The Monty Hall Paradox | mysimulator.uk

The setup, precisely

Three doors, one car, two goats. You pick a door. The host — who knows where the car is and will never open the door hiding it or the door you picked — opens one of the other two doors, always revealing a goat. You are then offered the choice to switch to the remaining unopened door or stay with your original pick. The question, posed to Marilyn vos Savant in a 1990 Parade column and infamous for the flood of mathematicians who wrote in insisting she was wrong, is: does switching help?

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It does, and by a lot: switching wins 2/3 of the time, staying wins only 1/3. The intuition that trips people up is treating the two remaining doors as suddenly equally likely once one is revealed — they are not, because the host's action was not random.

Why your first guess is frozen at 1/3

At the moment you pick a door, the probability the car is behind it is exactly 1/3, and the probability it is behind one of the other two doors combined is 2/3. Crucially, nothing the host does afterward can change the probability attached to your original door, because the host's choice of which goat-door to open is not evidence about your door — he was always going to be able to open a goat door no matter what you picked, since at least one of the two remaining doors is guaranteed to hide a goat. Your original door is stuck at 1/3 forever; the remaining 2/3 of probability, previously spread across the other two doors, collapses entirely onto whichever single door the host didn't open.

Bayes' theorem, worked through

For the sceptics, here is the calculation in full. Say you pick door 1 and the host opens door 3. Let C1, C2, C3 be the events the car is behind door 1, 2, 3 (each prior probability 1/3), and H3 the event the host opens door 3.

P(H3 | C1) = 1/2   (car behind your door — host picks either remaining goat door at random)
P(H3 | C2) = 1     (car behind door 2 — host MUST open door 3, it's the only goat left)
P(H3 | C3) = 0     (host never opens the door hiding the car)

Bayes:  P(C2 | H3) = P(H3|C2)·P(C2) / P(H3)
                    = (1 · 1/3) / [(1/2·1/3) + (1·1/3) + (0·1/3)]
                    = (1/3) / (1/2) = 2/3

So P(C1 | H3) = 1/3  (unchanged)   and   P(C2 | H3) = 2/3  (switch wins)

The host's forced move when the car is behind door 2 — he has literally no choice, door 3 is the only door he's allowed to open — is what pumps all of door 2's probability into your observation. That asymmetry between P(H3|C2)=1 and P(H3|C1)=1/2 is the entire paradox in one line.

The host's knowledge is the whole trick

Change one assumption and the answer changes completely. If the host opens a random door without knowing where the car is, and just happens to reveal a goat, the classic Monty Hall reasoning breaks: the two remaining doors really are equally likely, 50/50, because you have now conditioned on a host action that carried no information (it could have revealed the car and simply didn't, by luck). This variant is sometimes called the Monty Fall problem, and confusing it with the original — assuming the host's goat-reveal is uninformative — is the single most common source of the wrong 50/50 intuition.

Scaling it up makes it obvious

A common way to make the 2/3 result intuitive rather than merely provable: imagine 100 doors, one car. You pick one door (1% chance of the car). The host, who knows where the car is, then opens 98 of the other 99 doors, all goats, leaving your door and exactly one other door closed. Would you switch? Almost everyone says yes immediately — the host's forced elimination of 98 doors he knew were empty has concentrated essentially the entire 99% probability onto that one remaining door. The three-door case is the same mechanism, just with less dramatic numbers, which is exactly why it fools people the classic version doesn't.

Frequently asked questions

Why does switching win 2/3 of the time instead of 50/50?

Your first pick has a 1/3 chance of being right and that never changes. The host, who knows where the car is and always reveals a goat from the two doors you didn't pick, effectively concentrates the other 2/3 of the probability onto whichever single door he leaves closed. Switching claims that concentrated 2/3; staying keeps your original 1/3.

Does it still work if the host opens a door at random and just happens to reveal a goat?

No — that's a different problem (sometimes called Monty Fall). If the host doesn't know where the car is and could have revealed it by bad luck, then observing a goat really is uninformative and the remaining two doors become truly 50/50. The classic 2/3 result depends entirely on the host's guaranteed, knowledgeable goat-reveal.

Is there an easier way to see why 2/3 is right?

Scale it up: with 100 doors and a host who opens 98 known-goat doors after your pick, almost everyone immediately agrees you should switch to the one door the host pointedly left closed. The 3-door version uses the identical logic — it's just subtle enough with only 3 doors that intuition often gets it backwards.

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The Monty Hall Paradox: Why Switching Wins Two-Thirds of the Time