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/sci/ - Science & Math


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15542721 No.15542721 [Reply] [Original]

The monty hall problem is the best midwit filter.
I have never seen a person of subpar intelligence be able to comprehend why the Monty hall problem is true. No matter how much you explain it to them, they won't get it. This problem should be used as an offhand IQ test for schools and jobs.

>Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

Do you get it? or is it over for you?

>> No.15542727

>>15542721
It is my advantage to not switch. Goats are based. Fuck cars. But I want both goats, so what deal do I need to make with the host to get both goats?

>> No.15542733

Yea i mean cars break down and costs money..
With goat you can milk it, get some fur and like trade it or eat it.
Having car makes you fat and lazy, doesn't sound so intelligent to me

>> No.15542740

>>15542727
Technically speaking, the problem indeed doesn't specify whether the goat or the car is the price.

>> No.15542742

>>15542721
>the host, who knows what's behind the doors
I think this says it all. The syntax is resolved by the original problem. The only people who don't understand haven't read it.

>> No.15542741

>>15542740
The only factor curiously left out of the problem specification... the player's preference.

>> No.15542745

>>15542721
It's to my advantage to not switch my choice because I want a pet goat.

>> No.15542749

>>15542745
>>15542727
>>15542733
you can sell the car and buy a bunch of pet goats

>> No.15542750

>>15542749
That's not as special...

>> No.15542752

>>15542749
Its probably full of rust already and makes a sound

>> No.15542754

>>15542749
Makes sense. I don't think there's a country in the world where you can't trade a car for a goat.

>> No.15542759
File: 2.00 MB, 361x247, gorilla warfare.gif [View same] [iqdb] [saucenao] [google]
15542759

>>15542749
But I want THESE two goats. What if they are not for sale?

>> No.15542764

>>15542721
When I was in 8th grade, in a 10th grade math class cause I was smart, the teacher showed us this problem and I insisted that switching makes no difference. I insisted on this because I was unaware of the
>the host, who knows what's behind the doors
factor. I don't know if it's because I wasn't paying attention or because the video she showed us of the problem just didn't specify it, but the video seemed to me to imply that the host opened one of the two unchosen doors at random, not knowing who or what was behind the door, and it just so happened that a goat was behind the door.
It was never clarified to me that the host actually knew, so the teacher probably thought I was just dumb lol

>> No.15542767

>>15542764
Same here anon, frustrated me to no end because of how fucking ridiculous it is without that one qualifier

>> No.15542768

>>15542764
The host knowing is why switching makes no difference. It means one goat door is never in play, reducing the choice to two doors.

>> No.15542769

>>15542764
That doesn't matter if the host knows or not know ( if he could open the door with car its just over )
If you land your pick on car host has 2 choices to open a door
If you land on goat door the host has just 1 choice
So the game will play as if it wont matter what happens before because it 100% leads to 2 door game

>> No.15542770

>>15542764
>>15542767
Are you both under 50?

>> No.15542771

>>15542767
based

>> No.15542773

>>15542764
It makes no difference. All these ideas of the un-chosen door having a 2/3 probability just sound like schizo pseudoscience. At the start, each door has an equal 1/3 chance of being a winner. When the host opens a non-winning door then the remaining doors have an equal 1/2 chance of being a winner.

>> No.15542774

>>15542773
It assumes a non-real world in which unchosen doors influence the final probability. While this can be (erroneously) calculated using mathematics, it's impossible to replicate in a real world scenario.

>> No.15542775

>>15542768
You have it backwards. The host knowing is why switching DOES make a difference.
>>15542770
I'm 30 and I am >>15542764
>>15542769
It does matter if host knows, but in neither of the scenarios is the host going to knowingly open car door.
Either the host doesn't know which door hides the goat, and he simply opens one of the two non-chosen doors at random (which is what both me and >>15542767 interpreted the problem as being when we were first introduced to it), in which case if there is a goat behind the door, switching door choice makes no statistical difference
OR, the way the problem is actually intended, the host does know which door hides the car, and he opens one door knowing a goat is behind it (so if the player chose a goat door, he intentionally opens the other goat door instead of the car door), in which case the player switching doors doubles their probability of car.

>> No.15542780

>>15542773
If the host opens a door at random (as some of us first believed), correct.
However, if the host opens a door knowing it hides a goat, you are incorrect. Let me explain:
>you choose a door at random
>the host, who is omniscient, now opens a door of his choosing, NOT a random door
>the host will NEVER open the door that you chose
>the host will NEVER open the door hiding a car
>the host will ALWAYS have the option of opening a goat door, since there are two goat doors and you only get to choose one door
>your door has 1/3 chance of being car door
>given that the host always has the option to open goat door, and will never choose your door, and will never choose car door, the host opening a goat door says nothing about whether or not your door is car door
>what him opening that door DOES say is that, if your door is not the car door, then the car door is the unchosen door as opposed to the door he opened

>> No.15542782

>>15542775
>in which case the player switching doors doubles their probability of car
No, this is fucking retarded.

>> No.15542783
File: 70 KB, 742x463, monty.jpg [View same] [iqdb] [saucenao] [google]
15542783

>>15542775
There is 4 ways the game can play out
There is 100% chance the game leading to 2 door monty
Your prior choice then doesn't matter

>> No.15542784

>>15542749
Fuck that I want hollywood goats trained to stand behind a curtain.

>> No.15542786

>>15542780
Yeah and none of this changes the fact that all available doors have an equal probability.

>> No.15542792

>>15542783
This is why it's always 50/50. The setup for the question prevents any other outcomes from existing.

>> No.15542793

>>15542783
On the off chance you are not trolling, the first two options in your diagram are COLLECTIVELY as likely as EITHER the seond or third. The first possibility is 1/6, second is 1/6, third is 1/3, fourth is 1/3.
Reason being, your choice is made first. Host's choice is made second. If you choose car, then host will choose at random one goat door or the other goat door, whereas if you choose a goat door, then the host only has one door to choose from rather than two.

>> No.15542795

>>15542793
Determinism prevents your 1/3 vs 2/3 world from being real, because you can't have chosen otherwise.

>> No.15542796

>>15542782
>>15542783
>>15542786
>>15542792
genuinely cant tell if trolling or underageb&
>>15542784
based

>> No.15542799

>>15542795
>determinism
isnt real
God created the universe. Free will exists.

>> No.15542801

>>15542793
>Reason being, your choice is made first.
Why would your choice matter if the pre-existing values are that it will become 100% 2door monty whatever happens

>> No.15542804

>>15542793
THERE ARE TWO DOORS
ONE HAS A GOAT BEHIND IT
ONE HAS A CAR BEHIND IT
YOUR ODDS OF EITHER ARE 50%
That's the whole fucking scenario. Nothing that happens before then makes a single solitary difference, it's just theoretical math mumbo-jumbo that can't possibly map to reality, pure sophistry. The actual midwits are the people who buy into that number-cooking bullshit.

>> No.15542807

>>15542801
because if the player's choice is car then the host has two doors to choose from; otherwise the host only has one door. Therefore the four possibilities in the diagram are not equally likely. What ARE equally likely are whether the player chose door 1, door 2, or door 3. So you can just combine the top two possibilities.

>> No.15542810

>>15542804
This.

>> No.15542812

>>15542796
No trolling, not underage. There are 2 doors to choose from, they have equal probability of having a car behind them. Everything that happens before Monty opens his goat door is an irrelevant distraction.

>> No.15542815

>>15542807
Monty opening a goat door every time before you lock in your decision means that there are only 2 doors actually in play and you just don't know it yet. Which of the 2 remaining doors has the car is a literal coin flip.

>> No.15542817
File: 49 KB, 1154x154, 16547546755465.png [View same] [iqdb] [saucenao] [google]
15542817

>>15542775
>I'm 30 and I am
God damn you fuckers are young lol. Anyway yeah the problem solves itself as soon as you read it.

>> No.15542818

>>15542807
Yes and if you happen to choose car door there will be 2 different doors the host can choose from, thats an illusion from the players part that if you choose goat the host has "locked" to choose a certain door when the car picking scenario the host has "freedom" to choose between 2 doors

>> No.15542822

>>15542764
Literally me. I was about 13 in 10th grade. And I argued with the math teacher for almost the entire class and even after it for a long time. I doubt he thought I was dumb though, just that I was a nerd who cared too much.
Bugged me for so long, then about 3 years later I suddenly remembered it again, and it clicked and started to make sense. I had to think of it as the probability of choosing the incorrect door instead of thinking about it straight as the question asks it for it to make sense to me first time though, for all the ones here that still don't get it, try doing that.

>> No.15542836

>>15542822
See: >>15542804
You can set up as many arbitrary convoluted scenarios leading up to a coin flip as you want, the coin will still have a 50% chance of either turning up heads or tails.

>> No.15542839

>>15542836
>tells you to close eyes
>flips coin
>turns it over after it lands when your eyes are closed
heh heh heh nothing personnell kiddo ;)

>> No.15542844

>>15542812
>>15542804
But monty knows which one the car and goat is. So he wouldn't open the door with the car. That means the door he didn't open is more likely to be the car. Because he actively chose to not open it. so now there's 2 doors that he didn't open. Door 1, and door hence door 1 has 1/2 chance of being true. As opposed to the door you chose, door 2. Because you chose it when 3 options were available. So there's only 1/3rd chance of it having a car.

It might become more obvious if you think about a larger example. Imagine you were given 10,000,000 doors. And you selected a specific door. Now this being correct has a 1/10,000,000 th chance of being correct. Monty then goes on to open 9,999,999 of the doors. So now you are left with just 2 doors. The one you selected, and one which monty simply deliberately didn't open. Monty knows which the car was in, so now it is more likely that switching will give you a prize, since that is the door he deliverately didn't open. As opposed to your door which he didn't open because you had selected it. Your door still has 1/10,000,000 th chance of being correct, since you chose it when you had 10,000,000 options, Diminishing the number of doors doesn't change the probability of it being correct, because your knowledge when you chose it dictates that probability. But now door 2 has 1/2 chance of being correct, because that probability is dictated by Monty's choice based on his knowledge. So you should 100% in such a scenario choose door 2, the door that monty didn't open. Much more likely to win a car

If you still don't get it, I'm afraid OP might be right.

>> No.15542846

>>15542839
>flip coin and tell you to call it in the air
>you call Heads
>I catch the coin but keep it covered and peek at the result
>I inform you that the coin most definitely did NOT come up Quirglesmurf
>I ask if you want to stick with Heads or switch to Tails
What are your odds of winning if you stay or if you switch?

>> No.15542847

>>15542844
but the fact that he opened 9999999999 doors means that now it is 50-50 if your door has a ccar, it either does or it doesnt :)

>> No.15542849

>>15542844
My door could still have a car behind it. Monty opening one door over the other provides me with no information except that it informs me my choice was actually always 50/50 and the third door was a red herring. Assigning some greater value to the unchosen door makes no sense in reality, it's just a mathematical quirk that shows how probability calculations can sometimes be stupid and wrong.

>> No.15542850

>>15542849
This.

>> No.15542854

>>15542836
It's not a coin flip because it's not random. Monty has knowledge of the answers.
An equivalent would be playing dice with God, who already knows the result of the toss in the future. So of the 6 options, you choose, let's say 2 for the dice toss. God removes all other options except for 1. So now you should switch to 1. Because of God's knowledge of the future, that he knows that nothing from 3-6 could have been the result of the toss. It has now become more likely to be 1 than 2. Because 2 was based on your knowledge out of 6 options.

You're really missing the part where Monty hall knows what's behind the door.

>> No.15542856

>>15542849
>except that it informs me my choice was actually always 50/50
oh so if he doesnt open the door then your choice wasnt 50 50?

>> No.15542857

>>15542854
It doesn't matter that he knows what's behind the doors. All it means is that those choices (be it the third door or the 9 million doors or the 4 other sides of the die) were always invalid red herrings. The statistics were built on faulty information and the dataset must be corrected before proceeding.

>> No.15542858

>>15542844
>has a 1/10,000,000 th chance of being correct. Monty then goes on to open 9,999,999 of the doors.
Are you supposing monty will open 9,999,999 doors instead of 1 ?

>> No.15542861

>>15542856
If he doesn't open a door then you would have 2 alternatives to switch to instead of 1, making the total number of doors 3 instead of 2.

>> No.15542863

>>15542856
Correct, if he doesn't open the door then the choice is 1/3. All doors have an equal chance of hiding the car. When Monty opens a door he alters the premise, the remaining doors still have equal chances of hiding the car but that equal chance is now represented numerically as 1/2 because the door Monty opens is simply removed from the equation.

