A classic probability teaser
Marilyn vos Savant
In September 1990, Marilyn vos Savant, puzzle columnist for the U.S. magazine Parade, was sent a probability teaser by a reader. Its publication in her "Ask Marilyn" column together with her solution has produced much debate amongst mathematicians and laymen alike ever since. Apart from it's challenge to common intuition, one reason it has attracted so much interest is that Ms von Savant is listed in the Guinness Book of World Records as having the highest recorded score in an IQ test (228 : based on a score attained at 10 years old) and that proving her wrong, some respondents no doubt reasoned, might indicate that they too were blessed with such uncommon prowess. To date, Ms vos Savant has received over 10,000 letters on the puzzle, mostly disagreeable, and articles are still being written claiming new insights that show where Ms vos Savant got it wrong. The present article argues that the solution really depends on one's interpretation of the stated problem.

One of the earliest known appearances of the problem was in Joseph Bertrand's Calcul des probabilites (1889) where it was known as Bertrand's Box Paradox. It later reappeared in Martin Gardner's 1961 book, More Mathematical Puzzles and Diversions, as The Three Prisoner Problem and then resurfaced in 1975 - inspired by Monty Hall's U.S. gameshow "Let's Make a Deal" - in an article in The American Statistician by Steve Selvin entitled A Problem in Probability. The problem, as presented by Ms vos Savant in 1990 (apart from the change in notation) ran as follows. 'Suppose you're on a game show and you're given the choice of three doors. Behind one is a car, behind each of the others is a goat. You pick a door, say door A, and the host, who knows what's behind the other doors, opens another door, say B, which has a goat. He then says : "Do you want to switch to door C?" Is it to your advantage to take the switch?' Despite the fact that you, the contestant, have only doors A and C to choose from, and so it seems your probability of having the car is 1/2 each, Ms vos Savant answered "yes", switch to C. Initially, the contestant has a choice of three possible prize locations which, as far as he is concerned, are equally likely hence 1/3 probability each. His actual selection, door A, is shown in blue (in diagrams).

Now here's the first difficulty. There is no indication in the problem as to whether or not : (i) the gameshow host always reveals a goat from behind one of the other two doors (i.e. his choice follows a definite eliminating rule); (ii) he sometimes reveals a goat and sometimes a car, even though he knows where the car is (i.e. he manages to randomize his choice); or (iii) he only offers the switch when the contestant selects the prize door; or (iv) some combination of these strategies. It matters because it is his strategy that determines how the problem is analysed.
(i) MONTY NEVER REVEALS PRIZE : The diagram for this case is shown right. CASE 1: For the case where the car is behind the contestant's choice, door A (blue), we follow branch A on the diagram, and see that Monty cannot open door A (probability 0) since he opens another door.
Since a goat is behind each of B and C, Monty can randomly select which door to open (probability 1/2 each). CASE 2 : If the car were located behind door B, we follow branch B on the diagram and find that Monty still cannot open the contestant's door A (probability 0), cannot reveal the car behind B (probability 0) and so only has door C to choose from (probability 1). CASE 3 : The situation is symmetrical if the prize were behind C, for then Monty can open neither A nor C (probability 0) and only has the choice B (probability 1). CALCULATION : The final column on the Probability Tree Diagram shows the result of multiplying the probabilities horizontally to get the probability of combined events. For example, the probability that the car is behind B and Monty opens C is 1/3 (arising from 1/3 x 1). In our problem, the contestant chooses door A and Monty opens B and we want to know the probability that C has the car given that Monty opens B. Now, all the probabilities relevant to Monty opening B are shown inside a square in the diagram. These are the probabilities we can select from, and the one we need is the bottom one 1/3. So the calculation is, the probability that C has the car and Monty opens B (probability 1/3) divided by the sum of the probabilities that we can select from (1/6 + 0 + 1/3 = 1/2). When we calculate this, the answer is 2/3. This means (since probabilities total 1) the probability that door A has the car is 1/3 not 1/2 as common intuition might expect and it is better to switch to door C which has probability 2/3.
(ii) MONTY SOMETIMES REVEALS PRIZE. The second option discussed above, is that Monty randomly selects one of the remaining two doors. As far as running a game show is concerned, running the risk of exposing the prize runs the risk of ruining the suspense. However, it is conceivable that Monty might do this if he decides that if he reveals the prize he will move items around and run the whole scenario again until a goat is revealed (and recorded TV has the advantage of only broadcasting this moment). It is irrelevant, however, to discuss degrees of reality; the point is, the problem does not prevent this. In this case, the goat he reveals behind door B might occur by chance and with probability 1/2. The diagram (left) shows how Monty's probabilities on the second set of branches are thus modified.

CALCULATION : Following the previous method, the probability that the prize is behind C given that he opens B is now 1/6 divided by 1/2 (1/6 + 1/6 + 1/6) which is 1/3. The probabilities for A and for B are also 1/3 (lucky Monty missed the prize!) and so A and C have equal chances. This last case clearly illustrates the characteristic that originally produced the answer 2/3 for door C. The fact that Monty followed a rule for deciding which of the remaining two doors to open (i.e. avoid the car behind C by opening B) gave greater weight to the case that the car was behind door C. This can be seen by examining the probability in the bottom squared box in each diagram. In both cases, the denominator for the calculation is 1/2. However, the box in question for the first scenario is twice that for the second.

(iii) CONTESTANT CHOOSES PRIZE DOOR. If Monty offers the switch only when the contestant has correctly chosen the prize door A then there is zero probability that the car is behind door C. Here, there is definitely no advantage to the contestant in switching. It would be a generous gameshow that consistently adopted this strategy because, assuming the contestant is aware of it, by offering the switch Monty reveals the location of the prize.

CONCLUSION : As for the correct solution to the Monty Hall Problem, it really depends on one's interpretation of the way the game show is run. Those who have actually seen the real show have additional knowledge in this matter, but anyway this is irrelevant because the problem as stated allows several scenarios. Some commentators only consider cases (i) and (ii) and say that the probability that the prize is behind C is at least 1/2 so one might as well switch. However, this answer is not valid, because the problem asks "Is it to your advantage to take the switch?"and the possibility exists that Monty might sometimes reveal the prize, in which case the answer is "No", there is no advantage (nor disadvantage). If we allow (iii) then there is definitely no advantage. In other words, we cannot say that there is an advantage in all cases. The problem with The Monty Hall Problem is that the possibilities are not sufficiently narrowed for there to be a unique solution.

What Ms vos Savant actually said.  
Biography of Monty Hall