You’re on a game show, standing in front of three closed doors. The host urges you to pick *one* door, explaining that if you picked the right one, you would find behind it a coupon for a lifetime supply of coffee^{1}If coffee isn’t your cup of tea, feel free to creatively replace this coupon with a more desirable prize. that’s yours to keep. Open the wrong door and you get nothing.

You consider your options carefully, wondering which door makes the most sense to pick. But the truth is, you have nothing to go by, so a random guess is all you can cough up: you pick *Door A*. The host tells you to go ahead and open the door…but as you reach for the handle, he yells, “Hold on there! I’m going to spice this game up a bit.”

Unlike the hapless guests of the show, the host knows perfectly well which door has the prize behind it. So here’s what he does: he artfully walks around to one of the doors that you *hadn’t* chosen — *Door C* — opens it up, and shows you that there’s nothing behind it. Good thing you hadn’t chosen that one, eh? Phew!

“Now here’s the coffilicious question,” continues the host, “would you like to go ahead with opening *Door A*, or would you like to switch to *Door B*?” It’s a conundrum indeed…what would you do?

#### Bits of Mathematics

The answer to the Monty Hall problem may be somewhat counterintuitive for some, but it’s not difficult to calculate the answer. Let p be the probability that you picked the right door in the beginning (let’s call this event X). It’s easy to see that, given three doors to pick from, the probability of *Door A* being the right door is:

p= P(X) = 1/3

Let Y represent the complement of X, the event that one of the *other* doors has the prize behind it. With q as the probability of this event, it’s again easy to see that:

q = P(Y) = 1 - p = 2/3

Later on, the host reveals that there’s nothing behind *Door C*. This door isn’t chosen at random — the host knows perfectly well what’s behind each door, and he picks one that is guaranteed to be a dud. If *Door C* had the prize behind it, the host would have chosen to eliminate *Door B* instead. If *Door A* had the prize behind it, the host would be able to freely pick either *Door B* or *Door C* to eliminate. Of course, *you* have no idea how the host is making these choices, so from your perspective once *Door C* is eliminated, there’s some non-zero chance of the prize being behind either *Door A* or *Door B*.

##### A Red Herring

If you were to *now* pick one of these two doors at random (let’s call this event W), the probability w of the new pick being the right door would be:

w = P(W) = 1/2

At this point, you might think it makes no difference whether you switch or not, as both doors are equally likely to have the prize behind it — *but you would be wrong*. Arguably this is where many people get tripped up.

The reason is that we don’t really care about w. Why? Because you’re *not* randomly picking one of *Door A* or *Door B*. Rather, you’re trying to decide whether or not your *original* choice still makes sense to go with. Or to phrase it differently, we want to determine how the likelihood of your original choice being right stacks up against the new information available at your disposal that *Door C* doesn’t have anything behind it.

Here’s a quick and simple way to understand this. The likelihood p of your original choice being right still remains unchanged at 1/3, but whereas event Y originally represented either of *Door B* or *Door C* having the prize (with an equal likelihood of 1/3 each), the same event now represents just *Door B* having the prize, with the same total probability of 2/3.

In other words, if you stick to your original choice, you have a 1/3 chance of being right. If you decide to switch, you have a 2/3 chance of being right. And that’s why it makes complete mathematical sense to switch!

#### A Broader Perspective

If the mathematics makes sense but the solution does not quite “feel right”, here’s a handy tip. Try extending the puzzle to many more doors. For instance, suppose we were dealing with a hundred doors, of which you picked one, just as you did with the three doors earlier. Of the remaining ninety-nine, the host then jumps up and eliminates all but one. Would you have faith that you’d picked the right door out of hundred? Or would you be more likely to believe that the *one* door that the host cunningly failed to eliminate ultimately held the prize?

If you’re interested in an empirical answer to this question, try the simulator I’ve made available on GitHub.

-------------------------------------------------- Monty Hall Simulator --------------------------------------------------Monty Hall Problem-------------------------------------------------- Simulation with 3 doors, 1000000 iterations... Winning likelihood: → Original choice = 0.333568 → If you switched = 0.666432

That’s all for today, folks! 🖖