Monty Hall problem

I, apparently, am a mathematical "moron" to coin a term.

A local barfly recommended that I understand Bayes theorem to be a better handicapper.  That eventually led to a wikipedia entry regarding the Monty Hall problem.  I still don't fully understand the solution, but I think I am getting there.

Anyway, that led me to ponder a similar problem relating to a modern gameshow - the one with all the suitcases that Howie Mandel hosts.  I wonder if, when he gets down to the last 2 cases, would a contestant be better off by switching?  I assume he would, but I don't want to assume anyting that can be proven.  So, do we have any math guys in here that would care to discuss this?

You have to change your first decision. That's the right answer ;)

No, this could not be used in Deal or No Deal. The reason is because YOU decide which case gets eliminated, instead of the host who is aware which case yields which prize.