Monty Hall Problem: Should You Switch? Simulate and See

The Monty Hall problem simulator asks for two things: how many doors and how many trials. The classic setup is three doors — one hides a prize, the other two hide goats. After you pick a door, the host (who knows where the prize is) opens one of the remaining goat doors and offers you a switch. The simulator runs that scenario hundreds or thousands of times and tracks how often switching wins versus staying.

Start with 3 doors and 1,000 trials. Enable “simulate both strategies” so the tool runs a staying batch and a switching batch side by side. The chart that appears tells the whole story: the switch line settles near 67% and the stay line near 33%. Once you see it, bump the doors to 10 or 100 and watch the gap widen dramatically.

After the host opens a goat door, two doors remain. Most people look at those two doors and think “50/50.” It feels like a coin flip. But the two doors did not arrive at this moment by the same path, and that history matters.

When you first picked, you had a 1-in-3 chance of being right. That means there was a 2-in-3 chance the prize was behind one of the other doors. The host then opens a goat door — but the host already knew which door hid the prize and deliberately avoided it. That action doesn't split the 2/3 evenly; it funnels the entire 2/3 onto the single remaining door. Switching gives you access to that concentrated probability.

This is conditional probability in action. Your posterior odds after the host reveals a goat are not the same as your prior odds. The host's reveal is not random — it's informed — and that information is what makes switching the better bet. Formally, you're performing a Bayesian update: your prior (1/3 on your door, 2/3 on the field) gets updated by the evidence (host always avoids the prize), and the posterior concentrates on the switch door.

Walk through a single round with three doors. Suppose the prize is behind Door 2 and you pick Door 1:

Now imagine the prize is behind Door 1 (your original pick). The host opens either Door 2 or Door 3 — doesn't matter which. You switch and lose. That happens 1/3 of the time. The other 2/3 of the time, the prize is behind a door you didn't pick, and the host helpfully eliminates the other wrong door, leaving the prize sitting in plain sight for anyone willing to switch.

Run 1,000 trials in the simulator and the numbers land close to these predictions: roughly 333 stay-wins and 667 switch-wins. The running average chart shows both lines wobbling early and stabilizing as trials accumulate — a textbook illustration of the law of large numbers.

Theory says switching wins (D − 1) / D of the time, where D is the number of doors. For 3 doors that's 2/3 ≈ 66.7%. The simulator runs a Monte Carlo test: generate a random prize location, a random player pick, have the host open all but one non-chosen goat door, then record whether switching or staying would have won.

The 100-door version is the thought experiment that convinces most skeptics. You pick one door out of 100. The host — who knows where the car is — opens 98 goat doors, leaving yours and one other. Would you really stick with your original 1-in-100 guess? Almost nobody would, and that instinct is correct. The same logic applies to 3 doors; it's just harder to feel at smaller scale.

Thinking two doors means 50/50. The doors are not symmetric. Your door was chosen before the host acted; the other door survived because the host deliberately protected the prize. Equal count does not mean equal probability.

Forgetting the host knows. If the host opened a door at random and happened to reveal a goat, switching would genuinely be 50/50. The entire advantage comes from the host's knowledge. Remove it and the puzzle dissolves.

Confusing a single trial with a strategy. On any one round you might win by staying. That doesn't make staying a good strategy. Over many rounds, the 2/3 vs 1/3 split is ironclad. The simulator's running-average chart makes this viscerally clear: early wobbles even out into a stable gap.

Assuming the result doesn't scale. People accept the 3-door answer and then assume adding doors doesn't change much. It does — massively. Every extra door makes the initial pick less likely and the switch more powerful. At 100 doors, staying is a 1% strategy.

The ignorant host. If the host doesn't know where the prize is and opens a door at random, sometimes the prize is revealed and the game ends. Among games that continue (host happened to pick a goat), switching and staying are equally good. The information edge vanishes with the host's ignorance.

Multiple prizes. Hide two cars behind three doors and the calculus flips: staying wins 2/3, switching wins 1/3. The standard result depends on exactly one prize — extra prizes change the conditional probabilities entirely.

Many-door generalization. With D doors and one prize, the host opens D − 2 goat doors. Staying wins 1/D and switching wins (D − 1)/D. As D grows the ratio approaches 0% stay vs 100% switch — a clean limit that makes the underlying logic almost self-evident.

Using the simulator as a teaching demo. Run 10 trials live in front of a class and the results look noisy — someone will argue staying won more. Then run 10,000 and the lines converge. That contrast between small-sample noise and large-sample convergence is itself a lesson in statistical thinking, separate from the Monty Hall puzzle.