The Monty Hall Problem

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.

3 doorsThe 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.  Below is a link to a simulation of the problem.

[OC] I ran 100 Monte Carlo simulations of the Monty Hall problem. It’s better to switch than stay. from dataisbeautiful

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. It may help to imagine a different scenario with 100 doors. When you first choose one door out of 100, it is very unlikely that you happened to pick the car. Then, based on the door you chose, the host opens 98 other doors, revealing goats. It could be that you chose the car the first time and so the host just opened 98 other doors at random, but it is much more likely that you picked a goat, and the host revealed the other 98 goats, so if you switch, you will get the car.

3 thoughts on “The Monty Hall Problem”

  1. 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.

  2. 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.

    1. I’m not an expert, but I don’t think that the scenario you’ve proposed is comparable to the original. If you had amnesia, you’d lose the information you had previously, so you’d have a 50/50 chance of choosing correctly. But that wouldn’t change the fact that there’s still a one-third chance of the prize being behind the door you originally chose and a two-thirds chance of it being behind the other one, which you would know if you didn’t have amnesia.

      When you picked the first door, you had a one-third chance of guessing correctly. Nothing the host does to any of the other doors changes this, since the prize is never moved – the host opening a different door with no prize does not make it any more likely that the prize is behind your door.

      A different way to think about it is a situation where instead of the host opening a door with no prize, they combined both doors that you didn’t pick into one, so it’s now one door with any prizes from either of the other two behind it. You would want to switch in this case – there are still two doors, but the one you didn’t pick (which was previously two) has two doors worth of chances to win a prize.

      The original Monty Hall problem presents the exact same scenario, but instead of combining two doors, they just eliminate an empty one. This is functionally the same. Your options for the final decision aren’t actually just door A or door B, they are; a) your original door (1/3 chance), or b) not your original door (2/3 chance). Obviously one of the two doors you didn’t originally choose would still have no prize behind it no matter where the prize is, since it can’t be behind two of them – this would be obvious to you before you made your decision. It doesn’t actually matter which it is, you haven’t learned anything new. What the host is doing by opening a door with no prize is essentially the same as combining the two doors, since you’d now be getting the prize if it was originally behind either of them.

      The situation would be different if the host opened a door with no prize behind it completely at random, including your door in their options. But the host won’t ever open your door, regardless of where the prize is. So all they’re doing is distilling both other options into one option, which has no impact on your door’s probability at all.

      Since you would know this, you have a two-thirds chance of winning if you switched. If you had amnesia, you’d have a 50/50 chance of winning regardless of what you picked, but there’s still a two-thirds chance that it’s behind the door you originally picked – you just don’t know that any more, so your choice is random. Even if there’s a 99% chance of the prize being behind one door and a 1% chance of it being behind the other, or if the game was rigged and it was guaranteed to be behind one door, you’d still have a 50% chance of being correct if you didn’t know this and chose at random.

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