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/sci/ - Science & Math


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5106015 No.5106015 [Reply] [Original]

Ok is there someone here who can explain the Monty Hall problem simply?

>> No.5106016

Imagine there are one million doors, and the host opens every single door, all containing goats, except 2: one contains a goat and one contains a car.

>> No.5106018

>>5106016
I can understand that, since the probability of someone picking the 'right' door in your example is very small.

However let's say that there are 3 doors and the host DOESN'T know where the car is and he simply opens one with a goat. Should you still change door?

>> No.5106020

>>5106015
Probability means how believable something is, and this is why information changes probabilites. Knowing that a dice is weighed against some outcomes changes the chances, right?
The first time you choose a door you make an uninformed desicion, and when you change the door, you make an informed desicion.

>> No.5106021

First choice is 1/3
Second choice is 1/2

Probability is retarded.

>> No.5106022

>>5106021
but that's wrong you retard

>> No.5106023

>>5106018
Nope. It relies on the fact that the host opens a door non-randomly. He is actually giving you more information than just 'the door I opened was a goat'.

>> No.5106034

Lets see

On yout first choice, you have a 2/3 to choose the goat door and 1/3 to get the car door.

Lets focus on the first case
Assuming you got the goat door, two doors remain, one with a goat and one with a car, the guy opens up the car one, so by switching you get the car

Since it is more likely that you got a goat door on your first choice, this is the best course of action 2/3 vs 1/3

>> No.5106035

>>5106018
Yes, it makes no difference as long as the information gets to you

>> No.5106037

>>5106034
>the guy opens the car one
Meant the goat one, my apologies

>> No.5106045

Humans suck at probability so your instincts tell you wrong.

Which door the quizmaster opens depends on which door the prize is in. So which door you call "the other unopened door" depends on whether the goat is in there: "the other door" is *whichever of the two doors has the prize in it, if any of them does*. So if the prize is in either of the doors you didn't pick, it will be in the door you call "the other door". Your mind doesn't count the probability that you decide to give something a label into account when predicting what it'll do.

>> No.5106101

>>5106015
think of it with a billion doors and 999,999,998 are open after your first choice. would you really believe it's a 1/2 choice?

>> No.5106112 [DELETED] 
File: 34 KB, 393x364, montyhall.gif [View same] [iqdb] [saucenao] [google]
5106112

>>5106015

>> No.5106131

>>5106112

There's really no better way of explaining it than this. If you switch, you have 2/3 chance of winning, and only 1/3 if you don't.
It relies on the fact that the host knows which door the car is behind and MUST choose a goat to reveal.

>> No.5106133

maths msc here - it's 50:50. either there's a goat or there's not.

>> No.5106141

>>5106133

Congratulations on revealing that you have no qualifications in maths. I direct your attention to this >>5106112

>> No.5106143

>>5106112
This is the best explanation.

>> No.5106145

>>5106133
But 50/50 is 1 so you are guaranteed to win?

>> No.5106145,1 [INTERNAL] 

Okay since I am banned from 4chan I will explain this problem here.

Imagine an infinite row of doors. Only one door has a car behind it, all other doors have a goat behind it.

You pick 1.

All the doors open except the one you picked and one that you did not pick.

The reason the door you picked did not open is most probably because you picked it.

The reason the other door remains closed is most likely because it has a car behind it.

Get it? Simple. Now let's generalize it.

In any situation where there are more goats than cars, your first choice is likely to be wrong. After revealing all the other wrong choices, you are left with your likely wrong choice and the remaining choice which is likely correct.

>> No.5106149 [DELETED] 

>>5106141
You have been trolled.
That one is quite common.
https://archive.installgentoo.net/sci/thread/S5056419#p5056502

>> No.5106156

K bro here's how it goes down.
You got 3 doors.
2 have goats
1 has car
You want to car.
So you have 1 goat which you have 33% chance of picking it.
Then another goat, so you have 66% chance of picking a goat
There is 1 car, so you have 33% chance of picking that door.

You pick your door. So you had a 1/3 or 33% chance of getting a car and a 2/3 or 66% chance of getting a goat.
Then one door is opened. The door that is opened will always be goat, no matter what.
So now there is 1 goat and 1 car. You get the option to change because fuck you. So since one goat got eliminated, we have 66%-33%
so you have 33% of choosing the goat if you switch. So where the fuck did the other 33% from the other goat go? It goes to the car, so its 66% chance of getting the car.

I'm not a maths genius. But thats how shit goes down.

>> No.5106158
File: 43 KB, 344x517, justwait.png [View same] [iqdb] [saucenao] [google]
5106158

>>5106156
>You want to car

>> No.5106187

>>5106156
>>5106112
I do get the math behind it, but it strikes me as the same fallacy as with the coin toss: "Get a thousand times heads, how high is the chance for heads on the next throw?"

>> No.5106191 [DELETED] 

>>5106187
http://en.wikipedia.org/wiki/Gambler's_fallacy
Still 50%

The Monty Hall problem is different.
Switching yields 2/3 chance of winning.
Not switching means you only have 1/3 chance of winning.

I posted that image.

