My Math Forum Can someone explain the Monty Hall problem to me?

 Probability and Statistics Basic Probability and Statistics Math Forum

 July 22nd, 2016, 09:20 PM #21 Senior Member   Joined: Jan 2014 From: The backwoods of Northern Ontario Posts: 393 Thanks: 71 When you picked a door, the probability of the car being behind it was 1/3. The probability of the car being behind one of the other doors was 2/3. That probability does not change by the fact that Monty opens a door that does not have the car behind it. So the probability that the other door that Monty didn't open being the right one is still 2/3.
 July 23rd, 2016, 04:48 AM #22 Senior Member   Joined: Apr 2015 From: Planet Earth Posts: 141 Thanks: 25 When you picked a door, the probability of the car being behind it was 1/3. The probability of the car being behind one of the other doors was 2/3. But that CAN change when Monty opens a door, depending on how he chooses a door to open. Call the other two doors L, and R, for the one to the left of, and right of, the one you chose. (Wrapping around to the other side of the stage if necessary.) If the car is behind L, Monty must open R. If the car is behind R, Monty must open L. But if it is behind your door, Monty must choose between L and R. Say he chooses L with probability X, and R with probability 1-X. So there are two ways Monty can open L, one with probability 1/3, and one with probability X/3. If you see him open L, the probability the car is now behind your door is the probability the car is behind your door AND he opens L, divided by the probability he opens L. That's (X/3)/[(1/3)+(X/3)] = X/(1+X). Similarly, there are two ways Monty can open R, one with probability 1/3, and one with probability (1-X)/3. If you see him open R, the probability the car is now behind your door is the probability the car is behind your door AND he opens R, divided by the probability he opens R. That's ((1-X)/3)/[(1/3)+((1-X)/3)] = (1-X)/(2-X). Now, if we don't know how he decides, X=1/2 and both of these probabilities are 1/3. The point is that it doesn't have to be. "That probability does not change by the fact that Monty opens a door that does not have the car behind it" is incorrect logic that by coincidence happens to get the correct conclusion. Conditional probability is often non-intuitive, and the mistakes that get made are almost always based on assuming something can't change, when it can. Thanks from manus Last edited by skipjack; July 23rd, 2016 at 07:26 AM.
 July 23rd, 2016, 05:06 AM #23 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039 Agree...but this is by now quite well known: just google "Monty Hall problem"...

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