Re: The Monty Hall Problem....
FF they are all absolutely correct, even though the analysis seems counterintuitive.
The easiest way I have heard that explained is that the chosen door has a one in three chance of being correct, while the non-chosen doors have a two in three chance of being correct. This never changes. When one of the non-chosen doors is revealed, the remaining non-chosen door assumes all of the initial 2/3rds weighting, and is therefore twice as likely to be the correct door. This has been proven by computer models. The key is that the initial 1/3 2/3rds odds remain constant, and that new information revealed does nothing to change those initial odds.
Lies, damn lies and statistics....