The Three Door Problem
Kapitoly: The Three Door Problem, The Probabilistic Liar's Paradox, The question mark paradox, Simpson's Paradox, The medical paradox, The St. Petersburg Paradox, Non-transitive cubes
The three-door problem, sometimes called the Monty-Hall problem, is a probability problem that can confuse more than one person.
Assignment
Let's have three doors. The presenter, Monty Hall, places a car behind one door and a goat behind the other door. Your task is to find and select the door with the car behind it. At this point, he prompts you to choose one of three doors, let's label them A, B and C. You choose one door, say B. Next, the moderator enters the game, and from the remaining door, A, C, he opens the door behind which the goat is hiding - that is, he reveals a door that definitely does not lead to the goal. Let's say he opens the door A.
The point of the whole problem is in the next step - the moderator will offer that you can change your choice. You first chose the door B, then the moderator tells you that there is no car behind the door A and offers you that you can change your choice to the door C. But you can keep your original choice B.
How do you decide? Is it more likely that the car will be behind the door B, or behind the door C? Or does it not matter?
Solution
You have three doors. One door is out of the game, so you actually only have two doors left. There is definitely a car behind one of these doors. You don't know anything about the other two doors, you have no other clues. So the probability is obviously 50% for both doors.
That's roughly the mindset of a lot of people, so don't worry if that's what you thought. But of course it's a bad solution :-).
The correct solution
It's worth changing your choice to the door that the moderator didn't eliminate. To understand, we can break down the probabilities from the beginning. When you first choose a door, the probability that there is a car just behind it is 1/3. We have no other clue, so the probability is divided equally, 1/3 for each door.
So if you choose a door B, you have a chance 1/3 that there is a car behind it. At the same time, there is a 2/3 probability that the car is behind door A or C. So it is more likely that the car is behind the door we did not choose. This is important.
In the next step, the moderator enters the game and opens one of the doors that does not have a car behind it. The moderator opens the door A. So there is no car behind the door A. What happens to the 2/3 probability? We know that the probability that the car is behind the door A or C is 2/3 and we also know that it is not behind the door A. So the 2/3 probability spills over to the door C. The chance that the car is behind the door C is 2/3.
But our chance at the door B remains the same, it hasn't changed. So we have 1/3 the chance that the car is behind the door B that we have chosen, and 2/3 the chance that it is behind the door C that we can change our choice to. So it's reasonable to change your choice.
More doors
The example can perhaps be better understood if we increase the number of doors and modify the problem as follows: let us have ten doors. At the beginning, the contestant chooses one door and the moderator opens all the doors except the one we chose and except the one with the car behind it. The task is practically the same as in the previous example, only there, when the moderator opened all the doors, he opened only one door because he could not open more.
At the beginning you choose a door 1. The chance that there is a car behind it is 1/10. There is 9/10 a chance that there is a car behind another door. Next, the moderator removes 8 doors that you are sure there is no car behind and you are left with two doors again. The chance that the car is behind the door the moderator left is 9/10, while the chance that it is behind your chosen door is still 1/10.