/sci/ - Science & Math

[Anonymous No.11716580]

Can someone explain this Monty Hall bullshit to me?

Why in the fuck is "staying with the door you've chosen" not interpreted as "choosing one of the two remaining doors"?

Would this riddle change completely if Monty didn't just open a door to reveal a goat, but also explicitly revoked your initial choice so that you're back in a 'neutral' position?

Does it all hinge on pedantic semantics?

[Anonymous No.11716601]

scale up the problem to 1000 doors and 999 goats and the answer becomes more obvious

[Anonymous No.11716603]

>>11716601
It doesn't do much for me.

[Anonymous No.11716605]

>>11716580
>Would this riddle change completely if Monty didn't just open a door to reveal a goat, but also explicitly revoked your initial choice so that you're back in a 'neutral' position?
No, it's the same. You still have the same two choices - original door or new door. And picking new door is still better.

[Anonymous No.11716606]

>>11716580
no.
>you pick a door: 1/3 chance of being correct
>that means the odds are that it is a door you DIDN'T pick is 2/3
>Monty removes one of the other doors
>the odds are that it was one of the doors you didn't pick is still 2/3
>but you have been told it isn't the door that he showed you
>other door therefore has a 2/3 chance of being correct

this has been demonstrated in like 1000 different places, you really think a mongolian 5G conspiracy forum is where you're going to get a convincing answer?

[Anonymous No.11716607]

>>11716580
write a program and test it, it’s just a couple of lines

[Anonymous No.11716611]

>>11716603
Imagine there are a 1000 doors and you pick one. Then the host deliberately opens 998 out of the remaining 999 doors to reveal 998 goats. What is more likely, that the other door left is a random goat and you just happened to pick that 1 out of 1000 car, or that that is the car since the host must have avoided it in that case.

[Anonymous No.11716623]

>>11716603
you pick a door out of a thousand, now there is a 999/1000 chance you've picked a goat this probability never changes
Monty continues and opens 998 doors revealing 998 goats
the chance you picked the wrong door is still 999/1000.

[Anonymous No.11716624]

>>11716580
It's because Monty is guaranteed to open a door with a goat, and that door has to be one of the other doors. If Monty opened a random door (including the door you chose) just as you chose a random door, then it would be even.

[Anonymous No.11716882]

>>11716580

[Anonymous No.11716887]

>>11716624
>because Monty is guaranteed to
nah, for one round the odds are 2/3, even if Monty opens a goat door by coincidence

[Anonymous No.11716896]

>>11716580
What it comes down to is whether the guy opening the door is an omniscient observer who intentionally opens a door containing a goat (in which case changing door doubles chance of getting car), or if the guy opening the door picks a door at random which then just so happens to contain a goat (in which case switching door does not change likelihood of outcome).
In the omniscient observer scenario: You pick door. 1/3 chance it has car. Observer opens goat door, knowing that it contains goat (so assuming you picked a goat door, he is intentionally opening the door that is not the car door). So now if you did not pick car door, you know which if the other two is car door (since goat door is open)
That said, goats are cooler than cars so I would prefer a goat so I would not switch, and also it's 50/50 because it either happens or it doesn't.

[Anonymous No.11716910]

>>11716887
Wrong. If Monty is not omniscient and picks a door without knowing what is behind it, and it just so happens that behind that door is a goat, then you changing door will not alter your probability. The change in probability hinges on the fact that Monty knows which door the car is behind, and intentionally does not open that door.
Example with 1,000 doors:
You pick door 728. 1/1,000 change it is car. Monty opens all doors except 728 and 456, not knowing what is behind each. There are 1k possibilities for where goat is not hiding. 99.8% chance he opens a door with the car. 0.2% chance he does not. If he does not then well it's pretty likely your door has car but it's also pretty likely the other closed door has car. 50/50, it unironically either happens or it doesn't :^)
Alternatively, consider Monty knows which door has car. He opens 998 doors but not car door. On purpose. He leaves 292 and 728 close. Now it's 99.9% that switching to 292 will switch from goat to car.
tldr ur a tard

[Anonymous No.11716928]

>>11716910
wrong. hippie woowoo consciousness fag detected.

