12-17-2004, 04:52 PM
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Confirmed User
Join Date: Feb 2004
Location: Springfield, Illinois
Posts: 1,150
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ok found more info on it
it was Bayes' theorem
Quote:
found at http://argyll.epsb.ca/jreed/math9/strand4/4201.htm
The Monty Hall Problem
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is the Grand Prize; behind the others, Booby Prizes. You pick a door, say Door A, and the host, who knows what is behind each door, opens another door, say Door B, revealing a Booby Prize. The host then offers you the opportunity to change your selection to Door C. Should you stick with your original choice or switch? Does it make any difference? (This is similar to the routine on the TV game show Let's Make a Deal, hosted by Monty Hall, hence the name of the problem.)
Assuming that the host always chooses to open a door with a Booby Prize, and would never reveal the Grand Prize, the possibly surprising answer is that you should switch to the third door, which is now twice as likely as your original choice to be hiding the Grand Prize. This problem can be analyzed using Bayes' theorem or trees (see "You're the Expert" at the end of Chapter 6 of Finite Mathematics Applied to the Real World), but here is an intuitive argument. When you chose Door A, the probability that you chose the Grand Prize was 1/3 and the probability that it was behind one of the other doors was 2/3. By showing you which of Doors B and C does not hide the Grand Prize (Door B, say), the host is giving you quite a bit of information about those two doors. The probability is still 2/3 that one of them hides the Grand Prize, but now you know which of the two it would be: Door C. So, the probability is still only 1/3 that the Grand Prize is behind Door A, but 2/3 that it is behind Door C.
If you find this result counterintuitive (and even most mathematicians do), try running the simulation below. Choose a door by clicking on it. The host (your computer) will then open one of the other doors, revealing a pig. You may then, by clicking on the appropriate door, choose to stick with your choice or switch to the remaining door. After a moment the doors will close to allow you to try again. Below the doors are shown two running calculations: the experimental probability that you will win if you stay with your original choice and the experimental probability that you will win if you switch. After many tries, will these numbers be close to 1/2, or will they be close to 1/3 and 2/3 respectively?
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here is the link
http://argyll.epsb.ca/jreed/javaMath...tyHallSim.html |
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