If you guess on a 3-door game show, should you switch doors after a hint?
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Monty Hall Problem (best explanation) - Numberphile
Paper-cutout doors replay the game twice, showing why your 1-in-3 first pick and Monty's forced goat reveal park the whole 2-in-3 chance on the door you can switch to.
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What else makes you wonder?
What would this feel like with a hundred cupboards and the host opening ninety-eight of them?
Think about how much of the room your one blind point still covers when the row gets that long, and how much work the host has to do before he stops.
Where else does a hint from someone who already knows quietly change what you should do?
A friend who peeked at the answer sheet, a doctor ordering one particular test, a search engine choosing what to show you first — what did each of them have to avoid?
If a plan is the better bet, how many tries before you can actually feel it?
One round can go either way. Try imagining the same plan run five times, then fifty — when does the difference stop looking like luck?
More to explore
After you watchIf you guess on a 3-door game show, should you switch doors after a hint?
The short answer
Yes — switching is the better bet. On a 3-door game show where the host knows the layout and always opens a losing door, switching wins about 2 times out of 3 while staying wins about 1 time out of 3. It nearly doubles your chance, but it never promises a win.
Try this next
- What if the host had not peeked first? Set the tin helper to the one who is guessing and step through all six tries. Predict first: how often does a guessing helper uncover the key by accident, and what would that do to the cupboard game?
- What if the row had ten cupboards instead of three? Imagine your one blind point on a row of ten, with the host opening eight empty ones. Predict how much of the row your single point still covers before you work out the rest.
Now you — bend it
- What if What if you bumped the game up to 10 cupboards and the host opened 8 empty ones?Your one blind point now covers a much smaller slice of the row. Work out where the rest of the chance has to end up once the host has finished opening.
- What if What if the host opened a cupboard at random and sometimes showed the prize?Now his hint is not restricted anymore. Think about what that does to the two cupboards still closed in the rounds where he happened to miss.
Can you prove it?Across nine matched rounds the two fixed plans do not come out level. — Unseal all nine rounds one at a time and read the two star tracks. Because the rounds are fixed rather than random, anyone can replay the same nine and get the same two totals.
Design your own test:Predict how many rounds it takes before the gap between the two tracks stops looking like luck — then open two, then five, then all nine and check.
Explain it to a 6-year-old: Your first cup is usually the wrong one, and the helper who peeked is not allowed to lift the good one — so the other cup he left alone is a better place to look.
The whole story
How it works
Your first pick is just a 1-in-3 guess, so two times out of three the prize is behind a door you did not pick. The host knows where the prize is and always opens a losing door, never the prize. Because the host's choice is loaded and never lands on your door, your first pick stays at 1-in-3 — and the whole leftover 2-in-3 chance piles onto the single other closed door. Jumping to that door means you win whenever your first guess was wrong, which is two times out of three.
What people get wrong
It feels like once one losing door is open and two doors are left, it must be a fresh 50/50 coin flip, so switching cannot matter. That is the trap. The two doors are not equal: the host deliberately avoided the prize when choosing what to open, so the door you skipped carries the full 2-in-3 chance and your original door still only carries 1-in-3.
The catch
Switching is the better bet but not a guarantee. One time in three your first pick was already the prize, and switching walks you right off it — so staying still wins sometimes. Switching also only beats staying because the host knows the layout and always opens a losing door on purpose. If the host opened a door at random and might reveal the prize, the leftover game really would be 50/50 and switching would lose its edge.
Questions kids ask
If two doors are left, isn't it just 50/50?
No. The two closed doors are not equal, because the host did not open randomly — he knew where the prize was and made sure to open a losing door. That leaves your first pick at 1-in-3 and pushes the other 2-in-3 onto the single door you can switch to.
Does staying ever win?
Yes — staying wins whenever your very first guess happened to be right, which is about 1 time in 3. Switching just wins more often, about 2 times in 3, because your first guess is wrong twice as often as it is right.
What if the host opened a door without knowing where the prize was?
Then the trick disappears. A host opening a random door might accidentally show the prize, and in the rounds where he happens to reveal a goat, the two closed doors really are 50/50. Switching only wins 2-in-3 because the host knows and always avoids the prize on purpose.
Why does switching feel so wrong?
Because our brains treat the two leftover doors as a fresh, even choice and forget that the host's hint was loaded. Running the game hundreds of times shows the truth: the switch line settles near 2-in-3 and the stay line near 1-in-3, every time.
Talk about it
- Two doors are left and one prize — guess first: are they really equal, or is one a better bet?
- Why might a hint from someone who knows the answer be worth more than a hint from someone guessing?
- If you had to bet your dessert, would you stay or switch — and what makes you sure?
For grown-ups
This is the Monty Hall problem. The first pick wins with probability 1/3, so the prize is elsewhere with probability 2/3. Because the host knows the layout and always opens a losing door, no probability transfers back to the original pick; the entire 2/3 collapses onto the one remaining unopened door. Switching therefore wins 2/3 of the time and staying 1/3. The crux is the host's constraint — a host who opened a door uniformly at random, sometimes revealing the car, would leave a genuine 50/50 between the two closed doors.