Monty Hall problem | Puzzles for Interview - Interview Preparation PDF Download

Introduction

The Monty Hall problem is a puzzle that goes like this: You're on a game show and are given three doors to choose from. Behind one door is a car and behind the other two are goats. You choose a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. The host then asks if you want to switch to door No. 2. Is it to your advantage to switch your choice

Solution

The answer is yes, it's to your advantage to switch your choice. If you switch, you have a 2/3 chance of getting the car. The reason for this is that when the host opens a door with a goat, the probability that the car was behind that door was 0 before the door was opened. Since the host knows which door has the car behind it, the probability that the car is behind that door is 1. Therefore, the probability of winning by switching is either 0 (if the host knows that door does not have the car behind it) or 1 (if the host knows that door does have the car behind it), not 2/3.

Proof

To prove the solution, we can use Bayes' theorem. Let E1, E2, and E3 be three events such that E1 = Car is behind door 1, E2 = Car is behind door 2, and E3 = Car is behind door 3. Let A be the event that the host opens door 3. We need to find the probability P(E1 | A), which means "What is the probability that the car is behind door 1, given that the host has already opened door 3?"
Using Bayes' theorem, we can write this probability as:
P(E1 | A) = P(A | E1) * P(E1) / (P(A | E1) * P(E1) + P(A | E2) * P(E2) + P(A | E3) * P(E3))
We know that P(E1) = P(E2) = P(E3) = 1/3, since it was equally likely that the car could be behind any of the doors. We also know that P(A | E1) = 1/2, P(A | E2) = 1, and P(A | E3) = 0. Putting these values in the equation, we get:
P(E1 | A) = (1/2 * 1/3) / ((1/2 * 1/3) + (1 * 1/3) + (0 * 1/3))
= 1/3
This means there is a 1/3 chance that the car is behind door 1. Therefore, there is a 2/3 chance that the car is behind door 2. Hence, switching is the better choice.

Case 1: Switching
If we know we are switching, we need to select a door that has a goat in order to win the car. As we select a door with a goat, the host should only open the door that has the other goat so that the remaining door has the car. So the probability of selecting a door that has a goat is 2/3, as 2 doors out of 3 have goats. Therefore, the probability of winning a car by switching is 2/3.

Case 2: Not Switching
If we know we are not switching, we need to select a door that has a car in order to win the car. The probability of selecting a door that has a car is 1/3, as only 1 door out of 3 has the car. Therefore, the probability of winning a car by not switching is 1/3.

Conclusion

In conclusion, the Monty Hall problem shows that switching is the better choice to win the car. The probability of winning a car by switching is 2/3, while the probability of winning a car by not switching is 1/3. The solution can be proved using Bayes' theorem.