There is a classic problem, one that stirs up more debate among first-year statistics students, and other interested people, probably more than any other: the "Monty Hall" problem. As a little bit of background, Monty Hall was the host of a game show called "Let's Make a Deal", which was popular in the 60's and 70's. Contestants, often ridiculously dressed, would be given a small amount of money, or a small prize. Then, they were shown a door or a curtain. The contestant then had the choice of keeping the "sure thing", or trading their given prize for whatever may be behind the door or curtain. Sometimes, it would be a wonderful prize, sometims it would be a "zonk" -- a worthless joke prize, meaning the contestant lost their original stake.

In one game, Monty would produce three keys. One key would unlock a box (which contained a prize or large amount of money), or in some versions, would be a key to a car. The other two keys were "duds". After picking a key, Monty would show that one of the keys he was still holding was a "dud". He would then offer the contestant a "sure thing" prize, in exchange for their key, or, the contestant could stick with the key they are holding, in hopes it would unlock the treasure (or start the car). It was this show, and this particular game that gave birth to one of the most interesting exercises in statisics and probability that exists.

So here, then, is the Monty Hall problem: Say you are given a choice to pick door number 1, 2, or 3. You know nothing about what is behind any one specific door, only that behind one door is a prize (a car, for example), and behind the other two is nothing, or a "zonk" (a dud). After you have chosen a door, a different door is opened, always revealing that this other door has nothing behind it. You are then given the option to keep your choice, or switch to the remaining door. Which option gives you better odds of picking the car? Or are they the same?

The answer may shock you: The car is twice as likely behind the other door, so you should switch.

What?!?

First off, it should be noted that this situation never actually arose in "Let's Make a Deal". On the game show, the contestant could never switch keys or doors once a selection was made. Monty did show a "dud" key to build suspence, but this never changed the odds (because the contestant could not switch).

The problem was originally presented as the Three Prisoners problem, by Martin Gardner in his "Mathematical Games" column in 1959. Three prisoners are locked in a cell, with one being randomly chosen to be executed the next day. Prisoner "A" says to guard, "at least tell me one of my cellmates who will not be executed", claiming since he already knows at least one of the other prisoners will not be executed, this reveals no information to him. The guard says that if he tells the prisoner this information, it changes the prisoner's chance of being executed to 1/2. The question is then, is the guard correct?

The answer is, so long as the guard provides a truly random way to tell the prisoner this information, it does nothing to change the prisoner's odds. Of interesting note is, though, say Prisoner "B" overhears this conversation, and the guard tells A that C is not being executed. From B's perspective, this now increases his odds that he is the one who was chosen.

Let's say that the guard's chosen method is to put B and C's name in a hat, then, unseen to A, pick one out and read it. If A is being executed, he will simply say the name he draws out. If B or C is being executed, he will say the other name no matter what he draws out. In this scenario, there are six equally likely situations, which I will label with the person who is being executed first, and the name drawn out of the hat second:

AB, AC, BB, BC, CB, CC

Before the drawing of the name out of the hat, A knew his chances of being executed were 1/3. Now, being told that C is not being executed, that eliminates above options CB and CC. It also eliminates AB, since if that was the case, the guard would have said B was not being executed. So, with this new information, there are now 3 equally likely situations:

AC, BB, BC

For A, nothing has changed. His odds of execution are still 1/3. However, notice that if B overheard this conversation, he knows his odds are now 2/3! This certainly seems counterintuitive at first glance. First, the ultimate rule of probability is that all possible probabilities must add up to 1, or that is, 100%. If A's chances are 1/3, and C's chances are now 0, B's must be 2/3. However, a seemingly more intuitive solution to this dilemma would be to say that now A and B know they each have a 1/2 chance of being executed. This would be the case if the guard had come up to everyone and said "I'm going to randomly tell you one of the people not being executed". Then the guard would have the 2 people who weren't chosen in a hat, and pick one out and read it. For the 2 remaining prisoners, their chances that they were originally picked is now 1/2, given this new information. So why is it different?

Let's say that this second scenario happened. Again, there are six equally likely possibilities:

AB, AC, BA, BC, CA, CB

First, notice that by making this seemingly subtle change in the problem, the possible situations that arise are different. Now lets say the guard pulled out C's name. This only leaves two possibilities: AC and BC. Notice that for whatever name the guard calls, it only leaves two situations, one each involving the two prisoners who were not called being executed. The reason this scenario is different is because of the limited way the guard could reveal information in the first scenario. The guard could not include A's name in the possible names to draw, as to not give A any information that changed his chances. However, in doing so, this significantly changed things for B, who could now figure out his odds based on the information given. Another way to look at it is like this: if you look at the six possibilities for the first scenario; AB, AC, BB, BC, CB, CC -- you'll notice there's a 1/2 chance the guard will end up saying B's name, having no information beforehand as to who was actually chosen for execution. So, from B's perspective, there's a 50% chance the guard will say his name, meaning he's safe. There is also 50% chance the guard will say C's name, meaning B's chances of being executed are now 2/3. So B's chances of having been originally chosen are the 1/2 chance that C's name is said by the guard, times the 2/3 chance B was originally picked if C's name is said. So his chances of having been originally chosen to be executed are 1/2 * 2/3 = 2/6 = 1/3. This is actually the more "intuitive" solution. Notice if that C's name being called gave B a 1/2 chance of being the one chosen, and B's name being called (obviously) gave him a 0 chance of being the executee, then B's chances of being the one originally picked would be 1/2 * 1/2 =1/4. So as much as the 1/2 answer originally seems intuitive, the math does not back it up.

So what's all this about prisoners? I thought we were talking about a car. Well, it's actually the same problem, cleaned up a little bit so we don't have to talk about the chances of execution. The key to this problem is "Monty" is obligated to reveal a "dud" door. A simple way to think of it is this: there is a 1/3 chance you originally picked the car, so both of the other doors are duds. Switching in this scenario would cause you to end up with a dud 100% of the time. However, there is a 2/3 chance you originally picked a dud. Monty must show you the other dud. This means the remaining door must have the car. Switching in this scenario would cause you to end up with the car 100% or the time. So, switching will cause you to switch to a dud 1/3 of the time, but will cause you to switch to a car 2/3 of the time!

It would be different if, say, 3 different contestants were each assigned a door. If, in order to add suspense, Monty always opened a random "dud" door first, that person would now know their chances of having a car is zero. The other contestants would now each have a 1/2 chance of having the car. The reason this is different is because each of the contestants had an equal chance of their door being revealed if they picked a dud. In the original scenario, Monty could not reveal your door, and must reveal a dud. This limits Monty's choices if you originally picked a dud. In the second scenario, if you picked a dud, Monty could chose your door to be randomly revealed first, to build excitement and drama for the other two.

So, in the end, it all comes down to the information provided to you to determine the probability of an event. You need to consider the consequence of all information, even indirect information (Like B's knowledge that C wasn't chosen). In the end, make sure your problem works out forward and backwards mathematically (like we worked backward to figure B's original chance of being chosen as 1/3), and that all probabilities equal 1 (or 100%).

The amazement that comes from the seemingly senseless answer made completely sensible often causes this, and other probability "paradoxes", to be the topic of great debate and discussion. This problem has also remained famous for its association with a long-running TV show, and its very popular host. The question was also presented (with goats as the 'zonks' or duds) to Marilyn vos Savant, once listed in the Guiness Book of World Records for "Highest IQ", who writes a weekly column in Parade magazine. She presented the correct 2/3 answer, and the magazine was flooded with mail of people, some who were reputable mathematicians, convinced she was wrong. For this reason the problem is also known as "Marilyn and the goats"