The answer is you double your chance to win by switching

The answer is you double your chance to win by switching.
This is fact, no argument...

But you'll argue anyway, so argue away faggots.

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fuk off m8

Why you say that?
To dumb to figure it out?

> retarded Timmy
> Timmy say fuk off m8
> Timmy can't explain his retarded outburst

Good job Timmy...
Your very special and your carers love you very much....

And if i want the goat? Im gonna send your whole family ass to allah for being this stupid

OP is a faggot.

33% -> 50% is not doubling.

I.forgot about faggots like you..

Take your beastiality elsewhere...

Monty Hall Paradox

You don't double your chance to win by switching its an increase of 1 in 2 odds from 1 in 3 odds 50% as opposed to 33.333...%

IIRC, the logic behind this is that

>Odds of choosing the right door both times is (1/3) * (1/2) = (1/6)

>Odds of the other door being the right door after the change is (1/2)

>(1/2) obviously greater than (1/6) so you should switch

But that's fucking stupid, the choice you made in the first instance shouldn't have any effect on the probability of choosing the correct door at all once you knock one of the doors out so it should still be 50/50, no?

But you are the faggot...
It 33% to 66%

If you're so smart, explain how there can be a 50% outcome from a choice out of three

You won't, because you can't.

Monty Hall problem is poorly explained by Kevin Spacey. It's easier if you remove Monty and just think of it as an automated process.

1. 3 doors
2. You pick a door.
3. A door which you did not pick, and which is not the correct door, is opened.

> both times
Both?
Where did 2 choices come from?

The posterior probability of a correct choice is not 66% after switching. Lern2BayesRule

Sorry, I meant the first choice is your initial pick, the second is after one of the wrong doors is revealed and you have the option of changing your choice or not.

1/3, 1/3, 1/3
One is opened.
Now it must be on one of the two remaining. Your choice still has 1/3. Therefore the Other has 2/3 ( the rest) of having the prize.
Changing nets you 1/3 extra chance.

Fk dumb

I dont know who Kevin is, but your 1 2 3 steps are accurate.

no no,

The chance you choose the wrong door is 1/3. If you switch, the chance you get the wrong door is 1-(1/3)=2/3, namely you get the wrong door if you chose the correct door initially, which is 2/3 chance.

But thats wrong and you're fucking retarded

Almost everything you said was wrong, so first second or third, it's got nothing to do with me. ..
Except for giving me things to laugh about.

Bayes rule doesn't apply as the choice isn't unconditional. It's always a door with a goat which is opened.

The math is like this:
Initial chance of picking right door: 1/3
Game host will then remove one goat door from pool.
Since there's a 66% chance we also picked a goat door, the last door has a 66% chance to be the car.

But it is...
Learn to reality.

You start with a 2/3 chance of picking a goat

That means two out of three times, you pick one goat and Monty shows you the other goat

Therefore the car has a 2/3 chance of being behind the door neither of you selected, so you switch

Fucking retards

Also easier to see intutively if you imagine 10 doors. You choose one with 1/10 probability, then 8 incorrect doors are opened.

All the probability mass is shifted onto the closed door you didn't pick.

Math time.

Trust me, it only gets better from here
Sit back and enjoy the ride.

Kill yourself...
Right now.

Go derail something worth derailing you moron.

Babies first contour integral is just too much.

It's funny because you're wrong.
Switching doubles your chance to win.

Of course it does. I think you're responding to the wrong person.

Pls explain contour integral.
Sounds like something a bricklayer or plasterer may use, but doesn't sound relevant to this at all

It's 50/50 for me
So it doesn't matter if you change or not

You are correct, it's not relevant to the Montey Hall problem. I just put it here to see if anyone can do it, because I'm an asshole.

The basic idea of a contour integral is that it extends the Riemann integral over R to the complex plane. They are useful in calculating results in physics and computing inverse Laplace and Mellin transforms.

I responded to the post that had 2 replies saying dumb shit.
Both you I'm assuming.

Maybe you should go back and read the posts you responded to, because they can't be dumber than your responses.

