Chapter 12.5, Problem 63ES

Monty Hall Problem This is another explanation, using conditional probability, of the Monty Hall problem discussed in Exercise 54 of the last section. Let the door chosen by the contestant be labeled 1 and the other two doors be labeled 2 and 3. In the following exercises, we will use A to represent the event that the grand prize is behind door 1. B to represent the event that the prize is behind door 2, and C to represent the event that the prize is behind door 3. We will use $\bar{A}$ to represent that Monty Hall opens door 1, $\bar{B}$ to represent that he opens door 2, and $\bar{C}$ to represent that he opens door 3.

a. What is the probability that Monty Hall opens door 2 given that the grand prize is behind door 1? This is $P\left(\bar{B}|A\right)$.

b. What is the probability that Monty Hall opens door 2 given that the grand prize is behind door 2? This is $P\left(\bar{B}|\bar{B}\right)$.

c. What is the probability that Monty Hall opens door 2 given that the grand prize is behind door 3? This is $P\left(\bar{B}|C\right)$.

d. The probability that the grand prize is behind door I given that door 2 is opened (you have not switched choices) is given by Bayes’ Theorem in the following form.

$P\left(A|\bar{B}\right)=\frac{P\left(\bar{B}|B\right)P\left(A\right)}{P\left(\bar{B}|A\right)P\left(A\right)+P\left(\bar{B}|B\right)P\left(B\right)+P\left(\bar{B}|C\right)P\left(C\right)}$

What is the probability?

e. Find the probability of choosing the grand prize if you switch doors. That is, find

$P\left(C|\bar{B}\right)=\frac{P\left(\bar{B}|C\right)P\left(C\right)}{P\left(\bar{B}|A\right)P\left(A\right)+P\left(\bar{B}|B\right)P\left(B\right)+P\left(\bar{B}|C\right)P\left(C\right)}$

f. Is switching the better strategy?