Solving the Monty Hall Problem with Bayes Theorem: A Comprehensive Guide

Introduction to the Monty Hall Problem

The Monty Hall Problem is a famous probability puzzle that has captivated the minds of mathematicians, statisticians, and enthusiasts alike. This classic problem is based on a hypothetical game show scenario where contestants are faced with a choice that involves probabilities and strategic decisions. In this article, we will delve into the problem and its solution using Bayes Theorem.

Problem Setup and Initial Conditions

The Monty Hall Problem involves a scenario with three doors: behind one of these doors is a valuable car, while the other two conceal goats. The setup is straightforward:

You are given three doors: Door 1, Door 2, and Door 3.

Behind one door is the car, and behind the other two, there are goats.

You make an initial selection, let's say Door 1.

The game show host, who knows what's behind each door, opens one of the other two doors to reveal a goat.

You are then given the option to either stick with your original choice or switch to the remaining unopened door.

Understanding Bayes Theorem

Bayes Theorem is a powerful tool in probability theory that allows us to update the probability of a hypothesis as more evidence or information becomes available. In the context of the Monty Hall Problem, we can use Bayes Theorem to calculate the probability of winning the car based on our choices and the information provided by the host.

Defining Events and Calculating Probabilities

To apply Bayes Theorem, we need to clearly define the events involved and calculate relevant probabilities:

Event A: The car is behind Door 1 (your initial choice).

Event B: The host opens Door 3 to reveal a goat.

Define Prior Probabilities:

P(A) 1/3: The probability that the car is behind Door 1.

P(A^c) 2/3: The probability that the car is behind either Door 2 or Door 3.

Define Conditional Probabilities:

P(B|A) 1/2: If the car is behind Door 1, the host can open either Door 2 or Door 3, both of which have goats.

P(B|A^c) 1: If the car is behind Door 2, the host must open Door 3, and if the car is behind Door 3, the host cannot open it because it has the car.

Total Probability of B:

Using the law of total probability, we can determine the total probability that the host opens Door 3:

P(B) P(B|A)P(A) P(B|A^c)P(A^c) (1/2)(1/3) (1)(2/3) 1/6 4/6 5/6

Applying Bayes Theorem

Now that we have all the probabilities, we can apply Bayes Theorem to find the updated probabilities:

Probability that the car is behind Door 1 given the host opens Door 3:

P(A|B) frac{P(B|A)P(A)}{P(B)} frac{(1/2)(1/3)}{5/6} frac{1/6}{5/6} 1/5

Probability that the car is behind Door 2 given the host opens Door 3:

P(A^c|B) frac{P(B|A^c)P(A^c)}{P(B)} frac{(1)(2/3)}{5/6} frac{2/3}{5/6} 4/5

Conclusion and Optimal Strategy

The results from Bayes Theorem reveal:

If you stick with your original choice (Door 1), the probability of winning the car is P(A|B) 1/5.

If you switch to the remaining door (Door 2), the probability of winning the car is P(A^c|B) 4/5.

Thus, the optimal strategy is to always switch doors, as it gives you a higher probability (4/5) of winning the car.