Come forth and be confused!

[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]

Originally Posted by Lupin_the_3rd

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]

Originally Posted by Lupin_the_3rd

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

[aperson]

Hey dumbass, are you saying that if I have a lottery ticket, I have a 50% chance of winning? Because that's what your argument would state in that manner.

The removal of a box MODIFIES THE CHANCES OF THE ORIGINAL BOX BEING CORRECT.

[Lupin_the_3rd]

yes you would have a 50% of winning with the lottery ticket if all 32532343224 others didn't win
there is you and there is another guy
both have like a 99.9999% chance of winning
however there are two, each has an equal chance of winning

[aperson]

Look over the proof, learn math, look at more of my statements.

You have a 33% chance initially.

That chance doesn't just magically turn into 50% chance when a door disappears.


Edit: I'm writing a program to prove you wrong right now.

[Lupin_the_3rd]

if you are talking about you door compared to all the others, your math is correct. If you are talking about your current chance of winning a prize, it is 50%...both doors have a 66% chance of winning, regardless if its "your" door or not.

[aperson]

Sorry, the current chance is 66%, try again.

[Moogy]

Hey people, aperson obviously knows what he's talking about here. You guys do not. Stop trying to prove him wrong, because he IS correct, I will vouch for that.

[Lupin_the_3rd]

so you're telling me that because it's "your" lottery ticket, you have a better chance than another guy?
if 10000000 people have NOT won, there are two people left. Do you say, "i have a better chance than you because its my ticket"? No
you BOTH have an equal chance. there is no way that can be wrong....

[aperson]

No, I'm assuming the condition that YOUR ticket will NOT be removed.

[Lupin_the_3rd]

if thats the case, we are thinking of two separate problems

[jewpinthethird]

What is the prize? Because this is too much work.