Picking The Door In The Monty Hall Problem

The Monty Hall Problem is about what someone’s chances of winning a car if you had to pick the door that it is behind out of three doors. Behind two of the doors there is a goat and in the third door is the car. If you pick a door, the game show host opens another door to show a goat, should you swap doors or stick with the original door that you picked? The first thing that came to my mind is that it would not make a difference if you swap doors or not. You would still have a 50/50 chance of having the correct door. Since now you know that there are only two remaining doors, one with a goat and one with the car. It would only make sense that out of two doors one of them has the winning car. That is a 1 out of 2 probability of choosing the correct door. Well in all reality that response does make sense but according to the video that is not what the answer really is.
You should always swap doors; this will give you a better chance of getting the door with the car in it. In fact, Ron Clarke says that this increases your chance of picking the winning car by 2. If you choose not to swap doors then you have a 1/3 chance that you picked the correct door for the winning car and 2/3 chance that you picked a goat. If you do swap doors are one of the doors with the goat is revealed then your chances of winning the car are at better odds. The probability of picking the door with the car after swapping doors increases to 2/3 and the door with the goat decreases to 1/3. So it makes sense to always swap doors to increase your chances of