This problem is new to me, and fascinating. I've read most of the material relating to it on the excellent CTK site, and there seem to be numerous easy traps to fall into. Take for example Peter Stikker's explanation (www.cut-the-knot.com/peter.shtml). He draws a decision tree to demonstrate the solution - but omits some of the possibilities!

His first column shows the prize behind door A (1/3 probability); he chooses door A (1/3); the host opens door B (1/2) - but then what? Peter assumes that we MUST switch to door C, and does not provide for sticking with door A.

If we redraw his tree to show the sticking option in all cases, suddenly the probability of winning becomes 1/2, not 1/3. When the prize is behind door A, the eight possible outcomes are split 4 win and 4 lose. They are not all equally likely, but do fall into pairs of equal probability - so the win/lose outcomes are equally likely.

Will that be the end of it? I doubt it!
:-)

(PS: I tried to include the image in this posting - but couldn't get it to work. Sorry.)