>> No.15542865 [DELETED] 
File: 87 KB, 940x1387, Screenshot_20230704-214222_Samsung Notes.jpg [View same] [iqdb] [saucenao] [google]
15542865

>>15542844
But if your options have been reduced 2 doors as you say from 3 doors. And door 1 has 1/2 probability of having a car, and door 2 has 1/3 probability of having a car, it simply doesn't make any sense.
Think about it, if door 1 has 1/2 chance of being correct, then it also has 1/2 chance of being incorrect. And what does being incorrect mean here? Since door 3 is 100% incorrect, It means that Door 2 is correct. So that means door 2 has to have 1/2 chance of being correct as well. It simply can't be 1/3. The probabilities don't add up.

>> No.15542870

>>15542721
I figured you shouldn't switch automatically, but flip a coin to determine if you should switch or keep your first choice.

>> No.15542883

>>15542863
>>15542861
hmm.. what if the host knows which door hides the car, and he decides to avert his gaze from that to a goat door, and stare intently at the goat door ? but he doesnt open it?
does this reduce the chance?
what if you know he is gazing at a goat door but he is wearing sunglasses?
what if you think you can see through his sunglasses but you arent sure?
what if a pigeon swoops down and removes his sunglasses, so now you can see him gazing at a goat door with certainty?
and you know he is gazing at a goat door?
at what level of certainty that you know he is gazing at the door he is gazing at does it go from 1/3 to 1/2?
or does it stay 1/3 since he didnt open it?
what if he is gazing at the goat door, lunges for it with intent to open it, but then hitler comes back from the dead and shoots him before he is able to open the door?
i think this is where math crosses with quantum mechanics

>> No.15542894

>>15542883
All of these scenarios fall into the realm of psychology; the interplay between the host and the contestant and how one's actions (or perceived motives) influence the other's. None of it has numerical value.

>> No.15542895

>>15542883
the post that broke monty deniers

>> No.15542900

>>15542894
so all of those scenarios fall somewhere between 1/2 and 1/3? but that is a numerical value.
what if monty says he is gazing at a goat door, but he is wearing sunglasses so you cant quite tell? is that 1/2.5?

>> No.15542903

>>15542900
Nope, they have zero numerical value. You can't assign a numerical value to your personal perception of the possible intent or non-intent of Monty's gaze.

>> No.15542907

>>15542903
so when the host says he is gazing at a goat door, it is suddenly neither 1/2 nor 1/3 that you chose car door, and it suddenly becomes outside the realm of math? got it bro

>> No.15542924

>>15542907
The doors themselves have set probabilities. The rest of what you're describing can't change those probabilities. For everything you think about Monty's gaze, you could be wrong. For everything Monty tells you about his gaze, he could be lying. The only knowable and quantifiable probability is that there's a 50% chance that either valid door is hiding a car.

>> No.15542927

OP here. Why are you guys arguing that one door has 1/3 and the other has 1/2 chance? You do realize that the correct answer is supposed to be that the unchosen door has 2/3 chance, right? not 1/2? What you are saying with 1/2 doesn't even make basic mathematical sense lol.
I don't even need to say anything for the other side of the argument. They failed the IQ test straight up.

Both sides arguing here are retarded kek, this is an all-time low for /sci/

>> No.15542935

>>15542927
Of the 3 available doors, only 2 are valid, you simply aren't told at first. When the invalid door is revealed, you realize the only 2 valid doors each have a 1/2 chance of hiding the car.
See my coin flip example: >>15542846
Nothing that happens before the final choice matters, it's all a great big red herring and you always would end up with a 50/50 shot at the car.

>> No.15542950

>>15542927
I am arguing with the retard for the memes. I already know the chosen door has 1/3 chance of car whether or not host opens a door.

>> No.15542956

>>15542950
>and then host opens door
Tell me why the distribution lands on the other door and not your door
Seems paranormal

>> No.15542963

Monty wants you to chose a goat door, so he baits you and plays a trick on you. It also doesn't matter which door you pick, since you are always left with one car and one goat door in the end.
The fact he knows the correct answer doesn't magically change the odds for you, why would it?

>> No.15543008

>>15542956
it lands on the other doors collectively dumbass

>> No.15543015

>>15542775
The host knowing doesn't make a difference. The fact of the matter is he has revealed a goat. Whether or not he knew he was going to reveal a goat is irrelevant. I have no idea why the qualifier of him knowing what's behind each door is the only reason you came to understand the problem. You've seen a goat and you know you're more likely than not to have initially picked a goat yourself, and that's what's important. The host's knowledge doesn't mean shit.

>> No.15543024

>>15542935
This anon formulated it pretty simply in a rather different way that seems simpler than the typical explanation, and should be easier for you to understand if you begin thinking about it in the negative. >>15542822
Let me attempt to declassify what he was saying for you. When you choose the door, what are the chances that the door you chose was the car option?
1/3rd
Now think about it in terms of not having the car, or as the other anon said, of choosing the incorrect option.
That means that there is 2/3rd chance that the door you chose wasn't the car. Now no matter what else happens, no matter how many doors you open or close, it doesn't matter anymore. Your door will always have a 2/3rd chance of not having the car. Because the decision has already been made. And you had 2/3rd chances of choosing the 'incorrect option' when you made the decision. In laymen's terms, for every 3 times you would have made that decision, it sould have not had the car. That is unchanging. If you chose the door without the car, Monty opening or closing the door won't change the fact that you didn't initially choose the car. And every 2 out of 3 times, you would initially have not chosen the car given random odds.
So when you only have one other door left, and your door has 2/3rd chance of not having the car, it means that the other door automatically has 2/3rd chance of having the car. So you should switch, because the other door now has more odds of having the car.

I actually didn't think of it this way before either, but this makes it simple enough that even you should get it. Because really, the answer now simply becomes that you should switch, because the odds of having chosen the incorrect option were greater initially. Which is 2/3rd unchanging odds of not choosing the car.

But again, as I said in the original post, this is an IQ filter question. If you don't have the capability, you simply won't get it. No matter how much it is explained.

>> No.15543032

>>15543024
Your initial chances are 1/3 thats correct but that doesn't change the fact that the game is going to be 2door monty in the end
Whatever happens before the end game doesn't matter because it will lead to end game 100%

If the game was that monty could open randomly the car door and just happened to not do so, then your chances would be 66%
Because monty also had 1/3 of chance

>> No.15543035

>>15543024
Determinist physicalism disproved your hypothesis. You have no capacity to choose the alternative.

>> No.15543045

>>15543035
You were pre-determined to choose the incorrect option 2 out of 3 times on average. There, happy now?
>>15543032
So if your initial chances are one third. Just forget the end game even for a second. what does that mean for you to have a one third odds of choosing the car? Think about it if you can. That means that if you play the game 3 times, on average you will only select the car 1 time out of those 3 tries. That means that the other 2 times, the other unchosen door will have the car. End game doesn't even matter. Your decision is made when you choose the first door. Then and there it is concluded that you only have 1/3rd chance if it being correct on average. The end game only serves to narrow the other 2/3rd chance to one door instead of equally distributing it among two doors.
But OP out. You're either trolling at this point, or you genuinely failed the IQ test. Either ways it's not worth it for me to argue anymore.

>> No.15543048

>>15543024
You keep mistaking my disagreement with the conclusion for not understanding the explanation. I fully understand the explanation, I fully understand your conclusion, I also fully disagree. What's happening is that you're given false information at the outset, and when the false information (Monty's door choice) is revealed the numbers get screwed up unless you correct the false information. The real choice in the end is between 2 doors, and it always was going to be, you just weren't told at the start. The entire setup is a mind-game to trick you into thinking you're playing a game full of rules and twists and turns rather than choosing the result of a coin flip. It's done to add a false sense of tension, but when you examine the actual mechanics of the game you realize there was always going to be a completely invalid door that you'll never be allowed to choose in the final decision (the only one that matters).

>> No.15543060
File: 46 KB, 324x430, Erdos_budapest_fall_1992_(cropped).jpg [View same] [iqdb] [saucenao] [google]
15543060

>>15542721
pic related refused to believe that switch was the right answer until someone showed him a computer simulation of the problem. is erdos a midwit?

>> No.15543062

>>15543045
>forget the end game
>heres my way of playing it more times

We are only playing the game once, monty opens goat whatever happens
That distribution of the chance will spread evenly because it is 100% to becoming 2 door monty
If the game was that "monty opens goat whatever happens" was not true, monty could open a car door out of that 1/3 chance that still exists after you have locked down your choice (because 50% of 2/3 is 1/3) then you would have 66% chance

>> No.15543063

>>15543060
He's correct, and someone had to lie to him using a computer to make him forsake his own correct belief.

>> No.15543092
File: 157 KB, 625x755, Screenshot_20230704-234404_Samsung Notes.jpg [View same] [iqdb] [saucenao] [google]
15543092

>>15542721
The chances you pick a goat are 2/3.
If the host eliminates a door with a goat, that means if you picked a goat, the other door must contain a car.
So the chances of switching and the other door having a car are 2/3, the same probability as picking a goat initially.

>> No.15543095

>>15543060
He was too old to fully understand computers and was tricked by the air of infallibility given off by computer simulations. The midwits who believed their own bullshit about one door having a 2/3 probability created a simulation that didn't reflect reality; they wrote a program that assigned a 2/3 probability of winning to the un-chosen door (since that's how programs work) and this sleight of hand fooled him.

>> No.15543102
File: 47 KB, 474x605, 1686254388621.jpg [View same] [iqdb] [saucenao] [google]
15543102

Imagine falling for fake math used by self-felating retards without a shred of common sense in "statistics" """math""" classes.
>3 options
>1 in 3
>host removes 1
>2 options
>1 in 2
Bunch of midwit pseuds.
>b-b-but there were 3 options before therefore-
Retard.

>> No.15543107

Let's try this one on for size:
You are presented with 2 doors, 1 has a car behind it and 1 has a goat behind it. When you select 1 of the doors, the hosts asks, "Are you SURE that's the door you want? You can switch if you'd like?"
What is your probability of winning the car if you stay versus switching?

>> No.15543111

>>15543107
1/2

>> No.15543113

>>15543111
Exactly right, and that's why it's 1/2 in OP's example as well. Opening a goat door is nothing more than pageantry, all that's really happening is the host saying, "Are you SURE you didn't want that other door?"

>> No.15543128

>>15543113
nah it's 1/3, played it out on paper one afternoon

>> No.15543130

I actually thought reading these responses that this was embarrasing for /sci/. It is filled with actual midwits who fail a simple question yet act smug and smart in other threads. Makes sense for a board for obsessed with their IQ. OP is right in calling this an IQ filter.
But then I got to the bottom. and saw the number of posters. And realized that it's just one (or two idk) retard arguing over and over again. Now I'm just mega-embarrassed for him. Please stop. You are horrendously wrong brother. Quit it and work on getting your GED instead

>> No.15543133

>>15543130
If you think the answer is anything other than 50/50 or that one side of a coin flip can have a 2/3 probability then you're the midwit.

>> No.15543139

>>15543130
don't mind him, he's just exercising his narcissistic facade

>> No.15543141

>>15543133
Your argument is the same thing as saying I'll either die tonight or I won't. So there's a 50-50 chance, a coin flip, that I'll die tonight. This is ignoring everything else that makes it probablistic that I won't die tonight. If you actually think like this, you're not a midwit, you're a dimwit.

>> No.15543143

>>15543102
This. People are so deluded.

>> No.15543148

Terrible thread. If you can't see why it's 2/3 you're not thinking and being stubborn

>> No.15543155

>>15543141
No, you're the midwit who's distracted by basic misdirection and irrelevant factors. You were always going to have to choose between 2 doors, you just weren't told until the second round. You have 2 options, 1 door has been removed from play, 1 of the 2 remaining doors contains a prize, the odds are 1/2.

>> No.15543176

>>15542783
>assuming all 4 playouts have equal weight
bless your heart

>> No.15543196

>>15543148
If you think it's 2/3 you probably think stage magicians are actual sorcerers.

>> No.15543210

Anyone ever wondered that the answer could be 58.333% ?
>average of 50% and 66.666%

>> No.15543219

>>15542721
>and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat.
A second layer of IQ filtering is asking whether there's any difference between the host opening door 3 (to reveal a goat) KNOWING it has a goat and the host opening door 3 (to reveal a goat) NOT KNOWING it has a goat.
If so, why? If not, why not?

>> No.15543226

>>15542742
>The syntax is resolved by the original problem.
What a pseud way of stating a fact.

>> No.15543230

>>15543219
This. So many people here that are pretending to understand the problem keep saying the wrong stuff. Because they want to pretend they passed the filter. But they don't really get it. This question would essentially filter out 90% of this thread.

