Assume that a room is equipped with three doors. Behind two are goats, and behin

Question

Assume that a room is equipped with three doors. Behind two are goats, and behind the third is a shiny new car. You are asked to pick a door, and will win whatever is behind it. Let's say you pick door 1. Before the door is opened, however, someone who knows what's behind the doors (Monty Hall) opens one of the other two doors, revealing a goat, and asks you if you wish to change your selection to the third door(i.e the door which neither you picked nor he opened). Do you increase the probability of winning the car by keeping your original choice or by changing?

Explanation / Answer

Yes, you do increase your chances by switching doors. First, lets say you stick with your door that you initially picked. The chance that that door has the car is 1/3. That means that Monty Hall showing you the other door doesn't matter, because you are going to stick with your door, and you would expect to win the car 1/3 of the time if you played it this way many times. However, if you initially picked Door A, and he shows you that one of the other doors does not Have the car, then your door still has a 1/3 chance of having the car in it, but then where is the other 2/3 chance??? the answer is that the third door, the one you have the option of switching to, has that 2/3 chace of having the car in it, because it is a conditional probability situation; the probabilities have been altered when the information about another door is revealed. For an interesting simulation, go to http://people.hofstra.edu/steven_r_costenoble/MontyHall/MontyHall.html and http://people.hofstra.edu/steven_r_costenoble/MontyHall/MontyHallSim.html

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