The Monty Hall problem is one of the greatest brain teasers of all time. The correct answer is so counterintuitive that our brains want to reject it even after it is explained.

The problem was popularized by Marilyn vos Savant in a series of columns in Parade magazine in 1990 and 1991. She published the problem with the correct answer in the first column, only to have several mathematicians write in to tell her she was wrong, with many using a scornful tone and being sure to sign their names with “Ph.D.” appended. Their credentials did not make them any righter, and Savant patiently re-explained the correct answer in subsequent columns, finally calling on math classes across the country to actually perform the actions described in the brain teaser. Elementary school students were able to demonstrate that the mathematicians were incorrect. Although Savant referred only to a nameless game show, the scenario calls to mind the classic game show Let’s Make a Deal, as hosted by Monty Hall, and the puzzle came to be named after him.

Here is the scenario: You are on a game show, and the host directs you to pick one of three doors. You will win whatever is behind the door you pick. Behind one door is a shiny new car, and behind the other two doors are billy goats. You pick a door. However, the host, who knows what is behind each of the doors, purposely chooses a door that is not the door you picked and that has a billy goat behind it, and opens the door, revealing the goat. Now there are two doors remaining: your original choice, and the other unopened door. The host now asks if you would like to switch and choose the other door. Should you?

**Answer and Explanation**

Nearly everyone’s initial reaction is that it makes no difference whether you switch or not. There are two doors remaining, so the odds are fifty-fifty, or 1/2. However, this is incorrect. It is actually always in your best interest to switch.

The problem concerns conditional probability, and this means that in order to understand it, we must examine not just the result of two doors remaining, but the process of how we got there. Crucially, what you do affects what the host does, and that affects your odds of winning. One way to understand the odds is to think through all of the possible outcomes.

Let’s say that Door Number One is your initial choice. There are obviously three possibilities for where the car could be, so let’s examine them.

If the car is behind Door Number One, then Monty will reveal a goat behind either of the other two doors. If you stay, you win. If you switch, you lose.

If the car is behind Door Number Two, then when you choose One, Monty is forced to reveal the goat behind Three. If you stay, you lose. If you switch, you win.

If the car is behind Door Number Three, then when you choose One, Monty is forced to reveal the goat behind Two. If you stay, you lose. If you switch, you win.

So, switching results in a win two out of three times, and those are your actual odds if you switch: 2/3.

Note that your original odds of winning were 1/3, and that does not change: sticking with your original choice gives you a 1/3 chance of winning. But crucially, there is a 2/3 chance that your initial choice was incorrect, and in both those scenarios, the host is in essence being forced to tell you where the car is.

If you are not convinced, there are various ways of running a real-world experiment with a partner in order to demonstrate the odds. Even after an explanation or a demonstration, it can be difficult to understand that choosing between the two remaining doors is not in fact a fifty-fifty chance.

There actually are four cases, not three: (1A) The car is behind #1, and the host opens #2, (1B) The car is behind #1, and the host opens #3, (2) The car is behind #2, and the host opens #3, and (3) The car is behind #3, and the host opens #2.

The point of a conditional probability problem, which was ignored by both Marilyn vos Savant and those who criticised her, is to assign initial probabilities to all of the possibilities, and removing what is inconsistent with what is observed. In this problem, that means (1A) 1/6, (1B) 1/6, (2) 1/3, and (3) 1/3. When the host opens a door, two of these case are eliminated, leaving either (1A) and (3), or (1B) and (2).

Either way, there was a 1/6 chance of reaching the current state with the car behind #1, and a 1/3 chance if a goat is behind #1. You multiply these by 2 to make the sum of all possibilities are up to 1, making it a 2/3 chance that switching wins. And if the split between (1A) and (1B) is not even, the answer is not 2/3. What the above solution calculates, is the chance of winning if you must decide to switch before the host opens a door. And that’s a different problem.

Marilyn vos Savant has never provided an actual solution. Her first column was just a different way to use intuition (“If there were 1,000 doors, and the host opened 998, you’d switch pretty quick”). I won’t defend the answer provided by many (but not all, or as many as she implied) with PhDs, but they were right to call her out on this.

I despise this problem.

The mere act of making the new choice is, agreeably, a 50/50 choice. Why does that require a switch?

Imagine, for example, you choose door #2. The host opens door #1 to reveal a goat. Suddenly, a loose light falls on your head and you get knocked out. When you wake up, you have short term amnesia and can’t remember which door you chose in the first round (besides the obvious fact that it was not door #1).

That is to say, you forget the initial choice entirely and now, in choosing between door #2 and door #3, you happen to choose door #2. It’s not a choice to “stay” because both options are on the table; it’s making an entirely new choice (seemingly to you with amnesia) and choosing door #2.

With that logic in mind, what if you didn’t get amnesia? What if you merely set aside the previous decision, and create a whole new choice between door #2 and 3, and you choose #2 yet again? How is that different? Switching is unnecessary.