I'm not convinced

>guys what is the capital of France?
>"the capital of Spain is Madrid; you are fucking retarded if you can't deduce capital of France from that"

Troll brainlet ignored. Apply yourself.

>proven theorem

>concept of God was proven theorem
>flat earth was proven theorem
>geocentrism was proven theorem
>time and space is absolute was proven theorem

I'm not buying your social construct. Do you even know how to solve the problem?

I think you don't understand what a proof is.

I got argument that disproves Monty Hall theory.
Disprove my argument.

There's a big difference between all those things and mathematics

>comparing geography to math
kek

You pick the car, goat 1 gets removed, you switch and lose with goat 2.
You pick goat 1, goat 2 gets removed, you switch and win with the car.
You pick goat 2, goat 1 gets removed, you switch and win with the car.

You had a 66% chance to pick a goat. If you switch away from a goat, you win.

It's actually fifty/fifty even. Either you picked the right door, or you didn't.

Doesn't matter if you have an argument, you don't have a proof.

I'm with OP on this one.

OP BTFO

...

I've never heard of Monty Hall problem until this thread. Read the wiki and that led to three prisoners and bertrand's box. Then klingon-gamorrah's axioms.

Now I realize that statistics is fun. Thank you, Veeky Forums

...

You did your experiment wrong it you got 1/3 in both cases.

This is a great computer simulation exercise for kids learning to program.

>bertrand's box

first you choose 1 out of 3. by showing you one you don't want to choose it is made 1 out of 2 and you chances 50 percent. Say you are lost in a big building and have to choose between two doors to make your important meeting. Your chances are 50/50. The fact that in the whole building are 400 more doors does not change that. Makes sense?

Doesn't need experiment though.

Let's say I pick door 1 everytime and don't switch.
Now I randomize 30 numbers.

111111111122222222223333333333

So I won 10 out of 30 games, 1/3.

Let's say I pick door 1 everytime, and will switch to door what is left.
He will open door 2 or door 3, doesn't matter at all which.
- I lose everytime goat is in door 1 (1/3)
- If I switch, I will win half of the time (1/2*2/3)
But then we have to apply the correction factor ß.

So the result is:
- Stay: win 1/3 of time
- Switch: win 1/3 of time

Show this to Trump, I bet he'll select you to run NASA.

go to /pol/ you sjw

please be bait

Not to continue bumping this fucking thread again, but the easiest way to explain it is this:

Consider the same problem, but there are 100 doors. You pick one, Monty opens 98 other doors with goats behind them and leaves one closed.
Do you still think you have a 50% chance of winning if you stick with the door you picked initially?

No you understood it wrong, I'm afraid. You should always read instruction carefully before making a living embarrassment of yourself.

There is 3 doors. Not 100. 3 doors. Solve the problem using 3 doors. Not 100.

Just for the record:
If there were 100 doors, your chance of winning would be
- Stay: win 1/100 of the time
- Switch: win 1/100 of the time
Q.E.D.

lol so the chance of the car being behind the 2 remaining doors is 1/50, gotcha.

Thread lurker here. This didnt make the problem any clearer for me. After seeing 98/100 doors open how can you deduce any information about the 2 closed doors? All you know is that one of the doors is a winner. Saying that switching increases your chances makes it seem to me like my choice had some influence on which door was the winner.

just look at this and if you don't get it neck yourself.

I think i thought it out. With the 100 doors your first choice is 1/100 probability. Monty, knowing which door is the winner eliminates all other losing doors besides your door and one other. So your choice is to either remain with your 1/100 guess or use information that Monty had in that he knows which door is the winner. I can see how this scales down to 3 doors but i have a problem with the 2/3 odds when you switch. I know the odds of winning are increased when you switch but 2/3 seems way too high.

Imagine before you even play the game, you decide in advance what you intend to do; no matter what the host decides, you will either a.) Keep your first door no matter what or b.) Switch your door no matter what.

If you decide to keep your first door no matter what, then the only way to win is for you to pick the door with the car on your first guess. With three doors and one care, that means you only have a 1/3 chance if you decide to stay.

On the other hand, if you decide to switch, you will win only if you pick a goat on your first guess. Since there are two goats, there's a 2/3 chance of that happening, so if you decide to switch, you have a 2/3 chance of winning.

that's a pretty good explanation as well

It may be hard to believe, but yes, that's correct. It's called paradox for a reason! You just have to trust the math.

You picked door 1.
You have a 1/3 chance of having picked the car. This leaves a 2/3 chance the car is in one of the other two doors. Host removes a goat from the two remaining. This final door now carries the entire 2/3 chance of being car. Do you stay or switch?

The final choice is to either stick with your original 1/3 chance or to take the remaining 2/3 chance.

A scientific THEORY does not equal a mathematical THEOREM.

Plz let English not be your first language.

You gain additional knowledge about the doors. Your first pick was 1/3, but your second is 2/3 as one of the goats has been eliminated

But the experimental data disagrees with you.

Show your work, user.

I wanted the goat anyway.

There are 1000 doors

Behind 1 door is a car, behind the other 999 there are goats

You pick a door

The host reveals 998 of the goats, leaving only the door you picked and one other

Is it more likely that you picked the car on your first go out of 1000 doors or that the car is actually behind that other one?

2/3 is exactly right. It's the probability that you were wrong on your first pick since 2/3 of the doors conceal goats. Switching when you're wrong always gets you the right door and switching when you're right always gets you the wrong door so you have a 2/3 chance of being right if you switch because there is a 2/3 chance of you being wrong on your first try.

Both are equally likely.

Yes. you have a 50/50 chance of picking the car.

Either you picked the car or you didn't.

Flat earth was never a proven theorem.

How many times did you do it?

Does Monty only open a goat door if you've already chosen the winning one?

OFC it is, the higher-chance
-for-picking-again-meme was conceived by idiots who don't get that:
one cannot couple the initial chance with the new chance;
the game master ALWAYS reveals one door;
2 of 3 doors contain the losing condition by design.

Average people tend to confuse themselves with stats and likelihood, especially if MOVIE MATH is involved. Just hide the thread and move on.

You're stupid.

You actually don't realize that you initially picked a door with a 2/3 chance of a goat, and you actually think that switching is a 1/3 chance of the car, when it is trivial, basic knowledge, that it gives you a 2/3 chance of the car.

Watch this.
youtube.com/watch?v=WyoF-7rmakA

This.

Also this is a beautiful video youtube.com/watch?v=WyoF-7rmakA

youtube.com/watch?v=FAljAvR3L4s