goat in door 2. Stay = you loose, change - you win = 50/50 % chance. In 2 of the only 4 possible examples above you win if you change, and in the other 2 examples you loose if you stay = 50/50 %. These are the only 4 examples possible if the car is in door 2: 🐐🚗🐐 Example 1. Choose door 1, host reveals goat in door 3. Stay = you loose, change = you win = 50/50 % chance. Example 2. Choose door 2, host reveals goat in door 1. Stay = you win, change = you loose = 50/50 % chance. Example 3. Choose door 2, host reveals goat in door 3. Stay = you win, change = you loose = 50/50 % chance. Example 4. Choose door 3, host reveals goat in door 1. Stay = you loose, change - you win = 50/50 % chance. In 2 of the only 4 possible examples above you win if you change, and in the other 2 examples you loose if you stay = 50/50 %. And final example if the car is in door 3: 🐐🐐🚗 Example 1. Choose door 1, host reveals goat in door 2. Stay = you loose, change = you win = 50/50 % chance. Example 2. Choose door 2, host reveals goat in door 1. Stay = you loose, change = you win = 50/50 % chance. Example 3. Choose door 3, host reveals goat in door 1. Stay = you win, change = you loose = 50/50 % chance. Example 4. Choose door 3, host reveals goat in door 2. Stay = you win, change - you loose = 50/50 % chance. In the final possible scenario it’s all the same. 50/50 % chance to win or loose.
@@christerenstrom9798You're saying the probability of winning the lottery is 50% because you either win or lose, which is false. Do not confuse probability and possibility. You need to actually chose an option (stay or change) and calculate the chances. You'll see that changing doors is more likely to give you a win (google "monty hall table" and you see how you should make this table), but I'll post here a different way of thinking: Imagine 3 doors, behind 2 have goats and behind 1 have a car, let's say the car is behind the first door (🚗🐐🐐). We both agree that the car has a chance of 1/3 of being in the first door, 1/3 of being in the second and 1/3 of being in the third. Let's suppose you choose the last door, so the host NEEDS to reveal a door that HAVE A GOAT and it's not yours (because he knows which one has the car), so he reveals door number 2. Remember, that this information (that a goat was behind door number 2) you didn't have in the beggining, it's a NEW INFORMATION. Now we separate the doors into two groups, the group composed by the chosen door and the group composed by the other two doors, like this: 🚗🐐 and 🐐. What were the chances the car was in each group? The car had 2/3 the chances of being in the first group (door number 1 plus number 2) and 1/3 of being in the second group. Since now you know that the second door have a goat (that is 0% of having a car behind it), the other door needs to have all the probability of having a car. This is why changing doors always increase you chances. A=1/3 B=1/3 C=1/3; A+B+C=1; A+B=2/3, after reveals, B=0 so A+0=2/3>A=2/3; Now (A+B)+C=1>(A+0)+C=1>A+C=1>2/3+1/3=1
For people who still don't get it after watching the video & read comments (like me), here's one that is more intuitive: Situation 1: choose car -> swap and you get a goat Situation 2: choose a goat -> swap and get a car Situation 3: choose the other goat -> swap and also, get a car As you can see the chances of swapping and get a goat is 1/3, and no leak of information that changes the chances you'd picked the car at first
Diksha GiriIf your first pick is Goat A, you get Goat A. Swap = Car. If your first pick is Goat B, you get Goat B. Swap = Car. If your first pick is the car, you get the car. Swap = goat. So simple, 1/3 vs 2/3. No stupid 50/50 in this game.
Araqius oh right right so it is like rather than calculating the probability of getting a car second time we calculate the chances of picking up a goat or a car the first time. Since the chances of picking up the goat first time is 66% therefore it is better to swap the second time. I get it now, thanks.
True, and the explanation wasn't difficult to follow. Suggesting we should "swap" isn't the greatest advice since we will never be sure what is behind the door we chose. The odds could be against us or in our favor.
Araqius Hehe, Diksha Giri is still correct, as you are now using the Assumption that "swapping gets you the car" (which is ignoring the chance for the other goat). In Real Life physics, not only do you do not know what is behind the door, but by 'choosing the door' you don't actually 'have what is behind' the door, so probability is just a plaything here - until the outcome is set. Mathematicians, Statisticians et al would love to play with the probability of course, since it is 'after the fact' ;) It is a great example of Real Life Physics vs Theoretical Physics!
You basically have 0 % chance, because when you open the 1st door and the host asks you whether to swap... the crew is ready to swap the prizes while you make your decision... When you choose the host will talk long enough and get you under pressure, that the crew will swap the prizes for sure... except if they don't keep cars in small rooms
+Vladimir Dimov "You basically have 0 % chance" ... while simultaneously you basically have 100% chance because when you open the 1st door and the host asks you whether to swap ... the director is telling the crew "we need to increase ratings or the sponsors will leave the show - give away the car"
Michael beseler "except thats scamming, and they would be prosecuted" "... except that's not mentioned in the MHP question so is not material to the solution"
+O. Flake There is serious protection in place against cheating and collusion for every game show, as required by law. There would never be a situation where the crew is free to run around switching prizes without supervision.
For anyone that doesn’t understand: You start out with a 1/3 probability of getting the prize, but a 2/3 probability of not getting it. Monty will always reveal a non-prize door, so if you decide to switch your choice, you have a 2/3 chance that the original choice was not the prize, meaning that when you switch, the remaining door contains the prize - thus where the 2/3 probability of winning with a switch comes from.
*Easiest Explanation:* 1) If you pick a goat first and then switch, you will get a car 2) If you pick a car first and then switch, you will get a goat The odds of you getting a goat on the first pick are 2/3, so if you switch the odds of you getting a car are 2/3. Notes: ~The open door is not a red herring. When the door opens, it guarantees that switching will change from goat to car or vice versa. Making 1) & 2) above true. If you can understand this, you'll know why it is not 50-50. Essentially, the third door rigs the probability of the second pick. I know, It's pretty crazy, and I thought it was 50-50 too, but this makes perfect sense.
so it relies on shenanigans by the host to work? OK, say the host dies at the start of the game, then i pick a goat door. i have 2 doors left, how is it not 50%
Rona Kavanagh 'Fuck the car' is an imperative which may be taken literally to mean as stated, implying a second-person subject, or figuratively, to mean 'dismiss the car' where the principle actor may be first person in that it is their attitudinal stance which the command expresses. The result in either case is 'the car is fucked' which allows the adverb 'as well' to hang with a touch of irony due to the former equivocacy. Further, 'want' in the same sentence as 'fuck' is laden with double entendre, heightening the innuendo. So, you and the goat: not an item then?
Best explanation so far, you have a one in 3 chance, but if you initially picked either goat switching will make you win every time. Ironically you want your initial pick to be wrong forcing the host to eliminate the goat leaving only the car remaining.
The main reason this problem is counter-intuitive is because we tend to view each door as inherently having an equal probability of holding a certain item as each of the other doors. This is true in the initial conditions: we are told each door has the same chance of holding either a goat or a car. However, if we open a door it is clear to us that this probability distribution collapses for that given door. Nobody would say a door that has shown to have a goat behind it might still have a car behind it instead. We also acknowledge that the absolute probabilities of the remaining doors change as well. However, what we often fail to grasp is why this change is not evenly distributed among the two remaining closed doors. That comes from the fact that we did not _randomly_ remove any of the two goats. We specifically removed the goat from all the doors that we _did not select_. It's this piece of information that breaks the symmetry and gives rise to the counter-intuitive result that swapping doors is favorable. Hope that helps anyone understand it. Perhaps not :P
It's 50/50 still. Have tested it many times. It fluctuates on how many times you do it which is the issue and if you do the game 20 times vs 1-5 times you can see why. The amount of times is people messing with the problem if they choose to do it 100 times and even then it's still not more likely to be the other door. There is no rule or fact you need to do the test many times. Life and the universe doesn't work like that. You aren't promised 1 chance or 100. The only way it could be more than random or a higher chance for the other card is humans wanting to run the game so many times but that takes away from the actual probability. It's technically just random and no percent is accurate but the correct answers are you get it on the first try or it's around 50/50. But again it's random and no exact % is correct. Go take the tests yourself and you will see it's different Everytime and even the amount of times. Whoever said it's 66% didn't actually pay attention or is manipulating the game to their view which is incorrect just like the belief a god exists. Do the test yourself different days and different amounts and you will see, Seya little humans
This was a famous answer by Marilyn Vos Savant in Parade magazine yrs ago! She had all kinds of mathematicians and statisticians saying she was wrong! She was right!
Marilyn vos Savant was neither the person who invented the problem nor the first person to write about it nor was she the first to receive lots of negative feedback claiming that she was wrong. Don't tell me you came here from that idiotic Vox video.
Marilyn made it a national talking point in Parade magazine... before that no one but math nerds who read journals that have a few hundred circulation ever saw or discussed the 'problem'! My comment stands!
... *_Mr Manney_* Why Mathematicians get the MHP wrong… Their training requires them to believe the absolute truth of the scenario - the problem they set out to solve. They know instinctively that the answer is that a switching contestant will have a 2/3 chance to win the prize. They quite naturally resort to equations of probability because that’s what they know… Their equations - if they’re to work - need to have a zero substituted for what was previously a 1/3 chance - they make the claim that a goat door has zero chance of the car but there are two goat doors in the MHP so the unseen goat must by their definition also have a zero chance of the car. And the door with the car - well the mathematicians’ logic must also erase the car door’s 1/3 chance and make it a 100% chance. Mathematicians have it that all the doors in the MHP have a 1/3 chance of the car and at the same time No Door has a 1/3 chance - but they fail to recognise this dichotomy of reason. They operate on the basis of New Information - the revealed goat is for them new and significant - but they fail to accept that a bright child could forecast the goat as a predictable event because there will always be at least one goat between any two MHP doors - we do not need to see it to know of it - so it’s hardly new information. They drop a zero into their equation - crank the numbers and arrive with the answer that a single MHP door has a 2/3 chance of the prize - and that it must be so because the equation tells them that it’s the case. But to have a 2/3 chance of something requires there to be two objects of the same type - like goats for instance - but there is only one car! They have a door enjoying the 2/3 chance and not the contestant. The cynic has it easier - he or she does not believe the honesty of the MHP host - the cynic sees through his deception and reasons that the reveal of a goat is just a trick to make observers think that the goat and its door are removed from the overall set. The cynic reasons that the offer to swap to a single door is the second part of the host’s deception and that the actual but unstated offer is to surrender one 1/3 chance door for the other two 1/3 chance doors together. Mathematicians have it that chances change and that the host is honest - cynics have it that the host is a deceiver and that chances never change. And that actual content does not affect the chance of content.
Watching this helped at lot. More likely to pick the goat first time as 66% at that point. One goat gets eliminated straight away swap and the original odds flip to 66% in favour of getting car
NovaTime yes this is true but wouldn't you feel like a complete idiot if you picked the car first and swapped choices.. 33% is a little too high to be swapping imho.
mienbao I would interpret the unluckiness as actually luckiness because it's a definite win, but I understand what you're saying. It's just funny to interpret the words in a different sense. :)
correct simplification: -lets say you choose door 1 then decide to switch -there are 3 outcomes - car behind door 1= you lose because you switched - car behind door 2= you win because the host has to reveal 3 contains a goat - car behind door 3= you win because the host has to reveal 2 contains a goat so 2 out of 3 times you win according to this logic
For those who still are having trouble, let me draw it out: 1 = car, 0 = goat; Possible set: Door 1 | Door 2 | Door 3 1 0 0 0 1 0 0 0 1 Suppose you chose Door 1. Then the host will show you the goat. For each row, I will remove the goat which is either from Door 2 or Door 3. (stay) Door 1 | Door 2 | Door 3 Stay | Switch 1 0 0 1 0 0 1 0 -------> 0 1 0 0 1 0 1 So tell me. Is it better to stay or switch.
Your logic make no sense, it make no difference to swich/stay, your odds are still the same, you make it look like there is one extra chance to win if you swich but actually you also took away the chance to win, there forth you did nothing that could change the 33% chances to win. This video is just a joke ^_^
percentage is an illusion.When we remove one door, my choice becomes 50/50 from %33 when one door is removed.It doesn't stay %33 because now there are only two doors left.At the beginning i had %67 percent chance to choose a goat, now it is %50. because there are only two doors left.Get it? You're saying that you could win more if you swap by 2/3, but that's not true because we have 50/50 at the end.Watch the psychology of the guy it will take you to the car.if we had 9 goats and 1 car and when the host revealed one goat then swapping would be a better idea, but not in this situation.I get your thing, instead of staying with the %66 percent change of the choosing the goat because of the decision we made before we have to swap to increase the possibility, but no.When one door is removed, it is 50/50 from now on.I get it but it is hard to explain.If there were more goats swapping would be the best choice but not on 50/50 because it is 50/50 not %33.You forgot to remove one door when host removed a goat.
When there were 3 doors, swapping wins.But now there are only 2 doors left.I tell it again, if it was more than 2 doors at the end swapping would be the best idea but when there are only 2 doors at the end it becomes 50/50, it doesn't stay %33 when there are only two doors left.The possibility of choosing the one door of that %66 percent is high, i get it.Staying with that high possibility is wrong.But after one door is revealed, it is 50/50.
This is why I came here. Utterly stupid people who can't understand simple concepts even when they'e explained fully. Good Job Thunder Kat, now get outside. That extremely short bus will be here soon to pick you up.
+ᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚ They smell for a start
+ᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚwouldn't you want a car over a goat
And after pondering all day on this problem, and seeing it for myself proven with a sim program, I watched the video one more time and finally understood that indeed the chances of getting it right are 66% by swapping the original door and not 50/50 as I originally thought. It's so clear now that I feel dumb for not getting it the first time.
You only need to swap if you chose the wrong door at the start. Thus the probability of swapping and choosing the right door = probability of choosing the wrong door at the start = 66%
Ismael Mejia No what he's saying is if you do happen to choose a goat and swap you will have a car guaranteed. And you will most likely choose a goat the first time.
getbo i understand...most 50/50men understand, then rethink and go back to well theres 2 doors and one of each thing, and i have a choice to make in the moment, in the moment i say, it must be 50/50!
The simplest solution: 1. If contestant switches, he gets the "opposite" of what he first picked. 2. He has 2/3 chance of picking goat at first. 3......er, that's it.
I think this is the best explanation: I just wrote down a 5-digit number on a piece of paper (lets say it's number 34776 for example, but you don't know it yet).Can you guess which one is it?It's basically impossible with probability 1/99999, but try it anyway.Let's say you randomly pick number 22765.Than I remove all options except the number you pick (22765) and the number I actually wrote (34776).Than I ask you if you wanna stick with your number or change to the other.Do you still think you have 50% chance if you stay with your original choice?Of course not!Although there are only two options left, the probability that your original choice is the correct one stands still at 1/99999!!!The point is this - One number is still in the game JUST BECAUSE YOU COMPLETELY RANDOMLY PICK IT AND THEREBY FORCED ME TO LEAVE IT AS OPTION, while the other number is still in the game BECAUSE IT'S FORCED DUE TO ACTUALLY BEING CORRECT ANSWER.Just think about it, it's not that hard.
Still never could understand this. The way I see it, you only have two doors. The host is going to knock one out for ya. So there is a goat and a car. Two things, and he essentially asks you to pick a door again (swap or no-swap). That is 50/50. The third door doesn't matter to me anymore. If I could swap to the one the host revealed, sure. Overall, it is 1/3. But after he reveals the goat, I have two choices, and it becomes a matter of whether I have it or not. Heads or tails.
So let's play against each other 100 times. We both choose the same doors, and we both stay with our guesses. The difference between our games is that my rounds end as soon as we choose a door from three available, without you seeing what's behind it of course. Your rounds end after a goat is revealed. So explain to me how you would win 50 times while I win 33 when we have the same doors.
