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:
For the sake of simplicity, I chose an extreme example to illustrate the error in Marilyn's analysis. I did not intend to imply that I would treat every contestant this way.
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.
Sam Allen email@example.com wrote to question my claim that after discarding the third and sixth possibilities suggested by Patrick, the remaining four possibilities are equally likely. He suggested that the first two possibilities have a probability of 1/6 each, and the remaining two possibilities have a probability of 1/3 each. If the prize is equally likely to be behind each of the three doors, each of the three pairs of possibilities (first and second, third and fourth, fifth and sixth) has a combined probability of 1/3. Discarding the third and sixth possibility does not change this fact. Therefore, switching is still the best strategy even in this situation.
I'm periodically receive email telling me about computer simulations proving that Marilyn is correct. (I never claimed that her math was wrong. I only challenge her assumptions, not her math.) I also get requests to explain why Marilyn's math is correct. So, assuming that the Host always offers the contestant a chance to switch, this is why Marilyn is correct:
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.
Thank you to Rick Merrill Rick_Merrill@alum.mit.edu and Tony Pompa firstname.lastname@example.org who found Behind Monty Hall's Doors: Puzzle, Debate and Answer? John Tierney, Special to The New York Times, Sunday, July 21, 1991, Section 1, Part 1, Page 1, Column 5. Here are some quotes from Behind Monty Hall's Doors:
Which means, of course, that the only person who can answer this version of the Monty Hall Problem is Monty Hall himself. Here is what should be the last word on the subject:
"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."
Tim Mann email@example.com wrote to point out that Marilyn's analysis does not require that the game show host offer every contestant the opportunity to switch. All that is required is that the host's decision about whether to offer the contestant an opportunity to switch is independent of whether the contestant's initial choice is correct. For example, the host could make the decision whether to offer the switch before the contestant chooses a door, or the host could make the decision based on the flip of a coin, the day of the week, or some other variable independent of the contestant's initial choice.
Israeli Eitan firstname.lastname@example.org wrote to point out that the discussions above require the assumption that whenever the contestant's initial guess is correct, the host must randomly choose one of the other two doors.
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.
Andrew Goldish email@example.com has the following comments. If there's a flaw in his logic, I don't see it, but I'm sure my readers will find it quickly. 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.
Andrew Goldish firstname.lastname@example.org wonders why the following analysis is not valid:
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.
W. D. Scoones scoones@UVic.CA uses this argument to demonstrate that if the game show host always offers the opportunity to switch, Marilyn is correct. 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.
Posted at Mar 03/2008 06:09PM by Tommy:
There are infinitely many possible ways that the host could have been thinking but probably all to the disadvantage of the player, so less than 1/3 chance of winning no matter what.
Posted at Mar 11/2008 04:41AM by Perry Curling-Hope:
Solution to this probability statement assumes, as the 'brainteaser' does, that the behaviour of the showhost is constrained by simple logical rules. This being the case , the behaviour of the host is a 100% given, and the outcome is fixed and determined entirely by the contestants initial choice. Thereafter the outcome is certain.
Assuming the contestant takes the switch, which MUST be offered:-
If the contestant initially chooses the concealed prize, a 1 in 3 chance , he loses. If the contestant chooses a concealed goat, a 2 in 3 chance, he wins
Why? If he chooses the prize, (1 in 3) the host can't open this door, (against the 'rules' and defeats the object) He can only open one or other remaining doors, both with goats, it does not matter which. He then has to offer the switch to the only remaining door not yet selected, concealing the remaining goat, and the contestant loses.
If the contestant chooses a concealed goat (2 in 3) the host cannot open the door with the prize( again, against the 'rules' and defeats the object) and must open the door with the other goat. The only remaining door has the prize, and since the host must offer it, the contestant wins, 2 out of three times.
If the contestant refuses the switch, whatever happens after his initial choice is irrelevant....and he has a simple 2 in 3 chance of losing based on his initial choice. 1 prize, 2 goats!
