#why doesn’t choosing the same door you already chose have the same effect? that’s what I want to know#like does math not agree with the sage advice of ya authors that not choosing is also a choice?
The trick to it is that you're technically playing two games in a row, and the second one is the only one that you actually have to win.
In the first game, you have two chances to lose (picking a goat) and once chance to win (picking a car). Worse-than-even odds. But the important thing is, you don't actually get a prize for winning this first game. It's just set-up for the second one.
In the second game, sticking with your door is basically saying "I think I made a lucky guess in the first game, I'm sticking with that decision." Switching doors is saying "I don't think I got lucky in the first round, so I'm going to change my decision." You are gambling on whether you won or lost the first game, and what wins or loses you the prize is guessing correctly whether you were lucky in the first game. And because the odds of the first game were worse-than-even, guessing that you lost the first game is the safer bet, because you probably weren't lucky.
The really painful part of it is that our brains want to interpret it all as one game, where you've basically got 50/50 odds no matter what you do. That's what our every instinct is screaming at us should be happening, because the physical endgame is two closed doors, only one of them with something we want behind it, which has been there from the start. But it isn't one game with 50/50 odds. It's two games in a trenchcoat, and their combined odds are skewed.
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