A few posts back we told you about *Krazy Kaan’s Extravaganza*. It is a game where you first pick one of three doors, Krazy Kaan opens one of the doors you didn’t pick and shows you that there is no prize behind it, and then you have a chance to rechoose a door before accepting the prize (or lack thereof) behind the door you picked.

It turns out that Krazy Kaan “borrowed” his game from a famous problem in probability theory. It’s called the Monty Hall Problem, named after Monty Hall (the first host of the TV gameshow Let’s Make a Deal). In the original version of the Monty Hall Problem, two doors have goats behind them and one has a car — the idea being that you want to win the car, not a goat!

We asked if after Krazy Kaan opens one door and shows that there is a no prize behind it, should you switch? That is, does it increase, decrease, or not change your odds of winning if you switch?

As promised, the answer depends on how Krazy Kaan chose the door. You can work out the odds mathematically, but you can also do it using a little common sense:

- If you know Krazy Kaan chose a door at random from the two you didn’t choose, then you know no more and no less than the fact that the door Krazy Kaan opened doesn’t have the prize. So it is equally likely that the prize is behind the door you choose, and the door neither you nor Krazy Kaan chose. That is, when you have the opportunity to switch, all you know is that the prize is equally likely to be behind your door or the third, so far unselected, door. That is, you have a 1/2 chance that the prize is behind each door. Whether you switch or not, there is a 50% chance you’ll win. So switching doesn’t change your odds of winning.
- If you know Krazy Kaan knew which door has the prize behind it and deliberately chose to show you a door which doesn’t have the prize then, surprisingly, if you switch you’re twice as likely to win! Why? Let’s think about the possible scenarios. When you first pick a door, you have a 1/3 chance of being right and a 2/3 chance of being wrong. If you don’t switch, then it is as if Kaan never opened the door and you still have a 1/3 chance of winning and a 2/3 chance of losing. On the other hand, if you switch then you have a 2/3 chance of winning and a 1/3 chance of losing! Why? Because if you were wrong in the first place (there is a 2/3 chance of that), then the prize was behind one of the other two doors, but Kaan showed you the one which didn’t have the prize, so in switching you are now selecting the door which
*does*have the prize. So your 2/3 chance of losing has suddenly become a 2/3 chance of winning! Similarly, if you were right in the first place (which you had a 1/3 chance of being), then in switching you are choosing a door which*does not*have the prize. So your 1/3 chance of winning becomes a 1/3 chance of losing.

In short, if you know that Krazy Kaan uses his knowledge of where the prize is to pick a door, then by switching doors you also switch your chances of winning/losing!

It takes some thinking to realize that this is correct. Indeed, one of the great mathematicians of the 20th century, Paul Erdos, was initially stumped by the problem (at least according to his biography The Man Who Loved Only Numbers). A puzzle which is equivalent from the probability point of view is the Three Prisoner’s Problem introduced to the public by Martin Gardner.

To see a pictorial explanation, here is a good graphic (each possible game is given by one of the columns) which shows the outcome if you switch. You can use the same graphic in your head to see what happens if you don’t switch.

Alternatively, you can run play the Monty Hall Game a few dozen/hundred/thousand times yourself and keep track of the outcome of switching versus not switching.

The lo-tech way to do it is with a friend who has three cards (say two jokers and a queen), they deal them face down, and you play the game with the goal of winning the queen. The friend peeks and shows you one of the jokers you didn’t pick (hopefully they have a good poker face!).