Ask Marilyn ® by Marilyn vos Savant is a column in Parade Magazine, published by PARADE, 711 Third Avenue, New York, NY 10017, USA. According to Parade, Marilyn vos Savant is listed in the "Guinness Book of World Records Hall of Fame" for "Highest IQ."
In her Parade Magazine columns of September 9, 1990, February 17, 1991, and July 7, 1991, Marilyn discusses a gameshow in which the contestant is given a choice of one of three doors, behind one of which is a prize. The reader asks if, after the contestant chooses a door, the host opens a different door, revealing no prize, and offers the contestant the opportunity to switch doors, whether the contestant should switch or not. Marilyn replies that the contest should switch, because the second door has a 2/3 chance of winning.
Marilyn, there's nothing wrong with your math. As you noted, math answers aren't determined by votes. But TV ratings are! What could possibly have justified your assumption that the game show host offers every contestant the same choice? The initial question described only a single incident.
If I were the game show host, and you were the contestant, I'd offer you the option to switch only if you initially chose the correct door. In this case, the first door has a 100% chance of winning, the second door has a 0% chance, and switching would be a sure loser.
Unless you understand the motives and behavior of the game show host, all the mathematics in the world won't help you answer this question.
I assume that my letter to you explaining your incorrect assumption was lost in the noise of the thousands of letters you received.
I've received several comments and questions about my original answer. I'd therefore like to clarify my position.
Marilyn assumes that the gameshow host offers every contestant the opportunity to switch. In fact, there may be some game shows where this is the case. However, I believe that Marilyn was wrong to make such an assumption without stating it explicitly. If Marilyn had simply stated her answer was based upon the assumption that the game show host always offered the contestant the opportunity to switch, I would have been satisfied with her answer.
Assuming that the game show host does not offer this opportunity to every contestant, there are several possibilities:
Patrick J. LoPresti <firstname.lastname@example.org> wrote to point out an interesting alternative. Although the original question states that the host knows what's behind the doors, he nevertheless has the option of randomly choosing which door to open second. This means that sometimes (but not in the example cited by the original question), the host will open the door containing the prize. Regardless of which door the contestant chooses, there are six equally likely possibilities. However, since the question states that the door contains a goat, two of these possibilities are ruled out by this additional information. Assuming that you pick door number 1, the six possibilities are:
Since the contestant saw a goat, the third and sixth possibilities are eliminated. The remaining four possibilities are equally likely. In two of these remaining four possibilities, you have already chosen the correct door, and switching would be a mistake. In the other two possibilities, switching would be a winner. Thus, switching would be a winner in two of the four possibilities, or 50%.
This illustrates my original claim that unless you understand the motives and behavior of the game show host, there is no correct answer to this question.
It is interesting to analyze why the results suggested by Pat are different from Marilyn's results. Imagine that whenever the host randomly chooses the door containing the prize, before opening the door, he immediately chooses a different door. Now, the six equally likely possibilities are:
This makes switching a winner in four out of six possibilities, or 2 out of 3. This is the result reported by Marilyn. Although it appears as if there are only four possibilities (if one considers the third and fourth equivalent and the fifth and sixth equivalent), the four possibilities are not equally likely.
After you pick but before you open any doors, there's a 1/3 chance that you've picked correctly, and a 2/3 chance that you've picked wrong. Assuming that the host can open doors, but can not move prizes, nothing that the host does will change the probabilities described above.
Now the host opens one of the doors, and there's nothing behind it. There's still a 1/3 chance that you've picked correctly, and a 2/3 chance that you've picked wrong. This means that the remaining door has a 2/3 chance of being correct.
I hope I've done a better job of explaining this than Marilyn.
"If the host is required to open a door all the time and offer you a switch, then you should take the switch," he said. "But if he has the choice whether to allow a switch or not, beware. Caveat emptor. It all depends on his mood."
Let's look at an example when that assumption doesn't hold. First let's call the doors "left," "center," and "right". Now let the host's algorithm be as follows: Always scan from "left" to "right". In this algorithm, the host always looks first at "left" (if the contestant chooses "right" or "center") or "center" (if the contestant chooses "left"). If that door is not the winner, the host opens it. Otherwise the host opens the third door.
