When a probability table becomes a room

Miraia S. Chiou

Hilbert space · Quantum · Probability · Geometry

When a probability table becomes a room

Give each possible answer a direction, cast an arrow's shadow, and watch Hilbert space produce familiar beliefs. Then discover a prior that fits one question but cannot fit them all.

2026-10-29 · 14 min read

A friend might spend the weekend at the coast, in a forest, or in the mountains. You give the three destinations equal chances. Three numbers, each one third, seem to finish the story.

Now imagine a room with one direction for each destination. An arrow stretches through that room. Its shadow along each direction tells you the chance of that answer.

The room sounds extravagant. Why build geometry around three numbers?

Because you can do things to a room that a list of numbers barely hints at. You can group directions into larger spaces, turn the questions, and choose the arrow at random. Those operations connect ordinary probability, trial weights, and quantum state space. Some familiar statistical rules emerge as consequences of the room’s geometry. Others meet a surprisingly solid wall.

First, take the square roots

Let the three probabilities be $p_{1}, p_{2}, p_{3}$ . Each is nonnegative, and their sum is one. Form an arrow with components

ψ = (\sqrt{p_{1}}, \sqrt{p_{2}}, \sqrt{p_{3}}) .

The Greek letter $ψ$ , pronounced “psi,” names the arrow. Its squared length is

‖ ψ ‖^{2} = p_{1} + p_{2} + p_{3} = 1.

The double bars mean length. Our probability table has become a unit vector, an arrow of length one. For three equal chances it points along $(1, 1, 1) / \sqrt{3}$ .

A Hilbert space is a vector space equipped with an inner product, a rule for calculating lengths and angles, together with a completeness condition: every Cauchy sequence, whose vectors eventually become arbitrarily close to one another, converges to a vector in the space. Ordinary finite-dimensional Euclidean space is an example. Quantum mechanics usually uses complex coordinates, which can carry a phase as well as a magnitude.

A projection takes the part of an arrow that lies in a chosen direction or subspace. A subspace can contain several directions, much as a floor contains two directions inside a three-dimensional room. If $P_{k}$ projects onto the directions assigned to outcome $k$ , the corresponding share is

p_{k} = ‖ P_{k} ψ ‖^{2} .

Square the shadow’s length. That is the Born rule when the vector describes a quantum pure state and the projection describes an outcome. Here it also supplies an exact way to represent a probability table.

There are two different layers to keep in view:

Object	What it describes
One unit vector $ψ$	One probability table for the chosen questions
A distribution over vectors	Uncertainty about which probability table applies

Choosing a vector at random will take us to the second layer.

Give one answer more room

Suppose we build a six-dimensional complex Hilbert space. The coast gets one direction, the forest two, and the mountains three. The dimensions assigned to them are

(r_{1}, r_{2}, r_{3}) = (1, 2, 3), D = r_{1} + r_{2} + r_{3} = 6.

Here $D$ is the room’s total dimension. We cannot draw all six directions on a page, but their arithmetic is no harder than the three-dimensional case.

Choose a unit vector uniformly, meaning that its distribution favors no direction. In a complex space this is the rotation-invariant distribution often called Haar random. Project onto the three groups and square the lengths. The resulting shares obey

(p_{1}, p_{2}, p_{3}) \sim Dirichlet (r_{1}, r_{2}, r_{3}) .

The symbol $\sim$ means “has this distribution.” A Dirichlet distribution describes random nonnegative shares that sum to one. Its parameters act as prior trial weights, also called pseudo-counts. In this construction each complex direction contributes exactly one such weight.

The mean shares are the fractions of the room allotted to the answers:

E [p_{k}] = \frac{r_{k}}{D} .

The symbol $E$ means an average over the randomly chosen vectors. Our destinations therefore have mean shares $1 / 6$ , $2 / 6$ , and $3 / 6$ .

Predict before playing: if every destination gets four directions instead of one, will the mean shares change? Will the random tables spread out further, or gather nearer equal thirds?

