Friday, November 03, 2006

The Monty Hall Problem

Imagine yourself on a game show. You are the contestant and you have to pick a door out of the three doors in front of you. The host says that there is a prize behind only one of these doors. Now, you choose a door. The host then opens a door which is not the door you've chosen. Voila!! there is no prize behind that door. After all this drama, the host gives you a choice. you can keep your door, or you can change your choice to the other unopenned door. Faced with this seemingly ambiguous situation, what will you do? Which choice will give you a better chance of winning?

Now, for most people, the intiutive answer will be that it doesn't matter, because there are two doors and there is an equal probability that the prize is behind one particular or the other. So, the answer is 50% for each case.

As you may have guessed that this answer is wrong. And here is the explaination:
Suppose, instead of 3 doors there are a 100 doors. You select one of these 100 doors. The probability that you chose the 'prize' door is 1%. And the probability that you didn't choose the prize door is 99%. Out of the 99 unchosen doors, the host opens 98 doors, non of them containing the prize. With this fact in mind, we can speculate what is happening. The probability of the prize being contained in one of the 99 doors was 99%, but now we know that 98 of those doors do not contain the prize, but this does not change the fact that your chosen door's probability of being the 'prize' door is 1%. What has changed is the probability of the lone remaining door among the unchosen doors. Now its probability is 99% of being the 'prize' door. So, clearly, you'll be much better off changing your choice rather than staying with your current choice.
And what is true for 100 doors is true for 3 doors as well. Here, the probability of winning is 66.66% if you change your choice.

Some things are not what they appear at first.