Alright, so the infamous Monty Hall problem goes something like this!

The host of a game show, offers the guest a choice of three doors. Behind one is a expensive car, but behind the other two are goats. After you have chosen one door, he reveals one of the other two doors behind which is a goat (he wouldn’t reveal a car). Now he gives you the chance to switch to the other unrevealed door or stay at your initial choice. You will then get what is behind that door. You cannot hear the goats from behind the doors, or in any way know which door has the prize. Should you stay, or switch, or does it even matter??

The Solution of the Problem is like this :

Consider yourself to be in the game show Lets say you pick door 1 always. Its a randomn a choice as any. You can repeat the below solution with any door for that matter

As per the basics of math, three combinations can be made :

Assuming you denote a goat by “G” and a car by “C” , the combinations are :

G G C

G C G

C G G

Now consider the following the cases :

CASE 1 : G G C

You choose door 1. The host knows the position of the car. So he reveals the door 2. Now if you switch to door 3, then you will win the Car!

CASE 2 : G C G

You choose door 1. The host again knows the position of the car. So he reveals the door 3. Now if you switch to door 2, then you will win the Car!

CASE 3 : C G G

You choose door 1. Now the host can reveal any of the doors, since door 2 & 3 have goats behind them. Lets say he reveals door 2. Now if you switch you will get a goat! 😛

Hence as we see, by the basics of probability , the Switch will get you a car 2 out of every 3 times or 66.66% !