Monty Hall, But Make It Physics – Part 1 of 4

A short series that builds intuition for quantum mechanics (without asking you to become a monk)

The Monty Hall Problem – Series map:

  1. Post 1: Geometry, vectors, and the death of “50/50”
  2. Post 2: Information theory and the Bayesian perspective
  3. Post 3: Quantum mechanics and the power of projection
  4. Post 4: Wolfram’s multiway systems (and the Ruliad) in toy form

Post 1 — Geometry, Vectors, and the Death of “50/50”

  1. You’re on a game show. There are three rooms (or doors).
  2. Two hide goats. One hides a car.
  3. You pick one room.

Monty — who knows where the car is — opens one of the other rooms and shows you a goat. Then he smiles and asks:

Do you want to switch?

Most people feel the gravitational pull of a simple thought:

Two rooms left, one car… so it’s 50/50.

That thought is the bug.

Today we’re going to build the cleanest intuition I know, starting from one humble object:

the 45–45–90 triangle
(the geometry of splitting a unit amount into two perpendicular components)

By the end, you won’t just “know the answer.” You’ll see why switching is a kind of
probability concentration — like a fog bank collapsing into a single valley.

A promise (because readers are not psychic)

If I borrow a physics word, I will define it immediately in plain English.

In other words: you won’t be punished for being human.

Section 0 — The Monty Hall setup (the whole story in one breath)

The only “weird” thing here is Monty’s behavior.

That’s the entire paradox: a tiny rule-based measurement can beat a giant human intuition.

Keywords (the minimum vocabulary to stay sane)

If that last line feels like a lot: good. It’s the only line that matters.

Section 1 — The one distinction that explains everything

There are two universes people accidentally mix together.

  1. Universe A: Independent reveal (not Monty Hall)
    • If a door is opened at random regardless of where the car is, the reveal is independent of the hidden state.
    • In that universe, deleting one option can genuinely make things feel like 50/50 — because nothing “intelligent” happened.
  2. Universe B: Dependent reveal (actual Monty Hall)
    • Monty does not open a door at random. He follows a protocol:
      • He never opens your chosen door.
      • He never opens the car door.

So the reveal is dependent on the hidden state.

And here’s the whole moral, said plainly:

When evidence depends on the hidden truth, it reweights the remaining possibilities.

Now let’s turn “reweights” into something you can feel.

Section 1.5 — The sanity check (how to feel correlation in your bones)

Let’s stop pretending your brain is broken

It’s doing what it was designed to do: compress a situation into a snapshot and make a fast call.

Monty Hall punishes that instinct, because the snapshot (“two doors left”) does not contain what you need. It hides the valves that produced the snapshot.

Think of probability as conserved fluid in a three-chamber system.

At the start, the fluid is evenly distributed:

DoorProbability
$$\text Door 1$$$$\frac{1}{3}$$
$$\text Door 2$$$$\frac{1}{3}$$
$$\text Door 3$$$$\frac{1}{3}$$

Now Monty enters wearing a lab coat and a perfectly calm smile.

He is not “a guy opening a door.” He is a measurement apparatus: a machine with a rule book that routes the fluid.

And correlation means:

The valve position depends on the hidden truth.

No information yet, so your belief-fluid is uniform:

$$
P(A)=P(B)=P(C)=\frac{1}{3}
$$

Now you run a “diagnostic test” that is aggressively Monty-ish:

You get:

“B is not the problem.”

Your brain wants to do this:

Cool, B is eliminated. A or C remain. 50/50.

That’s the snapshot mistake.

A serious engineer asks:

Does this test’s output depend on the true cause?

Here’s the rule:

So:

$$
P(\text{“not B”}\mid A)=\frac{1}{2},\quad
P(\text{“not B”}\mid B)=0,\quad
P(\text{“not B”}\mid C)=1
$$

Now do the only law that matters:

Posterior $$\propto$$ likelihood × prior, then renormalize.