[Anonymous No.11716940]

>>11716928
Are you fucking retarded? What are you even trying to say? Please leave /sci/, double digit IQ fags don't belong here.

[Anonymous No.11716941]

>>11716910
true
>>11716887
false

[Anonymous No.11716945]

>>11716941
based and goatpilled

[Anonymous No.11716982]

>>11716940
Even if a random wind breeze opens a goat door, it's still a 2/3 chance if you switch.
Read a book.

[Anonymous No.11716985]

>>11716941
>>11716945
hillbillygoat idiots

[Anonymous No.11716987]

>>11716982
You have to go back.

[Anonymous No.11716988]

>>11716940

[Anonymous No.11716990]

>>11716987
>probabilities arise from m-muh feelings
have sex

[Anonymous No.11716994]

>>11716606
/thread

[Anonymous No.11717134]

>>11716928
shut the fuck up dumbass

[Anonymous No.11717140]

>>11716990
Yes, probabilities arise from knowledge.
Suck my cock you stupid fucking faggot. I'd say this to your bitch fuck face if I could but you're a faggot behind some screen so I cant

[Anonymous No.11717146]

Its only paradox because its a degenerate case of this entire problem
It doesnt work with less than 3 doors

[Anonymous No.11717261]

>>11717140
so opening the door doesn't change the probabilities? lol

[Anonymous No.11717263]

>>11716601
You are wrong.
>>11716623
No. Th problem changes. In the second option the 998 doors go away. It’s not 999/100
It’s 1/2, this is the part no one gets and everyone gets wrong, You get a new denominator after a door opens, it’s not 2/3 in the oj problem, it’s 1/2. It doesn’t matter if you change or not.
>>11716580
> Does it all hinge on pedantic semantics?
Yes.
>>11716882
This is stupid. The #2. Problem: door 1 is elimuted.
You know it is wrong. You would only pick door 2 or 3 and only 2 or 3 could win
ITS FUCKING 1/2 NOT 2/3 THE DENOMINATOR CHANGES
THIS IS SO FUCKING SIMPLE IT JUST SHOWS HOW MOST “SMART” PEOPE, ARE AUTISTIC RETARDS WITH NO LATARLT SKILLS

[Anonymous No.11717265]

>>11717146
rubbish. in what sense is it "degenerate"

[Anonymous No.11717270]

>>11717263
>The #2. Problem:
oh wow, you don't understand the picture at all

[Anonymous No.11717273]

>>11717263
you picking a door and monty opening the other is additional information which need to be taken into account. it's 1/2 only if you don't do it.

[Anonymous No.11717293]

>>11716910
> hiding. 99.8% chance he opens a door with the car. 0.2% chance he does not. If he does not then well it's pretty likely your door has car but it's also pretty likely the other closed door has car. 50/50, it unironically either happens or it doesn't :^)
Alternatively, consider Monty knows which door has car. He opens 998 doors but not car door. On purpose. He leaves 292 and 728 close. Now it's 99.9% that switching to 292 will switch from goat to car.
Not really, Monty hall wants you to lose remember, so even if you picked the right door he might still open all 1000 doors but one with a goat to bait you into changing.
Look here:
When you first pick, your odds are 1/1000. Your second pic is not (999/1000) vs (1/1000) like you claim. That is nonsensical your second pic is 1/2. There are two doors, one has a car one doesn’t. Period,

[Anonymous No.11717299]

>>11717270
Explain what I got wrong.
>>11717273
It’s only additional information if Monty hall acts the same way every time. Mr hall is a free actor. See here
>>11717293
He could be baiting you.

[Anonymous No.11717307]

>>11717299
>It’s only additional information if Monty hall acts the same way every time. Mr hall is a free actor. See here
he always opens a door with a goat. otherwise you're not talking about the monty hall problem.

