Hey guys, today i came across this interesting problem and i think i should share it with all of you. Actually it's a rather FAMOUS puzzle which, i think, many of you heard about. It's called

THE MONTY HALL PROBLEM!

For the benefit of those who haven't heard about it and those who have heard about it but forgot what it was about, the problem goes like this:

In the 1960s, there was a popular weekly US television quiz show called Let's Make a Deal. Each week, at a certain point in the program, the host, Monty Hall, would present the contestant with three doors. Behind one door was a substantial prize; behind the others there was nothing. Monty asked the contestant to pick a door. Clearly, the chance of the contestant choosing the door with the prize was 1 in 3.

So far so good?

Now comes the twist. Instead of simply opening the chosen door to reveal what lay behind, Monty would open one of the two doors the contestant had not chosen, revealing that it did not hide the prize. (Since Monty knew where the prize was, he could always do this.) He then offered the contestant the opportunity of either sticking with their original choice of door, or else switching it for the other unopened door.

The question now is, does it make any difference to the contestant's chances of winning to switch, or might they just as well stick with the door they have already chosen?

Well, most people think that it makes no difference if they switch. Here is their reason:

"There are two unopened doors. The prize is behind one of them. The probability that it is behind the one I picked is 1/2, the probability that it is behind the one I didn't is also 1/2, so it makes no difference if I switch."

HOWEVER THAT EXPLAINATION IS WRONG!!
Why is it so??

Try to think about it before you look at the full and correct explaination here

The hint is "Switching actually DOUBLES the contestant's chance of winning"
Interesting??



Definitely!! Enjoy :D