All this simulation isn't very much in the way of proof, is it? It's much quicker (and more formal and not empirical) to use conditional probability to get the result.
Initially, you select the correct door with probability 1/3. If you are correct, switching is guaranteed to fail, sticking is guaranteed to win. If you are incorrect, switching is guaranteed to succeed, sticking is guaranteed to fail. (Heuristically, in the less probable case switching fails and in the more probable case switching succeeds. We can be more formal as follows:) P(Win if switch) =P(Win if switch|first guess is correct)*P(first guess is correct)+P (Win if switch|first guess is incorrect)*P(first guess is incorrect) =0*1/3+1*2/3=2/3. P(Win if stick) =P(Win if stick|first guess is correct))*P(first guess is correct)+P (Win if stick|first guess is incorrect)*P(first guess is incorrect) =1*1/3+0*2/3=1/3. ------------------------ Yahoo! Groups Sponsor --------------------~--> Get fast access to your favorite Yahoo! Groups. Make Yahoo! your home page http://us.click.yahoo.com/dpRU5A/wUILAA/yQLSAA/MXMplB/TM --------------------------------------------------------------------~-> Yahoo! Groups Links <*> To visit your group on the web, go to: http://groups.yahoo.com/group/speedsolvingrubikscube/ <*> To unsubscribe from this group, send an email to: [EMAIL PROTECTED] <*> Your use of Yahoo! Groups is subject to: http://docs.yahoo.com/info/terms/
