>>>Nope. Think about this.
>>>
>>>He opens one door and asks you to choose again. You're not going to pick the open door so of the doors that you would choose from, there are only 999. Your odds change.
>>>
>>>The logic you use works for all the doors that EXIST, not the odds of the doors you will consider choosing. I'll take the odds for doors I will consider choosing because it increases my odds. But in the end, when there are only two doors left, the odds of you choosing the winning door of the two is 1/2 or 50/50 or 50%.
>>>
>>>>The point is that after you picked a door he opens 998 doors where there is no car ( - he knows where the car is). That makes the odds of the last door 999/1000
>>>>
>>>>Think about it
>>
>>
>>Ok, final attempt - I'm not trying to convince
>>
>>If I immediately open 998 doors and then you pick one - I agree it's 50%
>>
>>Now, all 1000 doors are closed and you pick one
>>
>>Your odds are 1/1000
>>
>>The odds that the car behind the other 999 doors are 999/1000 - correct ?
>>
>>That splits the doors in two parts
>>
>>Your door - odds 1/1000
>>The other 999 doors - odds 999/1000
>>
>>Of the other 999 doors, I open 998 where there is a goat behind
>>
>>We still have the two parts (1/1000 and 999/1000)
>>
>>Of the second part, 998 out of the 999 doors are open
>>
>>The second part is now one (closed) door.
>>
>>The odds of the second part are still 999/1000, but are now concentrated in that one closed door that you did not pick
>
>the rest is red-herring... if you have two to select from the odds are 1/2...... period. Odds are not dependent on prevous operations. if you have six babies all girls in a row, the odds of your next baby being a girl are stilll 1/2... not 1/7..

You had 1000 to select from.

Let's put this to rest - My name is not Rick