[Anonymous No.11717309]

>>11717270
You labeled it 2.
I fixed your retard problem.
It’s 1/2

[Anonymous No.11717319]

>>11717309
it's all one and the same problem retard
competitor chose door3, monty opened door1

[Anonymous No.11717324]

>>11717265
In a sense that its a border case which doesnt seem to represent the general trend
As many pointed out earlier, making it 100 doors makes the trend obvious, and with 2 doors the trend is not present at all

[Anonymous No.11717344]

>>11717309
the probability distribution is not uniform you retard. you're literally using the "it's 1/2 it either happens or not lol" logic.

[Anonymous No.11717351]

>>11717324
that makes it the first non-degenerate case

[Anonymous No.11717362]

>>11717344
Why is it not uniform? Are you saying Monty only shows a goat when I am right? Than it’s 0% chance. Are you saying Monty only shows a goat when I am wrong? Then it’s 100% chance if I switch.
Either way 2/3 is still wrong
>>11717319
It’s not the same problem, as I demonstrated, the denomanator changes, this is exactly the point we are arguing about. You get new odds and a a new choice

[Anonymous No.11717376]

>>11716882
nice logic

[Anonymous No.11717377]

The real issue is that 99% of the time the problem is stated incorrectly, which ends up confusing the probabilities.

[Anonymous No.11717388]

>>11717362
he always opens a door with a goat. the original distribution P is uniform:

P(car behind picked door) = 1/3
P(car behind first remaining door) = 1/3
P(car behind second remaining door) = 1/3

then he opens a door with goat. now the distribution is

P*(car behind picked) = P(car behind picked | goat behind Monty) = 2/3
P*(car behind remaining) = P(car behind remainig | goat behind Monty) = 1/3
P*(car behind Monty door) = P(car behind monty door | goat behind monty door) = 0

[Anonymous No.11717403]

why does two of the doors have goats anyway? why not just put nothing behind them?

[Anonymous No.11717417]

>>11717362
>denomanator changes blah blah
say that instead of doors we have boxes with a toy car and a stuffed toy goats

1) you select one box
2) the two other boxes are sent to china

the chance that the toy car is in china: 2/3

now one of the chinese boxes is opened

the chance that the toy car is in china: still 2/3

[Anonymous No.11717422]

>>11717388
This is wrong,
>>11717309
This is right. Either door is 1/2.
>>11717388
You can restate it as many times as you want, there is no 2/3, a door has been elimukayed, gone, taken, for all math purposes removed from the table.
The same is if you have 1000 doors.
First pick you pick door 200 (btw it’s the right door!) monty opens every door except door 200 and door 201.
There
Are
Two
Doors
One
Is
Winner
1/2.
The other doors do not matter unless you have some special knowledge about Monty.
In fact let’s remove Monty
You have a desk of cards face down, if you pick ace of spades you win $500 from me.
You pick a card, I turn over 50 cards (if I turn over the ace game is over no one wins or loses), none of them are ace of spades(darn). You can keep your pick or change. Your first odds were 1/52, now your odds are 1/2. Your odds are not 1/52 if you stay and 51/52 if you change, that’s stupid. You know those 50 are losers. In no case are the 50 cards still part of the problem
Do you take this bet? If you lose you pay me 52 to 1 odds.

[Anonymous No.11717424]

>>11717417
Wrong.
You changed the problem
> now one of the chinese boxes is opened
I Aussie you mean it was empty? The odds change BECUASE YOU NOW HAVE BETTER INFORMATION the odds go from 1/3 to 1/2

[Anonymous No.11717428]

>>11717422
nah
https://en.wikipedia.org/wiki/Conditional_probability

[Anonymous No.11717435]

>>11717424
>You changed the problem
nope, moving boxes around doesn't change probability

[Anonymous No.11717439]

>>11717422
>There
>Are
>Two
>Doors
>One
>Is
>Winner
>1/2.
"it's 1/2 it either happens or not lol" you're doing it again
>The other doors do not matter unless you have some special knowledge about Monty.
you DO have a special knowledge. you know that he ALWAYS opens a door with goat.

