The Monty Hall Problem

What this lecture covers

This recitation problem tackles the famous Monty Hall problem: you pick one of three doors hiding a prize, the host opens one of the other two doors (always an empty one), and you must decide whether to stay with your original pick or switch. The lecture answers which strategy is better and, importantly, why the exact rules of the game determine the answer.

It's a good exercise in translating a verbal puzzle into precise assumptions before computing, and in checking a counterintuitive numerical answer against multiple intuitive explanations.

Key ideas

• Rules matter: the host must always open an unpicked, empty door — this specific rule is what makes switching better, and changing it changes the answer.
• Staying wins only if your first guess was right: since the prize is equally likely behind any of the three doors, staying wins with probability 1/3.
• Switching wins whenever your first guess was wrong: since that happens with probability 2/3, switching wins with probability 2/3.
• "Doors as groups" intuition: switching effectively lets you claim both of the other two doors at once, since the host guarantees one of them is empty.
• Scaling up the intuition: with 1,000 doors and the host opening 998 empty ones, it becomes obvious that your first pick was very likely wrong, so switching is clearly better.
• Mixed strategies stay bounded: a strategy that switches only sometimes still produces a win probability between 1/3 and 2/3.

Walkthrough

Rules of the game (1:22)

The video lays out the exact setup: three closed doors, one prize, you pick a door, and a friend who knows where the prize is opens one of the remaining two doors, always choosing one that is empty. You then choose to stay with your original pick or switch to the other unopened door.

Why staying gives 1/3 and switching gives 2/3 (6:29)

The lecture argues that staying wins exactly when the original pick was correct — probability 1/3, since the prize is equally likely behind any door. Switching wins whenever the original pick was wrong, since the host is forced to reveal the other empty door, leaving the prize behind the door you'd switch to. This happens with probability 2/3. A full list of the six equally likely scenarios confirms the same counts.

Building intuition with 1,000 doors (9:28)

To make the result feel less surprising, the video scales the problem to 1,000 doors, where the host opens 998 empty ones after your pick. Since your first guess is right only 1 time in 1,000, switching to the one remaining unopened door is clearly the better bet — the same logic as the three-door case, just easier to see.

A mixed strategy and the final probability (13:35)

The lecture then examines a strategy that only switches depending on which door the host opens, introducing a free parameter p for the host's choice when either empty door could be opened. Working through the probability tree shows the win probability is p/3 + 1/3, which always lands between the pure-stay probability (1/3) and the pure-switch probability (2/3).

Before you watch

• Know the discrete uniform law and conditional probability trees, covered in the first two lectures of this course.
• No other background is needed; the problem is self-contained.

Check your understanding

1. Why does the host's rule of never opening the door with the prize matter so much to the answer?
2. Why does staying win with probability exactly 1/3?
3. How does the "doors as groups" explanation make the 2/3 switching probability intuitive?
4. What happens to the win probability of switching as the number of doors grows very large?
5. Why does a mixed strategy's win probability always fall between 1/3 and 2/3?