why is switching the preferred strategy in the Monty Hall problem?

I have a way of explaining without using fractions or any math crap. Just think of it this way:

I place 100 cups upside down on the floor, and I put a car key underneath 1 of the cups.

I say to you: "1 of these 100 cups has a car key underneath it. Choose one cup and if you get the correct cup you will get to keep the key to win a brand new car."

Then you choose 1 of the 100 cups. Before you pull the cup out of the floor to check if there are keys underneath it, I tell you: "WAIT! Before you pull that cup, I will remove 98 of these 100 cups, and neither of them has the car key underneath, so now there are only 2 cups left, the one you chose, and another one. Now I will give you the choice to either stay with your chosen cup, or to switch to the other cup that is left."

What would you do? You would OBVIOUSLY SWITCH to the other cup, because the keys are almost definitely underneath that cup, instead of being underneath the cup you first chose.

The Monty Hall problem is exactly like this, but instead of there being 100 cups and the presenter revealing 98 of them, there are only 3 cups and he reveals 1 of them. He still leaves 2 cups, the one you chose and another one, and that other one will be the cup with the prize most of the time.

The reason switching is the preferred strategy, is because the door that Monty opens depends on where the winning door is.

It’s easy to understand that your initial chance of picking the winning door is 1/3. It’s also easy to understand that the chance of at least one of the other two doors being the winning door is 2/3.

If Monty gave you the opportunity to switch WITHOUT opening a door, then your chance of picking a wining door by switching would be ½ of 2/3, which is 1/3 (same as your initial chance).

If HIS choice was random, then 1 out of 3 times he would open a winning door. The other 2 out of 3 times, your odds WOULD be 50/50.

Since his choice is NOT random, and since he will ALWAYS open a losing door, 2 out of 3 times, he will open a losing door and leave a winning door. The other 1 out of 3 times, both of the “other” two doors will be losers.

Because Monty ALWAYS opens as losing door, by switching, your chances increase from 1/3 to 2/3. It’s as if he’s saying do you want to stick with your initial choice (1/3)? Or pick both of the other doors (2/3), and I’ll open one of them to confirm that it is a loser.