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The Monty Hall problem is a puzzle in probability that is loosely based on the American game show Let's Make a Deal; the name comes from the show's host Monty Hall. In this puzzle a contestant is shown three closed doors; behind one is a car, and behind each of the other two is a goat. The contestant is allowed to open one door, and will win whatever is behind the door he opens; however, after the contestant has selected a door but before he actually opens it, the host (who knows what is behind each door) opens one of the other doors to show that there is a goat behind it, and asks the contestant whether they want to change their mind and switch to the other closed door. Does the contestant improve their chance of winning the car by switching or does it make no difference?The question has generated heated debate. As the solution appears to contradict elementary ideas of probability and common sense, it may be regarded as a paradox.
1 Problem and solution
1.1 The problem
Here is a famous statement of the problem, from a letter from Craig F. Whitaker to Marilyn vos Savant's column in Parade Magazine in 1990 (as quoted by Bohl, Liberatore, and Nydick).
- 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?
This is a restatement of the problem as given by Steve Selvin in a letter to the American Statistician (February, 1975). As stated, the problem is an extrapolation from the game show: contestants on Let's Make a Deal were not allowed to switch. As Monty Hall wrote to Selvin [1],
- And if you ever get on my show, the rules hold fast for you -- no trading boxes after the selection.
Selvin's subsequent letter to the American Statistician (August, 1975) appears to be the first use of the term "Monty Hall problem".
An essentially identical problem appeared as the "three prisoners problem" in Martin Gardner's Mathematical Games column in 1959. Gardner's version makes the selection procedure explicit, avoiding the unstated assumptions in the version given here.
1.2 The solution
The solution to the problem is yes: the chance of winning the car is doubled when the contestant switches to another door rather than sticking with the original choice. When the contestant chooses a door, there is a probability of 1/3 that they choose the door with the car: there is a probability of 2/3 that they do not choose the door with the car. When the host opens a door to reveal a goat, there is still a probability of 2/3 that the contestant has not chosen the door with the car (because when the host reveals a goat it does not affect this probability). Therefore if the contestant switches their choice, there is now a probability of 2/3 that they have chosen the door with the car.
2 Aids to understanding
- A pictorial explanation to help cement some of the explanations is given below. Here, the contestant chooses door 3.
- It may be easier for the reader to appreciate the result by considering a hundred doors instead of just three. In this case there are 99 doors with goats behind them and 1 door with a prize. The contestant picks a door; 99 out of 100 times the contestant will pick a door with a goat. Monty then opens 98 of the other doors revealing 98 goats and offers the contestant the chance to switch to the other unopened door. On 99 out of 100 occasions the door the contestant can switch to will contain the prize as 99 out of 100 times the contestant first picked a door with a goat. At this point a rational contestant should always switch.
- Another way of phrasing why the player should switch: By switching, the player is ensuring that he will win if he originally picked a goat. The probability of picking a goat was 2/3, so the player should switch.
- Instead of one door being opened and thus eliminated from the game, it may equivalently be regarded as combining two doors into one, as a door containing a goat is essentially the same as a door with nothing behind it. In essence, this means the player has the choice of either sticking with their original choice of door, or choosing the sum of the contents of the two other doors. Clearly, the chances of the prize being in the other two doors is twice as high. Notice how the above assumptions play a role here: The reason switching is equivalent to taking the combined contents is that Monty Hall is required to open a door with a goat.
- For the least reliance on verbiage and the most on formal mathematics, an approach using Bayes' theorem may be best. It also makes explicit the effect of the assumptions given earlier. Consider the position when door 1 has been chosen and no door has been opened. The probability that the car is behind door 2, p(C2), is plainly 1/3, as it may equally well be in any of the three places. The probability that Monty will open door 3, p(O3), is 1/2; if there can be any doubt, enumeration of cases will confirm this. But when the car is behind door 2, Monty will certainly open door 3, by the assumptions; that is, p(O3|C2) = 1. Hence the probability that the car is behind door 2 given that Monty opens door 3 is
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- Instead of attempting to calculate the exact probability of winning the car, we can execute a simulation of the game and count the number of times the contestant wins. This will give an approximation to the exact probability. See empirical solution of the Monty Hall problem for a Perl program which implements a simulation.
- Imagine a scenario in which Contestant A chooses a door. Monty then opens a goat door. And Contestant B opens the remaining door. Since the first contestant will choose the car door only 1 in 3 times, the second contestant will win the car 2 out of 3 times. Thus, the car is behind the remaining door 2 out of 3 times.
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