+Sib Pro Quo You are totally wrong. 2 doors left doesn't means it's 50/50. Let's say the game start with 1 door (1 car). After you make a pick, the host add a goat door. It's very clear that the winning chance of your door is 100%, right? Take a look at your logic. " So there is a goat and a car. Two things, and he essentially asks you to pick a door again (swap or no-swap). That is 50/50. " "Overall, it is *1/1*. But after he adds a goat, I have two choices, and it becomes a matter of whether I have it or not. Heads or tails." See? It's clearly wrong.
You're all talking to someone who cheated in calc and pre-calc in college because I knew I would never have to deal with the sine cosine bullshit in the real world. If the guy tells me I have three doors to choose from, but then tells me not to pick one of the three because it'll always be wrong, he has essentially told me to pick one of two doors. One has a car, another a goat. 50/50. If he told me to pick one of the three and never revealed a door, then sure, 1/3. But as soon as he reveals a door and tells me to swap or not, to me, he has essentially said, "Go ahead a pick one of the two doors again." Its like that second goat or third door never matters, cause he'll always throw it out of the game. In every case, it seems as if you either have a car or a goat behind the door. That is just 50/50 in my books. How is it 2/3 when he tells me one is bullshit? Like having to guess which of three pies is cherry, but your friend tells you not to pick the middle one, cause it is apple. The guy just narrowed it down for you.
***** The grammar in his comment is pretty off *shrugs* didn't get what he was saying. The other guy is then adding doors from two doors... Again - math is the pointless to me. Even you explaining how I would win 50 of 100 because I choose between two doesn't help. Sounds to me like you just agreed with me: because there are 2 doors left, chances are 50/50. "... their odds are 50/50 because there are two doors left."
***** Gonna be honest, its kinda funny to see you take this so personally. He kept saying "rounds" and "games" so I'm not sure how many times we are repeating that process and I didn't get what he was saying. I still don't get how it disproves my logic. I said having 2 doors to pick from is 50/50... and then you agreed by saying having two doors to pick from is 50/50... Instead of trying to explain it, you're getting your panties in a knot about it. If that's going to be the case, then just leave man. I'm asking for a clear explanation cause I don't understand, not an impatient douche.
Also, logically, if you always pick Door A, then Monty reveals Door B or C, and you stick. Do you win 50% of the time then? So that means the car was behind Door A 50% of the games. But what if you always pick Door B or Door C? Are those both 50% likely too? Obviously there would be a contradiction there: the car can't possibly always be behind your first pick with a 50% chance.
@@jasonl7651 the first time you pick, you have a 33% (1/3) chance of getting the car. When the host opens the other door, if you switch to the remaining door, your chances are 66% (2/3) of getting the car.
@@sheila6530 I'm not sure why that's a reply to me. You just stated the basics. My point was that the people who hold that it's 50/50 would lead to a contradiction, since the three doors must then add up to 150%
It's interesting to note that since the host knows where everything is, it cannot be a random opening of the door, and that alone excludes any possibility of there being 50:50 odds. A non-random process cannot have equal odds.
I am confused because probability can change, no? Yes originally it was 33% but now you only have 2 doors to choose from so that probability raises to 50%. It's not conditional because now you only have two doors to choose from. you're not going to choose the open door. Draw a tree diagram, it makes sense to me.
For me the easiest way to explain this is to think about what sticking or switching means in terms of the original guess. Sticking means the contestant thinks that their original guess was right Switching means the contestant thinks that their original guess was wrong Since the contestant has a 1 in 3 chance of being right the first time, they are more likely to be wrong the first time and win by switching.
But if you go to the internet, you'll find people that try to be correct all the time, so they end up losing most of the time as they thought they were correct initially. Can Monty Hall problem even applied onto internet arguments?
SIMPLE EXPERIMENT you can do which will PROVE this to be true, All you need is three coins and some method of recording results. It takes a couple of minutes, literally less time than it takes to argue about it; - Flip all three coins and lay them down on a table, if all are heads or all are tails then re-flip. If you get 1 head and 2 tails then head is the car and tails are the goats and vice versa. - Pick a coin in a predetermined and consistent position (e.g. the one on the right) to avoid selection bias (this represents your initial choice) - Mentally "reveal" one of the goats - Note whether the third coin is a car or a goat (this represents swapping) I did this 12 times and got a perfect 8:4 win-lose spread
37rainman I am aware of that. I just couldn't be bothered and i understood the logic behind it so what was the point? Peeps reading this can do it a hundred or a thousand times if they so choose, the result will be the same
martok2008 that site is flawed as fuck, if you choose the one on the left you lose, and if you pick the other two you win. I'd rather do my own probability and know it's accurate...
Important detail for everyone who thinks themselves a smartass and say it should be 50%. That would only be true if the gameshow host opened any random goat door, including the door you picked. If he did that then yes the chance would be 50% that you get it right. But it is specified that he ALWAYS opens one of the other doors. Therefore the solution shown here is correct. This problem is funny because people who are slightly more inteligent than average know that people who are less inteligent than average would switch, and then "overpredict", getting a wrong answer. It's a problem only dumb people or geniuses will get right the first time. Of course, some very average people will still not get it even after they think about it for a while and argue that the geniuses are dumb people.
Then again I already own a car, but don't already own a goat... which adds a suplementary layer of complexity to the problem as to deciding which one I actually want.
Tyler Holte Oh my GOD. Heavens thank you. Spending half an hour in the comment section brought me to a point where I thought mankind was so dumb that the only reasonable solution was to anihilate all life through a nuclear haulocost. But you made me realise some people get it.
Shane Warren If you can swap to get the other two doors you didn't chose, you get 2/3 to get the car and at least 1 goat. The host just take that goat away from you.
For people who aren't getting it, picture it this way: Imagine there are 1000 doors and behind 999 of those doors are goats, and behind only 1 is the car. You pick a door and there's 1/1000 chance it's a car, and a 999/1000 chance it's a goat. The host opens 998 doors--which takes a painstakingly long time--to reveal all the doors with goats except the door you chose and another door. If you switch doors your odds improve greatly: 999/1000 that it's the car, and 1/1000 that it's a goat. As the amount of doors approaches infinity, the odds of getting the car when you switch doors approaches 100 percent. Make sense now?
That's quite easy to understand, for me those who get confused by the 50/50 is because they think the host randomly opened the goat, but in Monty Hall Problem the host knows where is the car so he ALWAYS opens a door which doesn't have the car and that's the reason why we should always change. But is my first statement correct? Is the probabilities 50/50 if the host randomly opened the goat? So lets imagine that there are again 100 doors and there is one car. You choose a door and then your friend (who doesnt know where is the car) start opening the other doors at random and surprisingly there is no car found until two doors left (yours and another one). Now the host tells you to change the door if you want. What will you do? In this case many will think that the probabilities are reverse and now I have 99/100 the car to be in my door because that should be the reason my friend didn't manage to find the car in 98 doors opened. But, indeed the probabilities now are 50/50 (and pretty easy to prove) and whatever you do, change or not, is the same.
What if ur friend finds the car in his door?Thats the whole point of this problem its not 50/50 or what will change when 2 are left its of what u pick as first. In this video If he opened the doors randomly he would have a 1/3 chance of opening the car.But if he opened the car whats the point u would obviously swap..
@@wspang1993 If the game was done with 100 doors and 99 goats, and the host closes 98 doors, you'd have a 99% or 98% of winning by switching regardless.
I got into a big debate about this question with a friend who simply refused to believe/understand me even though I spelled it out as clearly as I could just like this video tries to. The only way I could explain it so he could understand is when I increased the number of doors. What I mean by this is, instead of only three doors with only one prize, imagine one billion doors with only one prize. You're allowed to pick a single door, just like in the initial setup. Explaining it this way, it's very easy to understand that the odds are not in your favor of picking the prize with that first pick and you probably picked a door without the prize, to which my friend agreed. The host, knowing where the prize is, opens up one of the doors that isn't the prize and asks if you'd like to change your answer. Once you decide, he opens another door and asks you again, and he'll repeat this until there are only two doors remaining. So, assuming you never change your answer until there are only two doors remaining, you'll be given one last opportunity. The host will have eliminated nearly a billion doors and you know that first door you picked is statistically unlikely to contain the prize and there's only one other door remaining. Should you swap? Of course he said yes. So when I brought it back down to three doors again, I explained AGAIN, when you pick a door, you only have a 33% chance as picking the prize right? So odds are the prize is in one of the other doors, RIGHT? I feel that was the moment my friend finally realized he was wrong. I could tell he was trying to figure out how to disprove me, by trying to some way that my example with a billion doors isn't the same as the one with only three, but even he eventually realized that wasn't the case and couldn't really refute what I was trying to say all along. And I know he wanted to. He loves to argue and hates being proven wrong but once he understood that this scenario holds true whether it's 3 doors/1 prize or 1,000,000,000 doors/1 prize, once every other door is eliminated with only 1 last door remaining, if you swap the odds will be in your favor.
also you can map it out assume you pick door 1 for example ( or door one is the first pick): 1 2 3 car goat goat goat car goat goat goat car okay so this shows that having door 1 representing you choice ( as the door order doesn't really matter in your choice) you have a 1 in three chance of picking the car off the start. now let's take that door away and then have the host pick a goat door ( as he never reveals the car under the rules). then we get this without door number left over: goat car car but wait there's now two chances to pick the car out of three total. you have 2/3 odds of following a path to victory if you swap.
I don't like increasing the numbers, because it does change the game drastically. The odds of choosing the right door out of a billion, is almost impossible. But the odds of choosing the right door, out of 3 choices, is very likely.
@John Bryan It doesn't change the game whatsoever. With 3 doors, you have a 66% chance to pick wrong, so the odds are, you will. All increasing the number of doors does is emphasize this. With 4 doors you have a 75% chance to choose wrong. At 5 doors, 80% to be wrong, 10 doors 90% etc. Since odds are that first choice is wrong (even at 3 doors), once you get down to only two doors remaining, the one you initially chose and the one the host hasn't eliminated, it's always better to switch because odds are, your initial choice was the wrong one.
If the odds were really 50-50 then you'd have to be able to refute the following: _For a dedicated strategy of either always switching or always staying, the initial pick determines the final outcome in 100% of cases._ In other words, once you've made the initial pick nothing else is left to chance. It is predetermined what your final result will be once you've selected a goat or a car. And since this is true, the odds can't be 50-50 as you are twice as likely to pick a goat at the outset than you are to pick the car.
There are so many people that don't quite get it. Let me simplify it for you. When you pick initially, you are most likely to get a goat. If you did get a goat, the host is forced to reveal the other goat you didn't pick. This means the final door must be the car. The chances of you getting a goat at the beginning is 66%, so because you switched, there is a 66% chance you get the car.
The number of people falling for this hoax theory is astounding… The fallacy here is believe that probably of the second attempt to open door is influenced by the probability of the first attempt, when in fact it’s not. As soon as you opened the 3rd door, the probability of the first door being the car increased to 50%, and not staying at 33%. Take a scenario with 2 doors for example. If I choose first door, host opens second door and reveals the goat, the probability of the first door being the car is now at 100% and not staying at 50%. If somehow the first attempt to open door influences probability in the second attempt, then this would be true.
@@robertdowney9017 surprisingly its not a hoax. there is a video on YT with the title " marilyn vox savant". i suggest you watch it. she even had mathematics professors apologizing to here after they eventually figured it out.
The problem with this whole debacle is that they are treating the first pick goats as actually two different goats which is correct but treating the second scenario with choosing the car first and switching to goat because ignoring which goat the host reveals and treating them as the exact same goat because in a 3 letter scenario there are 6 possibilities with no letters repeating and you have to take out 2 possibilities because host cannot reveal car so thus giving us 4 possibilities
When he opens the door with the goat...the car and the goat both have 50% chance. Because, the 2 doors that are left, either have a car, or a goat. The third door is completely irrelevant after this point. Simple elimination of fractions. Door 1...2/3 goat, Door 2...2/3 goat, Door 3....2/3 goat.... I pick door 3. The guy opens the first door to reveal a goat. NOW THERES ONLY 2 DOORS LEFT. In door 2, theres a 1/2 chance of a goat, same in door 3, odds are the same for the Car. Fuck this bullshit im out.
MyNameIsSimple72 You're wrong. This problem has been around for decades. Step one is realizing you have it wrong (it is a common thing - you're not alone; it is DESIGNED to make people think it is 50/50). Think of it like this. What if you NEVER switch? According to you, I have a 50/50 chance, so I should have just as good of a chance of winning the car by NOT switching, right? For me to actually win half the time, I would have to originally pick the car OUT OF THREE DOORS half the time. Does this make any logical sense to you? It shouldn't. Just play the game out in your head. Realize that no matter what you pick, if you choose to switch, you will ALWAYS end up with the opposite of what you originally picked. Go ahead and work out as many different scenarios in your head - - I challenge you to find a scenario where what I say is not true. You pick a goat, Monty will ALWAYS open the only other goat, and you will switch to a car. You pick the car, Monty will reveal one of the goats, and you will switch to the other goat. You pick a car and switch, you end up with a goat - - every time. You pick a goat and switch, you end up with the car - - every time. You will pick a goat first 2/3 of the time, so switching will result in you ending up with a car......2/3 of the time. The fact that the host reveals a goat door is IRRELEVANT because he will ALWAYS do it. He is REQUIRED to. It has absolutely ZERO impact on the odds. From the very beginning, you have a 1/3 chance that the car is in the door you originally picked, and a 2/3 chance that it is in one of the doors you did NOT originally pick. The only thing the host does when he reveals a goat is reduce the number of doors you did not pick from 2 to 1. There is STILL a 1/3 chance that the car is in the door you originally picked, and a 2/3 chance that it is in the only other door left that you did NOT originally pick. The problem's trick is convincing people that the action of revealing a goat changes the odds. It doesn't.
MyNameIsSimple72 you can also test this in real life. bigger sample size the better. take ~20 rows of 3 cups with the same ratio of hidden items and run through each row following the rules with a friend being the host. youll be correect roughly around 66% if you swap each time
MyNameIsSimple72 If your first pick is Goat A, it's Goat A. If your first pick is Goat B, it's Goat B. If your first pick is the car, it's the car. Your first pick gonna be the car 1 in 3 games (33%), not stupid 50%.