>> No.15543237

>>15542721
the problem does not address the psychological impact of how bad switching away from the car would feel. Humans feel loss harder than they feel equivalent gain. Some people could psychologically cope and do what is the best statistical chance of getting the car, and feel good about doing that, whereas other people might realize that if they switch away from the car they will feel so bad that they would rather risk the reduced chance.

>> No.15543240

>>15542742
but it never said he was avoiding the car. it said he happened to pick the goat. he could have picked the car afaik

>> No.15543266

>>15543237
Thats why gambling is so addicting, imagine finding a system to fool yourself infinite money
People only feel so good about the winning part of gambling because thats their aim.. House has greated a system where the % is always on their side, sure you win sometimes but its always from someone elses earlier loses
Technically you cannot ever win even a coinflip.. Only way to win a coinflip is to not play it.

>> No.15543274

>>15543266
>Technically you cannot ever win even a coinflip.. Only way to win a coinflip is to not play it.
No, I'm pretty sure the only way to win a coin flip is to guess which side will be facing up after it lands, which is possible.

>> No.15543275

>>15543219
That doesn't matter at all, if he knowingly opens the door to humiliate you.

>> No.15543284

>100 doors
>99 goats 1 prize
>I pick 1 door
>1/100
>98 doors are opened that are not prizes
>two doors remain, one with a prize one without
its still 1/100

>> No.15543286

>>15543274
Im talking about from the gambling point of view.. if you had 1000 coins and i had 1000 coins, we cannot win ever but only hope to get us where we started which is even
Winning that kind of coinflip would mean that the system has to have 2001 coins

>> No.15543295

>>15543286
Even though numerically the probability is 50/50, in reality you can have long strings of "luck" in the form of the same result coming up repeatedly. If the results are truly random, you could still get 1000 Heads in a row even though it would be exceedingly unlikely, so the odds of winning exist, however slim.
t. gambling addict

>> No.15543304

>>15543102
if we assume this second choice is new and independent then yes it is 1/2 but in the real world there were 3 doors, 2 goats and a car and the possibility of the car being behind one of those doors was and always will be 1/3. the door that had the goat still exists and there are still 3 doors in reality

>> No.15543306

>>15543304
>if we assume this second choice is new and independent
Which it is. There is no universe where you could have picked the goat door Monty opens.

>> No.15543310

>>15543306
it isnt independent because its still the same reality, the three doors still exist and now you can see through one of them. it is still 1/3

>> No.15543312

>>15543310
It is impossible for you to choose the third door. You cannot switch to the third door even if you want to.

>> No.15543314

>>15543295
I didnt get that gambling addict gene and im sorry if you did but im just interested about these things
Sure the hope is there to win jackpot from the lottery but those systems are already far beyond the 49% point (for harvest)
If you can luck yourself to 3 correct in a row, there has to be the 3 wrong choices in a row also because the system does not gain anything from outside and the best hope is to get even

>> No.15543315

So called "mathematicians" think the world works like stuff they scribble onto paper and rationalise themselves into insane positions. Its 1/3.

>> No.15543316

jesus christ this thread is giving me a tumor.
anyone who agrees with OP are the real idiots who just can't accept reality.

>> No.15543319

>>15543315
It's 1/2 after Monty removes a non-winning door from play and asks you to switch or stay, as you only have 2 doors to choose from now and only 1 of them has a prize.

>> No.15543325

>>15543319
t. Reality denier

>> No.15543330

>>15543325
The reality is that you were only ever going to choose between two doors. The first round is pure misdirection and pageantry. There's a 100% chance that Monty will show you a goat door and a 100% chance that you'll be able to select between 2 available doors for the only choice that actually matters, so the probability of winning with EITHER door is 1/2 because the third door is simply not in play and never was going to be.

>> No.15543349

>>15543330
The third door is right there and you make a choice based upon there being 1 car and two goats. You are simply denying reality.

>> No.15543354

>>15543330
yeah you don't "choose" the door until the round starts

>> No.15543356

>>15543349
A goat could be standing next to you on stage and it wouldn't impact the final choice between 2 doors.

>> No.15543363

>>15543356
Ok, do it with 1000 doors.
Pick a door. He opens the other 998 goat doors. now it's a 50% chance your door contains the car. Right? JUST STICK WITH IT, RIGHT????

>> No.15543364

>>15543349
>The third door is right there and you make a choice based upon there being 1 car and two goats.
In the first round, yes. At that point the probability of picking the door with the car is 1/3, or so you think. The second round changes the premise by revealing one of the goats, which removes that door from play. You are then asked to choose a door a second time, but this time there are only 2 doors to choose from and 1 of them has the car. The probability is now 1/2 because the number of valid available choices is 2 and the number of winning options is 1.

In truth, the host knew all along that you would eventually be reduced to picking between only 2 doors, so your probability was ALWAYS 1/2 but you simply weren't aware at the start that one door would be rendered an invalid choice. But again, that happens every time with 100% certainty and once you know that you realize that only the choice in round 2 matters in any way and choosing between 2 doors gives you an equal 50/50 shot at winning.

>> No.15543366

>>15543356
Naturally, since the choice is between three doors two of which have a goat behind them and another a car. The goat stood next to me is nothing to do with the choice, which is and will always be 1/3.

>> No.15543368

>>15543363
>JUST STICK WITH IT, RIGHT????
It doesn't matter if you stick or change, the probability is still 50%, you are choosing between two equally likely options. 1 car / 2 doors = 1/2 = 50%
All else is just psychological trickery.

>> No.15543369

>>15543363
No, because the choice was 1/1000, and when you remove 998 doors the choice is between two doors each of which have a 1/1000 chance of having the car behind it, since the other 998 doors don't disappear from reality.

>> No.15543371

>>15543366
No, if one goat was standing next to you there would only be 2 doors. Are you stupid? The door that Monty opens is an invalid option in all scenarios every time, invalid options cannot affect your probability. You are only actually choosing between 2 doors in the final round which is the only round that affects which door you actually open.

>> No.15543372

>>15543369
Your door was the 1/1000 set. Then he put the remaining doors through a donkey filter. His remaining set is filtered and has a different probability.

>> No.15543373

>>15543369
>the other 998 doors don't disappear from reality
They disappear from the equation as they are invalid answers. They were always red herrings and you were always going to be given a choice between only 2 doors in the end. You have to adjust the probability to reflect the new information.

>> No.15543374

>>15543371
>The host has 3 doors, two of which have a goat behind them and one has a car
>You, looking at these 3 options are stood next to a goat (this goat is unrelated to the two behind the doors)
Reality deniers can't even understand that in their newly concocted scenario where a goat is next to the contestant there are now three goats total, only two of which are related to the door problem.

>> No.15543378

>>15543363
You do it with 1000 doors and monty opens only 1 goat door
Its never implied that he opens the rest of the doors

>> No.15543382

>>15543374
You are standing in front of 3 doors. One door has a big red X on it and the host tells you that you're not allowed to choose that door under any circumstances and also assures you under sworn oath that the prize cannot be behind it. You are asked to choose a door, what is your probability of choosing the door with the prize?

>> No.15543384

>>15543382
I choose the door with the X

>> No.15543386

>>15543382
>believing what the host tells you when they have an interest in making you choose a door that doesn't have an expensive car behind it

>> No.15543389

>>15543386
If he lied, you can sue him.

>> No.15543391

>>15543382
You choice would be a 50/50 between two doors which each have a 1/3 chance of having a prize behind it, since there are three doors total

>> No.15543396

>>15543384
You are not allowed, select one of the doors without the X. All attempts at selecting the X door will be ignored and you will be prompted to select again. What is your probability of choosing correctly?

>> No.15543398

>>15543391
Yup, you're just stupid. Glad I could confirm.

>> No.15543399

>>15543396
I keep choosing the X until Monty gets angry and kicks me off the show for refusing to go along with his charade that the odds are 1/2 and not 1/3

>> No.15543402

>>15543398
>thinks 3=2
>calls other people stupid

>> No.15543405

For god's sake retards it's not 50%. Let me make it crystal clear for all of you.
When you first pick 1 out of 3 doors, the odds of picking the car is 1/3 while goat is 2/3 meaning it's more likely to pick a fucking goat. Got that retards?
Now suppose after you pick a door, the host makes it a lot easier for you and puts a poster saying 2/3 for goat doors and 1/3 for the car door. Then he asks you if you wanna switch or stay.
You look at your door and you see the poster say 2/3 because that's what would happen you unlucky bitch. The host opens the other 2/3 goat door and the only choice you see is the 1/3 car door so unless you're disabled, you switch.
End of lecture. Basically, the two choices u make are not independent. In other words, you're first choice affects the second choice retards.

>> No.15543411

>>15543405
If my door says 2/3 chance goat, then it is 1/3 chance car, same as the other door that says 1/3 chance car.

>> No.15543417

Picking at the beginning is 1/3 chance
Picking after the reveal is 1/2 chance
This much is certain.
So if you swap doors 1/2 of the time, essentially picking after the reveal, this should improve your chances to 1/2.

If you stick with your option all of the time, it's the same as picking before so it is 1/3. Since there are 2 choices (stay and swap), the other choice is 2/3.
So changing 1/2 of the time should have a probability of 1/2*1/3 + 1/2*2/3 = 1/2, which is consistent with what I said in the 1st part.

>> No.15543439

>>15543405
>the two choices u make are not independent
They are in every way that matters.
>you're first choice affects the second choice retards
It doesn't in any way that matters.

1 door and 1 goat have been REMOVED from the available choices and you are asked a second time to choose a door. Between the 2 doors remaining there is 1 car, you have a 1/2 chance of guessing right. If you account for a known door which is no longer an available option you're literally rigging the statistics using faulty information. The scenario has changed, the dataset must be updated.

>> No.15543446
File: 340 KB, 1169x1463, 5CC31DF5-EA82-4720-BBE3-EE50405B18FF.jpg [View same] [iqdb] [saucenao] [google]
15543446

>>15543405
You are wrong.
> the two choices u make are not independent
This is the gamblers fallacy, this is only true *before* you make the first choice, after the first choice, it has zero baring,
There are two choices, it is 50%
It’s 50%.

>> No.15543466
File: 37 KB, 539x488, moooontyyyyyy.jpg [View same] [iqdb] [saucenao] [google]
15543466

What would happen in 4 door monty ?

When you pick and host opens goat door, how do the chances go again ?

And how do you know which one would you choose ?

>> No.15543470

>>15543466
>No long math jargon
>Show me in fractions

Refute this and its proved forever

>> No.15543489
File: 8 KB, 649x359, monty_P.png [View same] [iqdb] [saucenao] [google]
15543489

>>15543466

>> No.15543494
File: 99 KB, 500x463, 8FB6EA21-279E-4134-8DE1-6E748B754670.png [View same] [iqdb] [saucenao] [google]
15543494

There are actually 18 outcomes if you want it to be a two step problem.
Becuase order matters, it’s not like throwing dice odds, where dice can be then put in any order
A,b,c. With either a,b,c being the winner.
If a is the winner and you picked a and switch you lose
If a is the winner and you picked a and stay you win
If a is the winner and you picked b and switch you win
If a is the winner and you picked b and stay you lose.
If a is the winner and you picked c and switch you win.
If a is the winner and you picked c and stay you lose.

If b is the winner and you picked a and switch you win.
If b is the winner and you picked a and stay you lose
If b is the winner and you picked b and switched you lose.
If b is the winner and you picked b and stayed you win
If b is the winner and you picked c and switched you win.
If b is the winner and you picked c and stayed you lose.

If c is the winner and you picked a and switch you win.
If c is the winner and you picked a and stay you lose
If c is the winner and you picked b and switched you win.
If c is the winner and you picked b and stayed you lose
If c is the winner and you picked c and switched you lose.
If c is the winner and you picked c and stayed you win
9 outcomes where you win, 9 outcomes where you lose.
9 outcomes where you switch and 9 outcomes where you stay.
3 outcomes where you stayed and won
6 outcomes where you stayed and lost
3 outcomes where you switch and lose
6 out comes where you switch and win
Staying is 3/9 win
Switching is 6/9 win

>> No.15543496

>>15543015
>>15543063

Based.

>> No.15543497

>>15542721
Just write a Python program nigga

>> No.15543501

If someone doesn’t understand it at first, I like to say to them, “imagine you have 1000 doors, there’s a .1% chance your right first time, the host opens EVERY door that you haven’t chosen except for the one with the prize, do you swap.

>> No.15543502

>>15543501
It’s still 50/50 once it resolves to the second choice, there are two doors left.

>> No.15543503

>>15542721

Real ones can suspect that the shilling for an answer that is NOT 1/2 somehow relates the occultism of money making.

>> No.15543506

>>15543501
>the host opens EVERY door that you haven’t chosen
Not true, could be just 1

>> No.15543510

>>15543502
No it's not schizo. The odds of you getting that door right is 1/1000. So when it comes down to two doors the odds are not 50/50. It's 1/1000 and 999/1000. If you still don't get it, you are beyond help.

