When I first read about the Monty Hall Problem I didn't pay much attention to it. Recently MK wrote something about this. A book I am reading has a chapter about it too. If you have not heard of it before, here it is :
You are in a game show on TV. In this show, the idea is to win a car as a prize. The host of the show shows you 3 doors. He tells you there is a car behind one of the doors and there are goats behind the other two doors. He asks you to pick a door. So you pick a door. Then the host, who knows what is behind each door, opens a door you didn't pick to reveal a goat. He says that you have a final chance to change your mind and ask you if you want to pick another door.
What should you do? Do you have a better chance if you change and pick the other door instead?
Most people think it doesn't matter. Whichever remaining door you choose the chance is 1 in 2 or 50:50. Not true. If you change your door, the chance is higher, 2 in 3.
It is quite intriguing and not easy to understand but it could be proved.
If you are interested, search for Monty Hall problem and read it up.
I tried out the game with 3 cards, one card with the word CAR written on it and the other two cards with the word GOAT. Every time I changed the original card I had picked. Results: 7 out of 10 times I got a car. That means if I didn't change the card I got a car only 3 out of 10 times.
3 comments:
This problem is documented in Wikipedia.
After reading it, I still don't understand the problem, especially the complicated mathematical formulae at the end of the article.
I am still of the opinion that it doesn't matter whether you switch or not.
Yes, the difficult cases and formulae are also beyond me.
But if I were in the game, I would switch since it doesn't matter or maybe I really got more chance.
The book I was reading gives quite a logical explanation. I think I'll put it up.
haha.. i read abt it recently too!
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