In the basic Monty Hall problem, why not the probability are 50-50? When reading the book of (Chapman & Hall_CRC Texts in Statistical Science) Joseph K. Blitzstein-Introduction to Probability Chapter 2. I have great puzzles about the second last paragraph.
To build correct intuition, let’s consider an extreme case. Suppose that there are a million doors, 999,999 of which contain goats and 1 of which has a car. After the contestant’s initial pick, Monty opens 999,998 doors with goats behind them and o↵ers the choice to switch. In this extreme case, it becomes clear that the probabilities are not 50-50 for the two unopened doors; very few people would stubbornly stick with their original choice. The same is true for the three-door case.
The author give a extreme case. After opening the $999,998$ doors. Why 'it becomes clear that the probabilities are not $50-50$ for the two unopened doors; very few people would stubbornly stick with their original choice'? I still can not abandon the idea that the probabilities are $50-50$.
How to think about it? Thanks for any help!  
 A: Forget the extreme case. Just think about the original case. 
Here is some intuition. Suppose you have not picked the door with the car. Then you have a goat, and Monty has no choice in which door to open (the one with the other goat). If you knew what was behind your door, you would then be certain that the unopened door contains the car. Now suppose you have picked the door with the car. Now Monty has a choice in which door to open, but in any case the unopened door contains a goat. 
Now the two scenarios I have described above are NOT equally likely. How often are you in the first scenario when you have not picked the car? $2/3$ of the time, because two out of three doors contain goats. How often are you in the second scenario, when you have picked the car? $1/3$ of the time because only one door contains a car. So if you imagine playing this game many, many times, $2/3$ of the time you will be in the first scenario where the unopened door contains a car. Only $1/3$ of the time will you be in the second scenario when the unopened door contains a goat.
If you want to think about the extreme case, the situation is exactly the same, except now there's a $999,999/1,000,000$ chance you are in the first scenario, and only a $1/1,000,000$ chance you are in the second.
A: Let's say you pick door number $1$. Monty opens all of the other doors (revealing goats), except door number $435,991$. Do you still think the car is behind door number $1$? Put another way, does door number $435,991$ also contain a goat?
It boils down to this: There are $999,999$ ways that you can be wrong on your first guess, and just $1$ way to be correct. The odds are overwhelmingly in your favor to choose the door that Monty didn't open. There's only one door he cannot open (the one with the car).
A: He opened 999,998 doors of which he knows they have a goat behind it. The door you originally selected was a random choice. So the other door that remains must almost certainly have the car.
Consider the case where Monty doesn't know where the car is.
When he randomly opens 999,998 doors it's extremely likely that he opens the door to the car somewhere and you loose.
In the very unlikely case where Monty randomly (without knowing) opens 999,998 doors and doesn't find the car the two remaining doors will have a 50-50 probability.
But that scenario is extremely unlikely , the chance of that is one in 500,000.
So in that case you could say you have a chance of $\frac{1}{500000}$ to get a 50-50 chance. Or a $\frac{499999}{500000}$ chance that the door to the car is opened before reaching the last 2 doors.
Now back to the case where Monty deliberately leaves the door to the car closed that chance transfers to the box you initially didn't choose because Monty more or less tells you that the car must be in the last box, unless by coincidence it was in the box you originally chose.