>> No.15543512

>>15543506
still pays off to switch

>> No.15543514

>>15543510
The odds WERE 1/1000
There is new information, now the odds are 1/2. It’s very very simple.
Your past choice does not effect your new choice in any way. There are two doors. You can choose door 1 or 2

>> No.15543515

>>15543512
show us >>15543466

>> No.15543531

These probability meme threads are pointless because any apparent retard is indistinguishable from a shitposting troll.

>> No.15543563

>>15543515
you select a
a:.25 b:.25 c:.25 d:.25

monty opens b & c
a:.25 b:0 c:0 d:.75

by switching from a to d you triple your odds

>> No.15543572

>>15543563
Thats not the question, monty opens B. not B and C

>> No.15543576

>>15542799
So you could do anything with your free will and you choose to live in a universe where you are subject to physical laws and god doesn't exist?

>> No.15543579
File: 159 KB, 713x469, 1688554492473.jpg [View same] [iqdb] [saucenao] [google]
15543579

>>15543024

>> No.15543581

>>15543572
you select a
a:.25 b:.25 c:.25 d:.25

monty opens b
>a:.25 b:0 c:.375 d:.375

by switching from a to c or a to d you x1.5 your odds

>> No.15543582

>>15542844
>10,000,000 dors
>you open 1
>monty opens 9,999,999
>u should chose the other dor
have a look at this retard giving intelligence advice wew lads

>> No.15543586

>>15543581
Show it in fractions how the 1/4 gets distributed

>> No.15543590

>>15543586
.25/1=.25
.75/2=.375

>> No.15543595

>>15543590
...

>> No.15543597

>>15543045
OP you are a raging faggot
suppose monty doesn't open a goat door, but instead says, "LAWDY ME! DERR BE A FOURTH DOR HER I NEVER DUN SEE BEFOR!!"
according to your jewpsych maths, then the door I've chosen has a 1/3 chance of being correct, two other doors also have a 1/3 chance of being correct and the new fourth door has a 1/4 chance.
but you are so retarded you might say my door is 1/3 and the others are all 1/4.
just depends what your kiked brain thinks

>> No.15543599
File: 10 KB, 676x523, montyhallsol.png [View same] [iqdb] [saucenao] [google]
15543599

here is the solution for anyone whos still confused

>> No.15543604

>>15543597
xD

>> No.15543612

>>15543494
retard
monty is going to open a goat door
i get to chose a door
read above as: i get to choose to open two doors, knowing that one of them will 100% be a goat door.
therefore, from the very start, it is a 50% one of my two doors will be the car door.

>> No.15543616

>>15543599
wrong. because you don't have a choice to open either 2 or 3 if you chose 1. You only have a choice to open whichever of 2 or 3 that monty didn't open YOU INSUFFERABLE FAGGOT

>> No.15543669

>>15542804
>>15542836
Some people get so mad about maths lol

OP is right, they come out of the woodwork every time to try to argue why the real world doesn't follow maths

>> No.15543674

>>15543196
Knowing the precise odds would be equivalent to seeing through a stage magician's tricks. Being fooled by your wrong assumptions like you would be thinking real magic is being performed, or at least not being able to tell how the magician pulls off his illusion.

>> No.15543679

>>15543364
>At that point the probability of picking the door with the car is 1/3, or so you think.
>or so you think
lol

The probability does not retroactively change, idiot.
>you simply weren't aware at the start that one door would be rendered an invalid choice
Of course you were aware, it's a premise of our fictional game show. But it wasn't always invalid. And in fact, had you chosen it at first, then necessarily, another door would've been rendered an invalid choice instead. Perhaps even the door you ended up picking this time. All of them were valid choices.

>> No.15543683

>>15543382
This is a different scenario, for the reason that the door with the red X was never an option to you and that the choice to eliminate it beforehand was never constrained by any choice you made.

>> No.15543686

>>15543446
>This is the gamblers fallacy, this is only true *before* you make the first choice
M8 the first choice is independent, your second choice is not, for the reason that Monty's choice is also dependent on your first choice

>> No.15543694

What makes it especially fun though is that if you are convinced the odds are 50-50 and therefore decide to let your choice of whether to stay or switch depend on a coin flip, your odds of getting the car will actually work out to 1/2.

>> No.15543696

Imagine there are 100 doors

You pick one, monty opens one
You swap to another door
Monty opens one door again
You swap to another door
You keep doing this untill there is only 2 doors left

What are the chances for the last 2 doors

>> No.15543708

>>15543466
The chances are always [number of prizes]/[available doors]. In this case, the chances are 1/3 once the number of available doors has been reduced to 3.

>> No.15543714

>>15543708
I think so too, the denominator is directly related to the doors
Not that it suddenly becomes 2/8, 3/8 and 3/8
That sounds just way too retarded

>> No.15543715

>>15543669
The real world does follow math, OP's math is not in line with reality because it doesn't account for the change in information and available choices. The removed door cannot influence the outcome of the remaining doors.

>>15543674
The idea that a coin flip could have a 2/3 chance of heads and a 1/3 chance of tails just because some third option was temporarily made available before being taken away prior to the coin flip is precisely analogous to being so enamored by a magician's misdirection and sleight of hand that you think reality is actually bending and changing through magic. Nothing that happens prior to the final choice between 2 doors has any bearing on the probability that there's 1 prize and 2 doors = 1/2 = 50% chance of winning no matter which door you choose.

>>15543683
It's no different from the second round, in which there are 2 closed doors to choose between and a third open door with a goat sticking out of it that you're not allowed to choose.

>> No.15543716

>>15542721
Doesn't matter, all choice is based on luck and is ultimately unpredictable unless initial conditions are always the same

>> No.15543717

>>15543696
Assuming there's still only 1 available prize, it's 1/2 in the final decision.

>> No.15543720

>>15542764
>>15542775
>I was smart
>The host knowing is why switching DOES make a difference.
This is embarrassing and you have misled all of these other posters here as well who have been arguing for the wrong thing this entire time. Defending your position.
The host knowing makes no difference at all. The host could be a bumbling retard or a toddler. Or a robot even. The host could know nothing at all. It doesn't effect the answer at all. The only thing that matters is that the host not open the door with the car, knowingly or unknowingly, because that would just render the question null and void and end the game. If the host does open the door with the car, that doesn't change the answer, it simply nullifies the question so that scenario doesn't matter at all. We can scrap it from our thoughts altogether.
The odds of you picking the car on your first choice will always be 1/3 no matter what the host knows, and hence the odds of the unchosen door will be 2/3. It doesn't even matter if there is a host at all. The odds remain the same nonetheless. The doors don't change what is behind them based on the host's decision. The host serves an almost cosmetic purpose in this problem. You can safely ignore his knowledge of anything.

>> No.15543728

>>15543715
>The removed door cannot influence the outcome of the remaining doors.
You are not entirely wrong. The location of the prizes was already determined. But Monty's choice was not. If you picked a goat door at first, then Monty's choice is constrained - he HAS to open the other goat door. You can't change where the car is, but you can (albeit unwittingly) change whether or not Monty has to open a specific door. And therefore you are also wrong to say
>Nothing that happens prior to the final choice between 2 doors has any bearing on the probability that there's 1 prize and 2 doors = 1/2 = 50% chance of winning no matter which door you choose.
And
>It's no different from the second round
It is, because in the original scenario, you influenced the setup of the second round.

>> No.15543729

3 doors, denominator 3
2 doors, denominator still 3

explain

>> No.15543731

>>15543597
>then the door I've chosen has a 1/3 chance of being correct, two other doors also have a 1/3 chance of being correct and the new fourth door has a 1/4 chance.
The math doesn't add up and you're a retard. Since you've added another variable here and not taken anything away from when you made the original choice, the odds of all the doors will be the same.
Let me simplify it for you. In the original scenario, you have only 1/3 odds of having the car behind the door you chose (and hence 2/3 of the unchosen door) because you have 3 possible outcomes with equal likelihood.
The car could either have been behind door 1
It could have been behind door 2
Or it could have been behind door 3

In your new scenario, all doors will have 1/4 odds because there's 4 possible outcomes with equal likelihood.
The car is behind door 1
It is behind door 2
It is behind door 3
Or it is behind none of them (Which means it is behind door 4, which at the time of your decision making, though unrevealed, still exists and still equally likely to have a car)
The odds of your door having 1/4 chance of having a car won't change here no matter what you do after (unless you add more doors, which just defies the premise of the scenario and adds another possible, equally likely outcome), just like the 1/3 odds won't change in the original example even if you open one of the doors.

>> No.15543732

>>15543720
Supposing that, firstly, the host could unwittingly open the car door when choosing completely at random, and secondly, the game does then continue and offer you a choice between the two remaining closed doors, you must account for those times when the game is 100% unwinnable at the second stage, which will naturally skew the odds. Granted, it is not a mystery at that point whether or not you still have a shot at the car, but still. The host knowing matters, because that means he has to indicate to you where the car is 2/3 of the time. If the open door randomly happens to be a goat to the host's own surprise, it means nothing.

>> No.15543734

>>15543729
Just because there are two possibilities that doesn't mean they're equally likely outcomes.

More specifically to the point, there are still three doors in the end. One of them is revealed to have 0/3 probability of having the car.

>> No.15543737

>>15543734
There is 2 doors though, the 1 door is out of the function all together
Tell me how the denominator can be 3 when there is only 2 doors in the function

>> No.15543744

>>15543737
>the 1 door is out of the function all together
No it's not. It's right there. There's a goat right on the other side. You can see it.

>> No.15543748

>>15543744
How is it still in the function when the choice is made inbetween 2 doors
Surely you cannot pick the already opened door again ?

>> No.15543752

>>15543732
If the host opens the door with the car, it doesn't need accounting for. Because, as you said, the game is 100% unwinnable. That means that your odds of winning are 0. At that point, it doesn't effect anything else, because a probability of zero doesn't matter.
So when the host unknowingly opens a random door, 3 things can follow.
The host opens a goat, and you choose to remain with your door. You have 1/3 odds of winning here.
The host opens a goat, and you choose to switch the doors. You have 2/3 odds of winning here.
The host opens a car, and you instantly lose the game. You have 0/3 odds of winning here.
The third scenario has 0 odds of winning so you can safely ignore it. It doesn't effect your odds of winning/losing if you choose to switch the door or not at all. The first two scenarios happening here are entirely independent of the third. So it is best to just ignore it.
The host doesn't indicate anything. The odds of the unchosen doors containing the door are always 2/3 . The host opening a goat door, knowingly or unknowingly, simply narrows the 2/3 odds of the unchosen doors having the car to a single unchosen door having 2/3 odds of having the car (Instead of both the unchosen doors having 1/3 odds each, adding up to 2/3 total). The odds of unchosen door(s) will always be 2/3 by virtue of the odds of the original choice always being 1/3, unchangingly.

>> No.15543754

>>15542721
no, this "problem" was invented by circlejerking narcissitic mathematicians like you

>> No.15543762

>>15543748
To determine the probability, you have to look at all the doors, not just the ones you choose between.

>> No.15543764

>>15543754
OP is a retard who doesn't understand simple probability. And you anon is a retard for even calling him a mathematician.

>> No.15543769

>>15543752
Surely you can't mean to say that your odds of winning remain the same if 1/3 of all games played are inherently unwinnable, vs a game that is always winnable?

>> No.15543783

>>15543176
Scenarios wont take sides, they are by default equal

>> No.15543784

>>15543769
No your total odds of winning the game are reduced. But the odds of winning with original premise in mind stay the same. That is to say, your odds of winning stay the same if Monty opens the door to a goat. If he doesn't do that, it is simply irrelevant to the original question.

>> No.15543788

>>15542721
The problem with it is that the second choice is not directly related to the first one. It is not the same setup as before, since new information was given.
So, conditional probability does not apply.

Not changing your door is a new choice, on the same level as changing it.

You made a choice with 1/3 probability of car.
Now you're making a new choice with 1/2 probability of car.
And picking the same door again is a choice you make concerning the second probability. It has nothing to do with the first one.


The second problem with all these math tasks is the gamblers fallacy.
In one-off happenings, the probability does not really help you. It will only become predictable over a (really) large set of repetitions. But if you get the car is just random either way.

>> No.15543795

>>15543788
>new information was given.
>So, conditional probability does not apply.
lol
lmao

>> No.15543797

>>15543728
The first round has no influence over the second round except in removing an option from the table. You now have 2 doors to choose from, the odds are 1/2. This whole "the odds of switching are 2/3" thing is just an artifact of bad data management that doesn't properly redefine the parameters of the question to match the available choices and outcomes.

