The Monty Hall Problem
The Monty Hall problem goes like this: You are presented with three identical doors. Behind one of them is a car and behind the other two are goats. You want the car. Monty Hall tells you to choose one of the doors. Regardless of which door you choose, at least one of the two remaining doors will have a goat behind it. Monty Hall, who knows where the car is, then opens one of the doors that has a goat behind it. He then gives you the option of either sticking with the first door you chose, or switching your choice to the other unopened door.
Question: What should you do? Should you stay where you are? Swtich? Does it make a difference?
This problem is a staple of courses in elementary probability theory. Virtually everyone, upon hearing this problem for the first time, reasons as follows: After Monty Hall opens one of the doors, there are only two doors remaining. Therefore, regardless of which door you choose you have a 50-50 chance of being right. So it doesn't matter whether you stay where you are or switch to the other door.
This argument is clear, convincing and wrong.
I suspect that every mathematician who has ever presented this problem to a lay person has had the following experience: Mathematician explains problem. Lay person reasons as above. Mathematician explains that, actually, you double your chances of winning by switching doors (more on this in a moment). Lay person gets annoyed, agitated and belligerent. Lectures mathematician on the subtleties of the problem. Repeats, ad nauseum, that after Monty Hall opens one door there are only two, equally likely, doors left!.
I was reminded of this when I came across this blog entry, from The Daily Howler, posted on March 31. The blogger, Bob Somerby, was commenting on this review of a new book about probability theory intended for nonmathematicians. The reviewer wrote:
Before scoffing, chew on the now famous Monty Hall problem, named after the host of “Let's Make a Deal.” A contestant knows that concealed behind three doors there are two goats and one new car. The contestant chooses Door No. 1. The beaming host opens Door No. 3 to reveal a goat, and then asks the contestant if he would like to change his choice to Door No. 2. Two doors add up to a 50-50 proposition, obviously. So why bother? Because the odds have actually shifted. The chances are now two out of three that changing to Door No. 2 will obtain the car. (Emphasis Added by Somerby)
Say what? We don’t know what the Kaplans wrote to provoke that highlighted sentence. But for the record: If the contestant changes to Door No. 2, he’ll obtain the car half the time—and “half the time” is not “two out of three.”
Ugh. That's totally wrong, I'm afraid.
The following day Somerby revisited the topic. Apparently some e-mailers informed Somerby that he was mistaken. Sadly, Somerby decided to dig in deeper:
Many e-mailers wrote to insist that there is a counterintuitive “Monty Hall problem” of the type we discussed yesterday (see THE DAILY HOWLER, 3/31/06). We haven’t had time to review this in detail, but:
We didn’t dispute that there’s some such effect—an effect which the Kaplans describe in their book. What we said is this: Whatever that counterintuitive “Hall effect” might be, the Times review doesn’t seem to describe it. We’ll persist in our statement about the situation as described in the Times review: In that situation, it just isn’t true that the contestant would gain an advantage from switching his guess. We’ll grudgingly try to sort through the matter. But what a bad time for this storm to reach land—on a weekend when tyrannical guvmint allows us just 47 hours.
Oh my. The description of the problem given in the Times may not be a model of clarity, but it is clear enough to make the point. Somerby is wrong and the reviewer is right.
I don't know about any counterintuitive “Hall effect”, but the mathematics of the situation are not especially complicated. When the contestant makes his initial choice he has a probability of 1/3 of being right. In other words, he will get it right 1/3 of the time and get it wrong 2/3 of the time. Nothing Monty Hall did in opening one of the doors changes that simple fact. By sticking with your initial choice you will be wrong 2/3 of the time. By switching, you will win 2/3 of the time.
Or think of it this way: Your initial choice is correct 1/3 of the time. Does it really make sense to say that when Monty Hall opens one of the remaining two doors your chances of having made the correct choice magically jump to 1/2?
Or try it this way: Suppose you choose door number one, and Monty Hall opens door number two. The choice Monty Hall then gives you is not really Door Number One vs. Door Number Three. Really the choice is Door Number One vs. Not Door Number One. Since door number one is only correct 1/3 of the time, surely it makes sense to swtich.
Still not convinced? Okay. Suppose you had 100 doors. The doors conceal one car and 99 goats. You choose door number one. Monty Hall then opens 98 goat-bearing doors. Are you seriously claiming that in this situation it makes no difference whether you switch or not? If you are not seriously claiming that, then explain to me how this situation differs from the original version. (In my experience, this way of putting it usually gets people to realize that things are not as simple as they originally thought. One time, though, I had a student in my office who was absolutely convinced that it made no difference whether or not you switched. No matter how I tried to explain it he woudn't give in. So I whipped out this example and smiled. Without missing a step he informed that even in this case it would make no difference whether you switched. I stopped smiling.)
Want more? Try this computer simulation. Do it many times and keep track of your statistics.
Think the computer simulation is rigged? Fine. Get out a pad and a pencil. Make a list of every possible scenario. (For example: Car behind door number one, you choose door number one, Monty Hall opens door number two.) It's a little tedious, but there aren't that many possibilities. Then put a little mark next to all the scenarios in which you will win by switching. I think you will find that by switching you will win 2/3 of the time.
Somerby's a smart guy, and I suspect he will eventually come to realize that he has made a mistake here. I've been a big fan of his blog, which is mostly devoted to exposing the blatant, jaw-dropping insanity that spews forth from our nation's press corps and political pundits, for a while now. But lately he's been annoying me a bit by hammering various liberal pundits for what strike me as pretty minor sins. So consider this brief essay a small measure of payback.
Let me close with two amusing variations on the Monty Hall problem.
Variation One: Monty Hall does not know which of the three doors contains the car. You choose one of the doors. Then Monty Hall chooses one of the remaining doors and opens it. It contains a goat. He then gives you the option of sticking with your original door or switching. What should you do?
Variation Two: This time there are five doors, concealing one car and four goats. You choose one of the doors. Monty Hall, who knows where the car is, opens one of the remaining goat-bearing doors. He then gives you the option of switching. You make your choice, after which Monty Hall again opens a goat-bearing door. Again you have the option of switching. This process continues until there are only two doors remaining. What is your best strategy?
Answers in a subsequent blog entry.