Thought Toys · Chance & inference · Exhibit 06

The Monty Hall problem

Pick a door. The host — who knows where the prize is — throws open a losing door and offers you the swap. Your gut says it's a coin flip now, so why bother. Your gut is wrong: switching wins twice as often. Play a few rounds, then run a thousand and watch two-thirds rise out of the noise.

Pick a door.

Three closed doors. Choose one to begin.

prize opened, empty your pick switch here
Always switch0 / 0
should approach 2/3 = 67%
Always stay0 / 0
should approach 1/3 = 33%
3100 (the host opens all the losers but one)

What you're seeing

At the start, your one door has a 1-in-N chance of hiding the prize, and everything you didn't pick holds the rest of the chance between them. That split is fixed the moment you choose — the prize doesn't move. Then the host, who knows what's where, deliberately opens losing doors. He never opens yours and never opens the prize. He's not adding luck; he's removing known losers from the pile you rejected.

So the whole leftover probability that used to be spread across all those other doors gets funnelled onto the single door he pointedly leaves shut. Your door is still stuck at its original 1-in-N. The other one now carries everything else. With three doors that's 1/3 against 2/3 — switch and you double your odds.

If that still feels like a trick, drag the slider to 100 doors. Pick one, and watch the host fling open ninety-eight empties, leaving just your door and one other. Now the question asks itself: did you really nail it first try at 1-in-100 — or is the prize behind the one door he was suspiciously careful to keep closed? Press Run 1000 rounds and the two bars settle onto their predicted marks: switching wins (N−1)/N of the time, staying just 1/N. Your intuition isn't bad at chance — it just forgot the host was giving away an answer.

The rule, exactly. N doors, one prize. You pick; the host (knowing all) opens every other door except one, always avoiding the prize and your door. Stay wins only if your first guess was right — probability 1/N. Switch wins exactly when your first guess was wrong — probability (N−1)/N, because the host has quietly concentrated all of that remaining probability onto the lone door he left closed. For N = 3: 1/3 vs 2/3; for N = 100: 1% vs 99%. (Checked offline: 300k-round Monte-Carlo simulations land on 0.333/0.667 at three doors and 0.010/0.990 at a hundred, matching the formula.) Counter-example, verified in node: the 2/3 edge depends on a host who knows where the prize is — if the host opens a random door, switching wins only 50%.

Also in Chance & inference: Bayes' theorem →

All 16 in Chance & inference
  1. 05The Galton board
  2. 06The Monty Hall problem — you are here
  3. 09Bayes' theorem
  4. 13Buffon's needle
  5. 14The central limit theorem
  6. 28Simpson's paradox
  7. 30Markov chains
  8. 31Averages that never settle
  9. 35The birthday paradox
  10. 38The drunkard's walk
  11. 39Zipf's law
  12. 40Benford's law
  13. 42The coupon collector's problem
  14. 48The wisdom of crowds
  15. 52Genetic drift
  16. 63The gaps are chaos. The count is law.

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