Unnormalized:

$$\text A$$$$\frac{1}{3}\cdot\frac{1}{2}=\frac{1}{6}$$
$$\text C$$$$\frac{1}{3}\cdot 1=\frac{1}{3}$$
$$\text Normalize (sum)$$ $$\frac{1}{2}$$
$$P(A\mid \text{“not B”})$$$$\frac{1}{3}$$
$$P(C\mid \text{“not B”})$$$$\frac{2}{3}$$

Not 50/50.

Because the evidence was generated by a rule that depends on the truth.

Translation back to Monty Hall

The moral is clean enough to engrave on steel:

Monty does not.
Monty is lawful.
Law creates bias.
Bias creates advantage.

Section 2 — Reduce the story to state space (tuples → vectors)

Think of the game as a three-chamber system.
The car’s location is the hidden plumbing state.
Your beliefs are conserved probability fluid being routed by rules.

Step 1: Enumerate the state space (be exhaustively boring)

Before any drama, exactly one room has a car:

WorldRoom 1Room 2Room 3
1CarGoatGoat
2GoatCarGoat
3GoatGoatCar

So we adopt the cleanest encoding:

Then each possible world becomes a vector (a one-hot basis vector):

WorldVector
World 1$$\vec{W}_1=(1,0,0)$$
World 2$$\vec{W}_2=(0,1,0)$$
World 3$$\vec{W}_3=(0,0,1)$$

Step 3: Your choice is also a vector

If you pick Room 1, define your choice vector:

$$
\vec{C}=(1,0,0)
$$

Section 3 — What is a state vector? Explain it to God (briefly, without flinching)

A vector is a geometric object you can add, scale, and project onto axes (coordinate directions).

When you represent states as vectors, you gain superpowers:

Here, “measurement” means: a rule that turns a hidden world into an observation, which lets you discard incompatible worlds and rescale what remains.

In the metaphor:

Monty Hall is secretly a measurement problem. And measurement problems love vectors.

Section 4 — Your first choice creates two subspaces: 1/3 vs 2/3

Suppose you pick Room 1.

Before anything is opened:

So your choice splits the universe into two chunks.
If you like the fluid picture: you just created two reservoirs.

Here, “subspace” just means: a bucket of possible worlds grouped together.

Section 5 — Probability concentration (projection + renormalization)

Caption: Your first choice isolates $$\frac{1}{3}$$. The other two rooms hold
$$\frac{2}{3}$$ as one reservoir. Monty’s reveal doesn’t split $$\frac{2}{3}$$ into
$$\frac{1}{3}+\frac{1}{3}$$; it routes the full $$\frac{2}{3}$$ onto the single remaining alternative.

Here’s what your brain tries to do incorrectly:

  1. “Two rooms left ⇒ split probability evenly.”
  2. But probability doesn’t split evenly just because two labels remain.
  3. Probability splits evenly only if the evidence treats the remaining options symmetrically.

If you picked Room 1, here’s the entire asymmetry in one table:

True world (car is in…)Monty can open…Forced or free?If you switch, you go to…
Room 1Room 2 or Room 3 (both goats)Free (2 choices)A goat (you lose)
Room 2Room 3 (must avoid the car + your door)Forced (1 choice)Room 2 (you win)
Room 3Room 2 (must avoid the car + your door)Forced (1 choice)Room 3 (you win)

So when Monty opens a goat door, he’s not “removing a random door.”
He’s performing a rule-based reveal that pushes probability weight around.

In the fluid picture: he’s opening a valve under constraints.

In plain English: you threw away the worlds that can’t match what Monty just showed you, then you rescaled the remaining probabilities so they add up to 1 again.

Section 6 — The 45–45–90 triangle: the geometry of “50/50”

Now we build the most robust intuition tool in probability: what “50/50” actually means.

The Triangle

A 45–45–90 triangle with legs equal corresponds to “equal projection” onto two perpendicular axes (at right angles).

In vector form, a unit vector (an arrow of length 1) at angle θ has components:

$$
(\cos\theta,\ \sin\theta)
$$

If you interpret those components as “weights,” then a 45–45–90 triangle is the special case where the weights are equal.