Here's another way to look at it: There are 2 unopened doors, so why does the extra 1/3 probability apply solely to the un-selected door instead of your door? Both doors are equally blind, there's no reason to favor one over the other. That's why the extra probability "spreads out" between the two and becomes 50/50, or in other words we recalculate the probability based on the new information that tells us the third door is simply no longer a valid option.

>> No.15543799

>>15543788
>In one-off happenings, the probability does not really help you.
Ironically, people like you are how lotteries stay in business. "Sure, probabilistically speaking, I got a one in a million chance, but hey, that's just if you play a million times! You either win or you don't, baby!"

>> No.15543804

>>15543797
>Here's another way to look at it: There are 2 unopened doors, so why does the extra 1/3 probability apply solely to the un-selected door instead of your door?
Yeah if you understood why you're wrong you'd obviously also understand why this is wrong lol. Monty Hall works precisely because there IS a reason to favour the door you didn't pick.

>> No.15543816
File: 780 KB, 1080x2185, 1682512397729.jpg [View same] [iqdb] [saucenao] [google]
15543816

>>15543795
See wikipedia

>>15543799
On the contrary. Most people believe if the play often enough they'll win eventually.
No one hopes he'll win the lottery when he only buys one ticket. And casinos don't make money from people that do it one time just for chance. They make money from people that think the HAVE TO win eventually.

>> No.15543822

>>15543816
>See wikipedia
Uhuh. Keep reading. You might learn something.
>No one hopes he'll win the lottery when he only buys one ticket.
You're right, of course. You have to buy two. It's 50-50. Duh. Like Monty Hall!

>> No.15543824

>>15542721
I made a math major so angry with this once. I already knew about it, and he was trying (poorly) to explain it to a dorm room full of people. At one point, he gave up explaining and just pulled up a simulator online to show the problem in action.
We ran it three times. I kept picking door 2 and sticking with it, and I happened to be right all three times. I said something like "statistics mean nothing to the individual" and he left.

>> No.15543836

>>15543822
>You're right, of course. You have to buy two. It's 50-50. Duh. Like Monty Hall!

How nice of you to demonstrate that you have room temperature iq, if you think 2 times 50% chance makes 100%.

>> No.15543837

>>15543836
I know sarcasm doesn't carry well through written text, but at some point you really have to consider the intellectual capacity of the receiving party

>> No.15543843
File: 53 KB, 900x916, Screenshot_20230705-112750_Samsung Notes.jpg [View same] [iqdb] [saucenao] [google]
15543843

>>15543797
>There are 2 unopened doors, so why does the extra 1/3 probability apply solely to the un-selected door instead of your door?
>That's why the extra probability "spreads out" between the two and becomes 50/50
Because the probability of the one unopened door wasn't 1/3 to begin with. The 1/3 probability is for the door you chose. That means that your door has 2/3 probability of being wrong. This is spread equally among the two unchosen doors, giving them each 1/3 probability of correctness initially. Once you open one of the doors, the probability of your door being wrong is still 2/3, so the 2/3 probability applies only to the sole unopened door now.
Basically, think about the unchosen doors as a collective instead of independent entities. The correctness of the unopened doors relies on the incorrectness of your selected door. And the probability of your initial choice being incorrect 2/3 times doesn't change. If it was incorrect to begin with (which it will be 2/3 of the times, because it's a random guess out of 3 choices), Monty opening one of the doors won't make it correct. In fact, nothing Monty or anybody does will make the incorrect choice correct. And since the incorrectness of your door doesn't change by Monty opening one of the doors, the unopened doors will continue to have 2/3 odds of correctness, no matter how many doors you open.
Even if you had 10 doors initially it doesn't change this fact, your choice would have 1/10 odds of correctness. because it would be incorrect 9/10 times since it's based on a random guess out of 10 options. That would mean that no matter how many other doors you open or close, the unchosen doors will be correct 9/10 times. If you had seven doors, five doors or three doors left, you would simply distribute the 9/10 odds of correctness evenly among them (since they are equally likely). And if you choose to open all 8 unchosen doors, the last remaining unchosen door will have 9/10 odds all by itself.

>> No.15543846

>>15543824
>and even Einstein clapped
sure bud, definitely happened

>> No.15543848

>>15543837
What do you expect me to consider, after great remarks like "UhUh."?
If you're shitposting, so will I

>> No.15543850
File: 4 KB, 478x211, monty.png [View same] [iqdb] [saucenao] [google]
15543850

>>15543788

>> No.15543851

>>15543846
It did. What exactly is unbelievable about it to you?

>> No.15543852

>>15543843
>Because the probability of the one unopened door wasn't 1/3 to begin with. The 1/3 probability is for the door you chose.
This falsely assumes that choosing the door metaphysically binds a certain probability to it before you open it. In reality, the door is only tentatively selected and you get a second chance to choose between that door and a second door. The odds don't care about what you plan to do or what you've tentatively declared, you're doing math using outdated irrelevant information because of some perception that you've "locked in" a probability by declaring a future intention under different circumstances. How do you not realize how stupid this sounds?

>> No.15543853

>>15543848
What did you expect me to tell you after you cherry-picked one paragraph from a wikipedia article because you felt the mention of "criticism" was meaningful, even though the criticism offered does not deny the solution of 2/3 explained in various ways throughout the entire article, including a whole section on conditional probability after the criticism section? You don't seem to be inclined to read much, so "uhuh" will have to do. But you are so averse to admitting you're wrong you're now even pretending to have understood that I was being sarcastic all along.

>> No.15543854

>>15543851
save it for your Canadian gf

>> No.15543855

>>15543852
OPs point: proven. The midwit brain just physically can not understand probability.

>> No.15543860

>>15543855
Don't mistake my disagreement with a lack of understanding. The only midwit is the person who sees 2 doors with 1 prize and says, "Actually, the probability must be out of 3 because in the past there were 3 doors to choose from!" It's just faulty logic using known invalid data points to cook up an incorrect probability that only "feels" right because of the perception of that declared intention from the past when the entire dataset was different.

>> No.15543868

>>15543860
>Don't mistake my disagreement with a lack of understanding.
That does not stem from your disagreement, your lack of understanding stands on its own

>> No.15543869

>>15543855
You probably "hehe" dont understand that i have been 66% for some 12 years already and switched back to 50-50

>> No.15543874

>>15543869
>this is your brain on drugs
filtered & scrambled

>> No.15543879

>>15543860
Do you have an autistic spectrum disorder? I'm asking seriously.

>> No.15543884

>>15543879
None I've ever been diagnosed with.

>> No.15543904

>>15543869
You never truly understood the problem, even when you were 66% for 12 years. Just look around in the thread. Most posters who are arguing for the 66% position don't REALLY understand it either, only some do. And it's painfully obvious who they are if you actually understand the reasoning behind the problem. They're just doing it to Larp as if they understand to seem like they passed the IQ filter. You were like them as well. You are at least better since you realized your inadequacy and stopped Larping. So you have some self recognition and executive functioning, which will do you good in life as opposed to those who are under the delusion they understand it, but don't really.

>> No.15543915

>>15543904
I never did Larp, i just found it reasonable.. im always willing to change my views (even now)
I was prolly the first who tried to show others it with the million door analogy but thats the place where i understood that it matters if the host opens rest of the doors or just 1

>> No.15543924

>>15542721
You always switch because your odds increase from 1/3 to 1/2, but you look like a retard when you switch into being wrong. I seriously doubt midwits are filtered by this.

>> No.15543925

>>15542742
>I think this says it all. The syntax is resolved by the original problem.
The sort of people who deliberately obfuscate story problems so they can yell AHA THE DOCTOR WAS THE CHILD'S MOTHER I AM SMARTER THAN YUO are themselves intellectually insecure and not genuinely interested in teaching critical thinking or problem solving

>> No.15543928

>>15543852
>>15543860
You midwits simply ignored these. This has been explained a million times itt, by a multitude of anons yet you ignore all of it >>15543843 >>15543850 >>15543494 >>15543405 >>15543092 >>15543045 >>15543024 >>15543489 >>15543501

Just some examples, there's probably more itt if you read. Read these fully, and look at the images. If you still don't get it, I would suggest give up. Either you are trolling or you are simply not smart enough to get it. Nobody here can force you to understand this, just like you can't force a retarded middle schooler to understand quantum physics. This has been explained in every way possible to you just itt, and you still don't get it. There's nothing more anybody else can say that won't be a regurgitation of that which has already been said. Just read the older posts. You're a retard and you should stop trying. You failed the IQ filter. You have been filtered.

>> No.15543932

>>15543095
>they wrote a program that assigned a 2/3 probability of winning to the un-chosen door (since that's how programs work)
Software developer here, coding simulations based off of our own expectations is contrary to the purpose of coding simulations
Show the source code for this particular sim or concede the point: any attempt to argue without providing the code is de facto a concession

>> No.15543939

>>15543915
>I was prolly the first who tried to show others it with the million door analogy
Hubris

>> No.15543941

>>15543915
That is appreciable that you are willing to change your views. Stubbornness is a killer of inquiry.
But, what are you talking about? what rest of the doors? the host only opens 1 door.
Honestly, forget the host even. You don't need him. Just that once you make a choice, one of the unchosen Goat doors opens automatically. By an AI or God or whatever. The statistics still don't change. The will, knowledge or the actions of the host don't matter, he is there for ease of understanding the problem and you let him confuse you.

>> No.15543958

>>15543928
ironic

>> No.15543961

>>15543941
>Stubbornness is a killer of inquiry.
Exactly, cannot even for the fun of it do anything else than whats been told
>what rest of the doors?
Dont worry about it, it was just when i explained it that the host opens 999 998 doors (remaining) when i figured that the whole system turns upside down if the premise was that host only opens 1 door
Then i experimented it with 4 doors, nothing special

>> No.15543965

>>15542721
Misdirection is useful for artistic purposes like stage magic or storytelling. For story problems it only serves the purpose of telling the reader you're smarter than they are which is debatable.
>This problem should be used as an offhand IQ test for schools and jobs.
The entire reason people fall for this trick question is specifically BECAUSE in real life scenarios, probability is about as relevant as calculus – unless you consider gambling a career instead of a game in which case you're already a mathematician of sorts. The question about coin flips (coin lands on heads 99 times, what's the probability of it landing heads on the 100th) is honestly about as far as most people outside mathematics or computer science need to be educated.
>This problem should be used as an offhand IQ test for schools and jobs.
Name five real-world situations unrelated to gambling where this particular question matters. Five. Abstract arguments about the value of critical thinking will be ignored.

>> No.15543970
File: 23 KB, 800x800, 800px-Raven_Matrix.svg[1].png [View same] [iqdb] [saucenao] [google]
15543970

>>15543965
What's the real-life practical applicability of pic related?

>> No.15543980

>>15543915
>it matters if the host opens rest of the doors or just 1
the odds are different, but in either case it still is better to switch

>> No.15543989

>>15543928
Your answer is wrong and the logic is bad no matter how many different ways you try to rationalize it.

>> No.15543990

>>15542812
No you’re missing the point. at first select it is 1/3 chance you pick the car. Then the host opens a goat door and offers the switch. Now if you choose to switch you have a 2/3 chance of getting the car because it’s either you picked the car at first (1/3) or you didn’t and switched which must be the opposite of that value or (2/3).

>> No.15543997

>>15543990
You're focusing too much on the idea of "switching," which means literally nothing because your first choice has been invalidated by new information. The question would be more accurately posed as, "Would you like to select the same door again or a different door?" You are making a new choice between 2 doors; the 3rd door has been removed from the equation entirely. The probability of winning is 1/2.

>> No.15544014

>>15543965
This is not misdirection. Nobody is misleading you. It is a simplistic math question that filters out everybody with an IQ below a certain level (not sure what that level is, but OP can pitch in. I would guess somewhere 115-125), as is obvious from this thread. Just like many other subjects; in college, for instance, intoductory classes like calc serve to filter out dimwits as well, even if they are humanities majors they still have to take it. Because the inability to pass means something, even though it's not relevant to their fields.
It isn't that it's relevant irl. It's that you should be able to figure it out. And if not, you should be able to at least understand it when explained to you. If you are unable to do so, it simply indicates something about your intelligence.
This question doesn't matter much in real life, you're right. Much like any standardized IQ test. Much like the SATs. or anything else that is used to measure an IQ. It has no other purpose than to filter out dimwits. If you say it is useless simply because it has no other purpose but IQ filtering, i say it is just as useless as any other IQ filter, so scrap the SATs, scrap the MCATs & LSATs, scrap elite colleges, scrap everything that are implicit IQ filters.