From random directions to random probabilities

Give each answer room

Imagine three possible destinations. Give each one some directions in an abstract room, then choose a unit vector at random. Its squared lengths inside those three groups give one probability table. Every dot below is a different possible table.

Build the three groups

Coast1 direction

Trial weight 1 · Mean share 33.3%

Forest1 direction

Trial weight 1 · Mean share 33.3%

Mountains1 direction

Trial weight 1 · Mean share 33.3%

Recording an outcome adds one trial weight: one complex direction, or two real directions. This represents the updated distribution of shares. It does not describe a physical system growing new dimensions.

A triangle of probability tables

Possible tableExact mean (ring)Selected table

Merge Forest and Mountains into “Inland”

The shares follow Dirichlet(1, 1, 1), with total trial weight 3. One complex direction per answer spreads the tables uniformly across the triangle; increasing all three counts gathers them nearer its center.

A seeded mathematical simulation, using independent Gaussian coordinates and normalization. The dots show the distribution of possible shares, not 240 observed destination choices. The triangle is share space, not a drawing of the full Hilbert space.

Open the room of possibilities in the playground.

The means stay at one third when all three groups grow equally. The spread decreases. More dimensions give each group’s total squared length more contributions to average together.

For a Dirichlet prior with weights $a_{k}$ and total $A = \sum_{k} a_{k}$ , the exact variance is

Var (p_{k}) = \frac{μ_{k} (1 - μ_{k})}{A + 1}, μ_{k} = \frac{a_{k}}{A} .

Variance measures spread around the mean. Here $a_{k} = r_{k}$ and $A = D$ . The geometric construction reproduces the familiar distinction between a belief’s mean and the trial amount that resists changing it.

Using real coordinates changes the result. Each real direction contributes half a trial weight:

\begin{aligned} complex room: & (p_{1}, p_{2}, p_{3}) \sim Dirichlet (r_{1}, r_{2}, r_{3}), \\ real room: & (p_{1}, p_{2}, p_{3}) \sim Dirichlet (r_{1} / 2, r_{2} / 2, r_{3} / 2) . \end{aligned}

You can see the reason by constructing a random arrow from independent Gaussian coordinates. A real coordinate supplies one squared Gaussian contribution; a complex coordinate has a real and an imaginary part, supplying two. Normalizing by their total produces the respective Dirichlet shares. These random-state distributions are developed in Życzkowski and Sommers’ study of induced measures.

Merge a destination, preserve its share

Suppose we only care whether the trip is coastal or inland. Forest and mountains become one answer.

Their projections occupy perpendicular subspaces. Consequently, their squared lengths add:

p_{i n l a n d} = p_{f o r e s t} + p_{m o u n t a i n s} .

Their dimensions add too:

r_{i n l a n d} = r_{f o r e s t} + r_{m o u n t a i n s} .

In our six-dimensional example, the three-way distribution $Dirichlet (1, 2, 3)$ becomes the two-way distribution $Dirichlet (1, 5)$ , or equivalently a $Beta (1, 5)$ distribution for the coastal share. A Beta distribution is the two-answer member of the Dirichlet family.

The demo’s merge switch keeps the selected vector fixed. It simply groups two shadows. This makes EP’s refinement rule tangible: a broad answer keeps the same weight when its more detailed names are joined back together.

Operation	Probability language	Hilbert-space language
Name an answer	Choose a category	Choose an outcome subspace
Merge answers	Add their shares and trial weights	Join their perpendicular subspaces
Predict an answer	Divide its trial weight by the total	Take its fraction of the total dimension
Record one outcome	Add one to its Dirichlet parameter	Represent the updated share law with one extra complex direction

The last row has a specific meaning. Starting from $Dirichlet (1, 2, 3)$ , observing “coast” changes the distribution of shares to $Dirichlet (2, 2, 3)$ . A fresh uniformly random vector in a room with grouped dimensions $(2, 2, 3)$ represents that posterior distribution exactly. This is an equivalence between distributions of shares; it does not say a physical measurement creates another spatial direction, or that Bayesian parameter learning is the same operation as quantum measurement collapse.

