Let's Make A Deal - The Monty Hall Problem
April 3, 2020
This problem has always baffled me. Given I have some extra time these days, I did some searching for an explanation of the solution that made sense to me. With some editing by me, I think I found three good explanations - although you initially may be left scratching your head! It is cited as an example of the use of Bayes' Theorem. The solutions I discuss below do not make use of the math of Bayes' Theorem but use the concept.
The Monty Hall problem is a brain teaser, in the form of a probability puzzle, loosely based on the American television game show Let's Make a Deal and named after its original host, Monty Hall. It became famous as a question from a reader's letter quoted in Marilyn vos Savant's "Ask Marilyn" column in Parade magazine in 1990:
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 Monty Hall, 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?
Vos Savant's response was that the contestant should switch to the other door. Under the given set of assumptions (see below), contestants who switch have a 2/3 chance of winning the car, while contestants who stick to their initial choice have only a 1/3 chance of winning the car. (I know, you think the probability should be 50% - keep reading!)
The set of assumptions about how the host and contestant choose their doors are:
The host must always open a door that was not picked by the contestant
The host must always open a door to reveal a goat and never the car.
The host must always offer the chance to switch between the time the contestant originally chooses a door and while the remaining two doors are closed.
The contestant choses the original door randomly.
The car is initially hidden randomly behind the doors
iI the player initially picks the car, then the host's choice of which goat-hiding door to open is random.
When any of these assumptions is varied, it can change the probability of winning by switching doors - in other words, it may change the strategy of switiching.
Many readers of vos Savant's column refused to believe switching is beneficial despite her explanation. After the problem appeared in Parade, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine, most of them claiming vos Savant was wrong. Even when given explanations, simulations, and formal mathematical proofs, many people still do not accept that switching is the best strategy. Paul Erdős, one of the most prolific mathematicians in history, remained unconvinced until he was shown a computer simulation demonstrating vos Savant’s predicted result).
The problem is a paradox of the veridical type, because the correct choice (that one should switch doors) is so counterintuitive it can seem absurd, but is nevertheless demonstrably true. The Monty Hall problem is mathematically closely related to the earlier Three Prisoners problem and to the much older Bertrand's box paradox.
Explanations of Solution
Most people come to the conclusion that switching does not matter because there are two unopened doors and one car and that it is a 50/50 choice. This would be true if Monty opens a door randomly, but that is not the case; the door he opens depends on the contestant's initial choice, so the assumption of independence does not hold.
Two key insights to the problem:
1. Under the given assumptions, there is more information about Doors 2 and 3 than was available at the beginning of the game when Door 1 was chosen by the player. Monty Hall's deliberate action adds value to the door he did not choose to eliminate, but not to the one chosen by the contestant originally.
2. Switching doors is a different action than choosing between the two remaining doors at random, as the first action uses the previous information and the latter does not.
There are three potential initial configurations of the game. In two of them, the player wins by switching away from the choice made before a door was opened. The contestant's chance of winning by switching doors is directly related to her chance of choosing the winning door in the first place: if she choose the correct door on her first try, then switching loses; if she chose a wrong door on her first try, then switching wins; her chance of choosing the correct door on your first try is 1/3, and the chance of choosing a wrong door is 2/3.
Explanation 2 The solution presented by vos Savant in Parade shows the three possible arrangements of one car and two goats behind three doors and the result of staying or switching after initially picking Door 1 in each case:
Another way to understand the solution is to consider the two original unchosen doors together. "Monty is saying in effect: you can keep your one door or you can have the other two doors." The 2/3 chance of finding the car has not been changed by the opening of one of these doors because Monty, knowing the location of the car, is certain to reveal a goat. So the player's choice after Monty opens a door is no different than if he offered the contestant the option to switch from the original chosen door to the set of both remaining doors. The switch in this case clearly gives the player a 2/3 probability of choosing the car.