For some reason the internet still has weird arguments about the Monty Hall "paradox". I think it might be because sometimes the way it is explained is a bit confusing to some people? Idk, but I do remember that when I first heard about it the explanation was kinda weird to me until I rephrased it in simpler terms.
Here is the thing. You have 3 doors, one of them has a car train behind, the others have goats. You pick a door at random, the host doesn't open it, but he eliminates one of the other doors by showing it has a goat behind it, and asks you if you want to change your choice. So perhaps you picked door A first, he eliminated door B, and now your choice is between door A and door C. If you change your initial choice, it turns out that you will be correct 2/3 times, whereas if you don't you will only be correct 1/3 times. This confuses some people because they expect that since it becomes a choice between 2 doors it should be 50-50.
The simple explanation is that if you picked the door with the train the first time, then by changing you will always lose. If you didn't pick the correct door the first time, then changing always wins. But you will only get it right the first time 1/3 times, whereas you will get it wrong the first time 2/3 times, so changing is advantageous. That's it. That's all there is to it. There doesn't need any more mystification.
If it is still not intuitive, imagine if you had a million doors, you picked one, and the host eliminated all the others except one. There is (almost) no way you picked the right one the first time, it is literally a one in a million chance. So if it is not that one, and it almost certainly isn't, it must be the other which the host practically hand picked for you.
EDIT: For the reasoning to work, the important assumptions are two. One, the host always eliminates a goat, never a train. Two, the host always reveals one of the other two doors, not the one you picked. Both of these are significant, without the first one you have only a 1/3 chance of winning regardless of whether you change or not, and without the second one it becomes a regular 50-50 choice.
what helped me get it was splitting the choices into 2 groups:
your door = 1/3 chance of winning total
all the other doors = 2/3 chance of winning total
say somebody gets to pick 'all the other doors' as an option; we know they have a 2/3 chance of winning
that chance will stay the same no matter what, since the goats/trains aren't getting reshuffled
so if doors get revealed as duds in this group, you would now just know that 2/3 can all be placed on whatever door(s) remain