Willoughby Krenzteinburg I discussed this with a friend as well, and isn't it so that the only reason it's not 50/50 is because of the host of the so called show? Just think about it
robin291292 What makes it work the way it does is the fact that the host is REQUIRED to reveal a goat door after your initial choice. This is what makes your odds increase from 1/3 to 2/3 if you choose to switch. The reason being, the hosts actions do not affect the probabilities when his actions are following this rule. No matter which door you choose, there will ALWAYS be a goat door available to open, and the host will ALWAYS open it. You know this going in, so the fact that the host does it is irrelevant. He is giving you no new information. From the beginning, there is a 1/3 chance you originally picked the car. There is a 2/3 chance the car is in a door you did NOT originally pick. All the host does by revealing a door is reduce the number of doors you did not originally pick. There is STILL a 1/3 chance you have a car in your door and a 2/3 chance the car is NOT in your door. Now, there is only 1 other door that is NOT your door - representing that 2/3 chance. In a variation of the game where the host does NOT know where all the prizes are and truly reveals doors at random (which means the host COULD potentially accidentally reveal the car door in some cases) - in the cases where the host reveals a goat, your odds of having the car are 50/50. This variation would be statistically similar to the game show, Deal or No Deal in that the contestant is randomly opening cases, and in the event that the contestant has revealed all but two cases (their original pick and one other case), then at that point, there is a 50/50 chance that each case contains one of the two remaining possible prizes. So yes - to answer your question. The fact that the host's actions are predetermined to an extent (the host can't reveal the goat door you originally picked, but would be forced to reveal the only other goat door) is what makes the statistics of the Monty Hall Problem work the way they do. Mathematically, it would look something like this : If you want to figure out the odds of BOTH event A and event B of occurring, you just multiply them. In the regular Monty Hall Problem : So, the odds that you pick a car (1/3) AND Monty reveals a goat (1/1 - because he is required to and has no choice) are : 1/3 * 1/1 = 1/3. The odds that you pick a goat (2/3) AND Monty reveals a goat (1/1 again - because he HAS to) are : 2/3 * 1/1 = 2/3. So these are the odds AFTER Monty has revealed the door. 1/3 that you have a car, and 2/3 that you don't. In the variation where Monty acts randomly : The odds that you pick a goat (2/3) AND Monty reveals a goat (1/2 - only two doors remaining - only one of which is a goat) are : 2/3 * 1/2 = 2/6 = 1/3 The odds that you pick a car (1/3) AND Monty reveals a goat (1/1 - there are only goat doors remaining for Monty, so he is guaranteed to reveal a goat) are : 1/3 * 1/1 = 1/3 The other 1/3 of the time, you will choose a goat, and Monty will reveal the car - and this game is unplayable. So we ignore these events. That leaves us with the remaining 2/3 of the games. 1/3 of the time you have the car, and 1/3 of the time you have the goat. So, in the cases where Monty has acted randomly and has revealed a goat, there is a 50/50 chance you have the car. Now, it is STILL in your interest to switch because you have no way of knowing which version of the game Monty is playing. At best, you double your chances - at worst, you take a lateral move that does not decrease your chances, so you might as well always switch.
If you picked the car and swap you lose. If you picked the goat and swap you win. 2 out 3 times you picked the goat. 2/3 times you swap, you win. *keep reading other user's explanations, and watching other videos. Once you understand it then, well, you'll understand it, yeah
Simplified: if you pick a goat, and the host reveals the other, and you swap, you win, cuz they are the only goats in the game. And there is initially 2/3rd (66%) chance to pick a goat.
Another way to overcome some of the counterintuitiveness of the improved odds is to think of it like this - the host CANNOT choose your door, and since he KNOWS which doors the car is behind, he HAS to pick the other door - besides your door and the car door. Your choice and his knowledge of the doors makes switching to the only other unchosen door makes it more likely - but not certain - that the other door contains the car. He is always going to open a goat door, regardless of whether you chose a car door or a goat door. He has eliminated the original possibility that a car was behind the door he picks. Doing that increases the remaining possibility that the other door has a car, because if it has a car - he could not choose it.
Another way to explain it is by making the example more extreme. Imagine there are not 3 doors but 100. You pick one at random, which is almost certainly wrong. The host then opens 98 doors to reveal 98 goats. Now you can see, he is pretty much showing you exactly where the car is.
@@aaroei7 replicate this experiment on real life or more realistically through a program. I can guarantee you you're wrong. It's a hard concept to understand.
Nicholas Davis I agree that if it’s with 100 doors, you should swap because he basically shows you the right door but if it’s with 3 doors it’s a 50/50
@Life Prof. B If I am lucky I get a goat..which means I am unlucky.. AND If I am unlucky I get a car..which means I am lucky So you mean I am lucky if I am unlucky and I am unlucky if I am lucky.. Enlighten me xD
you're twice as likely to land on a goat 1st time than you are on a car. by swapping you are guaranteed to get a different outcome to what you had started with. you're therefore twice as likely to get a car by swapping than sticking. makes sense :)
I finally get it, thank you. I think the issue of people not understanding is too much emphasis being on that second choice, which technically is a 50/50 but its not the only round. It makes perfect sense when you realize you should assume you're first door was a goat based on probability.
***** You are not calculating the odds properly. You've fallen for the trick. Intentionally or by mistake? Idk. Round 1: Is meaningless. You CANNOT win the prize in this round, period! Round 2: 2 doors, one prize. 50-50 chance if you stay OR switch doors. 1 2 C G G C See? Let me put it another way: What are the odds you'll win the car in the first (3 door) round? Zero. There always is a second round regardless of which door you choose. See it now?
5Cats That has to be the dumbest logic i've ever heard. Just because you don't open your door on round 1 doesn't mean its meaningless at all. From the start you have a 1/3 chance of being right. There is a 2/3 chances of the car being in the other two doors. That being said, one of the other two doors obviously has a goat, so just because he opens the goat door first does not make it a 50/50. There is still a 66% chance that either of those 2 doors has a car, and since he showed you the one with the goat, you know that if you switch, you will have the correct one (of the 66% chance) that has the car.
5Cats If YOU wish to look ignorant that is one thing, but to tell others they are wrong when they are absolutely correct is quite another. Perhaps I recommend to you to LISTEN to people who are telling you 100% factual information. This isn't a debated issue...it is math and factual math proven by computer simulation and logic. People are trying to educate you and you are just looking extremely uneducated to them right now.
5Cats Are you going for the record for the number of times you can present the same illogical and incorrect solution to the MHP in multiple threads? (And where's that Wiki reference to the MHP and the Gambler's Fallacy you keep talking about?)
There's no such thing as a 50% chance if you have 3 variables. Your first pick is of course 1/3 win. If host reveals goat, and if you stay, your odds are still the same, 1/3 win. Now by switching, you change your odds, gives 2/3 win. Switching ALWAYS gives you better odds. 50-50 chance does not make sense.
+The Absolute Zero the host NEVER opens the door with the car but he opens one AFTER you first choose one. you dont know whats behind yours but you deffinitly know the one the host opens after that is not the car. thats why it could never be 50-50 chance. This video explains it perfectly. why you call people dumb if you dont understand such a simple explanation like the maker of the video did?
+The Absolute Zero "the whole time you're really just playing with 2 doors" The three doors you are playing with the whole time are: (a) the door first chosen by the contestant, (b) the door revealed by the host (which cannot be (a) or any containing the car), and (c) the unaddressed door (which cannot be (a) or (b)).
+Freddie Orrell The thing is that you don't have three doors all the time. You start off with three closed doors and pick one. The host then opens a door, removing a variable. So you start off with three doors and end with 2 doors. P.S. probability of winning when you keep the door is 1/3 while swapping gives you a probability of 1/2.
The easiest way to interpret this is go over all possible outcomes: 1. You pick a goat. You DONT switch. You get a goat 2. You pick the other goat. You DONT switch. You get a goat 3. You pick the money. You DONT switch. You get the money 4. You pick a goat. You DO switch. You get money 5. You pick the other goat. You DO switch. You get the money 6. You pick the money. You DO switch. You get a goat As you can clearly see switching gives you a 2/3 chance of getting the money whereas not switching gives you a 1/3 chance
But your choice between switching or not is NOT between THREE doors, your choice is only between TWO doors. The one you originally picked, and the one that remains after Monty opened the third one that has a goat, and removed that door and goat from your choices. He moved that door and goat off the stage because it is no longer a choice you can make. Now you can only choose between two doors. Your chance of winning a car from one of those two doors is 50/50.
@@lxilind7567 I did figure it out, actually, lol I made a comment further down in these comments, and I think I just was not understanding that at the start of the game, with three doors, I was going to pick the wrong door 2/3 of the time. here it is: Ok, I figured it out. I think. 1$, 2G, 3G. I pick one (*), and a goat door is removed. *1$, 2G, __. Two choices. Stay/win, switch/lose. 1$, *2G, __. Stay/lose, switch/win 1$, __ , *3G. Stay/lose, switch/win One stay wins. Two switches win.
The thing is the game doesn't reset at the second part. The possible results in the second one depend on the results in the first one. If you hit the car in the first attempt, changing will make you lose. If you didn't hit it, changing will make you win. As you see, if changing or not make you win depend on your success at the beginning. It's not like flipping a coin, in which the results are all independent.
this is not a guarantee that you will win when swapping, it is just clearly stating that the probability of you choosing a goat at the beginning is higher than choosing a car therefore swapping will give you a better probability of winning
its easier to think of it from the HOST's point of view. 1- the HOST will ALWAYS avoid the correct door if he HAS to. 2- the percentage that the host DOES NOT HAVE to avoid the correct door = the percentage that the contestant chooses the CORRECT door. = 33% 3- the percentage that the host DOES HAVE to avoid the correct door = the percentage that the contestant chooses the WRONG door. = 66% 4- Since there is 66% chance that the HOST will HAVE TO AVOID the correct door, means the door he is leaving has a 66% chance of being the correct door.
The best way to understand the "swapping" problem is this (at least this is how I understood it the first place): Imagine there are 100 doors, and you are asked to pick 1 initially. So you choose 1. Then Mikroutsikos (greeks will get it) opens 98 doors, and now you are left with your initial choice and one closed door. The chance of having won from the beggining is 1% (1/100) whereas the chance of the car being behind the other door is simply 99% (99/100). Not swapping makes absolutely no chance. If you still don't get it, imagine 10.000 doors. Not swapping means you are certain that you have picked the car on your first choice :-)
astr1nos This is an effective extrapolation. The other is to ask what more they know about their first 33% choice when they say the new odds are 50/50. Most admit it still must be 33%, so the balance still must be 66%. Then, ask what more they know about the original 66% side, the one they didn’t pick, which is now just one door and must be twice as likely.
+HumptyDumptyOakland: True, but that changes once the 'goat' door is revealed, because you KNOW that one of the dreaded goats WAS NOT behind the door you picked. In plain language, there are no longer 2 goats available for the booby prize, but one. Your odds improved immediately, from 1 in 3 to 1 in 2.
+Georooney the odds of having chosen the car on the initial pick remain 1/3 even after a goat door is revealed but the odds of winning car if you switch has increased to 2/3 ... it's basic probabilities
Goat is 1, Car is 2, open door is 0 You pick middle each time 1 1 2 0 1 2 switch and win 1 2 1 0 2 1 switch and lose 2 1 1 0 1 1 This is why, the host would never open the door with the car after your first selection, he would open door 3, he has revealed new information the impacts you "50/50" choice on the final pick, the events are not independent of each other. 2 out of these 3 events you win if you switch.
I always liked to turn this intuitively by changing it to 100 doors. Suppose there are 100 doors, 99 goats and 1 car. You pick one door by chance an the host opens 98 goat doors. Now, do you swap or do you stay with that door you picked first? Now its obvious without doing the math.
The chart at the end has been the best this has been explained to me. I always hated this puzzle because everyone's explanation was kinda weak and didn't cover all possibilities well. This however did and now I feel confident in why it works as to before where ik the answer but couldn't explain why it was. Thank you.
If I’m interpreting it right, does the table show that if you don’t swap, you’re essentially locking in your chances of a 33% probability of winning. But if you DO swap, instead of a 50/50, you actually have a 66% chance when you still factor in the first door with the goat as a possible option that you’re just gonna “ignore”!!
+J Fleming The tricky part is in understanding that the elimination of a goat must be deliberate and non-random, not by accident. If the host (or the player, or anyone really) blindly picked one of the unselected doors and opened it, revealing a goat instead of a car *by chance* instead of knowing ahead of time that they're revealing a goat, surprisingly the advantage of swapping disappears. This is why you get people thinking that it's beneficial to swap cases at the end of a game of Deal or No Deal even though it's actually 50/50 either way.
J Fleming No, I have insight into it that you lack. If someone insisted that a player should swap at the end of Deal or No Deal for an overwhelmingly likely chance to win the grand prize, you would be unable to explain the difference between the two games to them, and therefore you lack a thorough understanding of why the game works the way it does, which is the entire point of a math-based probability puzzle. If Monty tripped and fell and accidentally pushed a door open to reveal a goat, there would be *no advantage* to swapping. If you don't understand exactly why that is, you haven't thought about it enough and are in no position to claim it's so simple.
Why may i ask is it called "Monty" hall problem ? I would take the goats over the car because i have some Welsh and Aberdeen mates who would pay top dollhairs for pampered tv goats.
It's the "Monty Hall" problem - which is a person, who was the host of the game show called "Let's Make a Deal" which featured contestants picking among three doors. This problem - while not exactly the same format as the game show - is certainly based on that game.
bro it's simple goat - 2/3 car - 1/3 monty opens goat door 100% of the time, leaving one goat and car if you pick one and switch, you WILL get the other option. you pick the goat and switch, you WILL get the car. and the probability of picking the goat? 2/3 you pick the car and switch, you WILL get the goat. and the probability of picking the car? 1/3 middle schooler explains 😎😎😎😎😎😎
The explaination is quiet simple : Only one car and TWO goats. So : 1/3 : you pick the car, then you must stay to win. 2/3 : you pick goat 1 or goat 2, then you must switch to win. Saying that the final choice is a 50/50 chance means that your initial choice is 1/2 to pick the car and 1/2 to pick a goat, which is mathematically wrong. Then : ALWAYS SWITCH. CQFD as we say in France :)
One minute ago, I was just like this can't be true! I supported people that disagreed with the fact that swapping increases the chance of winning. Now I understand it and think people that don't get it are stupid...
If there were only two options to begin with, yes, it would be 50/50. But since there were two goats and only one car, there's a 2/3 chance that the door you chose initially has one of the goats behind it, but only a 1/3 chance that you picked the door with the car. Now do you understand?
since the host is showing a goat behind a door no matter what, picking again is 50/50 odds , i understand how the odds increase from 33 to 50 but in reality its always a 50/50 pick since one door is being eliminated with no risk or reward
You are more likely to pick the goat (2goat out of 3 doors 66%) than the car(1 car out of 3 doors 33%) So when you pick the door with the goat(66%) they must switch it to other door with the Car, since they must reveal the one door with the goat when you pick the goat at first So the chance of you picking the goat and swap to the car is 66% And the chance of you picking the car at first and swap to goat is only 33% since it’s only 33% to pick the car at first
I understand the math behind it, and it actually makes sense when I see the numbers, but it's still so hard to wrap my head around it because in my mind it's still 50/50
That's because only two places the car can be presents the illusion of 50/50. What you fail to see is that the third door affects the odds for each door differently. A way to easily see this is that you are effectively trading TWO doors for one if you switch. How so? If the car is behind EITHER of the two doors.... you WIN! Think of the lottery. Would you trade your one ticket for two?
This makes sense. We pick a goat 66% of the time before the reveal of a door. But, if we swap after the door reveal, it would mean we have a higher chance of getting the car.
The number of people falling for this hoax theory is astounding… The fallacy here is believe that probably of the second attempt to open door is influenced by the probability of the first attempt, when in fact it’s not. As soon as you opened the 3rd door, the probability of the first door being the car increased to 50%, and not staying at 33%. Take a scenario with 2 doors for example. If I choose first door, host opens second door and reveals the goat, the probability of the first door being the car is now at 100% and not staying at 50%. If somehow the first attempt to open door influences probability in the second attempt, then this would be true.
@@robertdowney9017 But the first choice obviously influences the second choice? The game does not reset - the positions of the prizes does not change, and what you chose the first time is guaranteed to be one of two options on the second choice. For the two choices to be independent, the prize positions would need to randomize after the first choice but before the second.
for all yall that dont get it your first pick has a 66% chance of getting a goat right? so when he reveals the goat if you switch you have a 66% chance of winning bc (like i said earlier) you probably picked the goat and if you think its 50/50 i can see your logic and it would be true IF he revealed the goat before you picked the goat but since you pick the goat(66% chance) before he reveals the other goat you should swap bc you probably picked the goat in the first place. If you wanna debate im ready
This doesn't make any sense. At all. It doesnt matter what door you pick at all. Since as soon as tou pick the door 1 of the 3 doors is revealed to have a foat in it which makes It a 50/50 chance instantly. The 33%, 66% thing is utter bullshit since It is just the percentage of the past which becomes absolutely irrelevant after your pick. If you have an option to swap after your pick doesnt matter what you pick chances of winning are always 50/50
Okay for those who don’t understand: when the show first started, you had a 2/3 chance of picking a goat. Got it? Okay, so now we are revealed another door which has a goat. So either the door you chose or the 3rd door has the car. Still with me? So since there was a 2/3 chance of selecting a goat in the first place, and the second goat was revealed, the 3rd door is most likely the car. There was only a 1/3 chance of picking the car, so switching would be a good idea. I hope you understand and if you still don’t please feel free to ask. Thanks!
