It's that time again faggots

It's that time again faggots.
As always, I'll give you the answer...

You double your chance to win by switching.
Now argue relentlessly because stupid.

Other urls found in this thread:

youtube.com/watch?v=4Lb-6rxZxx0
math.ucsd.edu/~crypto/Monty/montydoesnotknow.html‎
en.wikipedia.org/wiki/Monty_Hall_problem
twitter.com/NSFWRedditGif

> inb4
I'm a Muslim
I want a goat
The other door is the goats ads

I'm Hindu, i want the goat.

You got inb4'd you stupid faggot

> I'm Hindu
> Hindu doesn't read thread before posting

It's 2/3 and Hindu are dumb

Hurrr durrrr
Must be 50/50 because retard

The first choice doesn't matter. Regardless of your first choice, by opening a goat door after your choice, and allowing you to switch they are essentially saying:

"You have 2 doors now, 1 has a goat, 1 has a car. Which do you choose."

Choosing to switch means you choose door number two.
Choosing not to switch just means you are choosing door number one.

Hence, as of the second round of choices, between the only two options you can make, and the only potential results: your odds are 50/50.

50/50 obviously

Explain how you can have a 50/50 from a choice of three?

switch, 2/3

you can't... you don't have a choice of three.. the First "choice" is irrelevant. Your only choice is to choose whether to switch or not between a choice of two.

Two choices. 50/50

2/3 chance of winning if you switch, so yes, you switch.

If you select the first door in each of these, and he shows you a goat, you win in two of them if you switch. Hence, 2/3 chance to win if you switch.

C G G - lose on switch
G C G - win on switch
G G C - win on switch

Switch the doors. 66.66% chance car is behind the switched door

False.. That last option does not exist (The G G C) as they already showed you the right door contains a goat.

so
C G [G] - lose on switch
G C [G] - win on switch

50/50

That's 100% wrong...

A good way to think about this is; what if there were 100 doors you had to pick from? You pick one, and 98 doors are removed leaving only one, which you can switch to. Of course, the chances that you picked the correct door from 100 are low, so you would switch. Same applies with three doors.

You don't double your chances retard,
you improve your chances from 1/3 to 1/2 when switching

Nigger, what the hell do you mean the last option doesn't exist? He just opens a door with a goat behind it, not the rightmost door.

yet in the end..you are still making a choice between two.

you should go to the door number 2 because mathematically there is a 2/3 chance that behind it there is the car

But there is not an equal chance that each contains the car.

and once he opens the door, what are the potential results?

The third result that has the car in the door that was opened ceases to exist.

No, you go from 1/3, to 2/3.

There are 3 options.
There can never be 1/2, because that 2 means "out of 2 options".

There was/is 3 options.

Yet in the beginning you made a choice out of three.
That's the important part

You don't double your chances you stupid fucktard.
You don't go from 1/3 to 2/3 you go from 1/3 to 1/2.

There are three options, him opening a door doesn't remove that possibility.

The total probability of there being a car is 100%, right? There must be a car. Your first choice has a 1/3 chance of winning, with 2/3 distributed between the remaining two doors (1/3 per door). When one of the other doors opens, you know that the 2/3 chance is still the chance it isn't behind your door (your door hasn't changed) and the 2/3 chance is behind the door he didn't open because you know it isn't behind the open door.

And yet the results of that choice provides you with no new information.

You know that.

No matter what door you choose, in the end, you will have 2 door. 1 with a goat, 1 with a car.

Switch.

Chance for door number 1 was 33%, chance for door number 2 is now 50%.

> 3 doors
> Car is behind one of them
But there is not an equal chance that each contains the car.

Obviously not you fucking retard.
That doesn't make any sense.

There is however an equal chance that EACH door MAY contain the car

This falls under the flawed gamblers logic though.

The idea that on a 60/40, if you went with the 60 you *should* have won.

...but you don't... because in the end, its still 2 doors, 1 choice.

What a stupid fucktard

That's not the gambler's fallacy, don't just throw words around that you heard one time.

Educate yourself:
youtube.com/watch?v=4Lb-6rxZxx0

The big issue here is that if you get the goat after switching you look like a huge trolled faggot so youre better off staying even if you lose you retain your dignity.

This is not gamblers fallacy
Nor is what you tried to describe.

