>>Christof (and others) are treating it as a "conditional" probablity problem, that is, "how do the odds change if something occurs?" That is what the original problem (as I've seen it posed) was. There is also an "independent" problem in which the past is irrelevant, as others interpreted it. So, the answer depends on a very exact statement of the problem, otherwise the solution can vary...
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>I don't know if I agree, but that's not really important. Quite possibly I have presented the problem in such a way that either answer could be correct. How would you state the problem that only the 2/3rd's would be the correct answer.

I guess just including the probability in the original question would do it, after reading it over. It's easy to get sidetracked (as some apparently did) and consider the goat/door to be already open when you're weighing the odds, and treat the problem as one simple independent probability (which has the 1/2 solution) - but in fact, you're supposed to be comparing two different odds, one based upon the other - hence the term: "conditional" probability.

I see Jim Booth stated it well near the start of the thread...

I see Ken Matson just posted a link to M. vos Savant's famous column on this problem, as well as the also famous "loophole," perhaps more interesting than the original problem :)