The Monty Hall problem

Why? Because an unintuitive answer always gives birth to controversy, and controversy leads to the use of reason. Simply put, because our brains need the exercise.

So, we have this probability-related problem loosely based on a TV-show (Let's make a deal -
the problem's name comes from that of the host - the dude from image2) that sounds like this:

"A thoroughly honest game-show host has placed a car behind one of three doors. There is a goat behind each of the other doors. You have no prior knowledge that allows you to distinguish among the doors. "First you point toward a door," he says. "Then I'll open one of the other doors to reveal a goat. After I've shown you the goat, you make your final choice whether to stick with your initial choice of doors, or to switch to the remaining door. You win whatever is behind the door." You begin by pointing to door number 1. The host shows you that door number 3 has a goat." The question is: do your chances to win the car change if you switch?

In one word, yes.

Although apparently absurd, your chances to win the car the moment you are left with two rooms and the option to change by choosing either door are not 50%.

Let me explain: the two options you have as a strategy are 1. do not switch and 2. switch. Being two complementary choices (there is no other one available: you either switch to the other room or stick with your original one), the sum of the possibilities of them gaining you the car is 100%. You are obviously going to think that their respective chances are equal, 50% and 50%. "After all, you only have two choices."

Well, no, not really.

Let's try and see what our chances to win are if we choose to stick with our choice.

The car lies behind one and only one of three identical doors. If we want to stick with our choice, then in order to win the car we must correctly guess the right door from the start. That means we have 1/3, or 33%, chance to win the car. Thus there is a 66% chance of winning by switching.

Think of it this way: you've got three doors. The possibility of guessing the right one is 1/3, while the other two doors have 2/3, together. Since the host eliminates one of the other doors, the 2/3 chances are redistributed to the remaining one, because this basically means: "you either choose one door, and have a 33% chance of winning, or choose the other two, and have 66%"(because, of the two left, he tells you which one COULD have the car by eliminating the other).

There is another, more clear way of proving that, if you switch, you have a 66% chance of winning. The probability of an event is equal to the ratio between the number of cases in which the event does occur and the total number of events. There are three cases, as shown in the following image (again, Wikipedia ftw!) - click for bigger picture - :


There are 3 cases that prove that switching results in a 66% win, while sticking with the original choice only gets us the car on 33% of the time.

P.S. A simple way to test the result is experimentally, with three playing cards. Just take three playing cards, one of which is an ace, for example, shuffle them, and take a pick. If you correctly guess the ace, then sticking with your choice would win you the car, so there's one "point" for it. If you do not, then switching to the other card would win, thus one "point" for switching. You'll definitely notice the difference and if you calculate the chances, they should come close to 66%. The more tries, the more accurate the chances.

P.S.2. I've tried the experimental way. After thirty cases, I had 18 for switching and 12 for "sticking". That is a 60%-40% ratio. After sixty cases and two hours of school, the chances got 63,33%-36,66%. Perhaps if I would continue, they would get closer to 66%-33%. Anyway, this confirmed the theory.

- article used: The Monty Hall problem -