**The Monty Hall problem** is a probability puzzle based on the American television game show whose host was Monty Hall.

Imagine you’re having a really great day and you’re feeling very lucky so you decide to go on the show. When you get there you see three doors.

These are the rules of the game: Behind two of these three doors are goats and behind the last one a brand new car (Let’s say you’re trying to get a car and not a goat).

Let’s say you already chose one of them, for example the red one. The host now, who knows what’s behind the doors, opens the green doors which have a goat.

Now the host offers you to change your decision. What do you do? Do you stick with the door you chose or do you change? It may seem odd, but the odds are not $ 50 – 50$. You should switch the door – by doing that you’ll have almost $ 67%$ chance of winning.

Why exactly should you change your decision?

First you have three choices. The chance of winning is $ 1 : 3$. If you stick with the door you chose your chance will remain $ 1 : 3$. This means that there must be $\frac{2}{3}$ chance the car is somewhere else. If we know that the car is not in the last door, this means that there is $\frac{2}{3}$ chance the car is behind the yellow door. It is not certain that the car will be behind the yellow doors, but there is twice as much chance he is behind the yellow door than it is behind the red door.

Now let’s imagine having 100 doors. Now you got yourself in a kind of a bad situation. You have only $\frac{1}{100}$ chance to get it right. Now what if the host opens 98 doors with all goats behind them? Now you are left with only 2 doors. Now it’s a bit clearer. The chance you got it right remained $\frac{1}{100}$, but the chance that the right door is somewhere among the other 99 door is now concentrated on the door you didn’t pick.

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