Also, a simpler method of explaining the Monty Hall problem is to think of it if there were more doors. Lets say there were a million (thats alot ["a lot" grammar nazis] of goats.) You pick one and the host elliminates every other door except one. The probability you picked the right door is one in a million, but he had to make sure that the door he left unopened was the one that had the car in it, unless you picked the one with a car in it, which is a one in a million chance.
That's awesome. I shall use it in the future. Wish I could multi upvote.
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Are you serious? Are you buying this? Ok - let me make this easy: There NEVER WAS a 33% chance. Ever. The 1-in-3 choice is a ruse. No matter what door you choose, Monty has at least one door with a goat behind it, and he opens it. At that point, you are presented with a 1-in-2 choice. The prior choice is completely irrelevant at this point! You have a 50% chance of being right, just as you would expect. Your first choice did absolutely nothing to influence the outcome! This argument reminds me of the time I bet $100 on black at a roulette table because it had come up red for like 20 consecutive times, and of course it came up red again and I lost my $$. A guy at the table said to me "you really think the little ball remembers what it previously did and avoids the red slots??". Don't focus on the first choice, just look at the second - there's two doors and you have to choose one (the one you already picked, or the other one). You got a 50% chance.
Fail.