The Monty Hall Problem is unintuitive. Even if you see an explanation or proof of it, it can sometimes still "feel wrong". Here we'll look at another way for it not to make sense, one that is simple but I don't see offered very often.

The Monty Hall Problem is as follows (as copied from Wikipedia):

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?

The optimal choice is to switch. You will win 2/3 times if you switch but only 1/3 times if you don't. It doesn't make too much sense, intuitively, because you have two options and our gut says that there should be a 50/50 chance of winning. Another factor is that it would feel much worse to lose if you had chosen the correct one originally but switched off of it.

There are many solutions offered all over the web, most of which appear on the wikipedia article. These range from running the game many times to see which outcome wins, to direct proofs, through use of conditional probability, appeals to the host "knowing" where the prize is, or appealing to intuition by making the game 100 doors rather than 3. But it is much simpler than that.

Consider the following variant on the game:

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 then says to you, "You have a choice, either win if the prize is behind Door No. 1, or win if it is not behind Door No. 1." Which choice gives you the greatest advantage to win?

It is immediately clear that, in this game, you should choose the latter option. There are twice as many chances to win than if you choose the former. The real question is then "What is the difference between these two games?"

The answer to this question is: Optics. Structurally, they are identical. Switching in the original Monty Hall game is totally equivalent to choosing to win if the prize is not behind the door you choose. To make this clear, imagine that you are playing the original game and choose to switch before the host open either door. You're then choosing to win exactly when the prize is not behind the door you originally chose. Making this choice before-or-after the host opens one of the doors is irrelevant. What the host is doing is dressing the the choice up in what is, essentially, clouds and mirrors making it seem more difficult than it actually is.

So the real problem with the Monty Hall problem is that it is a easy problem dressed up as a difficult choice through illusions. A Magic Trick.