The Monty Hall Problem

Origin

A version of the Monty Hall problem was published in 1959 by Martin Gardner in Scientific American, in 1965 by Fred Moseteller in an anthology of probability problems, and in 1968 by John Maynard Smith in Mathematical Ideas in Biology. Steve Selvin presented it for the first time in a game show format in The American Statistician in 1975. The problem attracted the most attention when Marilyn vos Savant wrote about its solution in her column of the Parade magazine in September 1990. Vos Savant was famous for having the world record for the highest IQ of 228, but over 1000 people with PhDs believed her solution was incorrect and wrote in to the magazine berating her correct explanation to the unintuitive problem. Paul Erdös was a particularly vehement opponent to her explanation, even denying her rigorous proof. He was only convinced after observing a computer simulation support vos Savant's solution.

Problem and Solutions

A slightly reworded version of the problem published in vos Savant's column is given by Mlodinow (2009):

Suppose the contestants on a game show are given the choice of three doors: Behind one door is a car; behind the others, goats. After a contestant picks a door, the host, who knows what's behind all the doors, opens one of the unchosen doors, which reveals a goat. He then says to the contestant, Do you want to switch to the other unopened door? Is it to the contestant's advantage to make the switch? (p. 43).

The intuitive, but incorrect, response is that it does not matter whether the contestant switches or not; since there are two doors left, the chance of finding the car behind either is $1/2$. This is the argument used by those who wrote in correcting Marilyn vos Savant. Vos Savant argued that it was better to switch, with a $2/3$ probability of finding the car with that strategy.

Consider the switch strategy. When a contestant originally chooses a door, they have a $2/3$ chance of choosing a goat. Assuming they choose a door with a goat, the host reveals the other goat, and when they switch, they will get the car. The only way to end up with the goat using the switch strategy, is to pick the car first, a $1/3$ chance, and then after a goat is revealed, switching to the other goat.

If the stay strategy is employed, when a contestant chooses a door, there is a $1/3$ chance of choosing the car. Since they are already determined to stay, seeing a goat behind a nother door adds no new information, and the probability that they already chose the car is still $1/3$.

For a simulation of the scenario, see the applet below:

This problem is often discussed in introductory probability courses.