You may have heard a bunch of hype lately about the Monty Hall problem, that keeps raising its head from time to time. The problem, stated unambiguously (and thus verbosely), is this:

Suppose you're on a game show and you're given the choice of three doors. Behind one door is a car; behind the others, goats. The car and the goats were placed randomly behind the doors before the show. The rules of the game show are as follows: After you have chosen a door, the door remains closed for the time being. The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it. If both remaining doors have goats behind them, he chooses one randomly. After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door. Imagine that you chose Door 1 and the host opens Door 3, which has a goat. He then asks you "Do you want to switch to Door Number 2?" Is it to your advantage to change your choice?

[http://en.wikipedia.org/wiki/Monty_Hall_problem]

Here is how I solved the problem, in my head:

- Since I know beforehand that Monty will eliminate one of the goat doors after I make my choice, I can imagine that the two goat doors are actually one.
- Because that "one door" is represented twice, I have a ⅔ chance of picking it. Therefore, I have a ⅓ chance of picking the car door.
- If that doesn't make sense, consider that behind each door is a big tube, and two of the tubes lead into a room containing a goat, and the other tube leads into a room containing a car. There is one goat, but I have a ⅔ chance of picking it.
- When one of the goat door options is then taken away, I can choose between two actions:
- stick with my original choice, which has a ⅓ chance of being the car
- swap my choice to the door that, since it is my only remaining option, has a ⅓ chance of
*not*being the car.. that is, it has a ⅔ chance of being the car

Said another way:

- If I picked the first goat door, the host has to reveal the second goat door. There is a ⅓ chance of this happening.
- If I picked the second goat door, the host has to reveal the first goat door. There is a ⅓ chance of this happening.
- If I picked the car (⅓ chance), the host can either:
- reveal the first goat door (⅓ × ½ = ⅙ chance of this happening)
- reveal the second goat door (⅓ × ½ = ⅙ chance of this happening)

- Of the four outcomes, I have a ⅓ + ⅓ = ⅔ chance of having the goat if I don't switch, and a ⅙ + ⅙ = ⅓ chance of having the car. That seems obvious, right? So then if I
*do*switch, I have a ⅔ chance of switching*away from*the goat,*to the car*. So I have a ⅔ chance of winning by switching.

Said yet another way:

Each door has a ⅓ chance of being the car. If we split the doors into: the one I picked, and the two I didn't pick, the "one I picked" has ⅓ chance of being the car, and the chance that the car is in the "two I didn't pick" is ⅔. One of the two I didn't pick is revealed as a goat door, so it has a 0 chance. Therefore the other door I didn't pick must have (⅔ - 0 = ⅔) chance of being the car.

There is another path to consider:

If I know beforehand that my first choice doesn't matter, because after I make it Monty will eliminate one of the doors, I can "throw away" my first choice. After Monty gets rid of one of the doors, I know that there is a ½ chance that I will pick the car door. By discarding the past, I now have a 50% chance of getting the car. That's true and correct and all, because if you eliminate past knowledge, you really do have a ½ chance of picking the right door. But why would you do that? There's a perfectly valid option that gives you a ⅔ chance, which is much better than ½. Why not *pay attention to past information* and use it to your advantage?

... Matty /<