[An.American.Expatriate]

Your initial chance of being correct does not change no matter how many doors he opened. It stays at essentially zero. I pick the other door because its now a certainty.

Can you explain that?

As I understand it, my initial chance of being correct 1 to n.

My chance of "still" being correct as doors open is 1 to n - doors opened.

What I am not undertstanding is your statment (and the contention of the author) that I have a HIGHER chance of being correct if I switch on the last "choice"

[Nateman]

Here is your error:

My chance of "still" being correct as doors open is 1 to n - doors opened.

Your initial chance of being correct is 1/n. Since n is infinity you have zero chance of making the correct first pick. No matter how many doors Monty opens your pick is still wrong. With two doors left you have your wrong pick and the winning pick remaining.