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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.

Problem and solution

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 ,

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.

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.

• See also Empirical solution of the Monty Hall problem

Aids to understanding

• A pictorial explanation to help cement some of the explanations is given below. Here, the contestant chooses door 3.

File:Monty2.gif

• 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

${\displaystyle P(C2|O3)={\frac {P(O3|C2)P(C2)}{P(O3)}}={\frac {1\times {\frac {1}{3}}}{\frac {1}{2}}}={\frac {2}{3}}}$

• 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.

• This analysis of the problem considers the player's options in terms of Staying or Switching, and the probabilities of each strategy paying off. Here's what the probabilities would be if Monty didn't open a door after the player's initial choice:
  • Staying: The player can only win by Staying if the car happens to be behind the door the player first picked. Since there's a ${\displaystyle {\begin{matrix}{\frac {1}{3}}\end{matrix}}}$ chance that the car is behind that door, the player's chance of winning by Staying is ${\displaystyle {\begin{matrix}{\frac {1}{3}}\end{matrix}}}$.
  • Switching: The player can only win by Switching is if the car isn't behind the door the player first picked, and if the player selects the correct door out of the two remaining when he/she decides to switch. There is a ${\displaystyle {\begin{matrix}{\frac {2}{3}}\end{matrix}}}$ chance that the car is behind one of those two doors, but only a ${\displaystyle {\begin{matrix}{\frac {1}{2}}\end{matrix}}}$ chance that the player will select the correct one; since both these conditions must be met, the chance of winning by Switching is ${\displaystyle \left({\begin{matrix}{\frac {2}{3}}\end{matrix}}*{\begin{matrix}{\frac {1}{2}}\end{matrix}}\right)}$, or ${\displaystyle {\begin{matrix}{\frac {1}{3}}\end{matrix}}}$.

  In this version, where Monty gives no additional information, the chances of winning by Staying or by Switching are the same. Here's what happens to the probabilities when Monty opens (and thus eliminates as a choice) a door that he knows contains a goat:

  • Staying: The player still can only win by Staying if the car happens to be behind the door the player first picked. Since there's still a ${\displaystyle {\begin{matrix}{\frac {1}{3}}\end{matrix}}}$ chance that the car is behind that door, the player's chance of winning by Staying is still ${\displaystyle {\begin{matrix}{\frac {1}{3}}\end{matrix}}}$.
  • Switching: The player still can only win by Switching is if the car isn't behind the door the player first picked. However, there are no longer two doors to pick from once the decision to switch is made; Monty has opened one of those two doors and eliminated it as a choice. The chance that the car was behind one of those two doors is now the chance that it is behind the only remaining one of those two doors, ${\displaystyle {\begin{matrix}{\frac {2}{3}}\end{matrix}}}$; now that Monty has eliminated the ${\displaystyle {\begin{matrix}{\frac {1}{2}}\end{matrix}}}$ chance of guessing wrongly after deciding to switch, ${\displaystyle {\begin{matrix}{\frac {2}{3}}\end{matrix}}}$ is the player's chance of winning by Switching.

Variants

With several minutes remaining in the game, Monty Hall chose two contestants for the "Big Deal". Behind one of three doors was the grand prize. Each contestant was allowed to choose a door (not the same one).

In this scenario, a variant of Selvin's problem can be stated. Monty eliminates a player with a goat behind their door (if both players had a goat, one is eliminated at random, without letting the players know about it), opens the door and then offers the remaining player a chance to switch. Should the remaining player switch?

The answer is no. The reason: a switcher in this game will lose if and only if either of two initial choices of the two contestants was correct. How likely is that? Two-thirds. A sticker will win in those 2/3 of the cases. So stickers will win twice as often as switchers.

Altenatively by enumerating possibilities (again you play 1 and the other player plays 2)

         Ejected Player  Probability    Switch Strategy     Stick Strategy
 1 2 3    
 C G G   2               1/3            Lose              
 G C G   1               1/3            Lose               
 G G C   1               1/6            Win                 Lose
         2               1/6            Lose                Lose

Player 1 wins 1/3 of the time with the stick strategy, or 1/6 of the time with the switching strategy. Half the time he is eliminated. Given that he is not eliminated there is 2/3 probability of winning with the sticking strategy.

There is a generalization of the original problem to n doors: in the first step, you choose a door. Monty then opens some other door that's a loser. If you want, you may then switch your allegiance to another door. Monty will then open an as yet unopened losing door, different from your current preference. Then you may switch again, and so on. This continues until there are only two unopened doors left: your current choice and another one. How many times should you switch, and when, if at all?

The best strategy is: stick with your first choice all the way through but then switch at the very end. With this strategy, the probability of winning is (n-1)/n. This was proved by Bapeswara Rao and Rao.

Origins

The game used in the Monty Hall problem is similar to three card monte, a gambling game in which the player has to find a single winning card among three face-down cards. As in the Monty Hall problem, the dealer knows where the winning card is, but here the dealer always tries to trick the player into picking the wrong card. As the card is often a Queen court card, it is also known as Find the Lady.

An older puzzle in probability theory involves three prisoners, one of whom (already chosen at random but unknown to the prisoners) is to be executed in the morning. The first prisoner begs the guard to tell him which of the other two will go free, arguing that this reveals no information about whether he will be the victim; the guard responds by claiming that if the prisoner knows that a specific one of the other two prisoners will go free it will raise the first prisoner's subjective chance of being executed from 1/3 to 1/2. The question is whether the analysis of the prisoner or the guard is correct. In the version given by Martin Gardner, the guard then performs a particular randomizing procedure for selecting which name to give the prisoner; this gives the equivalent of the Monty Hall problem without the usual ambiguities in its presentation.

Anecdotes

After this problem's solution was discussed in Marilyn vos Savant's "Ask Marilyn" question-and-answer column of Parade magazine in 1990, many readers including several math professors wrote in to declare that her solution was wrong. An equally contentious discussion of Marilyn's discussion took place in Cecil Adams's column The Straight Dope.

The Monty Hall problem is elegantly discussed, from the perspective of a boy with Asperger's syndrome, in The Curious Incident of the Dog in the Night-time, a 2003 novel by Mark Haddon.

References

• Bapeswara Rao, V. V. and Rao, M. Bhaskara (1992). "A three-door game show and some of its variants". The Mathematical Scientist 17, no. 2, pp. 89–94
• Bohl, Alan H.; Liberatore, Matthew J.; and Nudick, Robert L. (1995). "A Tale of Two Goats ... and a Car, or The Importance of Assumptions in Problem Solutions". Journal of Recreational Mathematics 1995, pp. 1–9.
• Gardner, Martin (1959). "Mathematical Games" column, Scientific American, October 1959, pp. 180–182.
• Selvin, Steve (1975a). "A problem in probability" (letter to the editor). American Statistician 29(1):67 (February 1975).
• Selvin, Steve (1975b). "On the Monty Hall problem" (letter to the editor). American Statistician 29(3):134 (August 1975).
• Tierney, John (1991). "Behind Monty Hall's Doors: Puzzle, Debate and Answer?", The New York Times July 21, 1991, Sunday, Section 1; Part 1; Page 1; Column 5
• vos Savant, Marilyn (1990). "Ask Marilyn" column, Parade Magazine p. 12 (Feb. 17, 1990).

External links

• THE MONTY HALL PROBLEM (at letsmakeadeal.com; quotes Monty's letter to Steve Selvin in full)
• Monty Hall Paradox (let's make a deal) (lengthy bibliography)
• Monty Hall Dilemma (Monty Hall simulation, discussions, and generalization)

• Mathematical recreations and puzzles
• Paradoxes