Here is a famous story called The Monty Hall Problem.

You are on a game show on television. On this game show the idea is to win a car as a prize. The game show host shows three doors. He says that there is a car behind one of the doors and there are goats behind the other two doors. He asks you to pick a door. You pick a door but the door is not opened. Then the game show host opens one of the doors you didn’t pick to show a goat (because he knows what is behind the doors). Then he says that you have one final chance to change your mind before the doors are opened and you get a car or a goat. So he asks you if you want to change your mind and pick the other unopened door instead. What should you do?

You should always change and pick the final door because the chances are 2 in 3 that there will be a car behind that door. But if you use your intuition you think that chance is 50-50 because you think there is an equal chance that the car is behind any door.

Explanation: Firstly you can do it by math like this

Let the doors be called X, Y, Z. Let Cx be the event that the car is behind door X and so on. Let Hx be the event that the host opens door X and so on. Supposing that you choose door X, the possibility that you win a car if you then switch your choice is given by the following formula

P(Hz^Cy)+P(Hy^Cz)=P(Cy).P(Hz|Cy)+P(Cz).P(Hy|Cz)=(1/3.1)+(1/3.1)=2/3

The second way you can work it out is by making a picture of all the possible outcomes like this:

So if you change, 2 times out of 3 you get a car. And if you stick, you only get a car 1 time out of 3.