A few points worth bearing in mind: 1. "The Monty Hall Problem" is the name of a logic puzzle. Whether it is the same as the TV show or not is irrelevant. 2. If the host opens a door at random and reveals a goat, then the probability is 50/50. 3. If a third person comes in after the opening of the door, and does not know what has happened, then for him then the choice of winning is 50/50. 4. Switching does not guarantee a win, but gives 2/3 chance of winning
Just to clarify a bit: Ad.2 If the host opens a door at random, his chances are 50/50, so half of the time he will screw the game (since he will reveal the car instead of goat), and half of the time he will reveal the goat and give option to a player to win a car by switching to an unopened door. That doesn't change the fact that first player, by choosing one of the doors gets 1/3 probability, while remaining 2 doors have 2/3 probability, hence, in all cases where host opens the door with a goat, player will have 2/3 possibility to win a car by switching. Ad.3 If a third person (second player) doesn't know the choice of first player, his chances of winning are 50/50 since there are only two doors, one less likely to have car behind it (the doors that the first player initially choose), and the one more likely to have car behind it (the one of two unopened doors in first round of first player), and he doesn't know which door has better probability.
@@max5250 If MH opens a door at random the chance of his revealing a goat is 2/3, of revealing the car 1/3. You are assuming that the contestant has picked the car,
@@rwb966 Correct, my bad. The host would open the door with a car in 33% of cases, and screw the game, while he will open the door with a goat 66% of time, and still give opportunity to the player to win by switching.
@@rwb966 May bad was I said that the host would open door with car 50% of the time, i my first reply, which is obviously wrong, he would open that door only 1/3 of time.
How dare you Detective Diaz, I am your superior officer.
*Bone!*
kevin is right
YOUR FIRED
99
@@vishnujeganmogan1750 WHAT. THE. ACTUAL. FUCK. I was right on season 4 ep 8 of b99 when I hit up this vid lol
Best scene from B99 ever?
If u guyz interested in maths riddles
I have one in my channel
what brand is the goat
Russell
honda
Jericho
suzuki
Mercedes
What if I want the goat? Huh, ever thought about that?!? Next time don't assume my transportation preferences.
This video is nearly ten years old, and about every fifth comment or so is about wanting the goat instead of the car. You're all so very clever and original.
TedManney First of all I know that and second of all I wrote this comment before looking at any other comments
RyoMiianPlayz No shit, Sherlock. You left the same comment a million other people did *because* just like them, you typed up the first obvious thought that popped into your head and sent it out into the universe. People like you clog up the internet with repetitive, unoriginal thoughts, and the worst part is that you can't be bothered to read what anyone else has to say and then you expect people to read your bullshit.
RyoMiianPlayz tl;dr
Ross Sadler I literally don't understand how anyone can be like you. I said I know that other people left the same comment. Just cause other people have the same thought as you doesn't mean it's an "unoriginal thought". I just literally can't rap my head around the fact that this is so important to you that I know that I'm "unoriginal"
This is a good example of why many people who study mathematics despise probability. It's countertuitive and it is also hard to visualize.
how is this countertuitive it makes perfect sense. probability is literally everything.
and poker proves this wrong every single hand.
I'm not aware of probability being despised by many that study mathematics. And in my experience, probability and statistics were by far the easiest higher level math courses I took (I majored in Mechanical Engineering, minored in math.)
I find the process of taking something that appears straightforward, and uncovering a surprising underlying principle to be a thing of beauty. And in many cases, I feel like a hero battling villains when I see data that has been presented in a misleading way and I get to identify it and call it out.
@@whitey6317Making sense and being counterintuitive are not mutually exclusive nor is making sense and being intuitive the same thing.
When some thing "makes sense" means that knowing the "solution" or "the way a X works", you can reason it out.
Being counterintuitive means that just via intuition you will have a hard time arriving at the answer.
Intuition usually does not examine all the possibilities but reasoning, i.e. the part of your brain that can make use of logic to make sense of things does (if done properly). Intuition on the other hand makes decisions based on the amount of data that you receive at a glance (you can train it of course to start inferring the answers you'd arrive at with reasoning, but that requires building experience).
The perfect example is why the regular person (which includes me) will think (or rather intuit) its a 50/50 chance. People don't consider that the chance of winning the car is bigger because you can make the choice. They assume that having one or two choices does not make an impact on the probability and even if they notice the fact that the host will always go for a door with the goat in the first reveal they only think they understand the implications.
Regular person thinks like this: "Its one in three, but because the a goat will be revealed on the first reveal its really 50/50, so because I want the car therefore I want to hit it on first try and keep the door on the 2nd change thus producing less anxiety." Upon thinking of it a little more, its easy to observe that the anxiety level will not change (it might even rise if the person doubts their initial choice - which is a likely occurrence) and its easy to observe that its wishful thinking to hit the lower probability.
A person who can flip their intuition may think: "Well, since the first thing to be revealed will be a goat, and definitely not the door I chose, I should be aiming to hit a goat so the host is forced to reveal the other goat, and really hitting the goat is the more likely reality, then I should just embrace the goat and then release it." This shows a surprising level of acceptance towards the possibility failure which stems from understanding and acceptance of the limitations of the way this choice works (under the assumption its not rigged) as opposed to the wishful thinking of the regular person's reaction to this.
I hope the above makes sense.
The Monty Hall problem only gets the answer it does if you add another hidden rule: that Monty always has to make the offer of another door. If the likelihood of his making the offer differs depending on the subject's first choice in an unknown way then you cannot determine the correct choice. If Monty for instance, likes screwing with you and only offers the possibility of another door when the person makes a good first choice, then you reduce your chance of winning by switching.
If you want to win a goat the best choice is to pick the door the host opened.
A surprisingly astute comment - many gainsayers wrongly assume the revealed door is eliminated from the probability equation when in fact, as you illustrate, it remains and has 3/3 chance of goat and 0/3 chance of car. While it may be eliminated from the set of 'doors a contestant who wishes to win would choose' or even from the set of 'doors offered to the contestant to switch to', it remains as the door whose conditional probability is 0/3 car following the reveal, the other doors now having 1/3 and 2/3.
Well, I wanted to leave a unique brain fart, just point something out that I assume nobody or hardly anyone has pointed out so far. Thanks for congratulating me on that. :-)
Senjiu Kanuba Your comment merits being taken seriously. Many people simply counter the advice to switch with 'But I want a goat', in a limited attempt to subvert the expectation that the car is desirable without respecting the core purpose of the MHP as a math problem. However the theoretical chance of trying to win the goat is an effective counter to misguided solutions such as the Ablestmage Error, where the doors are reduced to two, and Buxtonian Logic, where the remaining two doors' chances are not divisible in a different way to before the reveal.
To be honest, I found the problem too simple to take it seriously but I can see people having trouble with it.
It's the same problem with "You have 3 similar looking keys, one of them opens the door, you remember which ones you tried already. Which one is most likely to be the key, the first, the second or the third one you try?", people don't try to figure out the likelyhood of each event, they just answer on gut level (or with system 1 as opposed to using system 2). That's why you learn this stuff in school.
Or a more popular problem because it might be actually important to someone: "You doctor has a diagnosis tool that returns positive in 99% of the cases when someone has the illness and negative in 1% in that case. It returns positive in 10% of the cases when you do not have the illness and negative in 90% of the cases then. One in ten thousand has the illness. You are diagnosed positive. Is it more likely that you do have the illness or not?"
What's your first guess? And what do you think is the actual answer?
Take a moment and think about it.
The solution is that in a million people 100 have the illness, of those 99 are diagnosed positive and 1 is negative. Of the 999900 people that don't have the illness 899910 are diagnosed negative and 99990 are diagnosed positive. So if you are diagnosed positive you are either one of the 99990 who don't have the illness or one of the 99 who do have it. So your chance of actually having it is 99/100089 which is about 0.1%. That means, in this setup, if you are diagnosed positive you have about 99.9% of not having the illness and 0.1% of having it. If you are diagnosed negative you have a one in 899911 chance of having the illness and about a 99.9999% chance of not having it. So it's a relatively reliable test if it returns negative but if it's positive you should probably take another test, unless you can live with a one in thousand chance of having that illness.
So I agree with you in that many people just don't think enough. :)
Senjiu Kanuba Surprisingly if anyone wants to win a goat their best bet is to open the door of your house, there's a 77% chance your mum is home.
The short version:
1.) Your more likely to pick a goat on your first pick. (2/3 times more likely)
2.) So switch, the second time round.
this is actually quite well simplified
exactly
I disagree with your simplification. The entire logic of the problem relies on the fact that they are revealing a certain door to you. With your simplification, you don't know what is behind the doors on your first pick, so you have a 2/3 chance of being wrong. Then you say that you are probably wrong so you should switch. Well, you still don't know if you are wrong, and you don't know what is behind any of the other doors, so you are effectively eliminating your own choice making it a 50/50.
3) Unless you feel like winning a goat😂
@Xx BigBoss xX no, that's just logical. Probability is theoretical, whereas in reality it is still 50/50
Here are all possible outcomes of swapping vs. all possible outcomes of not swapping. I've numbered the goats to make the illustration a bit easier to understand.
Swapping:
player picks goat 1, host reveals goat 2, player swaps to car - player wins
player picks goat 2, host reveals goat 1, player swaps to car - player wins
player picks car, host reveals goat 1 or 2, player swaps to goat - player loses
Swapping odds - 2/3 for player
Staying:
Player picks goat 1, host reveals goat 2, player stays - player loses
Player picks goat 2, host reveals goat 1, player stays - player loses
Player picks car, host reveals goat 1 or 2, player stays - player wins
Staying odds - 1/3 for player
If anyone can't see it after that explanation, it's because God doesn't want them to have nice things.
it make more sense that way.
NOW I get it. I was doing mental gymnastics to try and figure it out but somehow this worked better than the video did
I came here from BoOoOoOoOoOoOoOoOoNnNnNnNnNnEeEeEeEeEeEeE?!?!??!?!?
Anonymous Person you mean Rosa Diaz
BOOOOOOOOONNNNNNEEEEEEE
BOoOOOOOOOOOOnnnnnnnnnneeee?!?!?!?!?!?!?
@Batman Fan what me and my husband do on our personal time is none of your concern
True that
The real question is whether the car comes with insurance or not
Maz Amd no drive it on your driveway
Ooooooh
And if I can save 15% if I switch to GEICO.
I’d just sell the car
Does the goat come with insurance
Say there are 1000 doors, and only 1 car still. Once you choose a door, the host opens 998 other doors with goats behind them, leaving a single door unopened, plus your initial door. What are the chances if you switch, vs if you don't? Would that not make for a very, very, very easy game in which you are almost assured to win?
+David G Yes, it would. As you might expect, swapping will win 99.9% of the time while staying only wins the other 0.1% of the time. Examples using more doors are often used to help people who don't "get it" because it becomes more clear that swapping is simply betting on our first choice being wrong when we start with a near-impossible chance of selecting the car off the bat. Counterintuitively, the more doors we start with the easier it is to win the game!
+David G Nice explanation!
+Somayajula Padmavathi How is this a nice explanation? Changing 3 doors to a thousand doesn't make any difference in explaining.
Drieks It shouldn't, but it does (to some people). It's a lot more obvious when you're only 0.1% likely to pick the grand prize from the start that swapping is almost certainly the only way to win. When the chances are 1/3 against 2/3, lots of people give themselves permission to regard the chances as "close enough" to 50/50 as to not make a difference. Using a thousand doors is a way to force certain people not to ignore the diminished probability of selecting the grand prize off the bat.
+TedManney Thanks, Ted. I couldn't have offered a better way to convey what I was actually trying to say.
And thanks Drieks, I see your point ~ the 1000 door explanation does not in any way help a person who is attempting to solve this problem mathematically.
why don't these people want a goat?
Looney Rose Maybe they think they are not worthy of owning a goat. 😊
Dude, do you know how many goats you can buy, if you sell the car?😂
lol. goats are better that cars.
'Americans'
Vegetarians smh :-/
Captain Holt: you're fired.(Amy)
Rosa: Haha
Shane Goes
Yep I came from this
what kinda game show is this
An imaginary one created for the purpose of illustrating an interesting phenomenon in conditional probability.
julianpalca lets make a deal did this
The name is based off of Monty Hall, the host for Let’s Make a Deal, so probably that, haha
what kind of person are you.lol
Think of it this way.....When you PICK initially, you have a 33% chance of guessing correctly, and a 66% chance of guessing incorrectly. Now, if you SWITCH doors, after asked by Monty, you ONLY LOSE if you picked the CORRECT door initially! If you DON'T switch however, the door that Monty opens (with the goat, after your selection) is just a INDEPENDENT event, and has NO impact on your initial selection! This is a GREAT problem....I have taken stats classes, advanced mathematics, and have even studied this specific problem in particular, yet every time it continues to amaze me!
Now I understand it! Thank you
@Steve It's just the contrary. The contestant chooses a goat door in 2 out of 3 times on average. On the other hand, the host knows the positions and cannot reveal the contestant’s selection and neither the door that hides the car, which means that everytime the contestant has chosen a goat door (2 out of 3 games), the other door the host leaves closed is which has the car. So, we will always end with two doors, but the switching one will have the prize in 2 out of 3 times, not in 1 out of 2.
Steve, you can't have a 1/2 chance of having the car behind any of the three doors you pick unless you change the total probability from 1 to 1.5 in your explanation. That makes it wrong no matter how you word it.
The total probability in the original question is 1 because there is one car among three doors. The total probability in your explanation must also equal 1...but it doesn't. In order to have all three doors with a 1/2 chance of having the prize if selected would require a total probability of 3x(1/2)=1.5 which is 1/3 more than there are cars.Two of the three doors can have a 50/50 chance if, and only if, the other door that was opened had a 1/3 chance of having the prize but didn't. Therefore your explanation must be wrong.
@Steve "except your chances change to 1/2 when they open a door containing nothing. there are only two possible solutions left. your selection now has a 50% chance to contain the prize."
That logic must then apply to this as well....except your chances change from 1/3 to 1/4 when they add a door containing nothing. There are four possible solutions left. Your selection now has a 25% chance to contain the prize.
Lol!!!!!!
One way to see this problem intuitively is to tweak the conditions of the game: Instead of three doors, imagine a million doors, with all goats and one car. After the contestant chooses a door, the host opens all remaining doors except one. Now it is obvious that the odds are overwhelming that the car must be behind the door he did not open.
I always was confused about why they switched even though it was explained to me. Now I understand. You are basically guaranteed to pick a goat, and the host reveals everything else you didn't pick except for one door. The host obviously isn't going to reveal the car, so you should switch to the one he didn't reveal.
@@TheSpacePlaceYT The car still might be behind your door (which he also didn't reveal) It's always two probabilities and the first one doesn't matter. It's the Monte Carlo fallacy in disguise.
Holyshit, i got it😂
@@johnrobertson93 yeah no. This is like saying the lottery is 50/50 because you win or lose. At first you have 1/,1000,000 chance of picking a car. Or 999,999/1,000,000 of picking a goat. So when he opens all other doors(which doesn’t change what your choice was) it’s more likely you picked the goat and swapping will leave you with a goat one time every million. It’s not a debate lol it’s literally solved
Um if you had a million options and you cut it down to two choices that's still 50 50...
