I will demonstrate here that the usual resolution of the Monty Hall problem is incorrect. The problem is stated as follows. In a game show, the player is shown three identical doors. Behind two of the doors, there are goats, and behind the third, there is a car. The game host asks the player to choose one door. After he has chosen one, the host opens one of the remaining doors and reveals a goat (since <I>he</I> knows which contains what). Now, the player is given the option of switching to the other door. The questions are: *Does switching improve his chances of getting the car? *Does it decrease his chances? *Or will the chances be unchanged whether he switches or not? The obvious answer is that it does not make any difference. Since there are two doors (the unopened ones) from which he can choose, and since one door contains a goat and the other the car, there's a 50% chance of either door containing the car. But the answer which is claimed to be correct says that if he switches, the chance of winning is 2/3, and therefore, it is better to switch. This, allegedly, shows that statisitics can sometimes be counter-intutive. The justification is given thus: There are three possibilities: *The player has chosen Goat 1, the game host shows Goat 2. Switching wins. *The player has chosen Goat 2, the game host shows Goat 1. Switching wins. *The player has chosen the car, the game host shows either of the goats. Switching loses. Since switching wins two out of three times, the probability is 2/3 in favor of it. This is the standard solution. I propose to show that this is flawed. For, the events haven't been completely listed. The complete list is as follows: *The player has chosen Goat 1, the game host shows Goat 2. Switching wins. *The player has chosen Goat 2, the game host shows Goat 1. Switching wins. *The player has chosen the car, the game host shows Goat 1. Switching loses. *The player has chosen the car, the game host shows Goat 2. Switching loses. So the third event from the previous case really consists of two possibilities because the game host can choose either of the goats. So both these cases should be listed, and since there is no preference for any goat, both of the last events have equal probability. So now there are two cases where switching loses. Which means that now there's only 50% chance of winning when you switch. Of course, you don't have to go to all this trouble if you address the problem more directly. Forget that the player has chosen any door. The question whether to switch or not, is the same as the question which door should he choose. there are two doors. One contains a goat. The other contains a car. The probability that a given door contains the car is 1/2.