The Door You Should Not Keep

Three doors are closed. One hides a car and the other two hide goats, and the arrangement was fixed before you arrived. You point at a door without opening it. The host, who knows where the car is, then opens one of the two doors you did not choose and shows you a goat behind it. Two doors remain closed: the one you pointed at and one other. You may keep your door or move to the other one.

The stage shows one such game, finished. Step replays it from your first pick: point, open, decide, reveal. Your door keeps its outline the whole way through, so you can see which door was yours after the doors have been opened.

The tempting reading is that two doors are closed, so each is worth one half. That reading treats the two doors as though they arrived in the same way, and they did not. You chose one of them, blind, with a one in three chance of being right. The host chose the other one, having looked, under a rule that forbids opening the car.

So ask what each door had to survive. Your door survived nothing: the host was never going to open it. The other closed door survived a filter. If the car was behind either of the two doors you did not pick, the host was forced to leave the car closed and open the goat. That door is carrying the whole two thirds of the time your first guess was wrong.

Here is the sentence the rest of the page is about. Against a knowing host, switching wins if and only if your original door was not the car. There is no third case. If you pointed at the car, the door left closed is a goat and switching loses. If you pointed at a goat, the host had to leave the car closed and switching wins.

Your first pick was wrong two times in three, because you were choosing one door out of three with no information. So switching wins two times in three. The plate above is that argument written out as a list rather than as an equation, and the list is complete: there are exactly three places the car can be.

The hundred-door setting in the laboratory is the same rule at a scale where it stops feeling strange. You point at one door, the host opens ninety-eight goats, and one door is left. Your door is still the one in a hundred you guessed at. The other one is everything else.

Let the doors be numbered $1$ to $n$. The car is placed uniformly at random, and independently you choose a door uniformly at random. The host then opens every door other than yours and one other door, never opening the car; where the rule leaves a free choice, which happens exactly when your door is the car, the host chooses uniformly among the permitted doors.

Write $W$ for the event that your first pick is the car, so $P(W)=1/n$. Sticking wins on $W$ and switching wins on its complement, giving

At $n=3$ these are $1/3$ and $2/3$. The laboratory's kernel walks the whole case space of car position, first pick, host choice and switcher's choice for every door count it can hold, and counts the wins; the counts and the fractions agree exactly.

The case space here is finite and small, so the argument can be completed rather than approximated. Fix $n=3$ and let your first pick be door 1; the labelling costs nothing, since the three doors are interchangeable before anything is opened.

Three equally likely arrangements exhaust the possibilities. If the car is behind door 1, the host may open door 2 or door 3 and the door left closed is a goat, so sticking wins and switching loses. If the car is behind door 2, the host must open door 3, and switching reaches the car. If the car is behind door 3, the host must open door 2, and switching reaches the car again. Two of the three arrangements are switching wins.

That is an exhaustive check of a finite space, which is what a proof of this statement looks like. It is worth being precise about why: the space of arrangements really is finite, every case in it really was checked, and the three cases really are equally likely under the stated model. When any of those three things fails, a finite computation is evidence and nothing more, and other pages on this site are careful to say so. Here it is a proof, and the page says that instead.

The general $n$ follows the same way. Switching wins precisely on the complement of the event that your first pick was the car, an event of probability $1/n$, so switching wins with probability $(n-1)/n$.

Change one thing and the answer changes. Set the host to open a door at random, keep everything else, and look only at the games where the host happened to open a goat. Sticking and switching now win half the time each. The car has not moved, the doors look identical, and your first pick is still one in three. What went is the host's knowledge, and with it the guarantee that a surviving closed door survived a filter.

That is the sharpest thing this page has to say, and it is why the failure preset is on the page rather than in a footnote. The two-thirds advantage is not a property of doors, or of having two options, or of anything you can see on the stage. It is made of what the host knows and is obliged to do with it. An ignorant host who misses the car tells you that one particular door was a goat and nothing else, and both closed doors are then worth one half.

The race view is scenery for all of this rather than evidence of it. It plays thousands of seeded games and draws the running win rates against the exact values, which are drawn as horizontal rules because they were counted, not fitted. If the lines settle near the rules, that is a check that the simulation implements the rule the enumeration describes. It is not why the answer is two thirds, and no number of trials would make it so.

The genuinely open questions here are not mathematical. Why the answer is so widely rejected on first hearing, and what a person is actually assuming when they insist on one half, are questions about reasoning rather than about probability, and this page does not settle them.