What does integration have to do with this problem whatsoever

how is this not the correct answer? it doesn't matter how the problem started, you have 2 doors to choose from, 1 is good 1 is bad you have no indication of what is behind your door or the other unopened door.

he just said it wasn't relevant

Spouting irrelevant shit is how you fail your classes user.
Do you want to fail you classes?

Oh, your not learning any classes, your self educated on basement standard.
Good for you Timmy

I'm not sure what you're getting at and this discussion isn't really going anywhere.

He's retarded.
He posted it so he could say it isn't relevant which serves no purpose other than to make him look foolish.

You're trying too hard.

This.

Either you pick the goat initially:
(G) G C

i can it explain better than you all did.
theres a choice of 66% you get the wrong door at first choice.

then one door is opened, and your initial choice is wrong at a higher rate(66%) than having hit the right door(chance - 33%).

So when 1 wrong door is opened, you probable have chosen a wrong door initially, so you should switch to the other door now because you can only switch between 2 doors that are left now.

now you dont choose anymore 1 door of three, you choose 1 door out of three while one wrong is granted open. so you basically chose 1 out of 2 -> bettter chances.

when you consider the chances as a whole, you choose one door out of 3 witch 1 wrong door granted, so your chance is 66%

It matters

It actually does. See

I wasn't getting at anything.
I was saying your an idiot, because everything you said had nothing to do with anything that had been said.

Fuck I mean

It matters because the game show host opening the door with a goat behind it didn't open a random door. The game show host knows which door is the winner and opened a non-winning door.
Thus part of the randomness has been removed.

> trying too hard
You said things irrelevant to anything relevant.
There was no trying there, you actually did that

No for fucks sake, you can't fucking "remove part of the randomness" if you still have two doors to choose from at the end.

The easiest way to understand the problem is to imagine there are a million doors instead of 3.
999999 of them contain goats, 1 contains a car.

You pick door #120743, host then reveals goats behind every door except #120743 (the one you picked) and say, #555319.

Now it should be obvious to anyone with half a brain that it's more likely that #555319 is the winning door, than that you randomly picked #120743 out of a million and got it right.

The fact that the host eliminates all the goats (except one, if you happened to pick the right one) is the key to understanding this problem

That's so much better than..
You are twice as likely to choose wrong.
If you chose wrong, switching wins.

But feel free to wrote another novel about it.

If you pick the right door he still chooses randomly

i can it explain better than you all did.
theres a chance of 66% you get the wrong door at first choice.

so after 1st choice, one wrong door is opened. your initial choice was(and is) wrong at a higher rate(66%) than having hit the right door(chance : 33%).

So when 1 wrong door is opened, and you probably have chosen a wrong door initially, you should switch to the other door now because you can only choose between 2 doors that are left now.

now you dont choose anymore 1 door of three, you choose 1 door out of three with one wrong door being granted open. so you basically chose 1 out of 2 -> bettter chances.

when you consider the chances as a whole, you choose one door out of 3 with 1 wrong door granted, so the chance you hit the right door out of three is 66%

Trips speak truth, Fuck off.

There is no randomness.
You should probably go back to the top and read all those words that are in the OP image...

See this
You . definitely explained it better with that novel you wrote.

Seen this,change door

This user is a great user.
No arguing, no right/wrong convoluted math equation, no bullshit at all.

He knows what right, and states it.

>Pick 2 doors out of 99999
>You know that one of them hides a car
>Pick one of these 2
What is the chance you pick the right one?

This is absolutely wrong.
Your initial choice doesn't influence your chance of winning at all. Regardless of whether you've chosen right or wrong, one wrong door is eliminated, leaving you with one right and one wrong door.
If you look at the stay/switch problem not as changing your previous selection, but as an entirely new selection (which it is), you're either selecting a right or a wrong door, thus resulting in a probability of 50%.

thank you

But there was only ever a choice to pick one door.

Rethink your shit now that you know that.

thats basically exactly what i said

yea, haha, just like lottery m8. either you win or you dont, kek

What are you thanking me for?

The other one is picked by the host

I know you're excited, but everything you said is wrong.

for your agreement that my long explanation explained it very well and better.

Sorry, that was meant for

But you didn't choose that door, and it happened AFTER you chose yours...

no problem

It's not about choosing anything
In probability it's just a random number

you guys don't get basic mathematics.

