I first came across this riddle in Episode 7 of Brooklyn Nine-Nine Season 4. In the episode, the precinct’s captain, Raymond Holt, is in a dispute with his husband, Kevin, over the correct answer to the riddle below:

Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice?

— Parade magazine, 1990

Despite the absence of goats being behind 2 out of 3 doors in the TV show’s explanation of the problem, for the sake of understandability (yes, that is a word) and staying true to the original explanation, from this point we will use goats as the alternative to revealing a car behind a door.

The unequivocally correct choice is to switch doors and choose the door that wasn’t revealed by the host (whose name is Monty, hence the riddle’s title), which has a \(\bfrac{2}{3}\) chance of having a car behind it, therefore leaving the original door (which the contestant intended to open) having the remaining \(\bfrac{1}{3}\) chance.

At first, one might incorrectly assume (I know I certainly considered it) that of the two remaining doors (since the host in effect removes one of the choices) the probability is evenly split between two. This is simply **incorrect**.

Let’s say the contestant chooses Door 1. The host cannot reveal a car and therefore must choose to reveal the door with the goat. As there is only one car, no matter which door it is behind, the host can always safely reveal a goat. For the entirety of the game, Door 1 has \(\bfrac{1}{3}\) chance of revealing a car (and a \(\bfrac{2}{3}\) chance of revealing a goat). Focusing now on just cars, the first door has a \(\bfrac{1}{3}\) of revealing it, leaving the remaining two to have a \(\bfrac{2}{3}\) chance of one of the doors (any door) revealing the car. When the host reveals the goat in let’s say Door 2, we only gain more information. The probability does not change, which is crucial to understanding the probability outcome. Whilst the contestant cannot choose Door 2 (as it has been revealed), it still has a goat behind it and therefore the denominator (3) of the fraction representing the probability must remain constant, and cannot change to 2, thereby creating that incorrectly first assumed 50-50 chance (\(\bfrac{1}{2}\)). When the goat is revealed behind Door 2, Door 3 automatically has the probability of containing the car as the probabilities of \(\bfrac{1}{3}\) and \(\bfrac{1}{3}\) (for each respective door) has changed to \(\bfrac{0}{3}\) and \(\bfrac{2}{3}\). The simple act of gaining more information does not alter the probability.

If that’s too confusing, let’s look at the problem another way. Let’s propose a simple change: instead of revealing one of the two remaining doors, the host instead allows the contestant to either stay with Door 1, or open both Door 2 and Door 3 and if one has a car behind it, they may have the car. This relates to Doors 2 & 3 having the combined probability of \(\bfrac{2}{3}\) of one of the doors containing a car. Again, whether the host reveals one fo the remaining doors in the original scenario, or the contestant effectively does when they smartly choose Doors 2 & 3 in our new scenario, they only gain new information and the probability is maintained.

It is important to remember that the whole notion of having a percentage chance of a certain probability is tethered to scenario needing to be carried out a number of times. Specifically in this scenario and in a perfect world the scenario (or game) needs to be played out \(n = 3\) times, where n is the number of doors. Let’s use a coin to illustrate this idea. Let’s flip it once and say that the result was tails. For one trial, we would be somewhat correct to say that it has a 100% chance of landing tails up. It would be more correct to say that for that particular trial in this particular circumstance, the result of one flip was tails. We of course know the probability of landing tails up is 50-50 (assuming a perfectly evenly weighted coin, ignoring air resistance, etc), however this is only confirmed over a larger amount of trials. For 10 trials, we may have 4 tails and 6 heads tosses, however for a higher amount of trials, the probability approaches a perfect \(\bfrac{1}{2}\) chance of landing on either side.

Going back to the riddle at hand, consider the following table which represents the 3 different scenarios (regarding the placement of the car) and the outcome of the game, should the contestant originally choose Door 1 and stick with it.

Door 1 | Door 2 | Door 3 | Result |
---|---|---|---|

Car |
Goat | Goat | Wins car |

Goat | Car |
Goat | Wins goat |

Goat | Goat | Car |
Wins goat |

The final explanation I will propose is if there were a larger amount of doors. For argument’s sake, let’s say there are 1,000,000 doors. The contestant chooses Door 1 and the host reveals 999,998 of them, leaving Door 314,159 and Door 1. Since the host again may only reveal goats, there is an extremely high chance that Door 314,159 is the one with a car behind it and that Door 1 is the incorrect choice. Thinking from a different point of view, the contestant may consider how likely they were to correctly choose the correct door on their first try.

Still don’t get it? Have a look at the very comprehensive Wikipedia article for the problem.