The Monty Hall problem involves a classical game show situation and is named after Monty Hall, the long-time host of the TV game show Let's Make a Deal. There are three doors labeled 1, 2, and 3. A car is behind one of the doors, while goats are behind the other two:
The rules are as follows:
The Monty Hall problem became the subject of intense controversy because of several articles by Marilyn Vos Savant in the Ask Marilyn column of Parade magazine, a popular Sunday newspaper supplement. The controversy began when a reader posed the problem in the following way:
Suppose you're on a game show, and you're given a 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?
Marilyn's response was that the contestant should switch doors, claiming that there is a
Think about the problem. Do you agree with Marilyn or with her critics, or do you think that neither solution is correct?
In the Monty Hall game, set the host strategy to standard (the meaning of this strategy will be explained in the below). Play the Monty Hall game 50 times with each of the following strategies. Do you want to reconsider your answer to question in [2]?
In the Monty Hall game, set the host strategy to blind (the meaning of this strategy will be explained below). Play the Monty Hall game 50 times with each of the following strategies. Do you want to reconsider your answer to question in ?
When we begin to think carefully about the Monty Hall problem, we realize that the statement of the problem by Marilyn's reader is so vague that a meaningful discussion is not possible without clarifying assumptions about the strategies of the host and player. Indeed, we will see that misunderstandings about these strategies are the cause of the controversy.
Let us try to formulate the problem mathematically. In general, the actions of the host and player can vary from game to game, but if we are to have a random experiment in the classical sense, we must assume that the same probability distributions govern the host and player on each game and that the games are independent.
There are four basic random variables for a game:
Each of these random variables has the possible values 1, 2, and 3. However, because of the rules of the game, the door opened by the host cannot be either of the doors selected by the player, so
The Monty Hall experiment will be completely defined mathematically once the joint distribution of the basic variables is specified. This joint distribution in turn depends on the strategies of the host and player, which we will consider next.
In the Monty Hall experiment, note that the host determines the probability density function of the door containing the car, namely
The host also determines the conditional density function of the door he opens, given knowledge of the door containing the car and the first door selected by the player, namely
So the distribution of
In most real game shows, the host would always open a door with a goat behind it. If the player's first choice is incorrect, then the host has no choice; he cannot open the door with the car or the player's choice and must therefore open the only remaining door. On the other hand, if the player's first choice is correct, then the host can open either of the remaining doors, since goats are behind both. Thus, he might naturally pick one of these doors at random.
This strategy leads to the following conditional distribution for
This distribution, along with the uniform distribution for
In the Monty Hall game, set the host strategy to standard. Play the game 50 times with each of the following player strategies. Which works better?
Another possible second-stage strategy is for the host to always open a door chosen at random from the two possibilities. So the host might well open the door containing the car.
This strategy leads to the following conditional distribution for
This distribution, together with the uniform distribution for
In the Monty Hall game, set the host strategy to blind. Play the game 50 times with each of the following player strategies. Which works better?
The player, on the other hand, determines the probability density function of her first choice, namely
The player also determines the conditional density function of her second choice, given knowledge of her first choice and the door opened by the host, namely
Suppose that the player switches with probability
In particular, if
We are almost ready to analyze the Monty Hall problem mathematically. But first we must make some independence assumptions to incorporate the lack of knowledge that the host and player have about each other's actions. First, the player has no knowledge of the door containing the car, so we assume that
The host and player strategies form the basic data for the Monty Hall problem. Because of the independence assumptions, the joint distribution of the basic random variables is completely determined by these strategies.
The joint probability density function of
This follows from the independence assumptions and the multiplication rule of conditional probability.
The probability of any event defined in terms of the Monty Hall problem can be computed by summing the joint density over the appropriate values of
With either of the basic host strategies,
Suppose that the player switches with probability
In the Monty Hall experiment, set the host strategy to standard. For each of the following values of
In the Monty Hall experiment, set the host strategy to blind. For each of the following values of
The event that the player wins a game is
Suppose that the host follows the standard strategy and that the player switches with probability
In particular, if the player always switches, the probability that she wins is
In the Monty Hall experiment, set the host strategy to standard. For each of the following values of
Suppose that the host follows the blind strategy. Then for any player strategy, the probability that the player wins is
In the Monty Hall experiment, set the host strategy to blind. For each of the following values of
For a complete solution of the Monty Hall problem, we want to compute the conditional probability that the player wins, given that the host opens a door with a goat behind it:
If the host follows the standard strategy and the player switches with probability
This follows from [15].
Once again, the probability increases from
If the host follows the blind strategy, then for any player strategy,
In the Monty Hall experiment, set the host strategy to blind. For each of the following values of
The confusion between the conditional probability of winning for these two strategies has been the source of much controversy in the Monty Hall problem. Marilyn was probably thinking of the standard host strategy, while some of her critics were thinking of the blind strategy. This problem points out the importance of careful modeling, of the careful statement of assumptions. Marilyn is correct if the host follows the standard strategy; the critics are correct if the host follows the blind strategy; any number of other answers could be correct if the host follows other strategies.
The mathematical formulation we have used is fairly complete. However, if we just want to solve Marilyn's problem, there is a much simpler analysis (which you may have discovered yourself). Suppose that the host follows the standard strategy, and thus always opens a door with a goat. If the player's first door is incorrect (contains a goat), then the host has no choice and must open the other door with a goat. Then, if the player switches, she wins. On the other hand, if the player's first door is correct and she switches, then of course she loses. Thus, we see that if the player always switches, then she wins if and only if her first choice is incorrect, an event that obviously has probability