[Anonymous No.11717449]

>>11717439
>know that he ALWAYS opens
irrelevant. it is enough to know that the opened door has a goat - doesn't matter how/why the goat door opened, the new info is purely in the fact that there is a bleating goat looking at you

[Anonymous No.11717450]

>>11717428
But the events are independent. The event being your choice.
You posted a link, but do you understand it? Because it doesn’t apply here. All Monty does opening a door is eliminate that door. It doesn’t move the car. The second choice is completely independent and distrecte
https://en.wikipedia.org/wiki/Gambler%27s_fallacy

[Anonymous No.11717453]

>>11717435
No. The odds were 1/3 now the odds are 1/2

[Anonymous No.11717463]

>>11717453
ooh, magic. does moving the boxes 3 feet change the odds? 4 feet? 0.00005 feet?

[Anonymous No.11717464]

using the 3 door example is open to confusion, use higher numbers of doors since it makes the logic more visible

on a related note what if there are infinite doors and infinity - one goats?

[Anonymous No.11717469]

>>11717450
https://en.wikipedia.org/wiki/Monty_Hall_problem#Simple_solutions

[Anonymous No.11717470]

>>11717463
No, opening the box changes the odds dumbass. It elimutaes a box

[Anonymous No.11717482]

>>11717464
*infinity minus one goats

[Anonymous No.11717483]

>>11717470
>opening the box changes the odds
yes, among the chinese boxes oc
but china as a whole stays at 2/3
>>11716882

[Anonymous No.11717492]

>>11717469
It’s wrong.
It’s telling you can’t explain
>>11717376
And have to result to links, becuase you don’t understand it

[Anonymous No.11717501]

>>11717483
>but china as a whole stays at 2/3
No, Becuase there is only one mystery box in China now. And one at home. China goes to 1/2 chance of having a toy.
Opening the box doesn’t make it more or less lie,oh the other Chinese box has the toy

[Anonymous No.11717503]

>>11717492
you just elimutaeted yourself, ESL

[Anonymous No.11717510]

>>11717503
Insults just mean you know you are backed into a corner

[Anonymous No.11717518]

Why does sticking with the door you chose originally not count as a new choice in a new situation?

[Anonymous No.11717520]

>>11717510
not when you're right

[Anonymous No.11717525]

>>11717518
finally someone asking a rational question

[Anonymous No.11717573]

>>11717492
>And have to result to links
in english, please

[Anonymous No.11717587]

>>11717261
Correct

[Anonymous No.11717667]

over half an hour in MS Paint 3D

[Anonymous No.11717844]

>>11717587

[Anonymous No.11717849]

>>11717518
it does. the choice has probability 1/3

[Anonymous No.11717861]

>>11717449
that's true. still the probability is 2/3 and not 1/2.

[Anonymous No.11717870]

>>11717449
>>11717861
yes oc

[Anonymous No.11717879]

>play game
>monty says, you can choose one door
>now monty tells you, before opening any doors, that you can switch and open up the other TWO doors if you'd like
>you take the switch
>he opens one of your doors with a goat
>lets you open the other door
same shit, and it's obvious you should switch.

[Anonymous No.11717895]

>>11716580
I never understood this problem until just now.
The way people look at it and frame it is strange. So, if we expand this to a thousand doors, you pick a door. There are two possibilities now. If you picked the right door, Monty has to remove all of the wrong doors except one. If you picked the wrong door, Monty has to remove all of the wrong doors. It is much more likely you picked the wrong door, so it is much more likely Monty removed all of the wrong doors, leaving you to switch to a correct door 999/1000 times. If you were to draw this out it would make more sense.
I am not absolutely certain of my answer though. But, idk, if we had two identical universes, and we pick the leftmost of three doors in both. In our first world, Monty removes one wrong door before we make our choice. In the sexond world, he removes it after our choice. Monty, no matter your choice, removes all doors other than the good door and a wrong door. It is equally likely you picked the good door in both universes. It doesnt really make sense that when you know two options are left, one good, one bad, that the odds arent 50% because some other doors dont have the goat.
Basically, it was a 1/3 chance, but Monty has made it a 50% chance by removing a wrong door 100% of the time. No matter what door you pick, right or wrong, all of the wrong choices except for one disappear. Okay, that realization has actually made it clear to me. Switching doors is the right option because it is far more likely you chose the wrong door and thusly that the other last door is the good one.

