Three Doors, Two Thirds, Zero Excuses

Three doors. One prize. A game show host who knows everything. And millions of people, some with actual math degrees, who still insist the answer is a coin flip. I find this delightful. Let me fix it for you, because someone has to.

Where the puzzle comes from

The puzzle borrows its name from Monty Hall, who hosted Let's Make a Deal for nearly 30 years after the show first aired in 1963, according to the Data Science Discovery write-up. Some people call it the Monty Hall Paradox, as Harvey Mudd's math page does. It is not a real paradox. Nothing contradicts itself. Your intuition simply loses a fight it should never have picked.

The setup is easy. You face three closed doors. One hides the prize, the other two hide losers. The sources can't agree on the prize: one says $1,000,000, the other a luxury car. I'll allow either. The math doesn't care. It only needs exactly one winner among three doors.

The rules that drive the answer

You pick a door. The host then opens one of the other doors, always a loser, and offers you a deal: stay, or switch to the last closed door.

Read the next part twice. The host knows where the prize is and never opens that door. Both sources build on this, and the whole answer hangs on it. Let the host open doors blindly and the situation changes. The sources don't work out that version, so I won't pretend they do. Everything here assumes the informed host.

Why switching wins two thirds of the time

Your first pick has a 1/3 chance of being the prize and a 2/3 chance of being a loser. That is the entire game. When the host opens a losing door, your original 1/3 does not budge, since nothing about your door changed. The remaining closed door inherits the other 2/3, as the Harvey Mudd page explains.

If you distrust me, and you shouldn't, list every outcome. Lay the three doors and three prize locations in a table. Staying wins in 3 of 9 equally likely combinations. Switching wins in 6 of 9. That is the sample-space approach, and it is not subtle.

Here is the shortcut for the impatient. Play 600 games. Roughly 200 first picks land on the prize and roughly 400 do not. In each of those 400, the host must reveal the other loser, which leaves the prize sitting behind the remaining door. Switch, and you win all 400. That's two thirds, and the same page uses this very argument.

The common mistake and how to test it

Most people see two closed doors and declare it 50-50. It feels airtight. It is also wrong, and the sources call it a fallacy. The host's choice was not random with respect to the prize. He steered around it, and that steering carries information.

Want it more obvious? Take 52 cards, with the ace of spades as the prize. You pick one card, a 1/52 shot. The dealer turns over 50 non-aces. Do you honestly believe your original card is now a coin flip? Switching wins 51 out of 52 times, per the card version. Same structure, just louder.

If you still resist, write a quick simulation. Play the game hundreds or thousands of times and count. Staying lands near one third, switching near two thirds. The Illinois page suggests exactly this, including checking the table and the code against each other. I did not need to. You might.

What the puzzle does not settle

The answer lives and dies by its rules. Change the host's behavior and you change the math. The sources don't cover the random-host version, so work that one out separately instead of borrowing my two thirds.

They also say nothing about applying this to real life. Fine. Use it as a training exercise instead. Before you trust any new information, ask what the source actually knew when it handed it to you. That habit alone beats most of the doors you'll ever face.