I have a doubt concerning a question about the Monty Hall Three-Door Puzzle, in probability. I found this problem in Rosen's "Discrete Mathematics and Its Applications".
The Monty Hall Three-Door Puzzle: Suppose you are a game show contestant. You have a chance to win a large prize. You are asked to select one of three doors to open; the large prize is behind one of the three doors and the other two doors are losers. Once you select a door, the game show host, who knows what is behind each door, does the following. First, whether or not you selected the winning door, he opens one of the other two doors that he knows is a losing door (selecting at random if both are losing doors). Then he asks you whether you would like to switch doors. Which strategy should you use? Should you change doors or keep your original selection, or does it not matter?
First of all, before I ask my specific doubt:
I understand that the best strategy is switching doors, because the probability that the initially chosen door is incorrect is high (2/3); therefore, it is most probably not the winning door. So, after the host opens a door (which he knows is a losing door), the probability that the prize is in the other closed door (and not in the initially chosen one) is higher (2/3).
Now, the specific question which I want to ask (found in Rosen's book):
Explain what is wrong with the statement that in the Monty Hall Three-Door Puzzle the probability that the prize is behind the first door you select and the probability that the prize is behind the other of the two doors that Monty does not open are both 1/2, because there are two doors left.
When the contestant chooses one door (before the host opens a door), the probability that it has the prize is 1/3. But, when the host opens one door (that he knows is a loosing door), the possibilities of where the prize can be are reduced by one, because now the contestant knows that the prize can only be either on the chosen door, or on the closed door. So, it seems reasonable to think that the probability that the prize is in any one of the two remaining doors is 1/2.
But this reasoning is apparently wrong. Can any one help me understand why?
Thank you in advance.
Best Answer
The reasoning "because there are two doors, the probability that the prize is behind either is $1/2$" depends on an unspoken premise that there is no reason to distinguish between the two doors.
But the situation is not really that symmetric. One of the two doors were chosen by the contestant without any special information. The other was chosen by the host, under the particular restriction that it and the contestant's choice must not both be non-winning.
Because the two doors were not chosen under the same conditions, the situation is not symmetric between them, and hence the argument that they should be ascribed the same chance of winning is not valid.