Comments by "Hallands Menved" (@Hallands.) on "The Simple Question that Stumped Everyone Except Marilyn vos Savant" video.

No, this is not correct! The unopened (and unchosen) 2nd door does not somehow accumulate probabilities from the goat door – or from 98 other doors. Not from the chooser’s point of view. All unopened doors accrue equal amounts of increased likelihood of being the prize-door with each eliminated door – and the chosen door is unopened, too, remember? There can therefore be no advantage in switching doors – but also no disadvantage. I do not believe any mathematician could fall for this nonsense… This is a classical example of people conflating actual facts with probabilities. But the probabilities can never change where the actual prize is placed!
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@ImYourProblem It changes the second you’re asked to choose again in the new scenario. It was 1/3 but now it’s 1/2…
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I’ve heard this presented before, but this presentation and Savant’s conclusion are incorrect: The host does not magically raise the probability of just the one door he leaves out; he raises the probability of all doors he leaves closed! The first door likewise has no way of magically retaining the lower chance, arbitrarily ascribed to it initially. It is left out by the host, too! So in the end it’s merely two doors, each representing 0.5 chance of hiding the car. But why don’t you ask an AI? It can run the scenario a million times and determine the actual frequency of the changing of choice leading to a car… It must come out 0.5 I think.
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@kt22027 How about you read my previous comment explaining why it was always fifty-fifty because you de facto pick 2 doors in the first round? Your own pick and the goat you force the host to exclude (because he can’t exclude the car). At some point it must sink in that this is a 2-door game.
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@klaus7443 From the host’s point of view the probability remains the same, namely 1. He knows where the prize is…
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@ImYourProblem In the new scenario, there are but 2 doors. How on earth wound one of them have a probability less than 1/2, now? Both are still closed. The door doesn’t know it was chosen first. What is this perceived difference between a closed door someone like of and a closed door someone discarded. Are the doors moody, now?
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@klaus7443 This isn’t a valid argument since the host’s chance remans 1 in all cases. Instead, take the second example with the 100 doors and simply imagine that the contestant has chosen, but nobody knows which door, because he put the number in a sealed envelope. As the host begins opening goat-doors, would you still claim that some unknown door among the remaining is somehow stuck with probability 1/100 while all the other remaining door approach 1/2? Step by step I have refuted all your »arguments« if you still don’t get it, I doubt you ever will. C’ya!
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@ImYourProblem The original probability remains 1/3, yes, but when the contestant is asked to choose the second time the prize can only be behind one of two doors. This makes for a probability of 1/2 for both doors, since the fact that the prize wasn’t behind door three doesn’t supply any clue to its placement behind 1 or 2. So the second time the contestant chooses, the probability is 1/2. What would you have it be? 1/3 for door one and 2/3 for door two? The sum must be 1, you know. So can you explain why opening one goat-door somehow makes door 2 twice as probable as door 1? Nope, Miss Brainiac made a mistake due to her lack of schooling, and I hope and trust that nobody at MIT fell for the flawed assumption that there’s no second round. In the second round the scenario is new and you choose anew…
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@morbideddie 1/3 or 100, but it doesn’t matter, because I get to choose twice! And in the final setting there is only two doors, each with an equal probability of 1/2. The initial probability doesn’t »carry over2 from one scenario to the next. Say I roll a dice and it comes out 6. If I roll it again, is the probability of rolling another 6 reduced by the fact that I just rolled one already? No of course not…
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@ImYourProblem Yes, but you don’t know which one until the host reveals it. Until that moment it’s 1/3, but after he does, it’s 1/2. Look at it this way, then: If the host invariably opens one door, regardless whether you pick the winner or a goat, the entire event always gives you a fifty-fifty chance, always! If you owned the car dealership delivering the prize cars and the show manager said »look, we’ll run 99 shows and use 3 doors, so we’ll pay you a third of the car’s worth at each show, because we can use the same car again if the contestant draws a blank, and statistically he will loose out 2 times out of 3. Do you accept?« Hopefully not, because one of the doors is only a gimmick used for suspense-prolongation and is always eliminated, after which the contestant gets to choose freely between only two doors! This means that the car dealer would be approximately 1/3 of his money short after 99 shows, because on average one prize would be won every second show, not every third…
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@klaus7443 You’re projecting now. The host doesn’t matter, because he always opens one goat door and lets you choose freely a second time! The third door is only a suspense-gimmick. At the final choosing, you always have a fifty-fifty chance – 2 doors, 1 car
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@klaus7443 Now you’re just ranting. Nobody said anything of the sorts…
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@ImYourProblem Oh, it matters: That door is excluded when you’re given the second round of choosing – making it a two-door game!
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@ImYourProblem No, let's not pick two, because then there really is none to exclude. Look, you pick one door out of three. The host excludes one of the remaining two – and lets you choose AGAIN between the two remaining (the one you chose plus the one not excluded by the host)! Between those two he lets you choose freely. The prize is behind one of them! You still have not received any new information telling you which door is more likely to be the winner. Fifty-fifty two door game!