At $$\theta=45^\circ$$:

$$
\cos\theta=\sin\theta=\frac{1}{\sqrt{2}}
$$

So a 45–45–90 triangle is a geometric picture of “equal share.”

What it is not a picture of:

“Two options exist.”

Two options can exist and still be weighted $$\frac{1}{3}$$ vs $$\frac{2}{3}$$.
The geometry decides the weights. Not the count of labels.

Section 7 — Put the game in 3D: axes as rooms, planes as “walls”

Think of a 3D coordinate system:

This idea will become crucial later when we talk about “not lining up” with the information the game gives you.

Section 8 — A matrix view (optional, but clarifying)

This representation is almost offensively simple: three basis vectors, stacked into the identity matrix. And yet it already contains real structure. The identity matrix is perfectly symmetric—transpose it, nothing changes. Even if you rotate it as a geometric pattern by 180∘180^\circ, it returns to itself. That’s the quiet punchline: before any choices or reveals, the game treats all doors the same. The asymmetry isn’t “in the doors.” It’s injected later by a rule—your choice plus Monty’s constrained reveal. We haven’t said “group theory,” we haven’t touched trigonometry, we haven’t invoked complex numbers—and we’re already staring at invariance under transformations. The whole series is basically: start with symmetry, then watch measurement break it in a lawful way.

If we use the Car = 1 encoding from above, the three possible world vectors are:

Here are the three Monty Hall “world vectors” in goat-indicator encoding:

W⃗1=(011),W⃗2=(101),W⃗3=(110)\vec{W}_1=\begin{pmatrix}0\\1\\1\end{pmatrix},\qquad \vec{W}_2=\begin{pmatrix}1\\0\\1\end{pmatrix},\qquad \vec{W}_3=\begin{pmatrix}1\\1\\0\end{pmatrix}

We’re using the encoding 1 = goat, 0 = car, so each world has two goats, hence two 1s and one 0. Each vector “ignores” exactly one axis — the axis of the door containing the car.

Stack those vectors into a matrix:A=(011101110)A= \begin{pmatrix} 0 & 1 & 1\\ 1 & 0 & 1\\ 1 & 1 & 0 \end{pmatrix}

This is a hollow symmetric matrix:

The clean interpretation: the diagonal entries represent “self-links,” and here they’re all zero — no door links to itself. In graph terms, AA is the adjacency matrix of the simple undirected graph on 3 nodes where every node is connected to the other two (the complete graph K3K_3​ with no self-loops).

And here’s the philosophical punchline: this is an absurdly small representation. Three vectors, nine bits of structure—yet it already contains symmetry, invariance under transpose, and even a visual 180° rotational invariance (rotate the matrix twice!). We haven’t touched group theory, trig, or complex numbers… and the geometry is already awake. The only asymmetry comes later, when measurement (Monty’s constrained reveal) and your first choice act like valves that reroute the probability flow.

Section 9 — Dot products: “what do I experience from my axis?”

Here’s the geometry move.

$$
\vec{C}=(1,0,0)
$$

Each possible world (where the car is) is also a vector, for example:

$$
\vec{W}_1=(1,0,0),\quad \vec{W}_2=(0,1,0),\quad \vec{W}_3=(0,0,1)
$$

Now we can use the dot product as a simple “alignment detector”:

Dot (Inner) Product: $$\vec{C}\cdot\vec{W}$$
If your choice matches the car location (you win).$$\vec{C}\cdot\vec{W}=1$$
If your choice does not match (you lose). $$\vec{C}\cdot\vec{W}=0$$

If you picked Room 2 instead, your choice vector would be $$\vec{C}=(0,1,0)$$

Section 10 — The punchline, finally stated as geometry

When you pick a door, you create:

So after Monty opens:

That’s it.

Switching wins because you choose the door that inherited the mass of the larger subspace.

Heuristics (serious rules for weird worlds)