The Monty Hall problem persists in Internet mathematics discussions, as if its results are somehow spectacularly unique or mystifying. Here is the problem: *You are on a game show and are presented with three doors. Behind one door is some wonderful prize, and behind the other two is a goat (or something else of negligible value to you). You choose a door. The game show host, who knows where the prize is, then opens a different door to reveal a goat. You can then choose to take your current door or switch. Do you switch?*

Part of the mystique of the problem is that, apparently, when Marilyn vos Savant solved the problem, several august mathematicians insisted her solution was wrong. And so it has been established as some incredibly difficult, counterintuitive problem.

The reality is, it’s not. I’ve mused in the past about why some mathematicians might struggle with it; the most obvious reasons are that, by framing the problem in terms of a long-term game show, mathematicians think there must be a trick (there is: the rules of the problem aren’t the rules of the show), and that the mathematicians in question weren’t particularly adept at basic probability.

Recently, I saw another problem that demonstrates the counterintuitiveness of the Monty Hall Problem. I’ll change some details to bring it in line with the famous three-doors problem:

*Schrödinger and Einstein buy three bratwurst at a deli. Schrödinger likes his spicy, so he buys one of those, and the other two are mild. Unfortunately, the sausages are unmarked and they look the same, so when the men get home, they can’t tell which is which. Schrödinger randomly takes a sausage and goes off to work. Einstein then takes another sausage and bites into it. What is the probability that Einstein has bitten into the spicy one?*

Einstein is choosing between two sausages, so it seems intuitive to think that he has a 50% chance. But in reality, **because we don’t know what Schrödinger has,** he has a 1/3 chance.

At this point, we’re living in one of six possible universes. Call the sausages S, M1, and M2 (which are spicy, mild, and mild, respectively). These are the possible universes:

- Schrödinger has S, Einstein has M1.
- Schrödinger has S, Einstein has M2.
- Schrödinger has M1, Einstein has S.
- Schrödinger has M1, Einstein has M2.
- Schrödinger has M2, Einstein has S.
- Schrödinger has M2, Einstein has M1.

In four of these six universes, Einstein will be biting into a mild sausage; in two, he’ll be biting into a spicy one. Therefore, he has a 4/6 (that is, a 2/3) chance of having a mild one.

If Schrödinger bites into his sausage first, it will eliminate certain possible universes. If it’s spicy, then Einstein has a 100% chance of having a mild one. If it’s mild, then Einstein has a 50% chance of having a mild one. But until either Einstein or Schrödinger takes that first bite, we can’t eliminate any of the possibilities.

Since Schrödinger is somewhere other than Einstein, that means that Einstein’s chances of having a mild sausage vary based on some distant, unresolved event. One impulse might be to focus on the “distant” aspect (quantum probabilities!), but the key is the “unresolved” portion. Unless Schrödinger calls Einstein during his lunch, Einstein will still believe he has a 2/3 chance of getting the mild sausage regardless.

In general, probability works on the basis of possible universes (or, if that’s too metaphysical, possible outcomes), and the likelihood of each occurring. Each event that happens with a known outcome changes the palette of possible universes. Schrödinger taking a sausage does not change the palette because we don’t know what sort of sausage he took; Schrödinger biting into a sausage *does* change the palette.

But another key is that Schrödinger has a *chance* of taking the spicy sausage. Note that Einstein has either a 50% or a 100% chance of having a mild sausage once we know which one Schrödinger has.

Imagine we were to stop Einstein just before he takes a bite and suggest he switch to the other one. Schrödinger has just called us, having finished his lunch, but we don’t tell Einstein what he said. Should Einstein switch?

Now we’re wandering into Monty Hall territory, but a key is still missing. Because Schrödinger grabbed a sausage at random, Einstein gains nothing by switching. If Schrödinger’s brat is still unknown, Einstein has a 1/3 chance of having the spicy sausage now, and a 1/3 chance of getting it if he switches. His odds are the same. If Schrödinger’s brat is now known, Einstein has either a 100% chance of having a mild sausage, or a 50% chance… either way, he gains nothing by switching.

Even the incongruity between Einstein’s odds if Schrödinger’s choice is known or unknown is only a phantom incongruity. The reality is that we’re stating the same probability in different terms. In the “known” cases, we’re stating the probabilities of B given A. If A is the event that Schrödinger has S and B is the event that Einstein has S, then; \[P(A) = \frac{1}{3}; P(B|A) = \frac{0}{2} \\P(\neg A) = \frac{2}{3}; P(B|\neg A) = \frac{1}{2} \\P(B) = P(A) \cdot P(B|A) + P(\neg A) \cdot P(B|\neg A) \\ = \frac{1}{3} \cdot \frac{0}{2} + \frac{2}{3} \cdot \frac{1}{2} = \frac{0}{6} + \frac{2}{6} = \frac{1}{3}\]

Where the Monty Hall problem differs is that specific universes are deliberately removed from the initial set *after *a choice has been made. Perhaps this is one reason why people struggle with it.

If Schrödinger knew which sausage was which and decided he wanted a mild one today (save the spicy one for tomorrow), he could simply randomly take a mild one. Of the original six options, only these four remain:

- Schrödinger has M1, Einstein has S.
- Schrödinger has M1, Einstein has M2.
- Schrödinger has M2, Einstein has S.
- Schrödinger has M2, Einstein has M1.

… and so Einstein has an even chance of getting the spicy one.

In the Monty Hall problem, it’s the second chooser who is constrained from taking the lone item. So let’s say, instead, that Einstein (being smarter than Schrödinger) knows which one is the spicy one. After Schrödinger chooses and leaves, Einstein deliberately takes a mild one. Once again, we’re removing possibilities, and wind up with:

- Schrödinger has S, Einstein has M1.
- Schrödinger has S, Einstein has M2.
- Schrödinger has M1, Einstein has M2.
- Schrödinger has M2, Einstein has M1.

Now it looks like Schrödinger has an equal shot of having the spicy one. But he doesn’t. Why not?

The subterfuge lies in what the universe of possibilities is at the time the choice is made. In the case where Schrödinger deliberately takes a mild one, the possibility that Einstein takes a spicy one is 50% *at the time that Einstein chooses.* In other words, each of the four remaining possibilities have an equal chance of occurring, both at the initial point and at the time that Einstein chooses (that is, 1/4).

In the case where Schrödinger chooses randomly and then Einstein chooses deliberately, the possibilities don’t have an equal chance of occurring. Einstein changes the possibility set *after* Schrödinger has made his choice, even though he’s removing two possibilities that didn’t *truly* exist from the outset (since Einstein, in any case, wasn’t going to take the spicy one in the first place).

Another way, and I think the proper way, of thinking about this is: What are Einstein’s choices in the last scenario if he’s committed to not taking the spicy bratwurst? They are:

- Schrödinger has S, Einstein has M1/M2 (it’s irrelevant which).
- Schrödinger has M1, Einstein has M2.
- Schrödinger has M2, Einstein has M1.

In other words, Einstein only has a true choice when Schrödinger takes a mild sausage; otherwise, he can choose, but the outcome of that choice is irrelevant.

The Monty Hall problem is a bit of mathematical three card monte: If you go for an intuitive response, you’ll likely get it wrong. If you apply simple probability theory and don’t overthink it, you’ll likely get it right. If you overthink it and break it down, you may well convince yourself that the wrong answer is the right answer after all.