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@ImYourProblem Oh, come on! Think straight: You haven’t received any info on the two remaining doors, but one entire door has just been excluded completely. That certainly changes the odds! Now you choose between only two doors with nothing to indicate any preference. That’s fifty-fifty, period! I’m out...
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@ImYourProblem You keep changing the scenario, but the fact remains: You choose TWICE! and in the ultimate round end up with 2 doors! But you can still choose freely. Nothing carries over from first round beyond pure illusion. What would be truly different was if the host sometimes could eliminate the prize door, but he always eliminates a goat door. No! It most certainly becomes a fifty-fifty in the ultimate round. Bye!
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Oh, I just discovered that it was on this video I addressed the problem earlier. The YT algorithm has just forgotten that I already watched it. Well, I read through all comments and my rebuttals, and I still agree with me completely… 😂
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@max5250 Don’t quit your day-job…
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@Hank254 Only the 30 responses to my first comment. No, my explanation is the correct one. The video already explained the other view perfectly, so why would you try? Only problem is that probabilities aren’t inherited. Then some of you say that the car is placed behind 1 of 3 doors making the probability of picking it 1/3, but even that isn’t true, because in every game a goat door is taken out, so in reality there’s already only 2 doors to choose between in the first round! The game show host doesn’t exclude the door with the car! The probability is 0.5
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@Araqius Assume that reality prevails and you only know what can be known: The door you first pick already has a 50% of holding the car because the host does not open a random door. His door is always a goat! So no matter what door you pick initially, it already at that point is a 50/50 game. The third door is a decoy. If you want the probability to be one third initially, you need to let the host open a random door. The risk of him opening the car door must be real to give a 1/3 probability.
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@Hank254 No, I’m the one understanding what that means, because as I’ve mentioned a few hundred times already the host never opens the car door. You’re just parroting what you hear and evidently don’t understand probabilities.
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@Araqius 😂 Why complicate. There never was two goat doors, don’t you see? There’s a goat door and a decoy door and a car door. The host never opens the car door, you see. Already from the start it is a 50/50 game for Pete’s sake…
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@Araqius Strange how you constantly need to change the scenario to argue. What wrong with the real one?
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@max5250 By numbering the doors you assume a sequence of choosing, but that’s not the case and sequencing isn’t allowed when talking random probabilities. There’s one goat allowed to remain and one thrown out – always! You don’t know which goat is tossed, but you know the car isn’t! This means that you were only choosing between a goat door and a car door all the time!
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@gnlout7403 Yes, it really is 50/50 because the host never selects the car door. In this game there’s one goat door one extra goat door and one with the car. No matter if you chose a goat or the car, the surplus goat gets tossed. You are choosing between a car and a goat. The other goat is never available because the host doesn't pick at random. It a two door game with one decoy door…
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@max5250 You keep numbering the doors. You can’t do that. You also generalize from examples. Not allowed. And the only way you can make a 100 doors game comparable is if the host opens 98 goat doors. I’m so absolute right it astonishes me that you can’t see…
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@Hank254 You’re not just a little condescending punk, you’re also a liar: You haven’t tested it, or you’d already know I am right. But I understand your problem. On the level you function, you’ll inevitably tend to believe everyone else is as dense as you yourself, and lying to seem right is probably not a rare occurrence with you either. 😘
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@RonaldABG All your talk serves no purpose. Only if the host could open a random door, would the probability be 1/3, but then his action would be inconsequential. The fact remains that you start out chosing one door. Whether goat or car doesn’t matter, because one goat is always removed next. So you have two doors, one car, one goat, and no matter what you chose now, you still have the 50/50 chance you always had of getting wheels. Don’t forget that the sum of probabilities must always add up to 1! The matrix field can’t add up to 0.5+0.3333. It’s simply not possible. It is a given fact before you chose, that you’ll end up in a situation where you chose between two doors only and one gives you wheels…
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@max5250 Please read my latest response and suffer. Then repeat after me: „My God! Hallands was right again!"
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@max5250 OMG, that’s just so dumb.
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@kt22027 Sure, but that’s not the case here. Whether you pick the car or a goat the surplus goat is shipped off and you chose between one goat and one car – and you cannot not choose! That would only mean sticking with the first choice… You must remember that you’re a winner – hit or miss in the first round – because the host is not allowed to ship the car off, now is he? No! So hit or miss first round, it’s either goat of mustang! Does the host eliminate a goat? Yes, he always does! Does he chose which goat to eliminate? Only if you already picked the car. You feel that the door you first chose is slightly more goatish (2/3) than the door left by the host (1/2) but this is an illusion because your first pick can also only be goat or car (1/2) and even though you have 2 ways to pick a goat and only 1 way to pick the car, this doesn’t matter because 1 goat is always eliminated. It is neither to your advantage to change your mind nor stick with the first pick, because in both cases the chance of wheels is 50/50. You could formulate it this way: In the first round you’re choosing 2 doors, because your choice determines what goat-door the host must open. Hit or miss, it’s 50/50…
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@Araqius I already addressed this nonsensical argument.
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