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MONTY HALL PROBLEM

Imagine you are in a game show, the game show host shows you three doors. Behind one of them is the star prize, which is a wonderful car. However, behind the other two doors are booby prizes, two donkies. What door would you choose?


You have no way of knowing which door has which item and depending on the door you choose, you will receive the item behind it. You are asked to pick a door, but before it is opened the show host will open one of the other two doors. The host knows where the car is and she/he will open a door to reveal a donkey. You are then asked if you would like to swap your choosen door for one of the twor remaining closed doors. The questions are: "Should you swap?" or "Should you stick with the original choice?" or "Will not it suppose any difference what you do?" 
Which will it give you the greatest chance of winning the car?

The Monty Hall problem is about probability and the problem is simple to understand, but the answer is counterintuitive. What should you do? The answer is you should always swap because it gives to twice the chance of winning the car. Why? There are many ways to explain why, but perhaps the easiest is to examine the chances of winning the car for the two strategies: not swapping and swapping 

NOT SWAPPING
At the start of the game, you are asked to pick a door. Since there are three doors and only one hide the car, the probability of picking the car is 1/3, or about 33%. And since there are two donkeys, the probability of picking a donkey is 2/3, or about 66%. 

So, if you don not swap your door it does not matter which other donkey door the host opens because you are sticking with your first choice and the chance you have already picked the car is 33%. And the chance you have already picked the donkey is 66%. So by not swapping, you have a 33% chance of winning the car and a 66% chance of winning a donkey.

SWAPPING
Let's consider what would happen if, by luck, you picked the car first time, 33% of chance. It obvious that if you picked the car on your first move and then you swap, you will go ended up with a donkey. So if you swap, you are going to win a donkey at least 33% of the times.

What about if you picked a donkey the first time? Here is the KEY of the problem. This time there is only one donkey that the host can reveal. So the host opens the only other donkey door and then you swap to the remaining closed door, THE CAR. In fact, every time you pick a donkey door for the first time and then swap, you will win the car. And the chances of you have picked a donkey first time are 66%. So by swapping, you have 33% chance of winning a donkey, by picking the car for first time, and a 66% of winning the car, by picking the donkey for first time.


So you should always swap to the remaining door, why? Because if you do, you will have a 66% chance of winning the car and only a 33% chance if you do not. Remember ALWAYS SWAP

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