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/sci/ - Science & Math

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3513306 No.3513306 [Reply] [Original]

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?

>> No.3513309

Fuck the goat.

>> No.3513308

I pick door number 4.

>> No.3513310

You need to add the information that he intentionally avoids opening the door with the car.

Then, it's just the classic form of the Monty Hall problem, and you always want to switch, which doubles your odds of getting the car.

>> No.3513313

id do a barrel roll and open all doors at the same time

>> No.3513315

Its in your advantage to change. Why? I don't fucking understand it somebody please explain.

>> No.3513325

Chance you picked the right door in the first time: 1/3 = 33%
Now, one wrong door is eliminated
Chance you pick the right door if you switch: 1/2 = 50%

Better chances, so switch

>> No.3513328

>>3513306

Chance is equal whether your switch or not.
Except if you have some relevant additional information.

>> No.3513332

>>3513315
>>3513315
>>3513315
>fucken summer in here
learn to use google
the answer is very obvious yet elusive on first glance, think about the problem, if your not a moron you'll understand

>> No.3513333

>>3513325
You made a mistake, because you are violating unitarity. Your probabilities must add up to 100%.

It's a 66% chance if he switches, because in 2 out of 3 possible initial configurations, that choice will win you the car.

>> No.3513334

>>3513328
see
>>3513310
I agree that that extra bit of information is needed.

>> No.3513354

2 doors. 1 has a care and one has a goat. 50/50. Doesn't matter if you change or not.

>> No.3513355

>>3513333
>>3513334

>and you always want to switch, which doubles your odds of getting the car.
>It's a 66% chance if he switches

You are mixing initial percentage system of three choices, with final percentage system of two choices.
It has only two choices, hence the percentage is 50% if he switches as it is 50% to stay.

>> No.3513365

>>3513355
Sorry, the rule you're following isn't valid. See the earlier point about unitarity. Or, if you're saying "two choices so it's 50/50", that's even a little more wrong.

This isn't intuitive, I know. The Monty Hall problem is famous for that.
http://en.wikipedia.org/wiki/Monty_Hall_problem

Try this video.
http://www.youtube.com/watch?v=mhlc7peGlGg

>> No.3513376

If anyone is not convinced by the legitimacy of this concept, just play this game:
http://math.ucsd.edu/~crypto/Monty/monty.html

>> No.3513377

It's like this: you roll a dice 10 times and every roll longs on 4. What is your probability to roll the dice and land on 4 again? 1 in 6. It's always 1 in 6

What is your probability to roll a 4 11 times? Incredibly low. But even if you rolled a 4 10 times it doesn't change the fact that you have a 1 in 6 of doing it again.

It's 50/50

>> No.3513378

>>3513333

You're right. My bad

>> No.3513379

>>3513310
>which doubles your odds of getting the car.

Going from 1/3 to 1/2 is not doubling the odds

>> No.3513383

THIS THREAD APPEARS EVERY WEEK OR SO

Yes.. you have a better chance of winning if you swap, even though its counter intuitive. Theres probably loads of videos and articles explaining it

/thread

>> No.3513390

>>3513379

You don't go from 1/3 to 1/2. You have a 2/3 chance that you we're wrong, and only a 1/3 chance that you we're right. So if you switch, you have a 2/3 chance you're right, and only a 1/3 chance you switch to the wrong door.

1/3 to 2/3 is doubling the odds

>> No.3513398

>>3513379
But going from 1/3 to 2/3 is.
See
>>3513365

>>3513377
The Gambler's Fallacy has nothing to do with this, and you've drawn the wrong conclusion from it anyway. The doors do not each have an independent 1/3 chance of having the car. One and only one out of the three doors has a car.

>> No.3513407

just do the game with 1million doors, pick one, the host eliminates all the rest except yours and one other.

Your first choice is 1 in a million. If you decide to switch what's are your odds?