That's the best explanation I came across on RUclips because it employs simple logic without fleeing to mumbo-jumbo math explanations.
This is fun that ppl are still arguing it. Here's probably the simplest explanation (simplest I can think of anyhow). A random number between 1 and 3 (inclusive) is chosen, this is where the car is placed. Again another random number from 1 to 3 is chosen, this is the contestant's choice. Let's say the contestant will never switch, so the door the host reveals is immaterial, he's going to stick with his original choice.
There are 9 total cominations, 11,12,13,21,22,23,31,32,33 and as you can see, only 3 of them the contestant picked the car and will win. 3/9 = 1/3. So by not switching, 1/3 chance of winning (2/3 of losing) and by the converse, 2/3 chance of winning (1/3 of losing) if switching.
First of, you listed permutations, not combinations, just pointing that out there.
Second, you post out in one set of permutations (11, 12, 13) implying that there will always be three doors remaining at the second choice, which is not true since only two doors are left.
It would be like this: 11, 12, OR 11 13 since the host does not remove their door choice.
For the first set of permutations, the car remains in either door 1 or door 2.
So if it was in door 1, and you stay you win.
If it was in door 2, and you stay you lose.
Staying has a 50% chance of winning.
Consequentially, changing has a 50% chance of winning.
For the second set of permutations, the car remains in either door 1 or door 3.
So if it was in door 1, and you stay you win.
If it was in door 3, and you stay you lose.
Staying has a 50% of winning.
Consequentially, changing has a 50% chance of winning.
+TheGreatslyfer
You said:
"For the first set of permutations, the car remains in either door 1 or door 2.
So if it was in door 1, and you stay you win.
If it was in door 2, and you stay you lose.
Staying has a 50% chance of winning.
Consequentially, changing has a 50% chance of winning."
You concluded that staying had 50% because you were only focusing in the number of options, forgetting you have more information about one door than about the other, and this occurs because one door was selected randomly but the other on purpose with knowledge. (Remember the host knows the positions and must reveal a goat after your selection from the other non-selected two.) Let's see:
1) The number 1 is still an option because it was your selection and the host couldn't reveal it regardless of its content.
2) The number 2 could have been removed if it had a goat. It was not a guarantee for it to be in the second round. For example, if number 3 had the car and number two had a goat, the host would have been forced to reveal the 2. But he didn't, number 2 survived a possible elimination, which increases the likelyhood of it having the car.
The other door the host leaves closed will be the correct in all cases you failed at first, so it wins 2/3 of the time.
I think the easiest way to understand this problem is to just imagine the doors open the whole time. Then you can conceptualize why changing doors results in a win more often than not.
Say the truck is in the left door every time. You pick the left door. The host opens the middle. You switch to the right. You lose.
The truck is again in the left door. You pick the middle door. The host MUST show you the right door. You switch. You win.
The truck is in the left door again. You pick the right door. The host MUST show you the middle door. You switch. You win again. And that’s why you win 2 out of 3 times. Because there’s a 66% chance that the host is going to be forced to eliminate the other goat door.
Ah. It makes sense now.
OHHHHHH. THANK YOU
Actually, from my point of view, it's 100% - even better. Reason, the host has to pick the other goat to get a winner for the game which is his goal.
Yes, imagine the doors being open all the time. These are the only possible options:
The only 4 examples possible if the car is in door 1:
🚗🐐🐐
Example 1. Choose door 1, host reveals goat in door 2. Stay = you win, change = you loose = 50/50 % chance.
Example 2. Choose door 1, host reveals goat in door 3. Stay = you win, change = you loose = 50/50 % chance.
Example 3. Choose door 2, host reveals goat in door 3. Stay = you loose, change = you win = 50/50 % chance.
Example 4. Choose door 3, host reveals goat in door 2. Stay = you loose, change - you win = 50/50 % chance.
In 2 of the only 4 possible examples above you win if you change, and in the other 2 examples you loose if you stay = 50/50 %.
These are the only 4 examples possible if the car is in door 2:
🐐🚗🐐
Example 1. Choose door 1, host reveals goat in door 3. Stay = you loose, change = you win = 50/50 % chance.
Example 2. Choose door 2, host reveals goat in door 1. Stay = you win, change = you loose = 50/50 % chance.
Example 3. Choose door 2, host reveals goat in door 3. Stay = you win, change = you loose = 50/50 % chance.
Example 4. Choose door 3, host reveals goat in door 1. Stay = you loose, change - you win = 50/50 % chance.
In 2 of the only 4 possible examples above you win if you change, and in the other 2 examples you loose if you stay = 50/50 %.
And final example if the car is in door 3:
🐐🐐🚗
Example 1. Choose door 1, host reveals goat in door 2. Stay = you loose, change = you win = 50/50 % chance.
Example 2. Choose door 2, host reveals goat in door 1. Stay = you loose, change = you win = 50/50 % chance.
Example 3. Choose door 3, host reveals goat in door 1. Stay = you win, change = you loose = 50/50 % chance.
Example 4. Choose door 3, host reveals goat in door 2. Stay = you win, change - you loose = 50/50 % chance.
In the final possible scenario it’s all the same. 50/50 % chance to win or loose.
@@christerenstrom9798You're saying the probability of winning the lottery is 50% because you either win or lose, which is false. Do not confuse probability and possibility. You need to actually chose an option (stay or change) and calculate the chances. You'll see that changing doors is more likely to give you a win (google "monty hall table" and you see how you should make this table), but I'll post here a different way of thinking:
Imagine 3 doors, behind 2 have goats and behind 1 have a car, let's say the car is behind the first door (🚗🐐🐐). We both agree that the car has a chance of 1/3 of being in the first door, 1/3 of being in the second and 1/3 of being in the third. Let's suppose you choose the last door, so the host NEEDS to reveal a door that HAVE A GOAT and it's not yours (because he knows which one has the car), so he reveals door number 2. Remember, that this information (that a goat was behind door number 2) you didn't have in the beggining, it's a NEW INFORMATION. Now we separate the doors into two groups, the group composed by the chosen door and the group composed by the other two doors, like this: 🚗🐐 and 🐐. What were the chances the car was in each group? The car had 2/3 the chances of being in the first group (door number 1 plus number 2) and 1/3 of being in the second group. Since now you know that the second door have a goat (that is 0% of having a car behind it), the other door needs to have all the probability of having a car. This is why changing doors always increase you chances.
A=1/3 B=1/3 C=1/3; A+B+C=1; A+B=2/3, after reveals, B=0 so A+0=2/3>A=2/3; Now (A+B)+C=1>(A+0)+C=1>A+C=1>2/3+1/3=1
Me: confused how the car got through the door
For people who still don't get it after watching the video & read comments (like me), here's one that is more intuitive:
Situation 1: choose car -> swap and you get a goat
Situation 2: choose a goat -> swap and get a car
Situation 3: choose the other goat -> swap and also, get a car
As you can see the chances of swapping and get a goat is 1/3, and no leak of information that changes the chances you'd picked the car at first
Thank you
but how do u know that the door you have chosen has a goat behind it . How can u assume that without opening the door
gaurav m You do not. You know which one is more likely to have a goat.
So simple and very clear... you are dude man....
This guy explains it the best
It is based on assumption on your first choice. It is just making something simple into something complicated......
Diksha GiriIf your first pick is Goat A, you get Goat A. Swap = Car.
If your first pick is Goat B, you get Goat B. Swap = Car.
If your first pick is the car, you get the car. Swap = goat.
So simple, 1/3 vs 2/3.
No stupid 50/50 in this game.
Araqius oh right right so it is like rather than calculating the probability of getting a car second time we calculate the chances of picking up a goat or a car the first time. Since the chances of picking up the goat first time is 66% therefore it is better to swap the second time. I get it now, thanks.
Araqius Thanks. Your explanation is elegant.
True, and the explanation wasn't difficult to follow. Suggesting we should "swap" isn't the greatest advice since we will never be sure what is behind the door we chose. The odds could be against us or in our favor.
Araqius Hehe, Diksha Giri is still correct, as you are now using the Assumption that "swapping gets you the car" (which is ignoring the chance for the other goat). In Real Life physics, not only do you do not know what is behind the door, but by 'choosing the door' you don't actually 'have what is behind' the door, so probability is just a plaything here - until the outcome is set. Mathematicians, Statisticians et al would love to play with the probability of course, since it is 'after the fact' ;)
It is a great example of Real Life Physics vs Theoretical Physics!
"ill give you 10 seconds to think about it"
*pauses video*
Jajajaja...hahahaha... :D :D :D L.O.L
Lol....nyc 1..!
You basically have 0 % chance, because when you open the 1st door and the host asks you whether to swap... the crew is ready to swap the prizes while you make your decision... When you choose the host will talk long enough and get you under pressure, that the crew will swap the prizes for sure... except if they don't keep cars in small rooms
+Vladimir Dimov "You basically have 0 % chance"
... while simultaneously you basically have 100% chance because when you open the 1st door and the host asks you whether to swap ... the director is telling the crew "we need to increase ratings or the sponsors will leave the show - give away the car"
except thats scamming, and they would be prosecuted
Michael beseler "except thats scamming, and they would be prosecuted"
"... except that's not mentioned in the MHP question so is not material to the solution"
Freddie Orrell except that's scamming, and they would be prosecuted
+O. Flake There is serious protection in place against cheating and collusion for every game show, as required by law. There would never be a situation where the crew is free to run around switching prizes without supervision.
For anyone that doesn’t understand: You start out with a 1/3 probability of getting the prize, but a 2/3 probability of not getting it. Monty will always reveal a non-prize door, so if you decide to switch your choice, you have a 2/3 chance that the original choice was not the prize, meaning that when you switch, the remaining door contains the prize - thus where the 2/3 probability of winning with a switch comes from.
When I heard BOOBY Prizes... I immediately thought. "Nice! I have 2/3 chances of winning the grand prize!"
*Easiest Explanation:*
1) If you pick a goat first and then switch, you will get a car
2) If you pick a car first and then switch, you will get a goat
The odds of you getting a goat on the first pick are 2/3, so if you switch the odds of you getting a car are 2/3.
Notes: ~The open door is not a red herring. When the door opens, it guarantees that switching will change from goat to car or vice versa. Making 1) & 2) above true.
If you can understand this, you'll know why it is not 50-50. Essentially, the third door rigs the probability of the second pick. I know, It's pretty crazy, and I thought it was 50-50 too, but this makes perfect sense.
so it relies on shenanigans by the host to work? OK, say the host dies at the start of the game, then i pick a goat door. i have 2 doors left, how is it not 50%
The_Urgulerg Yes, if the door you picked was random, then it would be 50%, but the host doesn't pick a random door, he ALWAYS chooses a goat.
i hate the host
The_Urgulerg Game show hosts can put on a good smile while tricking you. Are my explanation makes senses, though?
yes
Fuck the car, I want my goat,
Rona Kavanagh Are you going to fuck that as well?
Why do you use 'as well' when the sentence I made implied I was ordering someone else to do it?
Next time you try and be a smart ass you should cover your tracks.
It couldn't have been seen as a list of things I was going to do because "I want my goat" negates that.
Rona Kavanagh 'Fuck the car' is an imperative which may be taken literally to mean as stated, implying a second-person subject, or figuratively, to mean 'dismiss the car' where the principle actor may be first person in that it is their attitudinal stance which the command expresses. The result in either case is 'the car is fucked' which allows the adverb 'as well' to hang with a touch of irony due to the former equivocacy. Further, 'want' in the same sentence as 'fuck' is laden with double entendre, heightening the innuendo.
So, you and the goat: not an item then?
Best explanation so far, you have a one in 3 chance, but if you initially picked either goat switching will make you win every time.
Ironically you want your initial pick to be wrong forcing the host to eliminate the goat leaving only the car remaining.
The main reason this problem is counter-intuitive is because we tend to view each door as inherently having an equal probability of holding a certain item as each of the other doors. This is true in the initial conditions: we are told each door has the same chance of holding either a goat or a car. However, if we open a door it is clear to us that this probability distribution collapses for that given door. Nobody would say a door that has shown to have a goat behind it might still have a car behind it instead. We also acknowledge that the absolute probabilities of the remaining doors change as well. However, what we often fail to grasp is why this change is not evenly distributed among the two remaining closed doors. That comes from the fact that we did not _randomly_ remove any of the two goats. We specifically removed the goat from all the doors that we _did not select_. It's this piece of information that breaks the symmetry and gives rise to the counter-intuitive result that swapping doors is favorable. Hope that helps anyone understand it. Perhaps not :P
yea like that makes more sense then anyone else on here
It's 50/50 still. Have tested it many times. It fluctuates on how many times you do it which is the issue and if you do the game 20 times vs 1-5 times you can see why. The amount of times is people messing with the problem if they choose to do it 100 times and even then it's still not more likely to be the other door. There is no rule or fact you need to do the test many times. Life and the universe doesn't work like that. You aren't promised 1 chance or 100. The only way it could be more than random or a higher chance for the other card is humans wanting to run the game so many times but that takes away from the actual probability. It's technically just random and no percent is accurate but the correct answers are you get it on the first try or it's around 50/50. But again it's random and no exact % is correct. Go take the tests yourself and you will see it's different Everytime and even the amount of times. Whoever said it's 66% didn't actually pay attention or is manipulating the game to their view which is incorrect just like the belief a god exists. Do the test yourself different days and different amounts and you will see, Seya little humans
counter-intuitive ....it's not what you think ,,,everybody thinks better to switch
@Joseph Is my middle name How much money somebody can win if He can resolve The monty hall problem? , Is there a prize for Monty Hall solution?
This is exactly the only part missing from all explanations. You touched on the exact part where it gets tricky or counter intuitive
This was a famous answer by Marilyn Vos Savant in Parade magazine yrs ago! She had all kinds of mathematicians and statisticians saying she was wrong! She was right!
Marilyn vos Savant was neither the person who invented the problem nor the first person to write about it nor was she the first to receive lots of negative feedback claiming that she was wrong. Don't tell me you came here from that idiotic Vox video.
Marilyn made it a national talking point in Parade magazine... before that no one but math nerds who read journals that have a few hundred circulation ever saw or discussed the 'problem'! My comment stands!
... *_Mr Manney_*
Why Mathematicians get the MHP wrong…
Their training requires them to believe the absolute truth of the scenario - the problem they set out to solve. They know instinctively that the answer is that a switching contestant will have a 2/3 chance to win the prize. They quite naturally resort to equations of probability because that’s what they know…
Their equations - if they’re to work - need to have a zero substituted for what was previously a 1/3 chance - they make the claim that a goat door has zero chance of the car but there are two goat doors in the MHP so the unseen goat must by their definition also have a zero chance of the car. And the door with the car - well the mathematicians’ logic must also erase the car door’s 1/3 chance and make it a 100% chance.
Mathematicians have it that all the doors in the MHP have a 1/3 chance of the car and at the same time No Door has a 1/3 chance - but they fail to recognise this dichotomy of reason.
They operate on the basis of New Information - the revealed goat is for them new and significant - but they fail to accept that a bright child could forecast the goat as a predictable event because there will always be at least one goat between any two MHP doors - we do not need to see it to know of it - so it’s hardly new information.
They drop a zero into their equation - crank the numbers and arrive with the answer that a single MHP door has a 2/3 chance of the prize - and that it must be so because the equation tells them that it’s the case.
But to have a 2/3 chance of something requires there to be two objects of the same type - like goats for instance - but there is only one car! They have a door enjoying the 2/3 chance and not the contestant.
The cynic has it easier - he or she does not believe the honesty of the MHP host - the cynic sees through his deception and reasons that the reveal of a goat is just a trick to make observers think that the goat and its door are removed from the overall set.
