Well, actually that's the page's title. The article is titled "And Behind Door No. 1, a Fatal Flaw", which really isn't as fun.
Anyway, the Monty Hall problem is well described by the writer here:
Here’s how Monty’s deal works, in the math problem, anyway. (On the real show it was a bit messier.) He shows you three closed doors, with a car behind one and a goat behind each of the others. If you open the one with the car, you win it. You start by picking a door, but before it’s opened Monty will always open another door to reveal a goat. Then he’ll let you open either remaining door.
Suppose you start by picking Door 1, and Monty opens Door 3 to reveal a goat. Now what should you do? Stick with Door 1 or switch to Door 2?
The odd thing is that the answer feels entirely counter-intuitive. It seems that it goes against all that "coins don't remember results" guff, because the answer is:
[W]hen you stick with Door 1, you’ll win only if your original choice was correct, which happens only 1 in 3 times on average. If you switch, you’ll win whenever your original choice was wrong, which happens 2 out of 3 times.
That is, if you switch you are twice as likely to win.
To see how this relates to monkeys and M&Ms read the rest of the article.