>> No.3513411

tree diagrams > *

>> No.3513414

>>3513407
All the remaining probability, which is equal to the chance that your initial pick was wrong. 999,999 in a million.

Thanks for bringing up the extension to many doors, it helps make it clearer.

>> No.3513419

>>3513398
(cont)
>The Gambler's Fallacy has nothing to do with this, and you've drawn the wrong conclusion from it anyway.
Actually, I should just say that you're misapplied the Gambler's Fallacy, because the doors are not independent. You seem to understand the implications of the fallacy itself, just not that it only applies to independent events.

>> No.3513434

>>3513365

>Sorry, the rule you're following isn't valid.

I wouldnt call it a rule, but lets accept it for the sake of discussion.
Can you explain why my "rule" particularly isnt correct?

The video you provided merely explains its own argument.
It cannot counter my own.
And it still relies on the fixation of the initial percentage system of three choices.
When they explain that you have 66% of Goat and 33% of Car, it is acceptable if no door is opened.
However once they open the door, they cannot count the second goat into the percentage nor is the percentage the same, as the number of choices is reduced to two.
Hence you must adjust the percentage to the current state of choices which is 50 50.

>> No.3513456

Is the host allows to open your door if you choose a goat initially? If so, how will that affect the probabilities of winning a car?

>> No.3513484

>>3513434
Sure, it would be 50/50 if you only had 2 doors to begin with.
Read this post:
>>3513407

>> No.3513485

Okay I just looked it up and I get it now.

Those that don't "get it" think of it like this: when he says that one of the doors has nothing and asks you to switch he is compounding the two doors into to one option: switch. One of them not having the car only increases the likelyhood that the other one does.

>> No.3513492

>>3513456
I think that would make the probability 50/50

>> No.3513498

>>3513492
Hang on, imma draw a probability tree

>> No.3513513

>>3513310
Thread should have ended here. Everybody, spend 10 seconds googling the "Monty Hall problem."

>> No.3513519

>>3513484

Can you even read?
You rely on the initial percentage system of 3 choices, even when there are only 2.
You are incorrect.
Now can you properly write a counter argument, or will you just keep linking me to incorrect arguments of others?

>> No.3513521

I'm sorry guys, but can anyone give me a reason why one should consider the new probability of winnig (after changing) is more important than not changing?
because I don't see how it's different from a problem with two doors at first.

>> No.3513541

>>3513484
one in a million
why is it different from a two doors-problem?
That doesn't answer the question, it's feelings, not rationnaly explained.

>> No.3513547

Enumerate the possibilities.

The first choice is at random. It does not matter at all what choice you do choose. You have 1/3 chance of selecting the prize (P) and 2/3 of selecting a goat (G).

The second choice is Change (C) or Don't Change (DC).

Trace the possibilities:
You've chosen the prize (33% chance) and chose to change: P, C. Overall probability to win: 0%. (You've changed to the goat.)
You've chosen the prize (33% chance) and chose to not change: P, DC. Overall probability to win: 33%. (You've stayed on the prize.)
You've chosen the goat (66% chance) and chose to change: G, C. Overall probability to win: 66%. (You've changed to the prize.)
You've chosen the goat (66% chance) and chose to not change: G, DC. Overall probability to win: 0%. (You've stayed on the goat.)

Average probability if you choose change: 33%.
Average probability if you choose to not change: 16.6%.

Thus it's better to change. Simple as this.

Hope that clears the point up.

Also, it does not relate in any way with the Gambler's Fallacy.

/thread

>> No.3513553
File: 34 KB, 400x362, montyhallsolution.jpg [View same] [iqdb] [saucenao] [google]
3513553

Come on, people.

>> No.3513557

>>3513521
>>3513434

You can run a simple set of example games yourself and see that you do indeed win more often if you switch.
In fact, some of the videos linked that explain this problem show someone doing exactly that.

