MONTY HALL PROBLEM IN LOGIC. Fredric M. Menger, Emory University, Atlanta, GA, USA
The Monty Hall problem is a classic exercise in logic that has bewildered even Nobel Prize winners. I myself kept defending an incorrect answer until finally my own sister ultimately persuaded me of my error. Here is the puzzle:
Suppose a person is shown three doors. Behind one door is a new car. Nothing lies behind the other two doors. A person, called the guest, is asked to select one door, but not open it. He selects Door #1. Then a second person, called the host, and who knows the location of the car, opens Door #3 to show that it is vacant. Now the guest is given a choice: He can open either Door #1 or Door #2. Which door should he select to maximize his chance of getting the car? Should he stick with Door #1, or should he move to Door #2?
Incorrect logic
After Door #3 is opened and found vacant, the car must be behind either closed Door #1 or Door #2. There is, after all, one car but two doors. Thus, the chance of the car being behind Door #1 is 50%, and the chance the car is behind Door #2 is also 50%, so that it makes no difference which door the guest chooses. This logic happens to be incorrect, as seen next.
Correct logic
Initially, the chance that the car is behind Door #1, Door #2, and Door #3 is 1/3 each. Thus, there is a 1/3 probability that the car is behind Door #1. There is a 2/3 probability that the car is behind Door #2 or Door #3. Now there are two possibilities, recognizing that the host can obviously not open a door that he knows has a car behind it. (a) The car is behind Door #1 in which case the host is free to select either Door # 2 or Door #3 to open. The probability of this happening is, as stated, 1/3. (b) The car is behind Door #2. The probability of this occurring (and this is key) is 2/3 because the probability for Door #2 or Door #3 is 2/3, but Door #3 has been eliminated from consideration. This leaves the 2/3 probability associated with only Door #2. In summary, there is a 1/3 probability for Door #1, but a 2/3 probability for Door #2. The guest has twice the chance of getting his car if he moves from Door #1 to Door #2.
Suppose a person is shown three doors. Behind one door is a new car. Nothing lies behind the other two doors. A person, called the guest, is asked to select one door, but not open it. He selects Door #1. Then a second person, called the host, and who knows the location of the car, opens Door #3 to show that it is vacant. Now the guest is given a choice: He can open either Door #1 or Door #2. Which door should he select to maximize his chance of getting the car? Should he stick with Door #1, or should he move to Door #2?
Incorrect logic
After Door #3 is opened and found vacant, the car must be behind either closed Door #1 or Door #2. There is, after all, one car but two doors. Thus, the chance of the car being behind Door #1 is 50%, and the chance the car is behind Door #2 is also 50%, so that it makes no difference which door the guest chooses. This logic happens to be incorrect, as seen next.
Correct logic
Initially, the chance that the car is behind Door #1, Door #2, and Door #3 is 1/3 each. Thus, there is a 1/3 probability that the car is behind Door #1. There is a 2/3 probability that the car is behind Door #2 or Door #3. Now there are two possibilities, recognizing that the host can obviously not open a door that he knows has a car behind it. (a) The car is behind Door #1 in which case the host is free to select either Door # 2 or Door #3 to open. The probability of this happening is, as stated, 1/3. (b) The car is behind Door #2. The probability of this occurring (and this is key) is 2/3 because the probability for Door #2 or Door #3 is 2/3, but Door #3 has been eliminated from consideration. This leaves the 2/3 probability associated with only Door #2. In summary, there is a 1/3 probability for Door #1, but a 2/3 probability for Door #2. The guest has twice the chance of getting his car if he moves from Door #1 to Door #2.
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