Key points of this article

  • Of three doors, one hides a brand-new car while the other two conceal goats as gag prizes.

  • The player first chooses one door.

  • The host—modeled after Monty Hall, the emcee of the American TV show Let’s Make a Deal, who knows what is behind every door—always opens exactly one losing door.

  • The player must then decide whether to keep the initial choice or switch to the only other closed door.

  • The conclusion is clear: switching doors is overwhelmingly advantageous (win rate = 2/3).

  • With the simulator introduced here, you can see and feel this counterintuitive probability until it makes sense.

Introduction

Have you ever felt “my gut tells me this answer, yet the math says otherwise”? The Monty Hall problem is the quintessential showdown between intuition and logic.

I too once believed “after the host opens a door, the chance must be 1/2 either way” and could not accept the result. I eventually understood only after running thousands of trials in Excel and watching the odds converge to 2/3. This article reconstructs that experience and goes further by publishing a browser-based simulator that anyone can interact with instantly.

The puzzle takes its name from Monty Hall himself—the Canadian-American host of Let’s Make a Deal—whose stagecraft of revealing goats inspired the scenario.

Rules of the problem (three-door version)

  • Behind the three doors, [one hides the new car while the other two hide goats].
  • The player starts by choosing any one door.
  • The host, who knows every outcome, always opens one of the remaining doors that hides a goat.
  • Two choices remain: the original door and one still-closed door.
  • The player must decide to “stay” or “switch.”

Why does intuition fail?

Most people reason, “once only two doors remain, the odds must be 1/2 each.” That is a major misunderstanding.

  • The chance of initially picking the car is only 1/3.
  • Therefore the chance of initially picking a goat is 2/3.
  • Because the host always reveals a goat, the unchosen closed door accumulates that 2/3 probability of hiding the car.
  • Staying yields a 1/3 win rate; switching yields 2/3. Switching is twice as good.

Understanding it with a decision tree

Breaking down the cases in a decision tree makes the logic even clearer.

  1. If you picked the car first (probability 1/3)

    • The host opens a goat door.
    • Switching lands you on a goat, so you lose.

  2. If you picked a goat first (probability 2/3)

    • The host deliberately keeps the car closed and opens a goat.
    • Switching moves you to the car, so you win.

In total, switching wins 2/3 of the time, staying wins 1/3.

Extending the idea to N doors

The logic does not depend on the number of doors.

  • With N doors:
    • Staying keeps the initial win rate of 1/N.
    • Switching at the end wins with probability (N − 1)/N.

Imagine 100 doors. The probability that you picked the car right away is only 1/100. The host—again, a Monty Hall–style figure who knows everything—opens 98 losing doors and leaves one closed door untouched. That remaining door now carries 99/100 odds of hiding the car. Switching is obviously the superior strategy.

“See it” in Excel and in the browser

Running thousands of trials in Excel shows the switching strategy converging to a 66.6% win rate, perfectly matching the theory.

Numbers alone can feel abstract, though. That is why I built a browser-only tool for this article.

Simulator highlights

  • Manual play mode lets you pick doors yourself and watch the host’s reveal animations.
  • Auto-simulation mode runs tens of thousands of trials in an instant and visualizes the convergence.
  • N-door settings (3–10 doors) generalize the scenario so you can grasp the pattern.
  • Charting compares the win rates of each strategy in real time.

Experiencing both the intuition and the statistics side by side makes the heart of the Monty Hall problem far easier to grasp.

Common objections and cautions

  • What if the host does not know the contents and opens doors at random?
    → Then the host might reveal the car. That becomes a different problem entirely.

  • What if the host follows a unique rule for which door to open?
    → If players know that rule, conditional probabilities can change.

  • “Once two doors remain, surely it’s 1/2 each?”
    → The host’s action provides information. The two doors are not symmetric: one hides the car with probability 1/3, the other with 2/3.

Summary

  • The Monty Hall problem is deceptively simple yet brutally counterintuitive.
  • Accounting for the host’s reveal shows that switching doors always dominates.
  • Excel experiments and the browser simulator introduced here let you internalize the math through both numbers and direct experience.

Try it yourself → Monty Hall Simulator

Appendix (original memo reprinted)

Back when I repeated the Excel experiments, I realized that “switching is equivalent to having picked two doors at the start”—that intuition finally stuck. This article is a revised and expanded version of that old memo.

It is fascinating that the puzzle went viral in the United States in the late 1980s. When Parade magazine featured it, the editors received over 10,000 letters insisting that “the odds must be 1/2 each.” Among them were hundreds of objections from university professors and statisticians—proof that even experts can be swayed by intuition when the framing feels like a TV game show.

This episode shows how the Monty Hall problem defies human intuition and why rigorous reasoning matters.