The cynic reasons that the offer to swap to a single door is the second part of the host’s deception and that the actual but unstated offer is to surrender one 1/3 chance door for the other two 1/3 chance doors together.
Mathematicians have it that chances change and that the host is honest - cynics have it that the host is a deceiver and that chances never change. And that actual content does not affect the chance of content.
did ya write this all out cuz then y'all need a life.
Timothy Dabest did ya reply to a 7 month old comment cuz then ya need a life
Watching this helped at lot. More likely to pick the goat first time as 66% at that point. One goat gets eliminated straight away swap and the original odds flip to 66% in favour of getting car
This is really helpful for this common, everyday issue we all deal with.
you are most likely to pick a goat first
NovaTime And, really, that's all there is to it.
James Nelson its what people forget to say to people who dont get it
NovaTime yes this is true but wouldn't you feel like a complete idiot if you picked the car first and swapped choices.. 33% is a little too high to be swapping imho.
33% < 66%. It's foolish not to swap.
James Nelson ok let's (for the sake of argument) say that i agree with you.. Would it be wise to swap twice if you have 4 doors then?
This is by far the best explanation on youtube
You're basically banking on your unluckiness of picking a goat first, and by swapping, you're flipping your chances over.
i like how you explained it! :]
Thank you ~
mienbao I would interpret the unluckiness as actually luckiness because it's a definite win, but I understand what you're saying. It's just funny to interpret the words in a different sense. :)
correct simplification:
-lets say you choose door 1 then decide to switch
-there are 3 outcomes
- car behind door 1= you lose because you switched
- car behind door 2= you win because the host has to reveal 3 contains a goat
- car behind door 3= you win because the host has to reveal 2 contains a goat
so 2 out of 3 times you win according to this logic
Samuel Kueber I already understood from the video at this point, but I also think this is a very effective way of explaining it. Nice one
For those who still are having trouble, let me draw it out:
1 = car, 0 = goat;
Possible set:
Door 1 | Door 2 | Door 3
1 0 0
0 1 0
0 0 1
Suppose you chose Door 1. Then the host will show you the goat. For each row, I will remove the goat which is either from Door 2 or Door 3.
(stay)
Door 1 | Door 2 | Door 3 Stay | Switch
1 0 0 1 0
0 1 0 -------> 0 1
0 0 1 0 1
So tell me. Is it better to stay or switch.
Your logic make no sense, it make no difference to swich/stay, your odds are still the same, you make it look like there is one extra chance to win if you swich but actually you also took away the chance to win, there forth you did nothing that could change the 33% chances to win. This video is just a joke ^_^
Yes, that was how I came to understand this initially.
percentage is an illusion.When we remove one door, my choice becomes 50/50 from %33 when one door is removed.It doesn't stay %33 because now there are only two doors left.At the beginning i had %67 percent chance to choose a goat, now it is %50. because there are only two doors left.Get it? You're saying that you could win more if you swap by 2/3, but that's not true because we have 50/50 at the end.Watch the psychology of the guy it will take you to the car.if we had 9 goats and 1 car and when the host revealed one goat then swapping would be a better idea, but not in this situation.I get your thing, instead of staying with the %66 percent change of the choosing the goat because of the decision we made before we have to swap to increase the possibility, but no.When one door is removed, it is 50/50 from now on.I get it but it is hard to explain.If there were more goats swapping would be the best choice but not on 50/50 because it is 50/50 not %33.You forgot to remove one door when host removed a goat.
When there were 3 doors, swapping wins.But now there are only 2 doors left.I tell it again, if it was more than 2 doors at the end swapping would be the best idea but when there are only 2 doors at the end it becomes 50/50, it doesn't stay %33 when there are only two doors left.The possibility of choosing the one door of that %66 percent is high, i get it.Staying with that high possibility is wrong.But after one door is revealed, it is 50/50.
This is why I came here. Utterly stupid people who can't understand simple concepts even when they'e explained fully.
Good Job Thunder Kat, now get outside. That extremely short bus will be here soon to pick you up.
What is the problem of winning a goat?
+ᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚ They smell for a start
They eat everything, I mean everything.
lol
+ for one you do not get to keep the goat... the rules are that you do not keep those prizes.
+ᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚᅚwouldn't you want a car over a goat
And after pondering all day on this problem, and seeing it for myself proven with a sim program, I watched the video one more time and finally understood that indeed the chances of getting it right are 66% by swapping the original door and not 50/50 as I originally thought. It's so clear now that I feel dumb for not getting it the first time.
You only need to swap if you chose the wrong door at the start. Thus the probability of swapping and choosing the right door = probability of choosing the wrong door at the start = 66%
Best explanation yet
Simple version is, you have better odds of picking the goat, so when you swap, it will be the car.
5 minutes and 48 seconds to explain this visually
no, it wont necessarily be the car, but it's most likely
Ismael Mejia No what he's saying is if you do happen to choose a goat and swap you will have a car guaranteed. And you will most likely choose a goat the first time.
AJ18555 You hit the nail on the head - This is the key to understanding this.
getbo i understand...most 50/50men understand, then rethink and go back to well theres 2 doors and one of each thing, and i have a choice to make in the moment, in the moment i say, it must be 50/50!
The simplest solution:
1. If contestant switches, he gets the "opposite" of what he first picked.
2. He has 2/3 chance of picking goat at first.
3......er, that's it.
That is brilliant! I even have a book devoted to MHP and I have not seen such a simple exposition. Masterful!
Thanks, Andremus.
Brilliant !
Thanks, WT.
may your coffin be made of gold, kind person!
I want a Goat so I can pet the Goat all day and have a ton of goat milk.
I want the goat so I should not swap.
+Shanka DaWanka lol mk, if u get a car, will u ask for a goat???
Yes.
Sell the car and buy lots of goats from the money!
+Pallando Rómestámo I only have enough room for two or three goats.
Kevin was right, damn :D i was with Holt on this one..I understand it but it's still all luck
It's probability.
I'd heard of this before watching the episode and I was just screaming at holt the whole time
I think this is the best explanation:
I just wrote down a 5-digit number on a piece of paper (lets say it's number 34776 for example, but you don't know it yet).Can you guess which one is it?It's basically impossible with probability 1/99999, but try it anyway.Let's say you randomly pick number 22765.Than I remove all options except the number you pick (22765) and the number I actually wrote (34776).Than I ask you if you wanna stick with your number or change to the other.Do you still think you have 50% chance if you stay with your original choice?Of course not!Although there are only two options left, the probability that your original choice is the correct one stands still at 1/99999!!!The point is this - One number is still in the game JUST BECAUSE YOU COMPLETELY RANDOMLY PICK IT AND THEREBY FORCED ME TO LEAVE IT AS OPTION, while the other number is still in the game BECAUSE IT'S FORCED DUE TO ACTUALLY BEING CORRECT ANSWER.Just think about it, it's not that hard.
+Fak Jea Assuming that 00000 is a possible option, the probability of guessing correctly would be 1 in 100000, not 1 in 99999.
+TedManney Yeah, you're right but it doesn't really make a difference.
+Fak Jea Now there's an explanation I could follow. Thanks.
But if there are two goats and a car, I can hear the goats through the door and choose the car directly
Hhahahaha you're right
goats are dead
But if there are two deadf Goats I can easily pick the car too because then the goats smell terrible
Still never could understand this. The way I see it, you only have two doors. The host is going to knock one out for ya. So there is a goat and a car. Two things, and he essentially asks you to pick a door again (swap or no-swap). That is 50/50. The third door doesn't matter to me anymore. If I could swap to the one the host revealed, sure.
Overall, it is 1/3. But after he reveals the goat, I have two choices, and it becomes a matter of whether I have it or not. Heads or tails.
So let's play against each other 100 times. We both choose the same doors, and we both stay with our guesses. The difference between our games is that my rounds end as soon as we choose a door from three available, without you seeing what's behind it of course. Your rounds end after a goat is revealed. So explain to me how you would win 50 times while I win 33 when we have the same doors.
+Sib Pro Quo
You are totally wrong.
2 doors left doesn't means it's 50/50.
Let's say the game start with 1 door (1 car).
After you make a pick, the host add a goat door.
It's very clear that the winning chance of your door is 100%, right?
Take a look at your logic.
" So there is a goat and a car. Two things, and he essentially asks you to pick a door again (swap or no-swap). That is 50/50. "
"Overall, it is *1/1*. But after he adds a goat, I have two choices, and it becomes a matter of whether I have it or not. Heads or tails."
See?
It's clearly wrong.
You're all talking to someone who cheated in calc and pre-calc in college because I knew I would never have to deal with the sine cosine bullshit in the real world.
If the guy tells me I have three doors to choose from, but then tells me not to pick one of the three because it'll always be wrong, he has essentially told me to pick one of two doors. One has a car, another a goat. 50/50.
If he told me to pick one of the three and never revealed a door, then sure, 1/3. But as soon as he reveals a door and tells me to swap or not, to me, he has essentially said, "Go ahead a pick one of the two doors again."
Its like that second goat or third door never matters, cause he'll always throw it out of the game. In every case, it seems as if you either have a car or a goat behind the door. That is just 50/50 in my books.
How is it 2/3 when he tells me one is bullshit? Like having to guess which of three pies is cherry, but your friend tells you not to pick the middle one, cause it is apple. The guy just narrowed it down for you.
***** The grammar in his comment is pretty off *shrugs* didn't get what he was saying.
The other guy is then adding doors from two doors... Again - math is the pointless to me. Even you explaining how I would win 50 of 100 because I choose between two doesn't help. Sounds to me like you just agreed with me: because there are 2 doors left, chances are 50/50.
"... their odds are 50/50 because there are two doors left."
***** Gonna be honest, its kinda funny to see you take this so personally. He kept saying "rounds" and "games" so I'm not sure how many times we are repeating that process and I didn't get what he was saying.
I still don't get how it disproves my logic. I said having 2 doors to pick from is 50/50... and then you agreed by saying having two doors to pick from is 50/50... Instead of trying to explain it, you're getting your panties in a knot about it. If that's going to be the case, then just leave man. I'm asking for a clear explanation cause I don't understand, not an impatient douche.
Those who maintain the 50/50 claim have obviously not even had the curiosity to test it by playing the game.
Also, logically, if you always pick Door A, then Monty reveals Door B or C, and you stick. Do you win 50% of the time then? So that means the car was behind Door A 50% of the games. But what if you always pick Door B or Door C? Are those both 50% likely too? Obviously there would be a contradiction there: the car can't possibly always be behind your first pick with a 50% chance.
@@jasonl7651 the first time you pick, you have a 33% (1/3) chance of getting the car. When the host opens the other door, if you switch to the remaining door, your chances are 66% (2/3) of getting the car.
@@sheila6530 I'm not sure why that's a reply to me. You just stated the basics. My point was that the people who hold that it's 50/50 would lead to a contradiction, since the three doors must then add up to 150%
It's interesting to note that since the host knows where everything is, it cannot be a random opening of the door, and that alone excludes any possibility of there being 50:50 odds. A non-random process cannot have equal odds.
I am confused because probability can change, no? Yes originally it was 33% but now you only have 2 doors to choose from so that probability raises to 50%. It's not conditional because now you only have two doors to choose from. you're not going to choose the open door. Draw a tree diagram, it makes sense to me.
For me the easiest way to explain this is to think about what sticking or switching means in terms of the original guess.
Sticking means the contestant thinks that their original guess was right
Switching means the contestant thinks that their original guess was wrong
Since the contestant has a 1 in 3 chance of being right the first time, they are more likely to be wrong the first time and win by switching.
But if you go to the internet, you'll find people that try to be correct all the time, so they end up losing most of the time as they thought they were correct initially.
Can Monty Hall problem even applied onto internet arguments?
SIMPLE EXPERIMENT you can do which will PROVE this to be true, All you need is three coins and some method of recording results. It takes a couple of minutes, literally less time than it takes to argue about it;
- Flip all three coins and lay them down on a table, if all are heads or all are tails then re-flip. If you get 1 head and 2 tails then head is the car and tails are the goats and vice versa.
- Pick a coin in a predetermined and consistent position (e.g. the one on the right) to avoid selection bias (this represents your initial choice)
- Mentally "reveal" one of the goats
- Note whether the third coin is a car or a goat (this represents swapping)
I did this 12 times and got a perfect 8:4 win-lose spread
37rainman No, I have. By third I meant the third coin you look at, the one left over after the original choice and the goat you "reveal"
37rainman I am aware of that. I just couldn't be bothered and i understood the logic behind it so what was the point? Peeps reading this can do it a hundred or a thousand times if they so choose, the result will be the same
37rainman this one only takes ten minutes. So they should do it and be proved wrong, no excuses that way :P
Or go here: www.grand-illusions.com/simulator/montysim.htm
Faster, and you get 1000 experiments.
martok2008 that site is flawed as fuck, if you choose the one on the left you lose, and if you pick the other two you win. I'd rather do my own probability and know it's accurate...
Important detail for everyone who thinks themselves a smartass and say it should be 50%.
That would only be true if the gameshow host opened any random goat door, including the door you picked. If he did that then yes the chance would be 50% that you get it right.
But it is specified that he ALWAYS opens one of the other doors. Therefore the solution shown here is correct.
This problem is funny because people who are slightly more inteligent than average know that people who are less inteligent than average would switch, and then "overpredict", getting a wrong answer.
It's a problem only dumb people or geniuses will get right the first time.
Of course, some very average people will still not get it even after they think about it for a while and argue that the geniuses are dumb people.
Then again I already own a car, but don't already own a goat... which adds a suplementary layer of complexity to the problem as to deciding which one I actually want.
Tyler Holte Oh my GOD.
Heavens thank you.
Spending half an hour in the comment section brought me to a point where I thought mankind was so dumb that the only reasonable solution was to anihilate all life through a nuclear haulocost.
But you made me realise some people get it.
Shane Warren
If you can swap to get the other two doors you didn't chose, you get 2/3 to get the car and at least 1 goat.
The host just take that goat away from you.
this type of problms is solved by "expectation".
in not swapping case it is 1/3*1 + 2/3*0= 1/3 while in swapping case it is 1/3*0 +2/3*1 =2/3
GregTom2 I actually find a goat has many advantages over a car
Swap, always, as the odds to win are improved from 1/3 to 2/3.
For people who aren't getting it, picture it this way: Imagine there are 1000 doors and behind 999 of those doors are goats, and behind only 1 is the car. You pick a door and there's 1/1000 chance it's a car, and a 999/1000 chance it's a goat. The host opens 998 doors--which takes a painstakingly long time--to reveal all the doors with goats except the door you chose and another door. If you switch doors your odds improve greatly: 999/1000 that it's the car, and 1/1000 that it's a goat. As the amount of doors approaches infinity, the odds of getting the car when you switch doors approaches 100 percent. Make sense now?
@Tyler Porciello you didn't read, did you?
at least the host has kill 998/999 its chance by itself
it can be 1/2
@@vishesh_music no
After reading your explanation, probability of the last two doors still looks 50/50 to me. One door is a goat, one door is a car.
Very good explanation, I don't understand how some people still believe you have a 50/50 chance after watching this video.
yeah.. !
That's quite easy to understand, for me those who get confused by the 50/50 is because they think the host randomly opened the goat, but in Monty Hall Problem the host knows where is the car so he ALWAYS opens a door which doesn't have the car and that's the reason why we should always change. But is my first statement correct? Is the probabilities 50/50 if the host randomly opened the goat? So lets imagine that there are again 100 doors and there is one car. You choose a door and then your friend (who doesnt know where is the car) start opening the other doors at random and surprisingly there is no car found until two doors left (yours and another one). Now the host tells you to change the door if you want. What will you do? In this case many will think that the probabilities are reverse and now I have 99/100 the car to be in my door because that should be the reason my friend didn't manage to find the car in 98 doors opened. But, indeed the probabilities now are 50/50 (and pretty easy to prove) and whatever you do, change or not, is the same.
