Solution to Monty Hall problem

After writing and running a simulation program for n number of times I realised that the switching actually increases the probability of winning. The winning probability after switching is 2/3. Obviously without switching it would be 1/3.

There are n internet resources giving solutions to this problem. My explanation for the problem is as below.

You choose a door.  Probability of having car behind it is 1/3.

The host has two doors to open now. He knows what each door contains. They might contain following combinations

goat goat

goat car

car goat

In all three cases the goat has to be there behind one of the doors. The host always invariably chooses to open the door that has a goat.  If you notice above three combinations you notice that out of three cases he has in two cases the closed door will have a car.

So if you switch the probability of winning is 2/3.

People with mathematical disability might complain that this may not happen in practical situation but IT DOES HAPPEN and it must happen.

Monty Hall Problem

Imagine you are playing in a game show. You are shown three doors behind one door there is a car, behind others there are goats.

Now you select a door. The host open either of the other two doors. The opened door contains a goat.  The host than gives you an option. To switch your choice.  You can either stick to your chosen door or switch to the other door.

Will switching the door improve your probability of winning the car???????????

Solution: The solution is well known. No prize for guessing the correct one but the one who calculates it first will get an Apple ipod. 🙂 kidding