Monty Hall Problem

When people think about math, they usually think about complex equations, algebra, geometry, and the like. However, some seemingly simple problems like the Monty Hall Problem can be very complex. Think about this problem. 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. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice? At first, it seems like it would not matter if you switched or not, from your perspective, there are 2 doors left, and there is a car behind either of them, a 50-50 chance either way. Right??

Unfortunately, that is not the case. If you wanted to win a car, you should always switch. It comes down to whether you think the second decision is dependent on the first, or whether the second decision is independent on the first. The second decision is, in fact, dependent on the first. If your first decision chose a goat (⅔ chance) always swapping will get you a car, because Monty opened the door with the other goat, and always keeping will get you the goat you originally chose. If your first decision chose a car (⅓ chance) always swapping will get you a goat, cause you already chose the car, and always keeping will get you the car you originally chose. If you stayed with your first choice, you would have a ⅓ chance of getting a car, but the other door, believe it or not, has a ⅔ chance of having a car behind it. It's easier if you start with the probabilities of your second decision first, and then the probabilities of your first decision second.