The Monty Hall problem is a famous probability puzzle that baffles many, even those with strong mathematical backgrounds. Named after the television game show host Monty Hall, this problem illustrates a counterintuitive concept in probability theory.
The Game’s Setup
- There are three doors: A, B, and C.
- Behind one of the doors is a car; behind the other two are goats.
- You choose a door, say Door A.
- Monty, who knows what’s behind each door, opens another door to reveal a goat. He never opens the door you chose first and never reveals the car at this step.
- You now have the option to stick with your original choice (Door A) or switch to the remaining unopened door.
- Monty now opens your selected door, and you either win the car or get a goat.
Given this setup, what would you do? Would you stick with your original choice or switch to the other door?
The Detailed Analysis
To find the optimal strategy, let’s break down the problem into scenarios based on where the car is located.
If the car is behind Door A:
- Choosing Door A first and sticking with it = win.
- Choosing Door B first (Monty will open Door C, revealing a goat; remember Monty never reveals the car) and switching to Door A = win.
- Choosing Door C first (Monty will open Door B, revealing a goat) and switching to Door A = win.
- Result: 2 chances to win by switching, 1 chance to win by sticking.
If the car is behind Door B:
- Choosing Door A first (Monty will open Door C, revealing a goat) and switching to Door B = win.
- Choosing Door B first and sticking with it = win.
- Choosing Door C first (Monty will open Door A, revealing a goat) and switching to Door B = win.
- Result: 2 chances to win by switching, 1 chance to win by sticking.
If the car is behind Door C:
- Choosing Door A first (Monty will open Door B, revealing a goat) and switching to Door C = win.
- Choosing Door B first (Monty will open Door A, revealing a goat) and switching to Door C = win.
- Choosing Door C first and sticking with it = win.
- Result: 2 chances to win by switching, 1 chance to win by sticking.
Conclusion
The results are clear: regardless of the initial door choice, the chance of winning the car by switching doors is 2 out of 3. Conversely, sticking with the original choice only offers a 1 out of 3 chance of winning. Thus, the optimal strategy is to switch doors.
So, why do many people find the Monty Hall problem counterintuitive or even get it wrong?
- Overlooking Monty’s Knowledge: People often fail to account for the fact that Monty knows where the car is. He doesn’t open a door at random. He always opens a door revealing a goat, and this deliberate choice skews the probabilities.
- Mistaken Equivalence: After one of the doors is revealed to have a goat, it’s common to think the car is equally likely to be behind the two remaining doors as if it’s a fresh 50-50 choice. However, the previous decisions and Monty’s actions mean it isn’t a new 2-door problem but a continuation of the original 3-door problem.
- Human Heuristics: Our brains often rely on shortcuts or heuristics to make quick decisions. One such heuristic is to stick with our original choice, often rooted in the fear of regret if we switch and lose. This can cloud our judgment and make it difficult to assess the situation logically.
By understanding the underlying probabilities and the mechanics of the game, we can see why the optimal strategy in the Monty Hall problem is always switching doors after Monty reveals a goat behind one of the other two. This detailed analysis demonstrates the importance of considering all possible outcomes when solving probability puzzles. It also reminds us that our instincts about probability, influenced by shortcuts we take in thinking, can sometimes mislead us.