The Monty Hall problem can indeed be counter-intuitive. Let's break it down:

You're on a game show, and there are three doors. Behind one door is a car (the grand prize), and behind the other two doors are goats (consolation prizes). You don't know what's behind each door.

You pick a door, say Door #1. The chance that the car is behind this door is 1 in 3, or about 33.33%.

Then, Monty, who knows what's behind each door, always opens one of the remaining doors that has a goat. This doesn't change the probability of the door you've already chosen: it's still 33.33%.

Now you have a choice: stick with your original choice (Door #1), or switch to the other unopened door.

Here's where it can get tricky, but think about it this way:

If the car was initially behind your chosen door (a 33.33% chance), and you switch, you will lose (because you'd be switching to a door with a goat).

However, if the car was not behind your chosen door (a 66.67% chance, because there are two other doors), and you switch, you will win (because you'd be switching from a goat to the car).

The reason the chance is 66.67% (or 2 in 3) is because there are 2 cases out of 3 where switching will let you win (the cases where the car was initially behind one of the two doors you did not pick). That's why it's generally a better strategy to switch: because more often than not, the car will be behind one of the other doors.
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