So here is the outline of the scenario: You are in a gameshow, let's say The Price is Right, and have the opportunity to win a brand new Range Rover. It is shiny black with chrome rims, and you really want it, unfortunately, you must choose the correct door, A/B/C, and Bob Barker has already revealed what's behind C. Behind C there is the undesirable presence of a goat. At this point, you're probably thinking, "hmmm I have a fifty/fifty chance of choosing the car because the goat was already behind one door, so the only options left are either another goat or the beautiful car," but here is where you are wrong.
Initially, I would say it makes no difference whether you decide to stick with A or change your door of choice to B. But, apparently this isn't right, even though it makes complete sense. Because a goat was already picked, this situation becomes Not Independent, and swapping your door of choice will double the likelihood of getting the car. At the start of the game, the probability of picking a goat is 2/3, because there are two goats and three doors. If you don't swap, the probability of picking the car remains 1/3, and that of picking a goat is still 2/3, and does not change to 1/2. Switching to B is a auspicious decision because you are going to win a goat at least 1/3 of the time if you switch from the winning door to the goat door. If you pick the goat one first and then change to the other door, the probability of getting the car turns into 2/3. Always swapping to the remaining door will invariably double your chances.
Before entering any game show, I strongly recommend that you brush up on your probability, and review the difference between independent and Non Independent events. I hear that Ms. Mariner is great at assisting with that! Here is a wonderful diagram that perfectly exemplifies this situation and is based upon the preconceived notion that a goat was chosen first.
