Changes

== Problem Statement ==

== 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 [[goat]]s.  The contestant chooses door 1.  The host must then 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?

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 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 ==

== Solution ==

There are three possible scenarios in the problem.

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 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 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.

#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.

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.

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.

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.

Let "Prize x" be the event that the prize is behind door x, and let "Open x" be the event that Monty Hall opens door x. Then before the doors are open, P(Prize 1) = P(Prize 2) = P(Prize 3) = 1/3. P(Open 2 | door 1) = 1/2, as if the prize is behind door 1, Monty Hall has two doors he can open, as he must reveal a [[goat]], not the prize behind door 1. P(Open 2) = P(Open 3) = 1/2, as there are two doors Monty Hall can open, both equally likely. Thus, using Bayes' theorem, we get:

Let "Prize x" be the event that the prize is behind door x, and let "Open x" be the event that Monty Hall opens door x. Then before the doors are open, P(Prize 1) = P(Prize 2) = P(Prize 3) = 1/3. P(Open 2 | door 1) = 1/2, as if the prize is behind door 1, Monty Hall has two doors he can open, as he must reveal a goat, not the prize behind door 1. P(Open 2) = P(Open 3) = 1/2, as there are two doors Monty Hall can open, both equally likely. Thus, using Bayes' theorem, we get:

<math>P(Prize 1| Open 2) = \frac{P(Open 2| Prize 1)P(Prize 1)}{P(Open 2)} = \frac{\frac{1}{2}*\frac{1}{3}}{\frac{1}{2}} = \frac{1}{3} </math>

<math>P(Prize 1| Open 2) = \frac{P(Open 2| Prize 1)P(Prize 1)}{P(Open 2)} = \frac{\frac{1}{2}*\frac{1}{3}}{\frac{1}{2}} = \frac{1}{3} </math>