What if ur friend finds the car in his door?Thats the whole point of this problem its not 50/50 or what will change when 2 are left its of what u pick as first.
In this video If he opened the doors randomly he would have a 1/3
chance of opening the car.But if he opened the car whats the point u would obviously swap..
Combinations: If u pick the car,chances he ll open a car are 0.
If u pick a goat,chances he opens a car are 1/2
TWrecks Yeah thats what he said.
What if he switched the doors where the goats were behind
Shout out to all the people who've been fighting the good fight for all these years and showing people that swapping is the correct strategy,
@BuggyAl statistics only work if you get to repeat the choice many times, it wont work if you only get to choose once
@@wspang1993 If the game was done with 100 doors and 99 goats, and the host closes 98 doors, you'd have a 99% or 98% of winning by switching regardless.
I got into a big debate about this question with a friend who simply refused to believe/understand me even though I spelled it out as clearly as I could just like this video tries to. The only way I could explain it so he could understand is when I increased the number of doors. What I mean by this is, instead of only three doors with only one prize, imagine one billion doors with only one prize. You're allowed to pick a single door, just like in the initial setup. Explaining it this way, it's very easy to understand that the odds are not in your favor of picking the prize with that first pick and you probably picked a door without the prize, to which my friend agreed. The host, knowing where the prize is, opens up one of the doors that isn't the prize and asks if you'd like to change your answer. Once you decide, he opens another door and asks you again, and he'll repeat this until there are only two doors remaining. So, assuming you never change your answer until there are only two doors remaining, you'll be given one last opportunity. The host will have eliminated nearly a billion doors and you know that first door you picked is statistically unlikely to contain the prize and there's only one other door remaining. Should you swap? Of course he said yes. So when I brought it back down to three doors again, I explained AGAIN, when you pick a door, you only have a 33% chance as picking the prize right? So odds are the prize is in one of the other doors, RIGHT? I feel that was the moment my friend finally realized he was wrong. I could tell he was trying to figure out how to disprove me, by trying to some way that my example with a billion doors isn't the same as the one with only three, but even he eventually realized that wasn't the case and couldn't really refute what I was trying to say all along. And I know he wanted to. He loves to argue and hates being proven wrong but once he understood that this scenario holds true whether it's 3 doors/1 prize or 1,000,000,000 doors/1 prize, once every other door is eliminated with only 1 last door remaining, if you swap the odds will be in your favor.
also you can map it out assume you pick door 1 for example ( or door one is the first pick):
1 2 3
car goat goat
goat car goat
goat goat car
okay so this shows that having door 1 representing you choice ( as the door order doesn't really matter in your choice) you have a 1 in three chance of picking the car off the start. now let's take that door away and then have the host pick a goat door ( as he never reveals the car under the rules). then we get this without door number left over:
goat
car
car
but wait there's now two chances to pick the car out of three total. you have 2/3 odds of following a path to victory if you swap.
Very good example. I still think its horseshit. lol
I think the host can always manipulate the game as he sees fit.
I don't like increasing the numbers, because it does change the game drastically. The odds of choosing the right door out of a billion, is almost impossible. But the odds of choosing the right door, out of 3 choices, is very likely.
@John Bryan It doesn't change the game whatsoever. With 3 doors, you have a 66% chance to pick wrong, so the odds are, you will. All increasing the number of doors does is emphasize this. With 4 doors you have a 75% chance to choose wrong. At 5 doors, 80% to be wrong, 10 doors 90% etc. Since odds are that first choice is wrong (even at 3 doors), once you get down to only two doors remaining, the one you initially chose and the one the host hasn't eliminated, it's always better to switch because odds are, your initial choice was the wrong one.
@@aldi9802 It's not "very likely." It's exactly 1 in 3. Very likely means well over 50%.
Wait sorry, who would not want a goat over a stupid car??
+w senkow every time because I'm just sad like that haha
You want the car so you can sell it to buy more than 1 goat.
+FuriousGaming Your fun at a party aren't you?
+w senkow ☺️☺️😂😂😁😁😁😁
+FuriousGaming haha love that, that's a good shout ;)
If the odds were really 50-50 then you'd have to be able to refute the following:
_For a dedicated strategy of either always switching or always staying, the initial pick determines the final outcome in 100% of cases._
In other words, once you've made the initial pick nothing else is left to chance. It is predetermined what your final result will be once you've selected a goat or a car. And since this is true, the odds can't be 50-50 as you are twice as likely to pick a goat at the outset than you are to pick the car.
Agreed.
There are so many people that don't quite get it. Let me simplify it for you.
When you pick initially, you are most likely to get a goat. If you did get a goat, the host is forced to reveal the other goat you didn't pick. This means the final door must be the car. The chances of you getting a goat at the beginning is 66%, so because you switched, there is a 66% chance you get the car.
The number of people falling for this hoax theory is astounding… The fallacy here is believe that probably of the second attempt to open door is influenced by the probability of the first attempt, when in fact it’s not. As soon as you opened the 3rd door, the probability of the first door being the car increased to 50%, and not staying at 33%.
Take a scenario with 2 doors for example. If I choose first door, host opens second door and reveals the goat, the probability of the first door being the car is now at 100% and not staying at 50%.
If somehow the first attempt to open door influences probability in the second attempt, then this would be true.
@@robertdowney9017 surprisingly its not a hoax. there is a video on YT with the title " marilyn vox savant". i suggest you watch it. she even had mathematics professors apologizing to here after they eventually figured it out.
@@robertdowney9017I think mythbusters did this and proved the theory. It’s on youtube
@@robertdowney9017 Absolutely corect!I called a big corruption for the Marilyn Mach Vos Savant and poor logic for the most of the people.
The problem with this whole debacle is that they are treating the first pick goats as actually two different goats which is correct but treating the second scenario with choosing the car first and switching to goat because ignoring which goat the host reveals and treating them as the exact same goat because in a 3 letter scenario there are 6 possibilities with no letters repeating and you have to take out 2 possibilities because host cannot reveal car so thus giving us 4 possibilities
When he opens the door with the goat...the car and the goat both have 50% chance. Because, the 2 doors that are left, either have a car, or a goat. The third door is completely irrelevant after this point. Simple elimination of fractions. Door 1...2/3 goat, Door 2...2/3 goat, Door 3....2/3 goat.... I pick door 3. The guy opens the first door to reveal a goat. NOW THERES ONLY 2 DOORS LEFT. In door 2, theres a 1/2 chance of a goat, same in door 3, odds are the same for the Car. Fuck this bullshit im out.
MyNameIsSimple72 You're wrong. This problem has been around for decades. Step one is realizing you have it wrong (it is a common thing - you're not alone; it is DESIGNED to make people think it is 50/50).
Think of it like this. What if you NEVER switch? According to you, I have a 50/50 chance, so I should have just as good of a chance of winning the car by NOT switching, right? For me to actually win half the time, I would have to originally pick the car OUT OF THREE DOORS half the time. Does this make any logical sense to you? It shouldn't.
Just play the game out in your head. Realize that no matter what you pick, if you choose to switch, you will ALWAYS end up with the opposite of what you originally picked. Go ahead and work out as many different scenarios in your head - - I challenge you to find a scenario where what I say is not true.
You pick a goat, Monty will ALWAYS open the only other goat, and you will switch to a car. You pick the car, Monty will reveal one of the goats, and you will switch to the other goat. You pick a car and switch, you end up with a goat - - every time. You pick a goat and switch, you end up with the car - - every time.
You will pick a goat first 2/3 of the time, so switching will result in you ending up with a car......2/3 of the time.
The fact that the host reveals a goat door is IRRELEVANT because he will ALWAYS do it. He is REQUIRED to. It has absolutely ZERO impact on the odds. From the very beginning, you have a 1/3 chance that the car is in the door you originally picked, and a 2/3 chance that it is in one of the doors you did NOT originally pick. The only thing the host does when he reveals a goat is reduce the number of doors you did not pick from 2 to 1. There is STILL a 1/3 chance that the car is in the door you originally picked, and a 2/3 chance that it is in the only other door left that you did NOT originally pick. The problem's trick is convincing people that the action of revealing a goat changes the odds. It doesn't.
MyNameIsSimple72 you can also test this in real life. bigger sample size the better. take ~20 rows of 3 cups with the same ratio of hidden items and run through each row following the rules with a friend being the host. youll be correect roughly around 66% if you swap each time
MyNameIsSimple72
If your first pick is Goat A, it's Goat A.
If your first pick is Goat B, it's Goat B.
If your first pick is the car, it's the car.
Your first pick gonna be the car 1 in 3 games (33%), not stupid 50%.
Willoughby Krenzteinburg I discussed this with a friend as well, and isn't it so that the only reason it's not 50/50 is because of the host of the so called show? Just think about it
robin291292
What makes it work the way it does is the fact that the host is REQUIRED to reveal a goat door after your initial choice. This is what makes your odds increase from 1/3 to 2/3 if you choose to switch. The reason being, the hosts actions do not affect the probabilities when his actions are following this rule.
No matter which door you choose, there will ALWAYS be a goat door available to open, and the host will ALWAYS open it. You know this going in, so the fact that the host does it is irrelevant. He is giving you no new information.
From the beginning, there is a 1/3 chance you originally picked the car. There is a 2/3 chance the car is in a door you did NOT originally pick. All the host does by revealing a door is reduce the number of doors you did not originally pick. There is STILL a 1/3 chance you have a car in your door and a 2/3 chance the car is NOT in your door. Now, there is only 1 other door that is NOT your door - representing that 2/3 chance.
In a variation of the game where the host does NOT know where all the prizes are and truly reveals doors at random (which means the host COULD potentially accidentally reveal the car door in some cases) - in the cases where the host reveals a goat, your odds of having the car are 50/50. This variation would be statistically similar to the game show, Deal or No Deal in that the contestant is randomly opening cases, and in the event that the contestant has revealed all but two cases (their original pick and one other case), then at that point, there is a 50/50 chance that each case contains one of the two remaining possible prizes.
So yes - to answer your question. The fact that the host's actions are predetermined to an extent (the host can't reveal the goat door you originally picked, but would be forced to reveal the only other goat door) is what makes the statistics of the Monty Hall Problem work the way they do.
Mathematically, it would look something like this :
If you want to figure out the odds of BOTH event A and event B of occurring, you just multiply them.
In the regular Monty Hall Problem :
So, the odds that you pick a car (1/3) AND Monty reveals a goat (1/1 - because he is required to and has no choice) are :
1/3 * 1/1 = 1/3.
The odds that you pick a goat (2/3) AND Monty reveals a goat (1/1 again - because he HAS to) are :
2/3 * 1/1 = 2/3.
So these are the odds AFTER Monty has revealed the door. 1/3 that you have a car, and 2/3 that you don't.
In the variation where Monty acts randomly :
The odds that you pick a goat (2/3) AND Monty reveals a goat (1/2 - only two doors remaining - only one of which is a goat) are :
2/3 * 1/2 = 2/6 = 1/3
The odds that you pick a car (1/3) AND Monty reveals a goat (1/1 - there are only goat doors remaining for Monty, so he is guaranteed to reveal a goat) are :
1/3 * 1/1 = 1/3
The other 1/3 of the time, you will choose a goat, and Monty will reveal the car - and this game is unplayable. So we ignore these events.
That leaves us with the remaining 2/3 of the games. 1/3 of the time you have the car, and 1/3 of the time you have the goat. So, in the cases where Monty has acted randomly and has revealed a goat, there is a 50/50 chance you have the car.
Now, it is STILL in your interest to switch because you have no way of knowing which version of the game Monty is playing. At best, you double your chances - at worst, you take a lateral move that does not decrease your chances, so you might as well always switch.
If you picked the car and swap you lose. If you picked the goat and swap you win. 2 out 3 times you picked the goat. 2/3 times you swap, you win.
*keep reading other user's explanations, and watching other videos. Once you understand it then, well, you'll understand it, yeah
Simplified: if you pick a goat, and the host reveals the other, and you swap, you win, cuz they are the only goats in the game. And there is initially 2/3rd (66%) chance to pick a goat.
Yup.
Another way to overcome some of the counterintuitiveness of the improved odds is to think of it like this - the host CANNOT choose your door, and since he KNOWS which doors the car is behind, he HAS to pick the other door - besides your door and the car door.
Your choice and his knowledge of the doors makes switching to the only other unchosen door makes it more likely - but not certain - that the other door contains the car.
He is always going to open a goat door, regardless of whether you chose a car door or a goat door. He has eliminated the original possibility that a car was behind the door he picks. Doing that increases the remaining possibility that the other door has a car, because if it has a car - he could not choose it.
Dude, ny logic:
If u dont pick the car:
1.knock on 1 door, no goat sound, its the car
*knock*
*beep beep*
No goat sound, must pick another door.
Another way to explain it is by making the example more extreme. Imagine there are not 3 doors but 100. You pick one at random, which is almost certainly wrong. The host then opens 98 doors to reveal 98 goats. Now you can see, he is pretty much showing you exactly where the car is.
But maybe you picked the car one, theoretically you are correct but in real it’s still 50/50
@@aaroei7 no you have a 99% chance of winning in this case if you always swap lol
Nicholas Davis no, you are wrong
@@aaroei7 replicate this experiment on real life or more realistically through a program. I can guarantee you you're wrong. It's a hard concept to understand.
Nicholas Davis
I agree that if it’s with 100 doors, you should swap because he basically shows you the right door but if it’s with 3 doors it’s a 50/50
In real life, I'd say the host is likely to give you the option when you picked the prize in the 1st turn.
Agreed.
So this solution is basically reverse your luck,
if you are lucky, you get a goat,
if you are unlucky, you get a car.
@Life Prof. B
If I am lucky I get a goat..which means I am unlucky.. AND
If I am unlucky I get a car..which means I am lucky
So you mean I am lucky if I am unlucky and I am unlucky if I am lucky..
Enlighten me xD
@@abhishekacharya588 Yeah dude
In my opinion you win either way
you're twice as likely to land on a goat 1st time than you are on a car. by swapping you are guaranteed to get a different outcome to what you had started with. you're therefore twice as likely to get a car by swapping than sticking. makes sense :)
Am I the only one who would try to get the goat instead of the car?
+Wilford Brimley Like 10% or more of the comments on this (nine-year-old) video are "I'd rather have a goat" or "what if I'd rather win a goat?"
+Wilford Brimley seems u r very rich
Ajay Prabhu
Indeed, I have a goat farm.
+Wilford Brimley great ....which place are you from
Ajay Prabhu
Not Wales...
I finally get it, thank you. I think the issue of people not understanding is too much emphasis being on that second choice, which technically is a 50/50 but its not the only round. It makes perfect sense when you realize you should assume you're first door was a goat based on probability.
***** Now I get it, thanks!
***** You are not calculating the odds properly. You've fallen for the trick. Intentionally or by mistake? Idk.
Round 1: Is meaningless. You CANNOT win the prize in this round, period!
Round 2: 2 doors, one prize. 50-50 chance if you stay OR switch doors.
1 2
C G
G C
See?
Let me put it another way: What are the odds you'll win the car in the first (3 door) round? Zero. There always is a second round regardless of which door you choose.
See it now?
5Cats That has to be the dumbest logic i've ever heard. Just because you don't open your door on round 1 doesn't mean its meaningless at all. From the start you have a 1/3 chance of being right. There is a 2/3 chances of the car being in the other two doors. That being said, one of the other two doors obviously has a goat, so just because he opens the goat door first does not make it a 50/50. There is still a 66% chance that either of those 2 doors has a car, and since he showed you the one with the goat, you know that if you switch, you will have the correct one (of the 66% chance) that has the car.
