I think the problem statement is clear, and is independent of how the TV show actually operated.
The problem posed is that you have three closed doors, behind one of which is a car, and behind the other two are goats. You get to pick one of the closed doors, and will win whatever is behind it (you want the car).
One of the doors you did not pick is now opened, revealing a goat. You therefore now know the car is either behind the door you picked or behind the other closed door.
In this setup, the question is: if offered the chance to switch your selection from the door you initially picked to the other closed door, should you switch? Does it make a difference?
Mariyln Vos Savant said switch. The peanut gallery (some statisticians who should have known better) said it makes no difference - with two doors left you've got a 50/50 chance of winning the car whether you switch or not.
First off, you can run multiple simulations of the game (with random positions of the goats and car), and prove for yourself that always switching is a better strategy, so it's sad that anyone would insist that it's not, even if that seems logical.
The simplest way to explain why switching is the correct strategy is this:
When you make your initial selection, you have:
a) A 1/3 chance of selecting the car, and
b) A 2/3 chance of selecting one of the two goats
So, 1/3 of the time you selected the car and will therefore lose if you switch, and 2/3 of the time you selected a goat and will therefore win if you switch. Switching will therefore be a winning strategy 2/3 of the time.