>> No.15544023

>>15543997
>your first choice has been invalidated by new information
How has it been invalidated? If you selected the goat, does the new information suddenly make your selection the car? Or does the new information not dictate whether or not your initial selection was the car or the goat?
Think about it.

>> No.15544024

>>15543852
>This falsely assumes that choosing the door metaphysically binds a certain probability to it before you open it.
Holy fucking shit.

>> No.15544029

>>15543961
>system turns upside down if the premise was that host only opens 1 door
Not really it doesn't. >>15543843

>> No.15544048

>>15542849
It's not a mathematical quirk. You can easily simulate it or play it as a game with your friend if you have one.

>> No.15544052

>>15544014
OP's question is an example of how math can give false answers that seem correct if you abuse it or misunderstand it. Anyone who blindly follows the 2/3 answer because the numbers technically work on paper will probably be unable to discern other forms of misleading statistics because they're more obsessed with following the numbers than understanding how those numbers map to reality.

Here's an example:
Given this set of variables:
A = 1
B = 1
C = 1

Solve this equation:
A + B + C = ?
1 + 1 + 1 = ?
1 + 1 + 1 = 3
A + B + C = 3

Now remove A from the equation, what does it equal?
"2/3" midwits will insist that A's value can't disappear, so it must be added to B. B now equals 2, so B + C = 3
"1/2" chads will point out that A's value is gone and B and C have not been altered, so B + C = 2

THE DOOR WHICH IS REMOVED FROM THE EQUATION CANNOT INFLUENCE THE RESULTS IN REALITY
INCLUDING THE REMOVED DOOR'S VALUE IN THE FINAL EQUATION IS INCORRECT

>> No.15544060

>>15544023
It changes nothing other than how many options you have, and the car is still not revealed. You recalculate the probability with the new information and it is 1/2 because the third door is an invalid option.

>>15544048
I can simulate anything. Garbage in, garbage out. The 2/3 switch is based on faulty principles.

>> No.15544062

>>15542857
It's actually Monty opening the door that is a red herring. You always knew that there's at least 1 goat behind the other two.
It's a red herring to distract you from the fact that you aren't actually choosing between 2 doors. The choice is between 1 door (the one you originally chose) or both of the other doors. You know that 1 of the 2 doors is a goat. And you know that 2 doors are twice as likely to contain the prize than 1 door so switching is always advantageous.

>> No.15544069

>>15544052
>Sure the maths show I'm wrong but that's not REALITY
cope

>> No.15544070

>>15543015
It does matter because it means that the host could choose the car door and make it an invalid game. If you did that and not count the invalid games the probability does turn to 1/2 instead of 2/3 for switching (because choosing a goat door in the beginning has 50% probability of ending the game prematurely). The host knowing or not changes the game and its rules.

>> No.15544073

>>15544060
>I can simulate anything
Then do it.

>> No.15544081

>>15544029
Play 4 door monty
Pick A
Host B
You change to C
Host D

The B distribution goes to C and D and that makes them 3/8 and 3/8, A is 2/8 because you picked it
The D distribution goes to only A making it 5/8 and leaving C with 3/8
Equal steps and movement, 4 doors, 2 opened and 2 movements

>> No.15544084

>>15542721
The easiest way to conceptualize it is that the doors unopened and not selected retain their whole probability.

You select 1 door with 1/3 prob. The other two doors have 1/3 each, so 2/3 total. You eliminate 1 door so now the remaining has the whole 2/3.

With 4 doors, when you eliminate one the other two have 3/8 each.

>> No.15544085

>>15544081
So its 1/2 1/2 and not 5/8 3/8

>> No.15544090

>>15543439
if you have a 1/2 chance of being right then there is no benefit to switching since each door has a 50/50 chance which means regardless of what door you pick there is no difference. the only way you can then cope that switching does make a difference is by invoking the previous gamestate which means that the previous choice does infact impact the game

>> No.15544106

>>15544060
>I can simulate anything. Garbage in, garbage out.
You could have played it with a friend as I suggested. Since you didn't even consider such an option that I proposed I can only assumed it means you don't have a friend to play it with.

>> No.15544111

>>15544052
It doesn't just work on paper, it works in reality as well. This >>15543092 is a simulation of what it would actually look like if you removed one of the goat doors. And it's with colorful emojis as well!
You are simply simulating it incorrectly.

>> No.15544123

>>15543092
It really is this simple isn't it? You pick a goat, you get the car. You pick the car, you get a goat. How many cars are there and how many goats are there?

>> No.15544165

>>15544123
1 goat and 1 car

>> No.15544180

>>15544165
Wow, you get the most basic of facts about the Monty Hall problem wrong, there is no hope for you

>> No.15544199
File: 7 KB, 298x498, Capture.png [View same] [iqdb] [saucenao] [google]
15544199

>>15544069
I just went to random.org and generated 100 random numbers from 0-1 and pic related was the result.
1 came up 53 times out of 100
0 came up 47 times out of 100
Therefore the probability of getting a 1 when randomly choosing between 0 and 1 is 53%. The math works out, so that means I'm right in reality about the overall probability as well. The probability of getting 0 or 1 is not 50/50, it is in fact 47/53 and if you disagree then you have to prove to me how the math I did was wrong and that my results were not in fact 47/53.

See the problem yet?

>>15544090
>if you have a 1/2 chance of being right then there is no benefit to switching
Correct, it is the same.

>>15544111
Nope, the removed door is an invalid choice and irrelevant information.

>> No.15544202
File: 138 KB, 984x1749, Screenshot_20230705-142323_Samsung Notes.jpg [View same] [iqdb] [saucenao] [google]
15544202

>>15544081
That is a really smart way to look at it. And in that case, yes, you should switch back to A if you have the option to do so because A has a larger probability, even though it initially didn't.
But what the main difference here as opposed to the original problem is that you make the switch in the middle of the game. Which changes the probabilities.
When you make the switch over to C, C has a 3/8 probability of having a car. That simply means that every 8 times you make that switch, C will have a car 3 times. The reason that D's probability distributes entirely to A instead of C as well is because you already made the decision. Whatever selection you made for C when switching over won't change. 5 out of 8 times, that is going to be a goat. And once a goat selection has been made, it can't change. It would just mean that A or D will have a car for every time you select a goat. And once you eliminate D, it's just A that would have the car when C has the goat(Since your selection will not change because of D opening, C's probability will remain the same. It seems there may be some anachronistic confusion, as WHEN a box is opened is essential).
A has different probabilities in the beginning vs the end. But that's because the switch was made. If the switch is not made, A would still have a 1/4 probability. That doesn't mean that it would necessarily not have it. It simply means that it will have a car only 1 times out of 4. If in 4 different trials, you just stuck with your choice in the middle of the game, and only switched at the end, you would win 3/4 times. Which is a higher probability of winning than 5/8 (which happens if you switch over both in the middle and the end). And you would, in reality, win more often that way. There was no guarantee in your example that Monty would open D, it could have easily been A that he opened.
Probabilities work over many trials & are not a guarantee, and a lower prob. coming true doesn't negate the whole probability.

>> No.15544207

>>15544111
>You are simply simulating it incorrectly.
See my example in >>15544199. Clearly the odds of a coin flip are 47/53 and not 50/50. My results prove this is true, you can't dispute the math. See the issue?

>>15544180
1 of the goats is always removed from the equation in the second round and becomes irrelevant to the result.

>> No.15544209

>>15544199
You are trolling. I refuse to believe somebody is this stupid AND this smug about it. Be more subtle next time.

>> No.15544215

>>15544207
What is the chance that you choose a door with a goat when you make the first selection?

>> No.15544228

>>15544209
Anon, the stupid one are especially smug, and the smug ones especially stupid. It is entirely in line with our expecations.

>> No.15544232

>>15544215
Okay, one more fucking time...
You start by selecting from 3 doors. There is a 2/3 chance of selecting ANY goat and a 1/3 chance of selecting the car. More accurately, there is a 1/3 chance of selecting Car, a 1/3 chance of selecting Goat A, and a 1/3 chance of selecting Goat B.

No matter which door you choose, the host ALWAYS reveals a goat from one of the unselected doors for the sake of argument, he reveals Goat B) That door and that goat are now NO LONGER IN PLAY.

You are asked to select FROM THE TWO REMAINING DOORS. One has a Car (1/3 chance) and one has Goat A (1/3 chance). The door that used to hide Goat B is now open and out of play, your only choice is between the two remaining doors. So there's a valid 1/3, a valid 1/3, and an INvalid 1/3. Removing the invalid 1/3 because it is no longer relevant, we are left with two valid 1/2 choices.

>> No.15544233

>>15544209
I'm neither trolling nor stupid. I simply do not accept your faulty premise.

>> No.15544235

>>15544199
>The math works out
You didn't do maths, you did a simulation. You are proving the exact opposite of your point. You are looking at an event in reality (you generated random numbers) and saying it is wrong because it doesn't follow the maths (a coin flip is 50-50 exactly). But you say the former is "math" and the latter is "reality".

The simple fact is that the odds of a coin flip are 50-50 if you look at the pure maths but that doesn't mean that you will get a neat 50-50 distribution every time you flip 100 coins in a row. Just like >>15543824 won the car three times by staying. It happens sometimes. Probability is, by definition, not certainty. It is a prediction about what is more or less likely, based on incomplete knowledge.

>> No.15544238

>>15544207
>1 of the goats is always removed from the equation in the second round and becomes irrelevant to the result.
Anon, you are getting ahead of yourself. Do you agree with me that, if your initial pick was a goat, then switching would get you the car and vice versa?

>> No.15544242

>>15544232
>So there's a valid 1/3, a valid 1/3, and an INvalid 1/3. Removing the invalid 1/3 because it is no longer relevant, we are left with two valid 1/2 choices.
That's not how fractions work.

>> No.15544258

>>15543514
staying with your choice is the same as picking from the beginning, so it has the same probability as it originally did 1/1000

>> No.15544260

>>15544235
Okay, now apply everything you just said to the Monty Hall problem and we'll agree completely! 2 doors, 1 winning choice = 1/2 chance of winning whether you stay or switch. Running a simulation and getting a different actual result is not proof that that round 1 magically changed the actual chances of round 2.

>>15544238
I agree.

>>15544242
It's exactly how it works. The 1/3 probability is only in effect when there are 3 unknown valid doors to choose from. When you make 1 door known and it becomes an invalid choice, you are left with only 2 unknown valid doors. Those two doors had equal probability before and they have equal probability now, but it is best represented as 1/2 since they are the only 2 doors in play and 1 door must have the prize.

>> No.15544262

>>15544232
>There is a 2/3 chance of selecting ANY goat
So what does that mean? When you make your initial selection, how likely are you to have picked the goat? 2/3 odds you said correct?
So if one door has the goat, and the other has the car (necessarily since you already revealed one goat), and the door you picked will have a goat 2 out of 3 times, what will the other door have those 2 out of 3 times? A car I presume? With 2/3 odds in the other door correct?

Ignore everything else you said. Just the odds of selecting a goat or a car on first pick is what matters. If you picked a goat first pick, switching will give you a car. And the odds of picking a goat first pick are 2/3. So switching will give you a car 2/3 of the times. Just as this >>15543092 shows, 2 chances of goats, one of a car on the initial selection, we agreed on that. Switching means 2 chances of a car one of a goat because it's inverse to the initial selection since a two goats and two cars can't be there in the endgame, as one goat has already been revealed. Just look at the picture once ffs. If you don't get it by words, maybe colorful emoji pictures will help you.

If you can't see it, you are genuinely too low IQ to get it.

>> No.15544263

>>15544260
>Running a simulation and getting a different actual result is not proof that that round 1 magically changed the actual chances of round 2.
It is, if the rules of the simulation conform to the conditions of the game. Just like if you did your coin flip a million times, it would tend towards 50-50. Statistical outliers become less meaningful when your sample increases.

>> No.15544265

>>15544260
>I agree.
Good. So if your initial pick is a goat, you get the car by switching.

Second question, do you also agree that your initial pick is twice as likely to be a goat than a car?

>> No.15544271

>>15544258
How about this:
You are presented with 3 doors:
Door A
Door B
Door C
1 of these doors hides a car, 2 of these doors hide a goat. The host IMMEDIATELY opens Door A, which he knows hides a goat, and reveals that goat to you. Now he asks you to choose either Door B or Door C, what are your odds of selecting the door that hides the car?

>> No.15544274

>>15544263
Do you know how statistics work? Do you know how means work? Do you know what a sample mean and a standard deviation is? Do you know what a normal distribution is?
If you know all of the above terms, you would know why you sound extremely stupid right now. If not, nobody here is liable for teaching you basic things. I suggest reading an introductory stats textbook.

>> No.15544275

>>15544271
This is equivalent to coming into the original Monty Hall problem after the reveal of the goat and being asked to pick a door, which also gives you 50-50 odds. Because you lack the knowledge of the initial pick.

