Monty Hall problem

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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.

Contents

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 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.

  1. 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.
  2. The contestant initially chooses the door hiding goat 1. The host must reveal goat 2. Switching wins the car.
  3. 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.

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 | Prize 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:

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}

That is, the probability that the prize is behind door 1, given that Monty opens door 2, is 1/3, so the probability that it is behind door 3 is 2/3. Thus, the contestant should switch. The logic applies equally if Monty opens door 3.

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