A couple of days ago, a friend discussed an interesting puzzle. The puzzle concerns probabilities and I found it very intriguing. I thought I’ll write about it. The puzzle is popularly known as the Monty Hall problem. The problem is described as follows.
A friend, Monty, invites you to play a game where you have a chance to gain big. He brings you to a large hall with three large doors named, say, A, B and C. He says that behind one of the doors, he has parked a gleaming new car, and behind each of the other two, he has managed to make a goat stand quietly. If you win the game, you take the car. He asks you to choose a door. Say you choose B. He now opens another door and shows you a goat behind it. Say, he opens C, and shows you a goat peacefully ruminating behind it. He then asks you whether you want to switch your choice from your original door B to the remaining door A. You can either opt to switch your choice, or retain your old choice. Whatever your decide in response to this offer defines your final choice. He opens the door of your final choice and if you find a car behind it, you win. If you find a goat behind it, you lose.
The puzzle is, if you were playing this game, would you switch your choice when offered, or retain the original choice?
My immediate answer, like most people, was that it does not matter. Since two doors remain, and one of them has a goat and the other a car behind it, there is no need to change my choice. The probability of a car being there behind my original choice B, must be equal to the probability of a car being there behind the other door A, which is equal to 1/2. However, on deeper thinking this turned out to be a fallacy.
If I do not switch my choice from B to A when offered, the goat that my friend shows me behind C makes no difference to me. I simply stick with my earlier choice of B. The only way in which I can win in this case, is if the car is behind B. The probability of this is 1/3, since the car could be behind A, B or C with equal likelihood.
Now, if I switch my choice from B to A, the situation is a little more interesting. The door C does not have a car as the friend has shown. I win if A, and not B, has a car behind it. In other words, the only way I win is if my original choice B does not have a car behind it. And the probability of this event is 2/3. So, the chances of my winning are doubled if I switch my choice from B to A!