THE GAME SHOW PARADOX THAT STUMPED THE WORLD

In 1975, Steve Selvin sent a letter to the American Statistician posing a probability brain teaser to his respondent. Little did he know, this one problem would be one of the most controversial problems to ever exist!

What is this problem that stumped most of the world?

Let’s say you are on a game show. You are given the choice of three doors: 1, 2, and 3. Behind one door is a brand new car and behind the other two are goats. Your goal is to attempt to choose the door leading to the new car.

Whichever door you choose to start, Monty Hall will always open a different door, one with a goat. Suppose you start by choosing Door 1. He will open either Door 2 or 3, whichever leads to a goat. If both do, he will pick one randomly.

Now two doors will remain, 1 and 2. The question remains, will you stick to your previous plan and choose Door 1 or switch your answer to the other closed Door 2? Which door would maximize your chances of picking the car?

Why was this problem so difficult, though?

Most people don’t use conditional probability, the chances that an outcome could occur after another event occurs, when approaching this problem. Instead, they attempt to formulate the answer without thinking it through step by step.

Visual Approach:

If you think about it visually, if you start by picking a goat, then Monty Hall will reveal the other door with the goat, so to win the car in that case, you have to switch to the other closed door. However if you start with a car, you must stay on that door to win. There is 2⁄3 of a chance you start with a goat and 1⁄3 chance you start with a car. Therefore, it is more likely you will win if you switch your answer to the other unopened door after Monty Hall opens one door.

Conditional Probability Approach:

To put this in a more mathematical way of thinking, let’s revisit Bayes' Theorem that we previously talked about in this Math Magazine, a very important topic in conditional probability.

As a reminder, Bayes’ Theorem is a mathematical formula that can be used to calculate conditional probability for a test result, including the likelihood of a result based on a past result with similar circumstances.

Formula: P(H|E) = P(E|H)P(E)P(H)

In this case, H (hypothesis) represents door 1 leading to the car and E (evidence) represents Monty Hall revealing a door with a goat. P(H|E) represents the conditional probability of H, given E. So, let’s break this down into separate steps to think through this process more efficiently and understand it properly.

1. P(H) - Since we still have no information or clues to predict which door will lead to the car, the probability of door 1 having the car will be the same as doors 2 and 3. So, P(H) = 13.

2. P(E) - Since Monty Hall always picks a door leading to a goat, we already know this outcome. So, P(E) = 1.

3. P(E|H) - For this variable, we can assume our H, door 1 having the car, is correct and when calculating the probability of E occurring, we know it will always happen. So, P(E|H) = 1.

4. P(H|E) = (1)(1)(13) = 13

   
   

By using this formula, it can be concluded that multiplying the probability of the evidence occurring had no impact or change on the original probability of H, door 1 having the car.

However, when looking at the other perspective of door 2 having the car, the probability will be double than that of door 1 since the car can only be in two doors after Monty Hall reveals one with the goat. The probability of door 2 will equal 26.

Therefore, switching your answer to the other closed door after Monty Hall reveals one door will maximize your potential to get the problem right!