There’s one thing that never gets brought up by people when explaining the odds: the door reveal is a red herring.
People explain that because it’s a choice of three doors the odds of picking the right one is 1/3 therefore swapping gives you 2/3 odds. Odds presented by rigidly examining the game over the course of many static trial runs. Whenever people try to explain these odds, they will use a macro scale of the game. Playing it 50 times, 100 times, 300 times, a 1000 times. Throw in more doors. Run through every possible scenario. And the math of those odds are based on that rigid viewing of various possibilities. And that is all, technically, statistically correct.
But if you’re actually playing this game as it is presented in the monty hall problem, then you’re playing it once. Just, once. And when you are playing it, no matter what door you choose, a door will be revealed. That revealed door is always going to be A) not your door and B) a wrong choice. Which means you know, before you even pick any door, that the probability of the door that’s going to be revealed being the correct choice is 0%. And the game is rigged to where the 0% door is -never- going to be your choice, because they get it out of the way before your final choice. That means no matter what door you choose, you know beforehand you will always be left with 1 winning door and 1 losing door, regardless of the initial choice.
I’m sure people are ready to bum-rush my post and defend the 33%/67% odds with a smug sense of superiority. Don’t bother; I ‘get’ it. I understand the statistics. I understand the logic within the explanation. But playing to the odds is just a great way to make a wrong decision with confidence.