Monty Hall problem

The Monty Hall problem, also known as the three doors problem or the quizmaster problem, became famous in 1990 with its presentation in a popular weekly column called "Ask Marilyn" in Parade magazine. The column's author, Marilyn vos Savant, was according to the Guiness Book of Records at the time, the person with the highest IQ in the world. The problem is named after the stage-name of an actual quizmaster, Monty Halperin (or Halparin, according to some sources), on a long-running 1960's TV show "Let's make a Deal", though the events related in the Monty Hall problem never actually took place [1], [2].

Rewriting in her own words a problem posed to her by a correspondent, a Mr. Craig Whitaker, Marilyn asked the following:

Suppose you’re on a game show, and you’re given the choice of three doors: behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

Almost everyone, on first hearing the problem, has the immediate and intuitive reaction that the two doors left closed, Door 1 and Door 2, must be equally likely to hide the car, and that therefore there is no point in switching. However, some thought shows that this immediate reaction must be wrong. Consider many, many repetitions of the game, and suppose (for the sake of argument) that the car is hidden each time anew completely at random behind one of the three doors, while the contestant always starts by picking Door No. 1. In the long run, the car will be behind his initially chosen door, Door 1, on one third of the repetitions of the game. If he never switches doors, he'll take the car home with him on precisely those repetitions of the show where his initial choice was actually right - that's one third of the time. On the other hand, if he always switches, he'll go home with the car two thirds of the time - he'll win the car by "switching" exactly on every occasion when he would not win it by "staying".

Note that we are assuming, as most readers do and as vos Savant later explained was her intention, that whatever choice is initially made by the contestant, the host is surely going to open a different door revealing a goat and offer the option to switch.

One could say that when the contestant initially chooses Door 1, the host is offering the contestant a choice between his initial choice Door 1, or Doors 2 and 3 together.

The previous solution used a frequentist picture: probability refers to relative frequency in many repetitions. Also, it didn't address the issue of whether the specific door opened by the host is relevant. Could it be that the decision to switch should depend on whether the host opens Door 2 or Door 3?

Here is an alternative solution, which addresses this possible complication explicitly. Moreover, in this solution "probability" is used in the ordinary daily-life Bayesian or subjectivist sense: that is to say, probability statements are supposed to reflect the state of knowledge of one person. To be specific, that person will be a contestant on the show who initially knows no more than the following: he'll choose a door; the quizmaster will thereupon certain to open a different door revealing a goat; and thereupon certainly make the offer that the contestant may switch to the remaining closed door. For our contestant, initially all doors are equally likely to hide the car. Moreover, if he chooses any particular door, and if the car happens to be behind that particular door, then as far as our contestant is concerned the host is equally likely to open either of the other two doors.

The contestant initially chooses door number 1. Initially, his odds that the car is behind this door are 2 to 1 against: it is two times as likely for him that his choice is wrong as that it is right.

The host opens one of the other two doors, revealing a goat. Let's suppose that for the moment, the contestant doesn't take any notice of which door was opened. Since the host is certain to open a door revealing a goat whether or not the car is behind Door 1, the information that an unspecified door is opened revealing a goat does not change the contestant's odds that the car is indeed behind Door 1; they are still 2 to 1 against.

Now here comes the further detail which we will take account of in this solution: the contestant also gets informed which specific door was opened by the host - let's say it was Door 3. Does this piece of information influence his odds that the car is behind Door 1? No: from his point of view, the chance that the car is behind Door 1 can't depend on whether the host opens Door 2 or Door 3 - the door numbers are arbitrary, exchangeable.

Therefore, also knowing that the host opened specifically Door 3 to reveal a goat, the contestant's odds on the car being behind his initially chosen Door 1 still remain 2 to 1 against. He had better switch to Door 2.

Students of probability theory might feel uneasy about the informality of the last step. To feel more secure, they might consider the mathematical concept of symmetry and use the law of total probability to show how symmetry leads to statistical independence between the events "Car is behind Door 1" and "Host opens Door 3", when it is given that the contestant chose Door 1. Alternatively, they might like to explicitily use Bayes' rule: posterior odds equals prior odds times likelihood ratio. They just have to check that under the two competing hypotheses (whether or not the car is behind the door chosen by the contestant, Door 1), the fact that it is Door 3 (rather than Door 2) which gets opened by the host has the same probability 1/2.

For some readers, numbers speak louder than words. The following table is self-explanatory.

Probability Scenarios (Door 1 picked) Initial Arrangement Open D1 Open D2 Open D3 Probability of Occurring Keep Door Win Prob. Switch Door Win Prob. Car Goat Goat (p=1/3) No Yes (p=1/2) No p=1/3 x 1/2 1/6 0 Car Goat Goat (p=1/3) No No Yes (p=1/2) p=1/3 x 1/2 1/6 0 Goat Car Goat (p=1/3) No No Yes (p=1) p=1/3 * 1 0 1/3 Goat Goat Car (p=1/3) No Yes (p=1) No p=1/3 x 1 0 1/3 Note that the host has limited choices when the contestant chooses incorrectly

We observe that the player who switches wins the car 2/3 of the time. We also see that Door 3 is opened by the host 1/2 = 1/6+1/3 of the time (row 2 plus row 3), while winning by switching in combination with Door 3 being opened occurs 1/3 of the time (row 3). The probability of winning by switching given Door 3 is opened is therefore (1/3)/(1/2)=2/3. Knowing the identity of the opened door doesn't change the chance of winning by switching.