All the wrangling about this problem confuses the brainteaser, in which the hosts behaviour is fixed by logic. with a real show.
In a real show the host may:
Decide to open the door revealing the prize, which rather defeats the object, and has nothing to do with the probabilites contained in the brainteaser.
Not offer the switch or convince the contestant not to accept it, which is not part of the brain teaser and has to do with 'cheating' or psychology, not the problem being posed.
Not be aware of which door hides the prize, which was also not the problem being posed.
The brain teaser makes the offer to switch part ot the game 'rules'
Posted at May 07/2008 10:19AM by Dave B:
The reason why Andrew Goldish's second analysis (probability depends on perspective) is incorrect lies in his labelling of the doors. He examines Door C, which is defined as either one of the doors not picked by the player, each of which has a 1/3 chance of containing a prize. Door B is specifically defined as the one that is opened, revealing no prize. The problem here is that as defined, Door C and Door B can represent the SAME door. Specifically, out of the 2/3 of the time the prize is not behind Door C, half the time it will be behind Door A, and the other half it will be behind the other unlabelled door, in which case Door C/B will be opened.
Here are the statistics using this door labelling:
Prize is behind Door C (1/3):Door B is opened, player should switch.
Prize is behind the "other" door not selected by the player (1/3): Door C = Door B, player should switch.
Prize is behind Door A (1/3): Player should not switch.
So 2/3 of the time, the player should switch.
Posted at Aug 24/2008 04:12AM by Jada Leppky:
Further examination of Andrew Goldish's "Let's Make a Deal", involving two contestants:
As I see it, whether the remaining contestant should switch in this case is 50-50.
Assume contestant 'A' chooses a door first. S/he has a 1/3 chance of choosing correctly, meaning s/he is probably wrong, and the prize is behind one of the 2 remaining doors. For contestant 'A', up to this point the problem is the same as the "classic" problem.
At first it might seem that if contestant 'B's door is shown to be empty, then contestant 'A', who had a 2/3 chance of being wrong, should switch. But wait, the Host CANNOT show the un-chosen door to be empty, even if it is. This changes things. What it means is that the Host is not giving us as much information as we might think, and is in fact basically removing 1/3 of the possibilities.
In other words, in the event that contestant 'A' is the one left standing, s/he either chose right in the first place (1/3) OR s/he chose wrong (2/3) AND THEN contestant 'B' chose wrong out of the two remaining doors(1/2), which amounts to (2/3 x 1/2), or (1/3). Thus, each possibility is equally as likely (1/3 vs 1/3). You may wonder where the other 1/3 is. The other 1/3 is excluded, as it would have resulted in the other contestant being the one left standing. So we do some probability scaling. Technically, the odds should be scaled to 1/2.
Alternatively, if contestant 'B' is the last one standing, then contestant 'A' HAD to choose wrong FIRST. This would have left contestant 'B' with a 1/2 chance of choosing the correct door. Thus, it is a 50-50 tossup.
Here is a breakdown of all the possible outcomes:
First, let's label the doors 1, 2, and 3.
Let's put the prize behind door 1.
Let's call the contestants A and B.
For simplicity's sake, "A = 1" means contestant A chooses door 1.
Also " 3" means door 3 is un-chosen...
As you can see of the 4 instances where B is eliminated, 2 show A with the correct door (2/4 or 1/2), and 2 show the correct door as the un-chosen door. The same is true for those instances where A is eliminated.
Posted at Aug 09/2009 11:45PM by JuanBecerril:
This question and debate is in the book "The Curious Incident Of The Dog In The Night-Time" It has a simple graph on how marilyn is right. She said that there is a 2 out of 3 chance that when you switch you win the car because 2 doors have goats and 1 has a car. There is only 1 chance that you chose, stick and win a car, but there is 2 chances that you pick a door, switch and win a car. Yes, her answer doesn't apply to the fact that you already know whats behind one door, but her math is still right. Just go check this book out. It will make perfect sense to you.