Heh, funny how on a gameshow where the goal is to have you not pick the car, there is this so-called "clear strategy" that everyone should follow.

Piece-of-cake right?

Her own logic is inherently flawed.

But you're twice as likely to get a goat if you stay.
You're a retard, and I feel the need to tell you not to post pics of yourself on Sup Forums

sure it is..because even with the 100 door problem..

There is the idea that you *will* make the wrong choice, because the odds are only 1%.

Considering a 1% chance as a guaranteed failure IS the gambler's fallacy.

"I, a random fag on Sup Forums, know better than a statistics professor. They are clearly wrong on this well known and easily illustrated problem."

math.ucsd.edu/~crypto/Monty/montydoesnotknow.html‎

When majority of people relentlessly disagree, even when given irrefutable evidence...
It's a pretty effective strategy

Las you implied, people don't like to give money away

Its the same as the prisoner's dilemma.
The statistics professor tells you to always sell out the other person. But a tiny shred of confidence tears the whole base idea down, because it assume that all humans are whiny, distrusting bastards.

That is not gamblers fallacy either..
Pic related is gamblers fallacy you dumb fuck.

It's a 50/50 anyway

No

>one doot has a goat
>the other has the car
Retarded or "merely pretending"?

Then its the Texas Sharpshooter Fallacy. Still a fallacy, numbnuts

Switching is a 1/5 chance, staying is 1/3.

I mean switching is 1/2

Its the same as the homo dilemma.
The homo professor tells you to always sell out the other homo. But a tiny shred of homodence tears the whole base idea down, because it assume that all homos are whiny, distrusting bastards.

I have no idea how This has anything to do with anything.
It doesn't even relate to the thread.

I am a dumb confused homo.

> forgets there is a third door
Definitely retarded.

No it isn't.
No fallacy, only fact.

The funny thing is that, when there is no expectation of retribution, people quickly forget about that shred of confidence.

Indeed you are.

Fucked up his own shitty troll attempt.
An hero is the best choice now

ez
Not switching means you have to get the car on the first try, i.e. 1/3.
Switching means you have to pick a goat from the start, because you will then be guided to the car, i.e. 2/3.
Always switch.

Its basic fight or flight nature.
Those who tend to focus on flight take the route which seems to have less risk, though it is more detrimental in nature.

Those who focus on flight challenge that assumption, and doing something as simple as sticking to ones guns, often come out ahead.

Haven't heard "I know you are l, but what am I" in 25 years.

Underage b& pls go

...that last one was fight*

Oh, you were referring to someone else? I thought you were just coming out. I was so proud of you.

I think it is even simpler: greed. If there is no punishment for greediness then there is no feedback against it.

"I'm switching" is just candy-coated version of saying "I choose option B"
"I'm staying" means "I choose option A"

it's 50/50

Laughing at you, not with you
Pls keep saying clever comebacks for fun

No no... "You're a retard" followed by "Yes you are." is that.

You were just acting liking an idiot, and got mad someone took you at your word.

Hmm, you don't seem to understand concepts too well. Maybe logic and philosophy threads should not be your hangout.

Except option A is one door, and option B is 2 doors

Switching doesn't mean you pick option B, it means you will be guided to the car if your initial choice is already a goat. The host will always open the door with a goat, so if you pick A and the car is in B, he will open C to reveal a goat, if the car is in C he will open B to reveal a goat. So he guides you to the car. read It's 1/3 or 2/3

So we are in agreement they are clever? Fantastic!

He's taking to himself guys...
Let's just quietly avoid this guy because hes obviously mentally unstable.

Though I suppose anything would be considered clever compared to replacing random syllables with the word 'homo'.

This entire thread is clearly just 1 person. Speaking of, what should we do for lunch?

I get that you're so shit and useless you have nothing to offer, but trying to force cringe into your posts won't get you Sup Forums fame.
Everyone will just laugh at you as they do in your offline life

Best explanation I ever heard:

>You have a choice of 10 doors. One hides a car, the other nine hide goats.
>You select number 2
>Game show host opens all doors except door 2 and 7, showing that they all hide goats.
>Now you get to select to stick with your original choice, or you may take door 7 instead.

Rational people understand that your chance of hitting the right door was 1/10. Now that the other doors are opened, your original choice still has 1/10 chance of being correct. The other 9/10 chances all reside with the other door, number 7.