I said your explanation was longwinded and kind of crap, the linked to a much better, simpler, fee line explanation.

I'm happy you understand the concept, but you should lay of the drugs user

oh well then
never mind

youtube.com/watch?v=dciSc2rPSjE
We done?

and you don't get practical statistics

You people suck (unless you got 66%).

Here's a simple generalization to make it clear why the probability increases when switching.

Lets say there are 99 doors instead of just 3. 1 door has a prize, the others don't. Everything else is the same. You pick a door and the host opens another. Do you switch? Are the odds of the prize being behind your door greater or smaller than the prize being behind one of the 97 other doors?

kek

>Make it 100 doors: 99 goats, one car.
>You pick one.
>Then the host opens 98 goat doors.
>Only two doors remain.
>Yours, and 1 of 99 that you didn't pick
>You switch, because there was only a 1/100 chance you were right on your original pick.
>There was a 99/100 chance you were wrong initially.
>You WIN if you SWITCH after having been wrong initially.
>Same logic applies to the three-door case.

Then again on Sup Forums the goat is probably the better prize since none of you leave your basements anyway.

I wasnt/didn't...
You understand the concept though,
That's rare. Don't accept their lies user.

What if you were right initially

If the door which the host opens is always a goat, then the first choice is irrelevant. No matter what door you choose, there will only ever be one car and one goat you choose between.

So in other words, it's a 50% chance right from the get go. And by changing doors, you still keep that 50% chance. The third door is irrelevant.

Then...bummer. In the 3-door case, that happens 1/3 of the time. That's why it's a good game.

all good user ;-)
cheers

Satan trips can't be wrong.

Wrong, idiot.

Think of it like this. There are three doors, and the chance of choosing the right one is 33%. If you've chosen the right door, and you choose to switch, you lose. If you choose the wrong door, and you then switch, you win, because the only other door is the car. There's a 66% chance of choosing the wrong door on that first move, so a 66% chance of winning if you switch.

This only applies if you look at the problem as a whole, though. If you -only- look at the two doors remaining, and not at what happened prior to that, you have a 50% chance whether you switch or not.

Except that when you made your original choice, you had more doors to choose from. Stay only works if you were right initially, which was a 1-in-3 chance. Switch only works if you were wrong initially, which was a 2-in-3 chance.

Then you lose.
With a 2/3 chance to win, theres also a 1/3 chance to lose.

Nobody said it was a guarantee

But goats and doors and stuff.

You were 100% correct

Then you wrote a second paragraph.

Your Seco d paragraph is 100% wrong.

This is what people fail to realize, Monty sets out a specific scenario where a door is opened without the host having any knowledge of what's behind it, and it happens to be a goat. That is the blooming scenario, so of course "void games" where he picks the car aren't considered. That would be a completely different scenario.

However in the case of void games where host chooses at random and can end up with the car door, the odds remain at 1/3 for the player upfront, and either increase to 1/2 (host reveals a goat) or drops to 0 (host reveals car). Understood?

Host asks you if you want to switch door, guess what, you get to "choose" again, regardless of your original choice. What is there to choose now, you ask? Well, there are only 2 doors to choose from this time (because you're not that retarded to choose a door already opened, I don't even think the game allows you to do that anyways), unopened door with a goat, and unopened door with a goat, what would that mean? 50% or 1/2 choice scenario, nothing more, nothing less.

The 66% math rule is correct, but you can't apply it to real life scenarios if you're going to be changing rules (hurr durr let me open this door real quick lmao!!)

Then you get a fucking goat. It's not that complicated, dipshit.

>and unopened door with a CAR*

My bad, typo.

You've just confused them all and made them think it can be 50/50.

It isn't, you know it isn't.
Why would you do this?
Are you. Hunting scalps again?

This is just what happens when you try to add math to a comprehension problem. The math will be correct, the real life scenario won't (you will have half chance for each).

If you still fail to realize this, may god have mercy on your brainless soul.

In order for the Monty Hall problem to work, there are certain conditions that must be met. It has to be that you pick one door to start, then the host knowingly reveals a wrong answer door, and so on. The probability can be different if the exact conditions aren't met.

You're an idiot lol
After one of the doors is eliminated its not 66% its 50% learn to math

3?