For real coordinates, adding one trial weight takes two new directions. Try “Record this outcome” in each mode and watch the bookkeeping agree.

A weighted average becomes an expectation value

Give the three destinations travel times of 10, 20, and 50 minutes. For a sampled probability table, the expected travel time is

M = 10 p_{1} + 20 p_{2} + 50 p_{3} .

Now define a linear operator $Q$ that multiplies each coast direction by 10, each forest direction by 20, and each mountain direction by 50. The expectation value of this operator in state $ψ$ is

⟨ ψ, Q ψ ⟩ = 10 ‖ P_{1} ψ ‖^{2} + 20 ‖ P_{2} ψ ‖^{2} + 50 ‖ P_{3} ψ ‖^{2} = M .

The angle brackets denote the inner product. The two sides calculate exactly the same number. One side reads a weighted average; the other reads an operator through an arrow.

When every recorded value gets one complex direction, choosing the arrow uniformly gives $Dirichlet (1, \dots, 1)$ weights. This is the weight law of Rubin’s Bayesian bootstrap, which represents uncertainty by assigning random normalized weights to recorded values. With other Dirichlet weights it becomes the corresponding weighted version. In matrix geometry, the distribution of these expectation values is called a numerical shadow; Dunkl and colleagues study that construction.

Every positive weighted average of 10, 20, and 50 stays between 10 and 50. Turning it into Hilbert-space notation preserves that boundary. The question of how to represent a population mean beyond the observed range needs another modeling step, explored in Could the true average lie beyond everything you measured?.

Turning the question reveals what a table omits

So far we have kept the outcome subspaces fixed. Complex Hilbert space also allows questions whose directions cut across the original ones.

Take two unit vectors:

ψ_{+} = \frac{(1, 1)}{\sqrt{2}}, ψ_{-} = \frac{(1, - 1)}{\sqrt{2}} .

Their squared coordinate magnitudes are identical: both give $(1 / 2, 1 / 2)$ in the original basis. The sign difference is a relative phase of half a turn.

Now ask whether the state lies along the diagonal direction $u = (1, 1) / \sqrt{2}$ :

\begin{aligned} | ⟨ u, ψ_{+} ⟩ |^{2} & = 1, \\ | ⟨ u, ψ_{-} ⟩ |^{2} & = 0. \end{aligned}

The same original table leads to opposite answers under the new question. A fixed table of squared magnitudes does not retain the relative phase.

The projection operators for an original coordinate and the diagonal direction do not commute: applying them in opposite orders can change the resulting vector. Such questions are called incompatible. That extra freedom will matter when we try to assign one prior across several questions.

This is where a matrix becomes useful. A density matrix $ρ$ records the information needed for all these projection questions. The predicted share for a question $P$ is $tr (ρ P)$ , where the trace, $tr$ , means adding a matrix’s diagonal entries. Density matrices have trace one and are positive: they assign nonnegative shares to every projection.

EP can carry an additional positive trial amount $τ > 0$ in an unnormalized positive matrix $R = τ ρ$ . The correspondence becomes

\begin{aligned} total trial amount: & τ = tr R, \\ share for question P : & p (P) = \frac{tr (R P)}{tr R} . \end{aligned}

The normalized state supplies the shares; the scale supplies the trial amount. Assigning that scale an evidential interpretation is part of the EP model. Quantum measurement probabilities alone do not determine it.

For a geometric introduction to these matrices and their phase-sensitive questions, continue with When evidence puts on sunglasses.

A prior that fits one question can fail to fit three

Here is a small surprise with a short proof.

For an ordinary binary chance $P$ , the symmetric prior $Beta (1 / 2, 1 / 2)$ is perfectly valid. It gives more weight near zero and one than near the middle. A uniformly random real two-component arrow produces exactly this distribution of squared shares.

Could one prior over complex qubit states, states in a two-dimensional complex Hilbert space, give that same Beta distribution for every possible question direction?

A qubit’s density matrix can be represented by a point $(x, y, z)$ inside the Bloch ball:

x^{2} + y^{2} + z^{2} \leq 1.