Now assume that the contestant chooses "left" and the host opens "right". The contestant can be sure that the car is in the "center", since the host would have opened "center" if it were not the winner. On the other hand, if the contestant chooses the left door, and the host opens the center door, then the chances are 50:50 and it doesn't really matter if the contestant changes doors.
Although one should indeed switch doors in the problem as given, in the actual game show you should not have switched because the problem the contestants actually faced in the show was a little more complicated.
In the original show, two contestants were selected. Each of them picked a door, and they could not choose the same door. One of the selected doors was opened, revealing an object which is not the big prize (more on that later). Hall dismissed that contestant, turned to the other one, and gave him or her the option of switching.
The existence of the second contestant changes the problem dramatically. Here is how the probabilities actually worked.
Contestant A selects door A and contestant B selects door B. Door C is not selected.
Case 1 (probability 2/3). The prize is hidden behind one of the two selected doors: for argument's sake call it A (if it's B reverse the roles of A and B in this paragraph). Hall must open door B to dismiss contestant B ("We're sorry, it's not behind door B") and presents contestant A with the choice. Since A has selected the door with the prize, he or she should not switch. This is a classic example of only giving the option to switch to "selected" contestants!
Case 2 (probability 1/3): Neither contestant has selected the prize. No matter which of the two contestants is presented with the choice, he or she should switch because they will be faced with a choice between their original selection and door C, which contains the prize.
We find that in the actual game show the contestant should not have switched. The probabilities are the reverse of those in the classical problem!
Note also that in actual show it was still a little more complicated. Since there were actually two prizes of different values (each show had a goat, a small prize, and a large prize - not two goats), we ignore the case where you switch a goat for the small prize or vice versa - not considered in any of the other examples. I'll leave this for a later time.
The classical solution (assuming one contestant) is as follows. The contestant picks door A. There is a 2/3 probability before the doors are opened that the prize is not behind door A. Once a different door, door B, is opened and reveals a goat, the 2/3 probability is compressed into the remaining door, door C. Hence the contestant should switch. All that is well and good. Now consider the following, equally valid argument.
The contestant selects door A. Before any of the doors are opened, there is a 2/3 probability the prize is not behind door C. Door B is opened and contains no prize. Since there's a 2/3 probability the prize is not behind door C, there is a 2/3 probability that it is behind door A. So the contestant should not switch. The contestant's chances, it seems, depend on how exactly he or she looks at the problem -- clear nonsense. It's as if the point in space which is the solution to an equation depends on which coordinate system you use! A similar argument holds for the 2/3 probability that the prize is not behind door B (in which case the contestant has not learned any more information about the remaining two doors). In that case, since the two remaining doors are more or less equal in the contestant's eyes, switching does not matter.
Here are the responses from the readers.
To convince students that "switching" is better, I use what is, I think, an unusual argument based on a series of similar games that start from one in which nobody would "stay" and lead to the standard formulation, claiming at each step that from the contestant's view all games are identical. The only modification is to what the host says and does.
In the first game, the host offers both other doors (say, B and C) in exchange for door the first choice (A). Nobody has ever argued for "staying" in this game.
In the second game, the host again offers both B and C in exchange for A, but adds as an aside that the prize is not behind B (for example). Since you still get both, surely you should still switch. But, assuming that Monty is trusted, why would anyone ever go to the effort of asking for B to be opened?
Next the offer is to get both doors, but B is actually opened to reveal no prize. Once again switching is the best plan, but now the issue of trust is gone, and clearly the contestant needs only to ask for one door to be opened.
Finally, the host shows the prize is not behind B, and offers to switch, but says you must choose only one door, B or C. This is the standard problem. Since you didn't need to open two doors anyway, this last modification is not really a restriction, and the solution is not changed: so again should switch.
In each case, the preference for switching relies on mechanical behavior by the host (the Nature Host assumption), and his acting on superior information as to where the prize lies.