>> No.15544276

>>15544262
What you select in the first round is irrelevant to the second round because you are given a new chance to select a door freely with new odds under these new circumstances where there are only 2 doors.

>>15544263
There is nothing in reality that prevents me from flipping heads 1 million times in a row. The likelihood of doing that is infinitesimally small, but it is a possibility that can happen. However, it doesn't change the 50/50 odds of flipping heads or tails on the million and first flip.

>>15544265
>Second question, do you also agree that your initial pick is twice as likely to be a goat than a car?
During round 1, yes.

>> No.15544278

>>15544274
Meant for >>15544260 I presume

>> No.15544281

>>15544275
Yes, okay, so you understand that picking between 2 doors is a 1/2 chance of winning even if there's an open, un-selectable door standing beside them. Now you just have to understand that your personal foreknowledge of the first round HAS NO BEARING ON THE REALITY OF YOUR CHANCES IN ROUND 2.

>> No.15544284

>>15544276
>a new chance to select a door freely
Not him but the only way you could select a door freely is by forgetting which door you selected first.

>> No.15544287

>>15544275
>During round 1, yes.
Well, then, you have all the pieces. You agree that, 2/3 times, you will select a goat initially, and you agree that if you selected a goat initially, switching will get you the car. That's it. That's Monty Hall. From these pieces of information it is impossible to conclude anything other than that switching will win you the car two out of three times.

>> No.15544288

>>15544287
for >>15544276

>> No.15544289

>>15544278
Yeah, my bad.

>> No.15544292

>>15544284
Your personal perception of your chances do not influence your actual chances in reality. If I flip a coin and it comes up heads 3 times in a row, I might engage in some level of gambling fallacy and feel like heads is "hot" but the fourth flip is JUST AS LIKELY to be tails as it always was.

>> No.15544293

>>15544281
>Now you just have to understand that your personal foreknowledge of the first round HAS NO BEARING ON THE REALITY OF YOUR CHANCES IN ROUND 2.
You just have to understand that it does. You do realise that your answer is the "intuitive" one, right? The one that people would give without thinking about it? Why would you assume that our stance is the result of a lack of understanding and not of a careful consideration of the problem, when any random clown would come to your conclusion after glancing at the problem for two seconds? That is because people intuitively expect that the opening of the door is irrelevant because it seems to tell them what they already know, and the actual trick is realising that you can in fact deduce some information from it.

>> No.15544294

>>15544292
If you chose the door by flipping a coin, then 1/2 would be correct.

>> No.15544295

>>15544287
You forgot to mention the step where 1 goat is revealed and removed from play, leaving a recalculated probability of 1/2 for winning in the second round (the only round that can actually cause a win because the first round was just pageantry to build tension for TV)

>> No.15544299

>>15544281
Your choice in the first round influences the doors he chooses for the second round, since he's not going to open your door. By choosing your door you've guaranteed he will not reveal that door.
So the first round DOES influence the second round.

>> No.15544303

>>15544295
>You forgot to mention...
No, it's implied but irrelevant. You already agreed that picking a goat initially means switching wins you the car. An event that has a probability of 2/3, as you also agreed. There is no other possible conclusion from these facts. You are contradicting yourself.

>> No.15544304
File: 153 KB, 1080x1043, Screenshot_20220701-023848_TikTok.jpg [View same] [iqdb] [saucenao] [google]
15544304

>>15544276
You have made it incessantly clear that you are trolling. Stop it now you bonobo

>STOP FALLING FOR THE BAIT PEOPLE

>> No.15544323

>>15542721
Any theoretical question weeds out a lot of people just because they can't stick to the framing of the question.
When I mentioned the Monty Hall problem to person I know his reply was that he would study Monty Hall's face to know what was behind each door. When I told him that in this scenario Monty Hall has a 100% perfect poker face that will never betray the correct door he just couldn't get the idea of this theoretical statement and started arguing about looking at other practical solutions instead of trying to solve the theoretical as it is stated.

>> No.15544326

>>15544304
Literally replying to trolls

>> No.15544340

>>15544299
It influences the second round in ways that do not change anything about the outcome. Which goat door the host opens changes nothing about the available options.

>>15544303
You can't just skip over information. I smelled your dishonest bait and switch a mile away, which is why I didn't answer you straight initially. It seems my instincts were spot on.

>> No.15544347

>>15544304
The only trolls are the retards who think declaring an intention in one round magically locks in probabilities at a metaphysical level for the second round.

>> No.15544371

>>15544323
But what about thinking outside the box?
At what point does it become subverting the question?

>> No.15544381

>>15544340
>You can't just skip over information.
You're the one who's skipping over information. But only explicitly. You implicitly follow it anyway when you agree with me on the premises. Indeed, we *know* that the /reason/ that switching will get you the car in the event that you selected a goat is because Monty has to open a door and he cannot reveal the car. Of course. We got all that and we boiled it down to its consequences. Which you agreed to. So you do follow the logic even if you pretend not to. You agree to all the relevant premises. Yet you deny the conclusion. There is no bait and switch here (on my end, anyway). This is approaching formal logic in its rigidity. You could almost put it in the form of a basic syllogism. If you agree to the premises, the conclusion must necessarily follow.

>> No.15544386

>>15544347
But your pick in round one does influence which door gets opened next. Not that it's relevant because you are given the choice between the current door or the 2 left over doors. In this framing the host revealing a goat provides no new information because you already knew that one of the left over doors is a goat. You are in fact the one who fell for the Monty's tv drama bait.

>> No.15544399
File: 184 KB, 1920x813, Monty_problem_monte_carlo.png [View same] [iqdb] [saucenao] [google]
15544399

It's super easy to simulate the scenario to see that it's not 1/2 even if you don't understand why. Just go do that.

>> No.15544402

>>15544381
>You implicitly follow it anyway when you agree with me on the premises.
No, you're using disingenuous tricks to get me to agree with certain premises and then jumping to a conclusion I don't agree with by leaving out crucial information that changes the entire premise of the problem. It's incredibly disingenuous, you're no better than a con man.

You asked if I agreed whether there's a 2/3 chance of selecting a goat in round 1, then you asked whether switching in round 2 would produce the opposite result of whatever I selected, then you jumped to the absolute absurd conclusion that round 2's probability MUST be 2/3 for switching. When I point out that I never agreed to that leap of logic you try to declare that my agreement on the first two points requires that I agree with your illogical bullshit and then try to shame me with accusations of contradicting myself.

So no, fuck you. And fuck me for engaging with you honestly when I knew you were just trying to play that bait and switch trick.

>> No.15544406

>>15544399
Only if you intentionally program the simulation to give you the wrong answer.

>> No.15544409

>>15544371
If asked what 2+2 is, answering 5 isn't thinking outside the box. It's just wrong. The Monty hall problem isn't a philosophical inquiry. It is a Mathematical question with an objective answer.

>> No.15544410

>>15544386
>you are given the choice between the current door or the 2 left over doors
No, the door that the host opens in round one is not available for selection in round two. You are given the choice between the current door or the ONE left over door.

>> No.15544412

>>15544406
Go do it fairly. You can use your favorite programming language or even just a sheet of paper and dice. There's no excuse for you for not doing that.

>> No.15544417

>>15544402
Anon, you are irrational. You are not being tricked. Again, if you agree to the premises, the conclusion necessarily follows. There is no leap in logic there. What you call "absurd" is the obvious entailment of the preceding. Consider that you are clinging to the wrong answer for reasons that are not entirely logical. Because if you have agreed with my premises (and you do) then you have already agreed that the answer is 2/3. You only stubbornly refuse to admit it. Very well. I cannot force you to admit anything. But you have revealed yourself impervious to logic and unwilling to change your mind.

>> No.15544418

>>15544402
point to exactly which premise you disagree with here. Because you are not making any point.

1. When making the initial selection, there is a 2/3 chance you will select a goat.
2. Anytime you select a goat in your initial selection, switching doors in round 2 will give you a car
3. Following that premise 1 & 2 are true, it must be true that 2/3 of the time, switching in round 2 will give you a car.

Give me the number of the premise you disagree with.

>> No.15544419

>>15544410
It's a selection between 1 door or 2 doors. Opening a door does nothing. The game is exactly the same whether Monty opens one door or if he offers you to open both doors (unless you consider goat to be a possible prize then it's a little different).
You bet whether you chose the correct door or not. The second phase is not independent of the first one. If you had a single class of probability theory or just probability in math class in middle school you should be able to draw a tree of choices and see the probabilities of each one.
It's really not that hard. You can derive it rigorously or using a stochastic model. Either one is fine yet you refuse to do it.

>> No.15544426

>>15544419
This is the gambler's fallacy in action.

>> No.15544428

>>15542721
>you are more likely to choose one of the two goats instead of the one car
>therefore it's reasonable to switch once the uncertainty about the distribution decreases.
That's literally it. It also doesn't matter whether the host knows what's behind the doors or not, as long as the door opened has a goat. Imagine there are 10 doors and the host opens 8, which all happen to be goats. Would you really not switch?
300+ replies lmao

>> No.15544434

>>15544426
Anon, people keep explaining to you exactly how there is a connection between the two events that influences probability, and you just keep repeating "gambler's fallacy".

>> No.15544442

>>15544417
See below. Premise 1 and 2 are agreeable, but premise 3 is an absurd leap of logic that fails to account for the change in the dataset when one goat is revealed and that door is made invalid as a choice.

>>15544418
>Give me the number of the premise you disagree with.
3, because when you remove one door from the equation your choice is binary and there's a 1/2 chance of picking the car or the goat from the remaining doors. All the set dressing regarding your initial choice and the third door is rendered irrelevant now that you can only choose between 2 doors.

>>15544419
>The second phase is not independent of the first one.
It is in every way that matters. Which specific door is opened in round 1 is merely set dressing and pageantry because it will always reveal a goat and always make your choice be between 2 valid remaining doors.
>you should be able to draw a tree of choices and see the probabilities of each one
In round 2, you pick one of two doors and receive one of two outcomes. Only one outcome is a winning one. That is literally a 1/2 chance.

>> No.15544446

>>15544426
You don't even know what gambler's fallacy is clearly.
You can prove your case very easily by performing the experiment. Write a simulation, share the code and show that you would actually come up with 1/2.

>> No.15544454

>>15544446
Not him but his simulation would be a single coin flip.

>> No.15544456

>>15542727
Well, it's not about whether you like cars or not idiot.
It simply can be replaced with something you do like or just be symboled as trophy or any other blissful thing

>> No.15544458

>>15544442
>Premise 1 and 2 are agreeable, but premise 3 is an absurd leap of logic
It literally directly follows. If you have a 2/3 chance of selecting a goat, and if switching when you select a goat gets you the care, then you have a 2/3 chance of getting the goat when you switch. You calling it absurd is what's absurd here, ironically. If you agree with the premises, you can keep going on and on about how the door renders everything irrelevant, but you've already agreed that it doesn't.

>> No.15544462

>>15544458
>then you have a 2/3 chance of getting the goat when you switch.
the car*
dammit you've worn me down, I need a drink

>> No.15544470
File: 50 KB, 800x600, monty.png [View same] [iqdb] [saucenao] [google]
15544470

>>15544442
>>you should be able to draw a tree of choices and see the probabilities of each one
>In round 2, you pick one of two doors and receive one of two outcomes. Only one outcome is a winning one. That is literally a 1/2 chance.

>> No.15544473

>>15544458
There is not a 2/3 chance of ANYTHING occurring after the choices have been narrowed to 2.

>> No.15544475

>>15544454
>if I change the game the outcome is different
You don't say

>> No.15544480

>>15544473
Prove it. Simulate a million games or something.

>> No.15544486

>>15544473
>If you have a 2/3 chance of selecting a goat, and if switching when you select a goat gets you the care, then you have a 2/3 chance of getting the goat when you switch.
If you can explain to me the flaw in this logic right here, THEN and only then do you have the right to start talking about how many doors there are, as we would then be in need of an alternative. If the logic checks out then this is the answer, simple as.

>> No.15544493

>>15544442
Imagine there are 100 doors, 1 car and 99 goats. You pick one and have a 1% chance of having picked the car. Then 98 doors are opened and all are goats. But the 99% chance that you have picked a goat still remains, that is, you're still 1% likely to have picked the car but switching turns it to 100-1=99%.

>> No.15544496
File: 96 KB, 792x924, 1687321407781734.jpg [View same] [iqdb] [saucenao] [google]
15544496

>>15544486
>>15544493
I don't know why these two posts out of all of them made this logic click suddenly but now I see it. Goddammit.