The three coordinates describe three perpendicular measurement axes. The chance of the positive answer along each is

P_{x} = \frac{1 + x}{2}, P_{y} = \frac{1 + y}{2}, P_{z} = \frac{1 + z}{2} .

Ask, more generally, for each of these probabilities to have the same $Beta (p, p)$ prior, where $p > 0$ is the trial weight per answer. This $p$ is a prior parameter; the capital $P_{x}, P_{y}, P_{z}$ are random probabilities.

The symmetric Beta distribution has mean $1 / 2$ and variance $1 / [4 (2 p + 1)]$ . Since $x = 2 P_{x} - 1$ , we obtain

E [x^{2}] = E [y^{2}] = E [z^{2}] = \frac{1}{2 p + 1} .

Adding gives a necessary demand on the average squared radius:

E [x^{2} + y^{2} + z^{2}] = \frac{3}{2 p + 1} \leq 1.

Thus $p \geq 1$ . Half a trial per answer would demand an average squared radius of $3 / 2$ , even though every allowed state has squared radius at most one. No choice of correlations among the three coordinates can repair that sum.

A small obstruction with a visible budget

Can three questions share this prior?

Ask for the same Beta(p, p) distribution along every axis of a complex qubit. The three perpendicular axes each demand a share of the Bloch vector's squared length. Every physical qubit state has a Bloch vector with squared length at most one.

x²y²z²

Maximum: 1 02

Each axis requires: 0.500
All three require: 1.500

Impossible: the Bloch vector's average squared length would be 1.500, greater than every allowed state's maximum of 1. Each individual Beta prior is valid; their joint demand fails.

The bar adds the expected values of x², y² and z². It is not a measurement of one particular state's radius. Here p is prior trial weight per answer, not the probability of an outcome.

Try the three-question prior budget in the playground.

At $p = 1$ , a uniformly random point on the Bloch sphere supplies an explicit solution: its projection on any axis is uniform between $- 1$ and $1$ , so each positive-answer probability is uniform between zero and one. At $p = 2$ , a uniformly random point inside the ball supplies another: its slices have areas proportional to $1 - x^{2}$ , giving $Beta (2, 2)$ shares after the coordinate change.

The half-count prior still works for a real two-component system, sometimes called a rebit. Its Bloch picture is a disk with two independent coordinates. The corresponding budget is $2 / (2 p + 1) \leq 1$ , and $p = 1 / 2$ fits exactly. A uniform point on the circle constructs it.

The requirement matters. We demanded one prior over states that supplies the same Beta family for all the stated questions. Many other priors over quantum states are possible. A classical Beta distribution for one fixed question remains valid for every positive parameter. It is the attempt to fit several incompatible questions into one shared state model that creates the obstruction.

A new kind of room to explore

Hilbert space brings together several things that initially look unrelated: names become subspaces, random directions become Dirichlet shares, grouped dimensions become trial weights, and weighted averages become expectation values. The same mathematics lets us ask exactly where a classical belief model stops fitting a larger collection of questions.

An intriguing next step for an evidence system is to retain more than one fixed list of answer shares. Could a learner keep a state that supports several overlapping questions, then choose which projection is useful for its next decision? Which relations must survive when the state is compressed? How should trial amount travel alongside the normalized state?

Those are modeling questions. The identities above give us instruments for investigating them. Whether a particular learner or physical system follows the resulting dynamics is a further question for experiments.

For now, return to the room, give the mountains another direction, and merge the two inland destinations. The statistical rule and the geometric rule will move together.

Reading map

A belief is more than a number introduces trial amount, mean shares, and malleability.
When evidence puts on sunglasses develops the positive-matrix and Bloch-ball picture.
Could the true average lie beyond everything you measured? investigates the range of weighted averages and a proposed extension for continuous means.
Życzkowski and Sommers, Induced measures in the space of mixed quantum states, develops the real and complex random-state measures behind the dimension correspondence.
Dunkl and colleagues, Numerical shadows: measures and densities on the numerical range, studies expectation values of random unit vectors and their distributions.