147
edits
Changes
Jump to navigation
Jump to search
New page: The '''Monty Hall Problem''' is a basic example problem in statistic and probability theory based on the premise of the television show ''Let's Make a Deal'', originally ho...
The '''Monty Hall Problem''' is a basic example problem in [[statistic]] and [[probability]] theory based on the premise of the [[television]] show [[''Let's Make a Deal'']], originally hosted by [[Monty Hall]].
== Problem Statement ==
A contestant on a game show is presented with three doors. Behind one of the doors is a car, and behind the other two doors are goats. The contestant chooses door 1. The host must then reveal open a door to reveal a goat; he opens door 3. The host then gives the contestant a chance to switch his choice to door 2. If the contestant is trying to win the car, is it to his advantage to switch his choice?
== Solution ==
It may be tempting to say that the contestant neither gains nor loses anything if he switches. Since there are two closed doors, and one of them is the winning door, it may appear that the probability of winning is 1/2 whether the contestant switches or not. Such reasoning is incorrect; the contestant always has a higher probability of winning if he switches.
=== Illustration using scenario outcomes ===
There are three possible scenarios in the problem.
#The contestant initially chooses the door hiding the car. The host reveals one goat, leaving the other goat behind the remaining door. In this case, switching loses.
#The contestant initially chooses the door hiding goat 1. The host must reveal goat 2. Switching wins the car.
#The contestant initially chooses the door hiding goat 2. The host must reveal goat 1. Switching wins the car.
If the contestant switches, two scenarios can lead to wins; the other option loses. Hence, the contestant has a 2/3 probability of success if he switches, but only a 1/3 probability of winning if he does not.
=== Solution using Bayes' theorem ===
The problem can also be solved by using [[Bayes' theorem]] to evaluate the [[posterior probability]] that the car is behind the initially chosen door, given that the host has opened another door.
== Problem Statement ==
A contestant on a game show is presented with three doors. Behind one of the doors is a car, and behind the other two doors are goats. The contestant chooses door 1. The host must then reveal open a door to reveal a goat; he opens door 3. The host then gives the contestant a chance to switch his choice to door 2. If the contestant is trying to win the car, is it to his advantage to switch his choice?
== Solution ==
It may be tempting to say that the contestant neither gains nor loses anything if he switches. Since there are two closed doors, and one of them is the winning door, it may appear that the probability of winning is 1/2 whether the contestant switches or not. Such reasoning is incorrect; the contestant always has a higher probability of winning if he switches.
=== Illustration using scenario outcomes ===
There are three possible scenarios in the problem.
#The contestant initially chooses the door hiding the car. The host reveals one goat, leaving the other goat behind the remaining door. In this case, switching loses.
#The contestant initially chooses the door hiding goat 1. The host must reveal goat 2. Switching wins the car.
#The contestant initially chooses the door hiding goat 2. The host must reveal goat 1. Switching wins the car.
If the contestant switches, two scenarios can lead to wins; the other option loses. Hence, the contestant has a 2/3 probability of success if he switches, but only a 1/3 probability of winning if he does not.
=== Solution using Bayes' theorem ===
The problem can also be solved by using [[Bayes' theorem]] to evaluate the [[posterior probability]] that the car is behind the initially chosen door, given that the host has opened another door.