In one of the poorer gambling movies, "21," Kevin Spacey plays a teacher who poses the scenario from "Let's Make a Deal" to one of his students:
(this is a synopsis)
A game show host tells you to pick one door out of three doors. He tells you behind one door is a new car and behind the other two are goats. After you select the door, he opens one of the doors you didn't select to reveal a goat. He asks now if you'd like to change the door you've selected. Should you? Does it make any difference and why?
In the movie, the soft-spoken protagonist meekly offers that he would change his door. When asked why, he mutters something about "variable change" (which doesn't really explain why) and the teacher grins at his newly discovered genius. The audience murmurs at the deduction displayed which must be high-level as it is beyond even their understanding. This sets the stage for the rest of the movie which regales "card-counting" as the nectar of geniuses. Really, card-counting requires the mathematical ability of a 5 year old but the focus of a 6 year old (which we lost when we turned 7 and found the internet or tv or a ninja turtles costume. . . but I digress).
The correct answer is to change from the original door, but the explanation is not as easily understood. It all stems, however, from the concept that the host will never open the door containing the car then ask if you want to change (obviously). So the door that he chooses to you is not as random as it seems.
Beginner's visualization: Imagine instead that there are 100 doors. You pick one then the game show host reveals every door except one other door to be a goat. It a seems a little easier to conceptually realize that there's something more peculiar about the door he left shut in this case.
The real explanation: Let's examine the multiple possibilities here to get an idea of the percentages- Your first choice is random:
2/3 of the time you will pick a goat. In these times, the host will reveal the other remaining goat because he cannot simply show you the car then offer that choice. If you choose to switch in these cases, you will get the car 100% of the time.
----so 2/3 of the time we get the car every single time AKA 100% * 2/3 = 66.66666%
1/3 of the time you will pick the car and the host can reveal either of the other doors which contain a goat. It doesn't matter which one he reveals. When you change here, you get the car 0% of the time.
----so 1/3 of the time we never get the car AKA 0% * 1/3 = 0%
Adding these percentages together, we get the car 66.66666% of the time with this strategy. This is improved MASSIVELY over our original 1/3 times and ALSO over the apparent 50/50 choice after the host opens one of the doors.
A lot of people get caught up in the psychology of the problem and think they're somehow being tricked, but the lesson is to first examine the information at hand then arrive at the most optimal solution . . . Also card-counting is for noobs. Don't do it.