However, when the answer was given, many of the students were still a bit confused as to why it worked or made sense. I too nearly scratched a hole in my head trying to figure this out, and it wasn't until I framed the problem a little differently that the solution began to appear. After sharing my discovery (which, by the way, I'm not claiming to be the only one who's thought of this...) with my classmates, a few lightbulbs started appearing over a few of their heads. Hopefully, this twist of the problem can help you see why the not-so-obvious solution is the correct one.
Let's start with the formal problem:
Monty Hall, the gameshow host on "Let's Make a Deal", shows you three closed doors. Before any of doors are opened, he tells you that behind two of these doors you'll find a goat. But behind one is a prize worth taking home. The point of the game is to pick the door hiding the prize - you pick it, you win it.
After you choose a door, he opens one of the two doors that you didn't choose and reveals a goat. Keep in mind that Monty will always reveal a goat. You are then given the option of switching to the other closed door that you didn't choose or staying with your original choice.
Should you switch, and why or why not?
Right off the bat, most of us (including me at first) think switching makes no difference because we believe there's a 50/50 chance of either our 1st door choice or the other closed door having a prize behind it - but that's incorrect!
Let's change the game a little bit, say, there's a 100 doors, behind which are 99 goats and one prize. You choose one, and Monty opens up 98 doors revealing 98 goats (remember that Monty only shows goats and gives you one closed door to switch to). Again we're left with a similar choice: stay or switch to the only closed door left. Knowing that almost all of the goats have been revealed, and the odds are that we probably picked a goat from the start (a 99-in-100 chance to be exact), does it still seem like there's a 50/50 chance that we'll find the prize by switching?
Say the game has 1,000,000 doors and after we pick one, Monty reveals 999,998 goats. Does it seem that the winner is more or less likely to be behind the door he didn't open? In this case, there's a 999,999-in-1,000,000 chance that we probably picked a goat to begin with.
How about 1,000,000,000,000,000 doors? See where this is going? Now imagine these doors being group into sets: those that hide goats, and those that don't (which is always just one door).
Since we're more likely to pick a door at random from the set of losing doors (because there's so many more of them), along with Monty revealing the remaining goats in the set, it should seem that switching moves us to the set containing the winning door.
This same logic applies to our original 3-door case. To start, we've a 2/3 chance of randomly picking a losing door and only a 1/3 chance of picking the winner. Since we're more likely to pick a loser, Monty showing us the remaining losers pretty much tells us where the winning door is, 2/3 of the time! Therefore, you've a 2/3 chance of winning the game by switching.
Conversely, you've a 1/3 chance of losing by switching which corresponds to the 1/3 chance you have of picking the winner from the start. If you do, the set of doors remaining is the losing set, and Monty will open all the doors but one (it doesn't matter which ones).
And there you have it. Hopefully this has shed some light and perhaps settled a few arguments out there about why switching gives you a probabilistically better chance of winning. If you want to see this in action, check out this interactive feature from the New York Times online where you can play the game repetitively. Here you can see your score of wins and losses over time, and how you really do in fact win by switching about 2/3 (or 66.67%) of the time.
Have fun!
