Pursuing this interesting problem a little further, when an initial choice is made by the contestant, there is a 2/3 probability that a losing door is chosen. The host always has a losing (excluded) door that may be revealed thereafter. Thus if there was a 2/3 probability that the chosen door is a losing door, then there must be a 2/3 probability that the remaining, unrevealed door is the winning door. Thus goes the argument for switching doors. More likely a losing door was chosen at the outset so a switch will result in a winning door. Seems to make sense!