>> No.15544505

>>15544473
So you pick a door (1/3 chance to pick car and 2/3 chance to pick goat). The host does its thing or whatever and you ignore it completely and stick with the chosen door. You don't care about any other opened doors or switching. Are you trying to convince everyone here that by picking randomly out of the 3 doors and keeping your choice until the end will somehow result in you getting the goat only 1/2 of the time? Do you really not see the problem here?

>> No.15544512

>>15544442
1. When making the initial selection, there is a 2/3 chance you will select a goat.
2. Anytime you select a goat in your initial selection, switching doors in round 2 will give you a car
3. When you remove one goat from the game by revealing one of the unchosen doors, it doesn't change whatever selection you had initially made. (e.g If you had initially selected a goat, it is still a goat. The object behind your door of selection will not change)
4. The probabilities for your initial selection of a goat cannot change contingent on the revelation of one of the goats since the revelation of the second goat occurs well after the initial selection.
5. Following from premise 3 through 4, when you move into the second round, you will necessarily have the exact same selection as your initial selection, every single time.
6. Following from premise 5, It is necessary that your odds of having a specific selection in round 2 are also the same as the initial selection odds when the decision was made. Because it's literally the same selection that was made initially.
7. Since premise 6 is true, the probabilities of the initial stage are still in effect for the chosen door even in round 2.
8. Following that premise 3 through 7 are true, it must be true by virtue of premise 1 & 2 that 2/3 of the time, switching in round 2 will give you a car.

Point to which one you disagree with.

>> No.15544514

>>15544505
>>15544512
See: >>15544496
No need to rub salt in the wound. I'll just see myself out now.

>> No.15544518

>>15544514
Sorry, I thought you were a 3rd poster.

>> No.15544627

>>15542721
Just checking in and it seems the bullshit arguments are still ongoing. I would suggest giving it up. As I said originally, it is simply impossible to explain to a person whose IQ is below a certain point, all the necessary explanations have already been made.
It seems my theory did not falter though, and this thread is prime proof of the truth of my post. Many seem to have been IQ filtered in this thread. And brutally so.
Remember, the longer it took you to get it, the lower your IQ. If you figured it out on your own, you are at the top, and if the first explanation convinced you and you figured it out, a little lower, and so on.

>> No.15544651

>>15544627
>muh IQ
No one cares about your masturbatory horoscope numbers.

>> No.15544676

>>15543102
Winning the lottery is 50/50 you either win or you don't.

>> No.15545012

there could be 9 trillion doors, it doesn't matter
the choice you make is 50/50 between two doors, the first round doesn't matter and switching makes no difference

>> No.15545017

>>15545012
Another wild retard appears. It's over. Thread is over. The other ones were convinced. Just read the old replies, no one is going to argue with you again now.

>> No.15545034

>>15545012
It isnt 50/50 its 1/9trillion

>> No.15545064

>>15545012
>>15545034
50/50 = 1

>> No.15545084

>>15545012
I was ITT arguing exactly like you for most of this thread. Here's the perspective shift that helped convince me: No matter which door you choose, in the first round it is more likely that you chose a door with a goat behind it than the door with the car behind it. Once all other options are eliminated, you are still more likely to be sitting on a goat door, so therefore switching doors is more likely to benefit you with a win.

>> No.15545175

So weird how this problem is shown in an incorrect light so many times

>> No.15545265

>>15545175
There are many variations of it.

>> No.15545424

>>15545034
Seems that again some people dont understand to diffrentiate between monty opening rest of the doors vs 1 door just because they both exist in 3 door monty at the same time

>> No.15545511

/sci/ is the lowest IQ board

>> No.15545514

>>15545424
Some people see reality, others obfuscate reality with 'mathematics'

>> No.15545621

>>15544434

Gamblers will likewise tell you how there are countless connections between the events they gamble on.

>> No.15545646

monty is unable to open the door that has been chosen. Had monty picked a random goated door and his pick happened to be one of the unchosen doors then it is 1/2. But it is implicitly assumed that monty would not open the chosen door. Heuristically one of the remaining doors have stayed unopened because of necessity and the other by chance or necessity. The fact that that door was able to stay unopened introduces bias in that doors favor. They have not been subjected to the same filter because of the preceding events. Thus arguments involving "independence" are wrong.

>> No.15545657

>>15542721
The reason this problem baffles people is that a lot of assumptions are implicit and Monty only has limited agency.

Monty opens a door in order to encourage you to pick another one, the natural assumption is that he wouldn't do that if you had picked a goat door. Thus anyone who sees Monty as a rational antagonist will naturally precommit to their choice, the fact that Monty is trying to get you to change your mind should increase your certitude that you picked the car.
But the problem doesn't specify that Monty apparently HAS to open a non-goat door no matter what.

>The host must always open a door that was not picked by the contestant.
>The host must always open a door to reveal a goat and never the car.
>The host must always offer the chance to switch between the originally chosen door and the remaining closed door.
These aren't spelled out properly whenever the problem is presented

>> No.15545675

>>15545657
To illustrate that the problem is with Monty agent, we can look at the "Deal Or No Deal" version.
Instead of Monty opening the door, it's the contestant who is given the choice of opening another door. If it's the car he loses, but if it's a goat he gets a chance to switch his choice. The contestant takes that deal, open the other door, it's a goat. Should he switch?

In that scenario the chance is 50/50 so it doesn't matter whether he switches or not.
And even in a scenario with 100 doors, where the contestant has opened 98 doors without stumbling upon a car, the chance of either of the last two doors containing the car remains 50/50.

It's Monty that confuses people.

>> No.15545680
File: 5 KB, 687x219, monty_portions.png [View same] [iqdb] [saucenao] [google]
15545680

>>15545657
>But the problem doesn't specify that Monty apparently HAS to open a non-goat door no matter what.
I thought about it some more and came to the conclusion that it doesn't matter what monty's intent was. As long as you see a goat, that door will have 2/3 probability.

Pic related explains why, think about it carefully!

>> No.15545687

>>15545680
The intent itself doesn't matter, the fact that Monty has to open a goat door (has to avoid the car) matters.

>> No.15545689

>>15545687
yeah and you're not selecting "a door", you're selecting a set of two doors where you were allowed to peek inside.

>> No.15545694

>>15545687
he doesn't have to open a goat door, it can be incidental.

>> No.15545710

>>15545694
No, he has to avoid the car. Otherwise it's the same as randomly opening a door yourself like in
>>15545675

>> No.15545718

>>15545680
That pic is useful sophistry, your arrival at the correct answer has been accidental. What ensures dependency in two events is the assumed intentions of the host. It is the hosts assumed avoidance of the chosen door and the car when opening a random door that introduces dependence.

>> No.15545723

>>15545718
Nope! This applies when the host accidentally chooses the donkey. In the alternate scenario where he chooses the car, you aren't playing.

>> No.15545733
File: 4 KB, 356x348, 2023-07-06_15-00.png [View same] [iqdb] [saucenao] [google]
15545733

>>15545723
Here is a simulation where monty can pick the car. In only half of the cases is switching a win.

>> No.15545748

Seems like people are still getting brutally filtered by probabilitiy.
It's completely irrelevant what the host wants or knows.
You have one decision situation with low information/high uncertainty, and then the same situation with higher information/lower uncertainty.
It's more likely you've made a wrong choice under higher uncertainty. Therefore you should change it.

>> No.15545751

>>15545748
>It's completely irrelevant what the host wants or knows.
It's not irrelevant at all, it's the whole point. It's what leads to more information. Opening more doors randomly adds nothing:
>>15545675

>> No.15545758

>>15545751
why make up alternative versions of the problem? We're presented with a completely clear formulation where:
1. the door opened by the host is a non-chosen one
2. the door opened contains a goat
All the other possibilities are already eliminated, whether the host knew or didn't know anything. The information the player has is enough.

>> No.15545800

>>15545758
No, you're actually retarded. If the host didn't know to avoid the car, it'd be no different from the door being opened randomly by the contestant or whatever, and then the odds would be 50/50.
It's the fact that the host knows what's behind each door and that he HAS to open a door with a goat that gives switching better odds.

>> No.15545841

>>15545800
This is the problem, intention doesn't change anything. In round 1 you are twice as likely to select a goat door as you are to select a car door. If the other goat door is removed it's to your benefit to switch because the higher likelihood that you selected a goat door in the first round means that there's a higher likelihood that the remaining door is the car door.

>> No.15545858

>>15545800
seems like you didn't read my post and are seething for being filtered once more.

>If the host didn't know to avoid the car, it'd be no different from the door being opened randomly by the contestant or whatever
yes, it would indeed be no different. Whether it's the host who knowingly opens a door or the contestant who randomly opens a door, the situation as stated is that a goat is revealed. It *might* have been a car but that's not what the problem gives you. Just two decisions of which one can be made with lesser uncertainty.

>> No.15545860

>>15545841
If the second door is opened randomly, then it let's you know you are not in the scenario where you picked a goat and then the car door was opened. That lets you readjust your original probability of picking the car from 1/3 to 1/2, and by extension, 1/2 to the remaining door. There is no advantage to switching.

If the second door is not opened randomly, then it is done independently of your original choice, and so it doesn't affect that original probability of 1/3 for that door, forcing the remaining door to have 2/3 probability. There is an advantage to switching.

>> No.15545864

>>15545860
It doesn't matter if it's opened randomly. If you see a goat then the car is still in play and it was still more likely in the first round for you to have selected a goat door.

>> No.15545869

>>15545864
If you picked the car, then the host is guaranteed to open a goat door at random, while if you picked a goat door, he'd only a goat door half the time.
The fact that he DID open a goat door thus means you're more likely to be in the "picked the car" scenario than you thought you were originally. 1/3 goes up to 1/2.

>> No.15545891

>>15543965
>Give me an example of when this question is useful in life and don't say "developing critical thinking skills" I want a REAL answer.

>> No.15545925

>>15543965
>irrelevant in real life
pretty sure that's what your peasant ancestors said when their kids were forced to go to school to learn reading and writing

>> No.15545995

>>15545869
If he opens the car door the game ends so that scenario is nullified.

>> No.15546030

>>15543494
Without loss of generality you can call b the winner. It's instructive to rewrite it with 6 cases.

>> No.15546048

If you swap and the car was behind your door you look like a bigger retard than if you don't swap and the car was behind the other door, so it would be better to never swap regardless of any so called 'mathematical' sophistry simply because the social consequences of being wrong are lesser in the case of not switching.
Also its a 1/2.

>> No.15546114

>>15543715
Random selection only happens once in this scenario, when there is 3 doors. The choice of the two remaining doors is not a random selection.

>> No.15546120

>>15543797
>second round
There is no second round, random selection only happens once.

>> No.15546187

>>15545995
Exactly. So when he does open the door to the goat, it is indicative that you should switch. Because it's twice as likely that if he opened up on the goat, that you selected a goat as well. Because it's the only door he could open, as opposed to the other scenario, where it's a random chance between the two unselected doors.

>> No.15546201

>>15546187
>Because it's twice as likely that if he opened up on the goat, that you selected a goat as well.
This type of confusing causal relation is what prevents people from understanding what's actually happening. You are ALWAYS more likely to have selected a goat door because there are twice as many goat doors as car doors. Saying, "IF the host opens a goat door THEN it's twice as likely you did, too" is fallacious.

>> No.15546255

>>15542721
To be fair, average mathematician IQ is ~130, and a lot of them missed it the first time.
But if someone doesn't get it after learning the answer, they're sub 95 IQ absolutely

>> No.15546272

>>15546201
Both ways are true. Just different ways of looking at it. If you are truly smart, you should be able to see both methods.

>> No.15546288

>>15546272
No, saying that the host's choice CAUSES your choice to be more likely one or the other is fallacious and presenting that kind of fallacious reasoning is exactly what gets people confused about the proposal. I know because it's literally what was preventing me from accepting it. You are always twice as likely to pick a goat door simply because there are twice as many goat doors, whether the host opens a goat door or a car door simply determines whether the game continues (since the host revealing the car would necessarily halt the game).

>> No.15546331

>>15545733
You're simulating a different scenario. Monty doesn't open the doors randomly (unless you picked a car). It may seem that this distinction doesn't matter but it does change the outcome of a simulation because you are no longer rejecting games.

>> No.15546362

>>15545748
It does matter. there are multiple variations of this problem and there are different solutions depending on the host behaviour. The most common version is the host opening the goat door if you picked a goat and choosing randomly from the remaining 2 goats if you picked a car. In that version switching gives you 2/3. But in certain versions the host's behaviour can be gamed for a better outcome.
In the variation where the host is ignorant of the contents of the doors, switching does nothing because the host opening a car door ends the game. That makes you lose 50% of the time when initially picking a goat. So in 1/3rd of cases you win by switching; in 1/3rd you win by staying and in 1/3rd you lose no matter what.