#44, RE: Monty Hall Problem
Posted by Ramsey_KJ on Oct-26-04 at 10:19 PM
In response to message #43
Thanks for your imput. Still the best explanation that I read was Kevin Devins column, the link to which Alex posted earlier. In that column, Devin gave both an explanation and a new twist. Suppose that the prize is behind one of seven doors and that you can select three of them. Your chances of winning are 3 in 7, and the chance of losing is 4 in seven. Now the host opens three of the doors that you did not pick and no prize is behind any of them. Since the host's action was limited in that he had to pick doors that you did not pick, this did not affect the original odds and you still have a 4 in 7 chance of losing if you do not switch from the three doors that you pick to the one other door that the host did not open. Thus if you act on the host's new information and pick the one other door that is left instead of your 3 doors you origionally picked you will have a 4 in 7 chance of winning. See Kevin Devin's column at http://www.maa.org/devin/devin_07_03.html Have a Good Day KJ Ramsey
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