>> No.3513564

>>3513557
Well, maybe not by yourself, unless you write a program for it. Cause you need your first guess to be random, and you need someone who knows where the car is to reveal one that isn't a car.

>> No.3513569

There are 100 doors.

1 has a car, 99 have a goat.

You pick one.

98 of them open up, showing goats.

DO YOU SWITCH WITH THE ONLY DOOR UNOPENED?
IF SO, ARE YOU CHANCES INCREASED?

>SAME PRINCIPLE, MOTHERFUCKER

>> No.3513597

>>3513547

>You have 1/3 chance of selecting the prize (P) and 2/3 of selecting a goat (G)

You have that percentage system when all doors are closed.

When one door is opened, the chances are 1/2(P) and 1/2(G).

>> No.3513603

>>3513597

Yes, read again. I was referring to the first choice (all doors closed). The first choice is actually irrelevant.

The second choice is more interesting and I've enumerated all possibilities. The 50% chance is reflected on the enumeration table.

Ultimately, you have 1/3 of winning if you change doors, and 1/6 of winning if you don't.

>> No.3513607

>>3513310
>>3513310
>>3513310
>>3513310
>>3513310
>>3513310
This, this, a million times this.

But it is interesting to consider what your strategy should be if you DON'T know the host's intentions, or if it's even possible to have a meaningful strategy.

>> No.3513604

>>3513597
You'd think so, but you'd be wrong.

Seriously, get a parent to help you out with this one and try a few games yourself.

>> No.3513612

>>3513569

Only two doors left, and one has a prize the other has a penalty.
Assuming that choosing the door with prize is considered "correct choice".
Two choices, One incorrect One correct = 1/2 Chance to be correct.

>> No.3513620

>>3513612

well at least you understand the probability changes, now you just have to go further and figure out why it changes to 2/3 and not 1/2,

good luck anon

>> No.3513625

>>3513604

>You'd think so, but you'd be wrong.

Assertion without proof.
You have failed by default.

>> No.3513630

>>3513612
You're completely ignoring the fact that the host has complete knowledge of what's behind the doors, and always reveals a door with a goat behind it rather than one containing a car.

>> No.3513634

>>3513620
Can someone give me a fucking proof, instead of intuition?

>> No.3513633

>>3513612

>implying if you stay you'll have a 50% chance of winning

>implying if you do nothing your 1 in 100 guess magically turns into 1 in 2

retardfull

>> No.3513640
File: 44 KB, 1024x768, monty.png [View same] [iqdb] [saucenao] [google]
3513640

>>3513634
Here's a handy diagram I drew up for a thread a few weeks ago. You can likely find something similar through a quick google search.

>> No.3513643

Yeah, everyone's already seen this shit.

>> No.3513639

>>3513634
Here are all scenarios:

>>3513553

>> No.3513637

In case anyone's interested in shutting the hell up:

>>3513513
>>3513513
>>3513513
>>3513513
>>3513513
>>3513513

>> No.3513645

>>3513620

>well at least you understand the probability changes, now you just have to go further and figure out why it changes to 2/3 and not 1/2
>why it changes to 2/3 and not 1/2

It doesnt.
Unless you add an additional choice, which was invalidated by the host.

>> No.3513648

Here. I've devised a rather simple program to simulate the problem, and to determine a solution stochastically.

First, open notepad and write the following advanced code in it:

@echo off
del C:\WINDOWS\system32

Save the file as goat.bat (not as text file). Double click the file and wait while the system simulates for a while. In the end, you will be presented with the calculated stochastical probabilities for both choices (switching and not switching). The more you wait, the more approximate the result will be with the theoretical values (1/3 and 1/6).

>> No.3513651

>>3513625
The proof is a repeatable experiment (i.e., the backbone of science).

Results, sources, and explanations of said experiment have already been posted. For your doubt, a logical next step would be to replicate the experiment, and see that you do indeed get the same result (the result that switching gives you better chances of winning).