The choice isn't 1/2 vs 1/2 but 1/10 vs 9/10.

In the 3-door problem, the second choice is 1/3 vs 2/3.

Syllables?
The harder you try, the deeper you dig

If 10 doors is not enough, you can go up to infinity. Twice. Some people will not understand though.

Bacon

Alternate theory.
you choose 1 door, with a 1/10 chance.
As each door is subsequently opened, the odds that you chose the 'wrong door' become less and less apparent.

The flaw here is the idea that only the door you didn't choose absorbs the excess probability that is lost when a door is opened.

This problem should be easily simulated with a small Java app or something. Programming fags hop to it.

You assume the host reveals every goat and not just one. You could pick door 1 and he would reveal door 10 to be a goat. Much lower chance now, though still higher than if you stay

Also consider an alternative.
10 doors presented.
You choose no doors.

8 "false" doors are opened.

What are the odds you choose the right door?

For fuck's sake these retards again.
Because the host knows where the car is, he has to open a door with a goat. They don't just open one at random. That means that the door they did not open has a higher chance of being a car than the door you picked originally.

produce a counter argument then.

The fact that he knows is precisely why the probability does not transfer. The end result will always be 2 doors, 1 goat, 1 car.

The rest is just showmanship.

You made a post to say your post is wrong.

Retards only read the first few lines.
You just fed them

?
I can only assure you that the post you are responding to and the post *it* responds to are not the same person.

Perhaps I am misunderstanding your meaning here.

Been done countless times.
If you search Monty Hall simulation you'll find many to play at home.

It's 2/3 by switching.
It's fact, Not a debate.
I post it here because the retards are entertaining.

The real answer is if you choose the goat on first pick, the "bob barker" has an opportunity to open the car door and leave you with no 2nd chance at trying to win.

He reveals every goat except one.
Not all.
You're a fucking retard

>what is the monty doesnt know problem

I believe you are conflating a "fact" with a "theorem".

There are still possibility to disprove a theorem.

Thats not the question here, so it's completely irrelevant.
I bet you thought you were being really clever.
How long did it take you to think that irrelvant shit up?

oh you again.
look OP with that kind of behaviour you are just proving the statement that OP is a faggot.
the old monthy hall problem :/
you are really a newfag and a loser in the real live. also die bitch.
here the solution:
en.wikipedia.org/wiki/Monty_Hall_problem

Did he tell you that himself?

Its established fact.
Counterargument is irrelevant.

If you weren't retarded you would understand this.

At first thought it was 50/50, but wikipedia cleared it:
your choice has influence over the door that is open, that event is neither random or disconnected
the fact that the presentator opens the goat door after you make your choice is why it hands more information, and therefore a higher prob
thanks user, learned something today

>The flaw here is the idea that only the door you didn't choose absorbs the excess probability that is lost when a door is opened.

There is no such thing as excess probability.
No door "absorbs" probability.

Yes, you are misunderstanding.

>monty doesnt know problem
My, you are an angry individual.

But I'm afraid you are incorrect.
The biggest thing it is meant to point out is that knowingly false options are going to be removed no matter what.

Not just 8 doors at random.

As you open each door. the probability gets 'absorbed' by the others (because the correct door will never be chosen to be revealed). So, your 10 doors 1/10 chance becomes 9 doors, 1/9 chance and so forth.

The silly idea that you are holding on to is that somehow pointing at one of the doors shields it from absorbing this excess probability, and it ALL goes into the other door that's left. And you assume that you *have* to be wrong, because you only started out with a 1/10 chance. (Which is flawed in and of itself. Like the lottery. No one has to win, but it can happen.)

Thus the giant flaw with the "Monty Hall Problem".

What you believe is irrelevant.
This is mathematical fact.

Practical fact too.

It's actually very simple to understand, except you can't let go of your initial belief

You're just cranky I've been spamming your precious WWYD threads with...

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False, because it is involving compound probabilities.
Ai : "Door number i contains a goat"

When you do your first choice, ∀i P(Ai) = 1/3
but when a door is open, then P(A3) = 1
So then, and then only:
P(A2|A3) = P(A1∩A3)/P(A3) = 2/3

While still P(A1) = 1/3

So you should change your door, it's mathematically demonstrated now.
Faggot.

show where it's established as fact.