Suppose you're on a game show, and you're given the choice of three doors [and will win what is behind the chosen door]. Behind one door is a car; behind the others, goats. The car and the goats were placed randomly behind the doors before the show. The rules of the game show are as follows: After you have chosen a door, the door remains closed for the time being. The game show host, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it. If both remaining doors have goats behind them, he chooses one [uniformly] at random. After the host opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door. Imagine that you chose Door 1, and the host opens Door 3, which has a goat. He then asks you "Do you want to switch to Door Number 2?" Is it to your advantage to change your choice?Baaah!
Please think about it some more. Better yet, imagine that there are a hundred doors (all with smelly goats behind them, except for one hiding a car), and the host opens all the doors except one. So now, only two doors are closed: your initial choice and the one the host didn't open. Would it improve your chances to switch doors? Ask yourself, when you chose that first door, what were the odds that it had a car behind it? Then ask yourself, what are the odds that the car is behind the door that the host left closed??? "To switch or not to switch?", that is the question.
Don't be hasty, but for the solution (and the source of the above problem statement), see http://en.wikipedia.org/wiki/Monty_Hall_problem
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