Refer back to the Monty Hall problem. The game is played as follows:Monty chooses one of three doors to hide a car behind, with each having probability 1/3 ofhiding the car. The other two have goats behind them. Then you, the contestant, choose adoor. Monty opens one of the other two doors, always revealing a goat. If your door has agoat, he must pick the other goat door, but if your door has the car, he chooses one of thetwo goat doors with either choice equally likely. Suppose that you chose door #1.(a) you can use Bayes’ rule to calculate the conditional probability of winning by switching given the door he chooses. Show that this method gives 2/3 by using it to calculate, given that Monty chooses door #2, the conditional probability of LOSING by switching, i.e. the probability that door #1 had the car allalong.(b) Suppose Monty decides to shake things up a bit: rather than choosing a doorto hide the car behind in a way that each door is equally likely to be picked, supposethat door #1 has probability p1 of having the car, door #2 has probability p2 of havingthe car, and door #3 has probability p3 of having the car. Note that p1 + p2 + p3 = 1,since the car must be behind one of the doors. Also assume that p1 ≥ p2 ≥ p3. Therest of the game is the same. Suppose again that you chose door #1.The question: given that Monty chooses door #2, what is the conditional probabilitythat you will win the car by switching? (c) Based on your answer to part B, give fully simplified criteria for when it wouldbe more advantageous for you to NOT switch away from door #1.

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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Refer back to the Monty Hall problem. The game is played as follows:
Monty chooses one of three doors to hide a car behind, with each having probability 1/3 of
hiding the car. The other two have goats behind them. Then you, the contestant, choose a
door. Monty opens one of the other two doors, always revealing a goat. If your door has a
goat, he must pick the other goat door, but if your door has the car, he chooses one of the
two goat doors with either choice equally likely. Suppose that you chose door #1.
(a) you can use Bayes’ rule to calculate the conditional probability of winning by switching given the door he chooses. Show that this method gives 2/3 by using it to calculate, given that Monty chooses door #2, the conditional probability of LOSING by switching, i.e. the probability that door #1 had the car all
along.
(b) Suppose Monty decides to shake things up a bit: rather than choosing a door
to hide the car behind in a way that each door is equally likely to be picked, suppose
that door #1 has probability p1 of having the car, door #2 has probability p2 of having
the car, and door #3 has probability p3 of having the car. Note that p1 + p2 + p3 = 1,
since the car must be behind one of the doors. Also assume that p1 ≥ p2 ≥ p3. The
rest of the game is the same. Suppose again that you chose door #1.
The question: given that Monty chooses door #2, what is the conditional probability
that you will win the car by switching?
(c) Based on your answer to part B, give fully simplified criteria for when it would
be more advantageous for you to NOT switch away from door #1.

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