To extend the above, it's as if Monty gives you the chance to keep your one door, or open all 999,999 of the other doors, of which he kindly opens the first 999,998 of them for you, leaving, deliberately, the one with the prize. Clearly, one would choose to open the other 999,999 doors rather than keep the one.
You see this is exactly why this solution is flawed, it assumes that switching is increasing the amount of doors that you're looking behind.
At the start there is. 1/100,000 chance that you pick the right door.
Now. If the host asked you do you wan't to keep your door, or take the other 99,999 doors then you'd dramatically increase your odds, you'd be a fool it two switch. It's a 1/100,000 chance vs 99,999/100,000 chance...
But that's not what happens, the host only opens doors that are incorrect, whether you switch or not you have equal odds because you've already been given the doors,
The idea that there is more chance that the prize is behind door nu
Bed two because you picked door number 1 and it wasn't behind door 3- door 100,000 is just crazy.
Those that use the statistical proof that you're getting more doors to check ignore the fact that regardless of whether you switch you've seen behind 99,998 doors, there are only two doors left, you have a new desecrate decision, the prize is bhind 1 of 2 doors and the odds are 50:50