5Cats If YOU wish to look ignorant that is one thing, but to tell others they are wrong when they are absolutely correct is quite another. Perhaps I recommend to you to LISTEN to people who are telling you 100% factual information. This isn't a debated issue...it is math and factual math proven by computer simulation and logic. People are trying to educate you and you are just looking extremely uneducated to them right now.
5Cats
Are you going for the record for the number of times you can present the same illogical and incorrect solution to the MHP in multiple threads?
(And where's that Wiki reference to the MHP and the Gambler's Fallacy you keep talking about?)
i already have a car but no goats, how to ensure i get goats?
There's no such thing as a 50% chance if you have 3 variables. Your first pick is of course 1/3 win. If host reveals goat, and if you stay, your odds are still the same, 1/3 win. Now by switching, you change your odds, gives 2/3 win. Switching ALWAYS gives you better odds. 50-50 chance does not make sense.
+The Absolute Zero the host NEVER opens the door with the car but he opens one AFTER you first choose one. you dont know whats behind yours but you deffinitly know the one the host opens after that is not the car. thats why it could never be 50-50 chance. This video explains it perfectly. why you call people dumb if you dont understand such a simple explanation like the maker of the video did?
+The Absolute Zero "the whole time you're really just playing with 2 doors"
The three doors you are playing with the whole time are:
(a) the door first chosen by the contestant,
(b) the door revealed by the host (which cannot be (a) or any containing the car), and
(c) the unaddressed door (which cannot be (a) or (b)).
+The Absolute Zero If you pick a goat with your first pick and switch you will win. So I ask you. What are the odds of picking a goat first?
+Freddie Orrell The thing is that you don't have three doors all the time. You start off with three closed doors and pick one. The host then opens a door, removing a variable. So you start off with three doors and end with 2 doors.
P.S. probability of winning when you keep the door is 1/3 while swapping gives you a probability of 1/2.
+Spruce Wayne What would you win if you were allowed to switch to the open door?
The easiest way to interpret this is go over all possible outcomes:
1. You pick a goat. You DONT switch. You get a goat
2. You pick the other goat. You DONT switch. You get a goat
3. You pick the money. You DONT switch. You get the money
4. You pick a goat. You DO switch. You get money
5. You pick the other goat. You DO switch. You get the money
6. You pick the money. You DO switch. You get a goat
As you can clearly see switching gives you a 2/3 chance of getting the money whereas not switching gives you a 1/3 chance
Who said anything about money? There are goats and a car. Also, your explanation is NOT the easiest.
But your choice between switching or not is NOT between THREE doors, your choice is only between TWO doors. The one you originally picked, and the one that remains after Monty opened the third one that has a goat, and removed that door and goat from your choices. He moved that door and goat off the stage because it is no longer a choice you can make. Now you can only choose between two doors. Your chance of winning a car from one of those two doors is 50/50.
@@OhJodi69 Holy shit, if this explanation doesn't convince you, you're beyond help my friend.
@@lxilind7567 I did figure it out, actually, lol I made a comment further down in these comments, and I think I just was not understanding that at the start of the game, with three doors, I was going to pick the wrong door 2/3 of the time. here it is:
Ok, I figured it out. I think.
1$, 2G, 3G.
I pick one (*), and a goat door is removed.
*1$, 2G, __. Two choices. Stay/win, switch/lose.
1$, *2G, __. Stay/lose, switch/win
1$, __ , *3G. Stay/lose, switch/win
One stay wins. Two switches win.
The thing is the game doesn't reset at the second part. The possible results in the second one depend on the results in the first one. If you hit the car in the first attempt, changing will make you lose. If you didn't hit it, changing will make you win. As you see, if changing or not make you win depend on your success at the beginning. It's not like flipping a coin, in which the results are all independent.
this is not a guarantee that you will win when swapping, it is just clearly stating that the probability of you choosing a goat at the beginning is higher than choosing a car therefore swapping will give you a better probability of winning
its easier to think of it from the HOST's point of view.
1- the HOST will ALWAYS avoid the correct door if he HAS to.
2- the percentage that the host DOES NOT HAVE to avoid the correct door = the percentage that the contestant chooses the CORRECT door. = 33%
3- the percentage that the host DOES HAVE to avoid the correct door = the percentage that the contestant chooses the WRONG door. = 66%
4- Since there is 66% chance that the HOST will HAVE TO AVOID the correct door, means the door he is leaving has a 66% chance of being the correct door.
The best way to understand the "swapping" problem is this (at least this is how I understood it the first place): Imagine there are 100 doors, and you are asked to pick 1 initially. So you choose 1. Then Mikroutsikos (greeks will get it) opens 98 doors, and now you are left with your initial choice and one closed door. The chance of having won from the beggining is 1% (1/100) whereas the chance of the car being behind the other door is simply 99% (99/100). Not swapping makes absolutely no chance.
If you still don't get it, imagine 10.000 doors. Not swapping means you are certain that you have picked the car on your first choice :-)
astr1nos
This is an effective extrapolation.
The other is to ask what more they know about their first 33% choice when they say the new odds are 50/50. Most admit it still must be 33%, so the balance still must be 66%. Then, ask what more they know about the original 66% side, the one they didn’t pick, which is now just one door and must be twice as likely.
Thanks now I get it
You are wrong about 1 thing. Not swapping still gives you a 1% chance of winning, not "no chance."
But the thing is, you don't know if you choose a goat or car in the first time...
+Eve Just a Player
You *do* know that you're twice as likely to pick a goat than the car though
+HumptyDumptyOakland: True, but that changes once the 'goat' door is revealed, because you KNOW that one of the dreaded goats WAS NOT behind the door you picked. In plain language, there are no longer 2 goats available for the booby prize, but one. Your odds improved immediately, from 1 in 3 to 1 in 2.
Georooney
Your odds improve to 2 in 3 if you SWITCH.
+Georooney the odds of having chosen the car on the initial pick remain 1/3 even after a goat door is revealed but the odds of winning car if you switch has increased to 2/3 ... it's basic probabilities
Nicolas Vallée Only because Monty knows he will open a door with a goat behind it. If he opened a door at random, the chances would still be 1 in 3.
what ever you choose, you don't get a car. They can switch the car to a goat at the back of the door lol.
You've worked at the fairground, I take it?
hmm sound reasonable 😂
Goat is 1, Car is 2, open door is 0
You pick middle each time
1 1 2
0 1 2 switch and win
1 2 1
0 2 1 switch and lose
2 1 1
0 1 1 This is why, the host would never open the door with the car after your first selection, he would open door 3, he has revealed new information the impacts you "50/50" choice on the final pick, the events are not independent of each other. 2 out of these 3 events you win if you switch.
If anyone ever watched the movie "21" this thing is in it
I always liked to turn this intuitively by changing it to 100 doors. Suppose there are 100 doors, 99 goats and 1 car. You pick one door by chance an the host opens 98 goat doors. Now, do you swap or do you stay with that door you picked first? Now its obvious without doing the math.
ABC=3/3
A=1/3 -> BC=2/3
B=0/3 -> C=2/3
The chart at the end has been the best this has been explained to me. I always hated this puzzle because everyone's explanation was kinda weak and didn't cover all possibilities well. This however did and now I feel confident in why it works as to before where ik the answer but couldn't explain why it was. Thank you.
If I’m interpreting it right, does the table show that if you don’t swap, you’re essentially locking in your chances of a 33% probability of winning. But if you DO swap, instead of a 50/50, you actually have a 66% chance when you still factor in the first door with the goat as a possible option that you’re just gonna “ignore”!!
It's pretty fucking simple actually. You're all overthinking it.
+J Fleming The tricky part is in understanding that the elimination of a goat must be deliberate and non-random, not by accident. If the host (or the player, or anyone really) blindly picked one of the unselected doors and opened it, revealing a goat instead of a car *by chance* instead of knowing ahead of time that they're revealing a goat, surprisingly the advantage of swapping disappears. This is why you get people thinking that it's beneficial to swap cases at the end of a game of Deal or No Deal even though it's actually 50/50 either way.
+TedManney You're overthinking it too lmao
J Fleming No, I have insight into it that you lack. If someone insisted that a player should swap at the end of Deal or No Deal for an overwhelmingly likely chance to win the grand prize, you would be unable to explain the difference between the two games to them, and therefore you lack a thorough understanding of why the game works the way it does, which is the entire point of a math-based probability puzzle. If Monty tripped and fell and accidentally pushed a door open to reveal a goat, there would be *no advantage* to swapping. If you don't understand exactly why that is, you haven't thought about it enough and are in no position to claim it's so simple.
I WANT THE GOAT
+TedManney And you're just assuming I don't already understand the problem. Also, still overthinking.
Why may i ask is it called "Monty" hall problem ?
I would take the goats over the car because i have some Welsh and Aberdeen mates who would pay top dollhairs for pampered tv goats.
It's the "Monty Hall" problem - which is a person, who was the host of the game show called "Let's Make a Deal" which featured contestants picking among three doors. This problem - while not exactly the same format as the game show - is certainly based on that game.
1. named after the first host to have it ever
2. so would the entire comments section
bro it's simple
goat - 2/3 car - 1/3
monty opens goat door 100% of the time, leaving one goat and car
if you pick one and switch, you WILL get the other option.
you pick the goat and switch, you WILL get the car.
and the probability of picking the goat? 2/3
you pick the car and switch, you WILL get the goat.
and the probability of picking the car? 1/3
middle schooler explains 😎😎😎😎😎😎
The explaination is quiet simple :
Only one car and TWO goats.
So :
1/3 : you pick the car, then you must stay to win.
2/3 : you pick goat 1 or goat 2, then you must switch to win.
Saying that the final choice is a 50/50 chance means that your initial choice is 1/2 to pick the car and 1/2 to pick a goat, which is mathematically wrong.
Then : ALWAYS SWITCH.
CQFD as we say in France :)
lol, I'm high right now and I still understood it because he explained it so good :)
Must be some weak as shit
One minute ago, I was just like this can't be true! I supported people that disagreed with the fact that swapping increases the chance of winning. Now I understand it and think people that don't get it are stupid...
It's a rite of passage!
Yes, this video did a really good job of explaining it clearly.
I don't understand it though. Wouldn't 2 options nock it down to 50/50?
If there were only two options to begin with, yes, it would be 50/50. But since there were two goats and only one car, there's a 2/3 chance that the door you chose initially has one of the goats behind it, but only a 1/3 chance that you picked the door with the car. Now do you understand?
since the host is showing a goat behind a door no matter what, picking again is 50/50 odds , i understand how the odds increase from 33 to 50 but in reality its always a 50/50 pick since one door is being eliminated with no risk or reward
I'd follow one rule with LMAD: Always take/keep the money, taking this scenario out of the equation.
You are more likely to pick the goat (2goat out of 3 doors 66%) than the car(1 car out of 3 doors 33%)
So when you pick the door with the goat(66%) they must switch it to other door with the Car, since they must reveal the one door with the goat when you pick the goat at first
So the chance of you picking the goat and swap to the car is 66%
And the chance of you picking the car at first and swap to goat is only 33% since it’s only 33% to pick the car at first
I understand the math behind it, and it actually makes sense when I see the numbers, but it's still so hard to wrap my head around it because in my mind it's still 50/50
That's because only two places the car can be presents the illusion of 50/50. What you fail to see is that the third door affects the odds for each door differently. A way to easily see this is that you are effectively trading TWO doors for one if you switch. How so? If the car is behind EITHER of the two doors.... you WIN!
Think of the lottery. Would you trade your one ticket for two?
This makes sense. We pick a goat 66% of the time before the reveal of a door. But, if we swap after the door reveal, it would mean we have a higher chance of getting the car.
The number of people falling for this hoax theory is astounding… The fallacy here is believe that probably of the second attempt to open door is influenced by the probability of the first attempt, when in fact it’s not. As soon as you opened the 3rd door, the probability of the first door being the car increased to 50%, and not staying at 33%.
Take a scenario with 2 doors for example. If I choose first door, host opens second door and reveals the goat, the probability of the first door being the car is now at 100% and not staying at 50%.
If somehow the first attempt to open door influences probability in the second attempt, then this would be true.
@@robertdowney9017 But the first choice obviously influences the second choice? The game does not reset - the positions of the prizes does not change, and what you chose the first time is guaranteed to be one of two options on the second choice. For the two choices to be independent, the prize positions would need to randomize after the first choice but before the second.
what if i want a goat?
dont swap hahaha
Thank you so much for making this video. My friend tried and failed to explain it to me and this cleared everything up for me.
for all yall that dont get it
your first pick has a 66% chance of getting a goat right?
so when he reveals the goat
if you switch you have a 66% chance of winning bc
(like i said earlier) you probably picked the goat
and if you think its 50/50 i can see your logic
and it would be true IF he revealed the goat before you picked the goat
but since you pick the goat(66% chance) before he reveals the other goat you should swap bc you probably picked the goat in the first place. If you wanna debate im ready
67% you dumb oaf!
66.66666666666666666666 ya bishhhhh
This doesn't make any sense. At all. It doesnt matter what door you pick at all. Since as soon as tou pick the door 1 of the 3 doors is revealed to have a foat in it which makes It a 50/50 chance instantly. The 33%, 66% thing is utter bullshit since It is just the percentage of the past which becomes absolutely irrelevant after your pick. If you have an option to swap after your pick doesnt matter what you pick chances of winning are always 50/50
Yoggeyash
Wrong. Idiot.
There's is a 2/3 chance to initially pick a goat.
If you switch, your odds flip to being 2/3 for the car.
+Yoggeyash
It doesn't make sense to you because you are not smart enough.
Okay for those who don’t understand: when the show first started, you had a 2/3 chance of picking a goat. Got it? Okay, so now we are revealed another door which has a goat. So either the door you chose or the 3rd door has the car. Still with me? So since there was a 2/3 chance of selecting a goat in the first place, and the second goat was revealed, the 3rd door is most likely the car. There was only a 1/3 chance of picking the car, so switching would be a good idea. I hope you understand and if you still don’t please feel free to ask. Thanks!
A few points worth bearing in mind:
1. "The Monty Hall Problem" is the name of a logic puzzle. Whether it is the same as the TV show or not is irrelevant.
2. If the host opens a door at random and reveals a goat, then the probability is 50/50.
3. If a third person comes in after the opening of the door, and does not know what has happened, then for him then the choice of winning is 50/50.
4. Switching does not guarantee a win, but gives 2/3 chance of winning
Just to clarify a bit:
Ad.2 If the host opens a door at random, his chances are 50/50, so half of the time he will screw the game (since he will reveal the car instead of goat), and half of the time he will reveal the goat and give option to a player to win a car by switching to an unopened door.
That doesn't change the fact that first player, by choosing one of the doors gets 1/3 probability, while remaining 2 doors have 2/3 probability, hence, in all cases where host opens the door with a goat, player will have 2/3 possibility to win a car by switching.
Ad.3 If a third person (second player) doesn't know the choice of first player, his chances of winning are 50/50 since there are only two doors, one less likely to have car behind it (the doors that the first player initially choose), and the one more likely to have car behind it (the one of two unopened doors in first round of first player), and he doesn't know which door has better probability.
@@max5250 If MH opens a door at random the chance of his revealing a goat is 2/3, of revealing the car 1/3. You are assuming that the contestant has picked the car,
@@rwb966
Correct, my bad.
The host would open the door with a car in 33% of cases, and screw the game, while he will open the door with a goat 66% of time, and still give opportunity to the player to win by switching.
@@max5250 That's right (there's no bad).
@@rwb966 May bad was I said that the host would open door with car 50% of the time, i my first reply, which is obviously wrong, he would open that door only 1/3 of time.
I am here, SIXTEEN FCKING YEARS LATER, because I was required to watch this by my instructor. HOLY CRAP!!!