Earlier quoted context omitted.
Interestingly, if the host picks randomly (and if he reveals the car, you... start over, or you get the car, or you get nothing, or ... it doesn't matter because it happens not to have happened in the time we're considering) then you are faced with a 50/50 chance.
I don't understand how this could be. If you start over when the host picks the car, then isn't that the same as the host picking the goat every time, i.e. the same as the host knowing.
Let's say you pick door 1. Let's go through all the possibilities: the states of the 3 doors, which one monty reveals, and what do we do, and what is the result?
1 | 2 | 3 || Monty Reveals | You switch | Result
---------------------------------------------------
G | G | C || 2 | Yes | Win
G | G | C || 2 | No | Lose
G | G | C || 3 | N/A | Start Over
G | C | G || 2 | N/A | Start Over
G | C | G || 3 | Yes | Win
G | C | G || 3 | No | Lose
C | G | G || 2 | Yes | Lose
C | G | G || 2 | No | Win
C | G | G || 3 | Yes | Lose
C | G | G || 3 | No | Win
Count 'em up: When you switch, 2 wins and 2 losses. When you don't, 2 wins and 2 losses. Of course, the situation is symmetrical for any starting guess you make.