The Myth of the Monty Hall Problem
Spoiler: it’s not that difficult.
At some point, you’ve probably heard of the Monty Hall Problem. It’s been referenced in books and movies, and it was briefly made famous by Parade magazine contributor Marilyn vos Savant in 1990.
It’s alleged to be an incredibly difficult and counterintuitive problem that befuddles common folk and experts alike. Here’s what economist Tim Harford wrote about it recently at the Financial Times:
Forget Fermat’s last theorem. The most vexing challenge in mathematics just might be the Monty Hall problem. Monty Hall — born Monte Halparin — presented nearly 5,000 episodes of Let’s Make a Deal, the US game show that inspired the puzzle. It is an onion of a conundrum; layer after layer, and guaranteed to make you cry.
The only thing about the problem that makes me want to cry is that people keep retelling this once mildly interesting riddle as though it was a deep, unfathomable mystery that confounds even statisticians.
The “problem” goes like this: you are on Monty Hall’s game show, and there is a new car behind one of three doors. Behind the other two are goats.
You pick a random door. The probability the car is behind that door is 1/3.
Therefore, the probability it is behind a different door is 2/3.
Monty knows where the car and the goats are, and he helpfully opens one of the other doors, revealing a goat.
Monty asks if you want to change your guess now.
Nothing about the scenario has changed. The odds of your original guess being right are still 1/3. The odds the car is behind a different door are still 2/3. And now, thanks to Monty, there’s only one other closed door.
So of course you should switch — by telling you which of the other doors it isn’t, Monty just doubled the chance it’s the last door.
People are (allegedly) supposed to inevitably, irretrievably, hopelessly fooled into thinking that the odds are 50/50 just because there are only two closed doors remaining. But if you look at it sequentially like this, the correct answer is intuitively obvious.
I think Fermat’s last theorem is safe, for now, as the most famously difficult math problem.
*This ignores whether Monty’s decision to offer the choice at all is random, or whether it’s skewed based on what he knows about the contestant’s first choice (i.e., sometimes Monty might be trying to psych out people who guessed correctly). Just how skewed the frequency of the offer would need to be to cancel out the large benefits of revealing the goat is also mathematically answerable, although I’m too tired to figure it out right now.
