Come forth and be confused!

[aperson]

Well wtf is the answer already? It's driving me nuts.

And if it is something stupid like 'It was actually in the door the guy opened but it was under an invisible sheet' I'm just gonna #$#ing snap.

It isn't, but you'll snap anyway because you wouldn't believe me.

[scorpio1690]

Then fucking tell us

[Moogy]

Great. My brain hurts now.

[aperson]

Okay.

You should switch doors, because the other door has a 66% chance of being the correct one.

[MystieRains]

Okay, crap. I guess I'll write and think at the same time. So, three doors. 33% chance you'll actually win whatever prize is in there. 66% you'll have a wonderful one way ticket to thin air. One door is opened, w()()t. Nothing there. You cleared that. You're not going to choose the one that's been opened of course, but still remains the fact there's still three doors, so it can't be a 50/50 thing, because you didn't start with 2 doors. You had 33% of choosing the right door, and 66% of choosing the wrong. Therefore, it'd be best if you switched doors because there's a better chance of you being right. Right?

To me that kinda makes sense, but then again, maybe I'm just an idiot... x.x

PS - Bloody fucking hell, You just couldn't wait till I was done, could you?!

PPS - I might as well just keep the crap up, cause someone might need it explained. I guess I should be happy I was right, but it's so much less satifying when I have my post after his ;-;

[aperson]

Yes, though it's not exactly correct, it is pretty close.

Here's an exhaustive proof:

There are three initial possibilities. Each have equal probability:

code:--------------------------------------------------------------------------------
Door 1   Door 2   Door 3   Probability
                  Prize      1/3
         Prize               1/3
Prize                        1/3
--------------------------------------------------------------------------------

You picked a door. Since you had no knowledge about what was where, this choice is essentially random. Thus, for each position of the prize there are three equally probably choices you could have made.

code:--------------------------------------------------------------------------------
X denotes your choice
Door 1   Door 2   Door 3   Probability
  X               Prize      1/9
           X      Prize      1/9
                  X, Prize   1/9
  X      Prize               1/9
         X, Prize            1/9
         Prize      X        1/9
X, Prize                     1/9
Prize      X                 1/9
Prize               X        1/9
--------------------------------------------------------------------------------

These are all possibilities and they are equally probable.
Now, when the host chooses which door to pick, if you chose wrong, there's only one door he can open that does not have the prize.
If you chose right, he can choose one of two doors to open; assuming he chooses randomly, two new cases are created, each with 1/2 the probability of the initial case:

code:--------------------------------------------------------------------------------
X denotes your choice
O denotes opened door
Door 1   Door 2   Door 3   Probability
  X        O      Prize      1/9
  O        X      Prize      1/9
  O               X, Prize   1/18
           O      X, Prize   1/18
  X      Prize      O        1/9
  O      X, Prize            1/18
         X, Prize   O        1/18
  O      Prize      X        1/9
X, Prize   O                 1/18
X, Prize            O        1/18
Prize      X        O        1/9
Prize      O        X        1/9
--------------------------------------------------------------------------------

Now count the total probabilities of the cases in which you should stay. They are 1/18 + 1/18 + 1/18 + 1/18 + 1/18 + 1/18 = 6 * 1/18 = 1/3.
The total probabilities of the cases where you should switch:
1/9 + 1/9 + 1/9 + 1/9 + 1/9 + 1/9 = 6 * 1/9 = 2/3.

[MystieRains]

Wow, never figured so much math work to be in it x.x But pretty close will do, cause I pretty much won the prize ^_^

Have any more?

[Lupin_the_3rd]

but it would seem like you would be creating a new problem, because the old formula has been used to determine the chance of it when there were three doors.
there is a 66% chance of it being your door when you consider all three doors, but if you see the two doors as a new problem, then it is a 50% chance...in real life you wouldn't have a better chance:

Door Door Door
X

so there are two doors. because its denoted that the third door is gone, you have a 50% chance of getting the prize at this point, not 66%. 66% comes from the fact that you actually "picked" two doors out of the three, getting you that number.

[aperson]

No, but really it requires no mathematical knowledge, that just PROVES it.

Think of it this way: You have 10000 boxes, only one contains a prize. You select a box, then a guy removes 9998 other boxes because none have the prize. Do you keep your box or do you switch boxes.

YOU SWITCH THE GODDAMN BOXES.

[Lupin_the_3rd]

but did your box get picked as empty?

[aperson]

so there are two doors. because its denoted that the third door is gone, you have a 50% chance of getting the prize at this point, not 66%. 66% comes from the fact that you actually "picked" two doors out of the three, getting you that number.

WRONG!

Look, there is a 2/3 chance the first door you picked is a loser. So if there's a 2/3 chance you're gonna lose and there's another door sitting beside you, I'd say that door only has a 1/3 chance of losing.

[aperson]

but did your box get picked as empty?

Since I'm expanding the equation from the original problem, I'll say 'HEY GENIUS, READ THE ORIGINAL RIDDLE, COULD HE PICK YOUR BOX THERE, NO.'

[dontcareaboutmyid]

I know about probabilities and what not, but the 33% if oyu stay you win is also the 50% of the new two boxes which is acually 83%, but that also makes the other box 115% right which is impossible, which makes your problem = teh lose

[aperson]

I'm sorry, you are 'teh lose' because the percents do not mathematically add in this manner. You sir, are a god damn idiot. A sixth grader could spot out how stupid your argument is.

[Lupin_the_3rd]

so then you have a 50% chance
there are two boxes--does one have a greater chance than the other? no it is an unknown. One box does not have a greater chance than the other
there are two boxes. One prize. 1 prize divided by two boxes =.50=50% chance, regardless if there were 4389527309 boxes picked before it