>> No.5106222

>>5106191
did you see the quotation marks? probably not. I simply didn't recall the name, thus described the fallacy.
I asked for in how far it differs.
Sure, the information reveal modulates the previous probability, but it doesn't change anything about the whole "two doors and two objects to choose from" situation afterwards.
the same way how statistically the previous coin toss modulates the coin toss sequence, in actuality it doesn't change anything about the coin toss itself.

>> No.5106229 [DELETED] 

>>5106222
It is still not the same.

Initially it is a straight up choice between 3 doors, so you have 1/3 chance of picking correctly.

The second choice is to switch or not. It is a choice only between 2 things.

If you do not switch, then you are gambling that your 1/3 chance of being correct before, is indeed correct. Which is still 1/3.
If you do switch, you are betting that you are incorrect.
Switching when incorrect on tthe 1/3 choice, automatically means that you definitely win.
Therefore 2/3

Switching is advantageous.

>> No.5106245

>>5106229
well, as I said, I never had any problem with the pure mathmetical concept. it simply struck me as odd in a real world scenario type of way.
I just don't want to accept it for some reason. Though this will serve well enough to make me shut up
>Which is still 1/3.

>> No.5106249 [DELETED] 

>>5106245
>I just don't want to accept it for some reason.
The maths is sound.
How can you not accept it?
The image I posted is very straightforward.

>> No.5106252

>>5106245
>I just don't want to accept facts for some reason.
Oh you, Americans.

>> No.5106302

>>5106249
>>5106252
I actually wanted to reply in a sensible manner, but 4chan's shitty algorithm (whoever the ass face retarded shitfucker was who wrote the damn thing is, I sincerely hope that person dies a horrible death) keeps interpreting my post as spam.

saging this shit

>> No.5106332

The idea that there are people here who cannot wrap their heads around such a simple problem is really rather depressing.

>> No.5106355

>>5106332
yeah, nah, you're just playing a smartass cause wikipedia spoon-feed you the answer

>> No.5106357

>>5106355
No, not really. I've never understood why people think the problem is difficult.

>> No.5106359

>>5106018
Yes, you should still change the door as the decision of the host gives you additional information.

The host would of course always know where the car is as it's pretty damn stupid if the host accidentally opens the car door
"oops, well, i guess you're stuck with a goat dear contestant." would be pretty damn anticlimatic

>> No.5106362

>>5106357
>>5106357
>I've never understood why people think the problem is difficult.
People are retarded, that's why.

>> No.5106390

>>5106359
If we assume the host does not know and picks a door at random, but only keep cases where the host opens a door containing a goat, then every program I have written to check suggests that switching and not switching are always the same [win half the time, lose half the time]. If the host has knowledge, then you get the 1/3 win without switching and the 2/3 chance with switching. If you don't believe these results, right your own program and post the results. Why people argue over this shit when it is so easy to test is beyond me.

with the host have no knowledge:
wins with switching: 1493/3000
wins without switching: 1507/3000

with the host having knowledge:
wins with switching: 981/3000
wins without switching: 2019/3000

>> No.5106394

>>5106390
I seem to have switched the last two sets of numbers. Oops.

>> No.5106447

People, people, this is one problem where just thinking about it won't go anywhere. Let's actually test the thing, rig up a ball& cup setup with two people, run the scenario a hundred times switching every time, and see what the results are.

>> No.5106472

>>5106447
I'm resisting the urge to attempt to name this process.

>> No.5106474

>>5106362
>>5106357
A lot of mathematicians and theo.physicists didn't believe it at first

>> No.5106497

>There is a formal mathematic solution on the wiki page
>says 2/3
>if you're too dumb to understand, leave /sci/ forever

Saging this shit.

>> No.5106533

>>5106474
And I've never understood why. I don't consider myself smart at math or anything, but I cannot understand what people find confusing about the problem. There are 3 possible situations; in 2 them, you win if you switch, and in the other one, you lose if you switch. Thus 2/3 chance of winning if you switch. I don't get how people can seriously argue about it, it just seems so simple. When I first saw people arguing about this problem on /b/ many years ago I thought they were trolling.

>> No.5106642

Okay, wow. I was firmly in the "choice is 50/50" crowd, but after running >>5106447 in a Python program (1000 tests), the results were 651 cars and 349 goats.

It really does seem wrong, but I stand by my results. Switching the choice DOES improve your chances.

Program if anyone cares:
https://www.dropbox.com/s/qe0p4ltw1ow163f/MontyHall.py

>> No.5106645
File: 240 KB, 1911x1038, hhhhhh.jpg [View same] [iqdb] [saucenao] [google]
5106645

Just made it,hope you like it.

>> No.5106682

>>5106645
i lol'd at the vrooom vroom and goat thingies

>> No.5106686

If you picked the goat he wouldn't be giving you the option.

>> No.5106687

this is the most stupid and misunderstood problem ever

the first time you pick the chance is 50%
problem solved

>> No.5106693

>>5106686
What.

>>5106687
What.

>> No.5106709

>>5106693
As far as >>5106686 goes, Monty the game show host wouldn't give you another choice if you picked goat. He would just reveal and give you your goat.

Giving away a car to 66% of contestants would bankrupt Lets Make a Deal pretty quickly.

>>5106687 is just stupid.

>> No.5106747

>>5106709
The game show is just an analogy. You don't have to go into details of whether it's sensible for the host to let you choose, just accept the situation.