How is it better to switch in the Monty Hall Problem?

Question

I'm not a mathematician so I'm probably getting something wrong here, but no matter how I think about it I can't accept that it is necessarily better to switch. Here is my best attempt at a proof:

Alternative Perspective on the Monty Hall Problem:

In the classic Monty Hall problem, a contestant is presented with three doors, one of which hides a car (the winning prize), while the other two conceal non-winning outcomes, such as goats. After the initial selection of a door by the contestant, the host, who knows the locations of the prizes, reveals one of the other doors to show a goat. The contestant is then given the option to stick with their original choice or switch to the remaining unopened door.

Key Assumptions:

• There is a single car and two goats, each initially equally likely to be behind any of the three doors.

Argument and Conclusion:

Random Removal of a Goat Door:

• In a scenario where the contestant does not initially choose a door, a goat door is randomly removed by the host, leaving the contestant with a binary choice between the two remaining doors: "Door A" or "Door B." This is the same as if the contestant randomly picked a door and and the remaining goat door is removed.

Equivalence of Choice Options:

• The question "Would you like to switch?" presents two possible answers: "Yes" (implying a switch) or "No" (implying no switch). The question "Which of these 2 remaining doors would you like to pick now?" also offers two possible answers: "Door A" or "Door B." Both questions ultimately lead to a binary choice between two doors, resulting in a 50/50 probability of selecting the car.

Conclusion:

Whether the contestant initially makes a choice or not, the probabilities associated with each remaining door are evenly distributed at 50/50. In both scenarios, the decision to switch or stick with the original choice does not affect the probability of winning the car.

Answer 1

The crucial point in the Monty Hall problem is that the host's action of revealing a goat door changes the probabilities. The key assumption you made is the source of the problem. Let me break down why your argument is incorrect:

Key Assumption: You assumed that there is a single car and two goats, each initially equally likely to be behind any of the three doors. This is where the problem arises. In the classic Monty Hall problem, the location of the car is fixed at the beginning, and the host knows where it is. The initial choice by the contestant is crucial.

Random Removal of a Goat Door: In the classic Monty Hall problem, the host doesn't randomly remove a goat door. The host specifically opens a door that (a) the contestant did not initially choose and (b) has a goat behind it. The host's action is not random; it's based on knowledge.

Equivalence of Choice Options: Your argument attempts to equate two scenarios, one where the contestant initially chooses a door and one where they do not. However, this equivalence is not valid because the initial choice does matter. When the contestant initially chooses a door, there is a $\frac{1}{3}$ chance they picked the car and a $\frac{2}{3}$ chance they picked a goat. When the host reveals a goat door, the probabilities shift.

Conclusion: Your conclusion that the probabilities are evenly distributed at 50/50 is incorrect. The probabilities are not evenly distributed, and they do change when the host reveals a goat door. The Monty Hall problem is famous because it defies our intuition. In fact, if the contestant switches doors after the host reveals a goat, their chance of winning the car is $\frac{2}{3}$, not $\frac{50}{50}$.

To fully grasp the Monty Hall problem, it's essential to understand that the initial choice matters, and the host's action of revealing a goat door is not random but strategic. This is why switching doors is the statistically better choice.