The movie 21
There is one thing I like in this movie, Trick math and card counting was one of the fascinating thing in the movie and the introduction of the principle of variable change (The Monty hall problem and conditional probability). A game show host tells a contestant to choose between three doors. Two doors have a goat behind them…and the other has a brand new car. He tells the contestant to choose a door. The contestant chooses door #1. Then, the host, knowing what is behind each door, reveals what is behind door #3-a goat. He then asks the contestant if he would like to change his choice or stick with door #1. My initial thought would be to stick with door #1 as it was the original hunch and the probability is 50/50. But this is the wrong answer as the probability is not 50/50, which surprised me. The right answer? You should change to door #2 because the probability that the car is behind door #2 is greater: 66.6% to 33.3%. Why is this? Well, they explained that, at the beginning the probability that the car is behind the door choosen is 33.3% or 1/3. But then, then the host reveals that there is a goat behind door #3 and we have just two choices, which most of us assume have an equal chance of having a car behind them. But they do not However, after door # 3 is eliminated, you are left with doors one and two. Than door #2 and #3 are combined, than door #2 get 66.6% chance and door #1 remains at 33.3%: Of course, you still might get it wrong, but at least your odds are better.
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