Why Switching Pays in the Monty Hall Problem

Whether it is in the classic Monty Hall scenario or a variation with a larger number of doors, switching doors significantly increases your chances of winning the prize. This article explores the mathematical reasoning behind this phenomenon, which is crucial for understanding the correct strategy in these types of probability puzzles.

Understanding the Monty Hall Problem

The Monty Hall problem presents a simple yet counterintuitive situation. Imagine a game show where you face three doors—behind one lies a grand prize, while the other two hide goats. You are given the choice to pick a door. After your initial choice, the host, who knows what's behind each door, opens another door to reveal a goat. The question is, should you stick with your initial decision or switch to the remaining unopened door?

Initial Choices and Probabilities

When you first choose a door, you have a 1/3 chance of selecting the grand prize. This initial probability remains unchanged by the host opening one of the other doors. This is because the host's action of revealing a goat does not alter the fact that your initial choice was random and unbiased. Conversely, the probability that the grand prize is behind one of the other two doors is 2/3. When the host reveals a goat behind one of these doors, the probability of the prize being behind the chosen door remains 1/3, while the combined probability of it being behind the other closed door becomes 2/3, as there are now just two unopened doors, one of which must contain the prize if the initial 2/3 probability holds true.

Increasing Your Chances by Switching

One of the key principles in the Monty Hall problem is the concept of conditional probability. By revealing a goat behind one of the doors, the host provides new information that influences the probability of the outcome behind the remaining doors. If you initially chose a door with a goat (which has a 2/3 chance), the host will reveal the second goat, leaving the prize behind the remaining unopened door. Conversely, if the door you initially chose has the prize (which has a 1/3 chance), only one unopened door contains a goat. Thus, switching to the remaining unopened door will result in a win 2/3 of the time, while sticking with the initial choice will only win 1/3 of the time.

The Larger Door Scenario

What if the number of doors is not three but 100? Let's consider a similar setup with 100 doors and concentrate on Door 43. After your initial choice, the host opens 98 doors, revealing goats behind each one. The mathematical principles underlying the Monty Hall problem remain the same, but the scale of the scenario emphasizes the significance of switching.

Probability Analysis in the 100-Door Scenario

In the 100-door scenario, your initial choice of Door 43 has a 1/100 chance of being correct, while the combined probability of the prize being behind one of the other 99 doors is 99/100. After the host eliminates 98 of these 99 doors, the remaining unopened door still represents the 1/100 chance of concealing the prize. Therefore, the probability that the prize is behind Door 43 remains 1/100 when you do not switch, but the probability that the prize is behind the remaining unopened door becomes 99/100 when you do switch. This demonstrates that switching doors is a consistently effective strategy, even in scenarios with a large number of options.

Conclusion

In both the classic Monty Hall problem and its larger-scale variations, the strategy of switching to the remaining unopened door increases your chances of winning. This is not just a matter of logic but also a demonstration of how conditional probability redistributes probabilities based on new information revealed through the process of the game. By understanding the principles of conditional probability and the behavior of the host in the Monty Hall problem, you can maximize your chances of success in similar probability puzzles.