Talk:Monty Hall problem

I dont beleive everyone else, This is my theory:

alright, now lets say each O is a door-

door #'s: 1 2 3

 doors:      O    O    O

probability: 1/3 1/3 1/3

okay each door has a 1/3 chance; but the two on the left have 66% chance.correct, So now say we pick one for example, door 3; awesome possum, so we say that it has a 33% chance of it being a car. So now we open door 2 and it's a goat. Now some think that door one still has a 66% chance and door 3 has 33% chance, but thats not true because now we only have 2 numbers; therefore it's 50/50 chance between the two

and I was looking at this and they were using larger numbers to try and explain it, this is what I think:

door #'s: 1 2 3 4 5 6 7 8 9 10 doors: O O O O O O O O O O

So each door has a 10% chance now say we predict door 9; at this point, you only have a 1 in 10 chance of getting it right. now next step:

lets say we open all the doors except for door 7 and door 9. Now you would think it's in door 7, BUT thats not true now we just have two doors and TWO doors only not 10 but Two:

door #'s: 7 9 Doors: O O

okay and now they both have a 50/50 chance. PROBLEM?

Well then,This is what I think, my friend tells me I'm crazy but I just don't know.