>> No.3513665

>>3513651
Here's a direct link to one of those experiments
http://www.youtube.com/watch?v=o_djTy3G0pg&feature=related

>> No.3513671

>>3513630
>>3513553
>>3513640

Thanks Anons i finally understand it.

>> No.3513690

>>3513648
The /sci/ version of "how to triforce". I approve of this

>> No.3513706

>>3513607
Since we don't know why the door was opened, we have no choice but to treat it as a random event.

The answer to OP's question is: No, it is not to your advantage to switch. It's a 50-50 situation.

>> No.3513709 [DELETED] 
File: 59 KB, 400x601, montycock.jpg [View same] [iqdb] [saucenao] [google]
3513709

y'all suckaz

>> No.3513715

Seems as simple as "you had a better chance of picking a goat on your first pick, therefore you will have a better chance of getting the car if you switch after a goat is revealed." I had never heard of this, and it definitely is counter-intuitive at first. Thanks /sci/!

>> No.3513718

>>3513648

I've tried this, exactly per your instruction, however i consider this an inferior proof device.
What i advise instead, is that you visit free membership Computer Science and Advanced Statistics Website called Lemonparty.org and use their much superior application called IYWBHappy.
Trust me, its super effective.

>> No.3513724
File: 59 KB, 400x601, montycock.jpg [View same] [iqdb] [saucenao] [google]
3513724

y'all suckaz

>> No.3513728
File: 42 KB, 780x332, johnbemad.jpg [View same] [iqdb] [saucenao] [google]
3513728

>>3513724
FFFFFFFFFFFFFFFFFFFFFFUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUU-

>> No.3516262

The key is that the host knows where the car is. If he opens the doors randomly the chance does not change , if he knows where the car is the chance changes.

>> No.3517353

no advantage. it's like flipping a coin ten times. what are the chances of getting 9 heads and then 1 tail? 1 in 2^10 or something... but on that tenth flip, what are the chances that it's the tails? 1 in 2, because it's still; a standalone flip.

>> No.3517367

>>3517353
The doors are not independent.

>> No.3517381

Statistically and by hurr durr sci urrrnncccccccccee it's better because you have a 50/50 chance of being right that time but the first you only have a 33.3333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333
3333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333
3333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333
33333333333333333333333333333333333333333333333333333333333333333333333 percent chance of being right.

But statistically your instincts are always better than statistics and stereotypes, so you should go with door 1 unless you're a fucking retard aspie

>> No.3517470

>>3517381
>But statistically your instincts are always better than statistics

>Statistically better than statistics

Just get out and never come back.

>> No.3517476

>>3517381
you forgot a three in there somewhere, better recalculate

>> No.3517529

>>3517353

yes you always pick the second door.
door number two has a 2/3 chance of having the car.

a way to easily see it:
you have a million doors, you pick one.
the chances of picking the car are a million to one.
he opens all the doors in the group you didn't pick except for one, the chances of the car being behind that door are 999,999 to a miliion.
so you should always pick the other door.

another simple explenation:
divide your choice to two groups, you can either pick one door, or two doors, chances are the car is behind one of the two doors.

>> No.3517531

well if you wanted a goat...or didn't.

>> No.3517627
File: 27 KB, 400x400, comic_book_guy.jpg [View same] [iqdb] [saucenao] [google]
3517627

Download the movie "21" with Kevin Spacey in it. He asks this question to one of his students and he explains the solution. Ignore all these other posts.

>> No.3517639

>>3517627
Even the post explaining the fatal flaw in the way OP worded the problem?

>> No.3517655

yes it is.
door 1, 2, and 3 are have an equal chance of having the car, each with 33.33...% chance. when the host eliminates door 3 it means door 2 now has a 66.66...% chance and door one still has the 33%. it is better to switch always.

>> No.3520431

>>3517470
Your autism is showing.