Monty Hallin'

Two days ago I saw the movie 21. Its based on a true story. Its a story about five MIT students and how they are recruited by a university professor to count cards in Black Jack. The plot is absolutely predictable but I definitely enjoyed it.

They referred to the Monty Hall Problem in the beginning of the movie. This sent nostalgia spinning up my spine. I never did pay attention in that class on probability, maybe cause the room was always so jam-packed and the professor was sooo irritatingly annoying. Thus, I never did get the hang of it and I think that was my lowest grade in all my years at uni. So I obviously couldn't remember much about the problem except its name.

The Monty Hall Problem is a problem of probability based on a tv show. The host asks you to pick one door out of three knowing that behind two of them is a goat and behind just one of them there is a bran new car. The objective obviously is to pick the door with the car behind it.



Once you pick one of the doors which has a probability of 1/3 to have the car behind it, the host then goes and opens one of the other two doors. It reveals a goat. The host then asks you whether or not you want to change your choice. So you offered to keep to your first choice or to switch to the other closed door.

The simple scenario is that you pick one door and stay with it. The probability in this case is 1/3 that the car is behind it. But actually it would be wiser or mathematically wiser to switch. Why?

The probability of your chosen door is 1/3 which makes the probability that the car is behind one of the other two doors 2/3 (shared across both doors). When the host reveals one of the two other doors, he has made the 2/3 probability rest solely on the closed door of the pair. Therefore your chosen door has a probability of 1/3 (same as before) and the other remaining door has a probability of 2/3. This is why it is wiser to switch.



You can look at it from various angles. You can analyze it using Bayesian analysis or map out all scenarios and see what happens. Another way to look at it is that the probability of choosing a goat in the first place is 2/3. In all cases if you choose a goat in the first place, and switch, you will win the car. This is because the host will reveal the other goat and thus the remaining door has the car behind it. But if you stick to the choice you only have a 1/3 chance of winning the car.

An easier way to look at this is to increase the number of doors to say; a 100. You choose one and the host reveals 98 of the other doors. Will you switch? The probability of choosing a goat first is 99/100 which is 99%. It is almost certain that you choose a goat door. Then the passes along the other 99 doors and opens them all except for one somewhere along the line. A rational decision would be to switch to that door knowing that there is a 99% chance you choose a goat to start with and a 1% chance you choose the car from your first choice. Pretty cool huh?

Its a really cool problem that sparked confusion and debate in the world. The problem and solution were published in the Parade magazine. About 10,000 readers (including several hundred mathematics professors) wrote to the magazine claiming the solution was wrong. People saw the problem of switching as a bran new problem independent of the first choice. In actual fact it is because depending on which door you choose at the beginning the host will reveal one of the other doors which has a goat. So the host's choice is based on your choice. So therefore there is a relationship between which door the host reveals and the door you choose.