[Anonymous No.11717908]

>>11717895
Yes, you get it now.

[Anonymous No.11717960]

>>11717879
that's not the same thing

[Anonymous No.11718026]

>>11716580
No it would not change anything.

[Anonymous No.11722443]

>go to bed, thread at ~5 posts
>thinking it'll die out because there have been numerous good and intuitive solutions
>wake up to 77 posts with 1-2 retards vigorously defending their position and refusing to follow well-accepted solutions
maybe the IQ threads aren't so bad after all

[Anonymous No.11723388]

>>11722443
to be fair most of the solutions are just appeal to intuition, which isn't a proof

[Anonymous No.11725466]

>>11723388
I have literally never seen a proof posted

[Anonymous No.11725494]

I think part of the issue is people are overly attached to the door they picked, and think it is special.
The thing is, if there were 100 doors and you picked any other door, then the door you initially picked is incredibly likely to be among the 98 doors that monty opens.
You only feel it is special because you picked it initially, and it is one of two doors left after monty is done opening.

>>11716896
>or if the guy opening the door picks a door at random which then just so happens to contain a goat (in which case switching door does not change likelihood of outcome).
it changes the likelihood of the outcome because you observe that it is not a car.
it is highly likely he will open the door with the car if he worked like this, but if he opened the door with the car you would know, and there would be no point swapping.
As long as you are in a universe where he opens a goat door, you are better of swapping
OF COURSE, this 100% contingent on that you the player is ALWAYS given a choice; if monty is able to decide whether or not to give you a choice, then he could refuse to give you a choice whenever swapping would land you a car.
For the case where he always gives you the choice AND he does not know, the likelihood of getting a car by swapping only matters if you discard the universes where the door he opened accidentally had a car (because obviously swapping here is pointless).

>>11717261
it does not change the priors, but conditonal on the car not being behind the doors monty opened then the belief you as a rational person should place in the car being behind the remaining door should be higher.

[Anonymous No.11725502]

>>11717293
>Not really, Monty hall wants you to lose remember, so even if you picked the right door he might still open all 1000 doors but one with a goat to bait you into changing.
This is why it is vital to state that in every instance of the problem, monty gives you a choice of swapping, regardless of what is behind the door you picked.
(and it is also obviously implied that there is no shuffle-machinery behind the doors that dynamically change where the car is, and also that there actually is a car behind one of the doors)
As long as the above is true,even if monty is opening doors at random, swapping will be to your benefit as long as one of the opened doors did not have the car.

[Anonymous No.11725534]

>>11716580
>Does it all hinge on pedantic semantics?

It hinges on Baysian statistics, which are notoriously unintuitive. It might make more sense if you think of it backwards, your chances of picking wrong are 2/3, the host always picks a dummy door to open so you know you either switch from the wrong door to the right door or vice versa.

So, your chance of choosing the first door correctly is 1/3 meaning you have a 1/3 chance to fuck up by not switching and a 2/3 chance to win.

[Anonymous No.11725536]

G1 = Goat 1, G2= Goat 2.

When you choose a door, three possible states exist, you landed on car, G1, or G2. Below that are decision branches that lead to outcomes. Switching has 2/3 chance win.

[Anonymous No.11725539]

>>11725536
Sorry, didn't mean for the picture to post sideways.

[Anonymous No.11725565]

>>11717667
I agree with this interpretation.