I can accept the fact that on a Roulette wheel (as long as there are no defects or imbalances in the wheel or ball) that the odds are the same each spin and previous spin outcomes have no influence over the current spin. However, if I see black come up 32 times in a row I am betting on red for the next spin.
You think that even in the hundred-door case? Test it. Hell, even test it in the 3 door case. It is empirically not 50%.
If the host had an even chance to show you either door, you’d be right, but since the host always shows you a goat, the two events (picking a door and choosing whether to switch) are no longer independent, since if you pick a goat it forces the host to pick the other goat.