I was stubborn about this for so long, and I’m still not entirely sure I understand it, but here is a perspective that made me doubt my belief.
Imagine the Monty Hall Problem, but with 100 doors and only one grand prize. You pick one; it obviously has a 1/100 chance of being a grand prize. Then Monty reveals 98 doors without grand prizes in them such that the only doors left are the one you chose and one that Monty left unopened. Monty obviously arranged for one of those two doors to have the grand prize behind it. The “choice to switch” is really just a second round of the game, but with a 1/2 chance of winning (wrong, your odds change only if you “participate” in round two).
If you stick with your door, you are relying on your initial 1/100 chance of winning. If you switch, you are getting the 1/2 odds of the “second round”.
Apparently with three doors, switching gives you a 2/3 chance of winning, but I don’t understand the math of how to get that answer and I wouldn’t be able to calculate the odds of the 100 door version. I just know intuitivey that switching is better.