Sleeping Beauty problem is an inspiration to understand the relativistic behavior of probabilities. Here, after developing a precise transformation between observers of this problem we will apply it to another known problem, the CHSH violation, and there are good news for people who are interested to understand!

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Above wallpaper reference¹

The following is the result of applying the mathematical tools^{2 3} that I developed years ago on the Bell's theorem. The idea, argument, and years of thinking are mine, and I am accountable for the result, but I used Grok 4.6 and Claude Opus 5 for the calculations, proofs, and edits. Good news is that it is rigorous enough to weaken the Bell's theorem sufficiently to give room for locality and realism back to the reality! I am very thankful to Bell who found this deep problem inside the probability theory, without knowing it's relativistic! Amazing job!

By the way, for completeness let's mention that I consider my posts as full-fledged scientific articles, so feel free to comment or even cite. Every algebraic step below is also formalised in Lean 4, and the script is in the geo-prob-bell repository.

Let's read it. Hope you enjoy!

What this post does

This is a physics post, and the arguments in it are meant to be checked line by line. But it is written to be read by anyone who is willing to follow a little algebra, not only by people who already speak the jargon. Where a technical word is needed it gets explained the first time it appears.

Here is the idea in one paragraph.

A probability is never just a property of a thing out in the world. It is a property of a thing plus a list of the outcomes you are keeping track of. Change the list and the numbers change, even though nothing about the thing has changed. That was the argument of Probability is Relativistic², and Geometrical Probability³ added the second half: probabilities are tensors on that list, and a conditional probability is not a single number but a rank-two one, a whole table with one index for each thing being conditioned. Call the list of outcomes a chart, and call the descartian package of events with different types a counting space.

This post points the tool at Bell's theorem⁴ — the result usually summarised as "the world cannot be both local and deterministic." Everything is built from scratch; the only thing carried over from the earlier posts is the habit of saying out loud which outcomes are being counted, and the rule that says which tensor rank each object has.

The plan, in order:

1. Warm up on the Sleeping Beauty puzzle, where there is only one index and no conditional probability at all, and watch the answer move when we transfer the list of outcomes for different observers. The maps that do the transferring turn out to form a group, which is worth pinning down before anything else.
2. Say why the classical picture and the quantum picture are not on an equal footing here — why one of them can grow its list of outcomes and the other cannot.
3. Write down what a Bell experiment actually records. This is where the tensor rank matters: what the experiment records is a joint distribution, which is one probability vector on the product chart — a rank-one object, exactly like Sleeping Beauty's — and the conditional table is what you get by dividing a marginal back out of it.
4. Turn that vector into a correlation number, \(E(x,y)\), by pairing it with an assignment of a value to each outcome.
5. Combine four of those numbers into the quantity \(S\) that Bell's inequality is about.
6. Work out what happens to the vector when the observer changes.
7. Leverage the possibility of adding extra event-types in a non-deterministic system and mix, exactly as in step 1.
8. Let the extra event-types carry a value of their own, and read off the interval of chart parameters on which two valid observers disagree.
9. Do it again for the case where each wing is re-charted on its own, which is a different map and gives a different pair of numbers.

The thread running through all of it is a single quantity. A chart says what an observer can tell apart, so what a counting space naturally measures is what her chart leaves out: an outcome the apparatus really produces that her list does not have a slot for. That is the hidden knowledge of the system. Steps 7 to 9 put a price on it.

What is new here is the insight the relativistic property of probability brought to our attention. And the fact that we cannot argue about absolute statements using relativistic variables will hit us hard! You can derive an invariant variable calculatable for each observer independently by only relativistic variables, then make an absolute statement from there, but, you cannot directly use relativistic variables to prove any absolute statement. This is what we will witness in this post.

1. Warm-up: Sleeping Beauty, one index

Start with the simplest case: one list of outcomes, no conditional probabilities, and a number that visibly moves when the list changes.

Sleeping Beauty⁵ is put to sleep, a fair coin is tossed, and she is woken once on Monday if it lands Heads and twice, Monday and Tuesday, if it lands Tails, with her memory erased in between. On waking, what is her credence in Heads?

The famous disagreement is really a disagreement about what to count. One camp counts three kinds of waking — Heads-Monday, Tails-Monday, Tails-Tuesday — and answers \(\tfrac13\). The other counts two, because for them the coin toss is the event, and answers \(\tfrac12\). Neither side has made an arithmetic mistake. They have written down different lists. Both halfer and thirder are valid observers of the system. This is relativity.

To compare them, put both lists on the same footing. Take the two-slot answer and pad it out to \(n\) slots with zeros, one zero for each kind of waking this observer does not distinguish:

\[ \overrightarrow{d} = \Bigl(\tfrac12,\;\tfrac12,\;0,\;\underbrace{0,\ldots,0}_{n-3}\Bigr)^{\top}, \qquad \mathbf{1}_{n}^{\top}\overrightarrow{d}=1 . \]

Now: what is a legal way to change the list? Whatever it is, it has to turn a probability distribution into a probability distribution — nothing negative, and the total still one. For a one-index object the family that does this is the stochastic mixer

\[ M_{n}(\mu) = \mu\, I_{n} + \frac{1-\mu}{n}\,\mathbf{1}_{n}\mathbf{1}_{n}^{\top}, \]

Read it as a dial. At \(\mu=1\) it does nothing: the observer's own list, fully resolved. At \(\mu=0\) it blends everything: every slot equally likely, all structure gone. In between it leaves a fraction \(\mu\) of the distribution where it was and spreads the rest evenly across all \(n\) slots. Notice, here we don't claim \(M_{n}(\mu) \) is all possible transformations in the counting space. But it's just enough for arguing about the sleeping beauty and the Bell's theorem.

Applying it,

\[ M_{n}(\mu)\overrightarrow{d} = \mu,\overrightarrow{d} + \frac{1-\mu}{n}\,\mathbf{1}_{n}, \]

which has two kinds of slot:

\[ P_{\text{H}}(\mu) = \frac{\mu}{2}+\frac{1-\mu}{n} = \frac{\mu(n-2)+2}{2n}, \qquad \bigl(M_{n}(\mu)\overrightarrow{d}\bigr)_{\text{padded}} = \frac{1-\mu}{n}. \]

So

\[ P_{\text{H}}(1) = \frac12, \qquad P_{\text{H}}(0) = \frac1n . \]

\(n\)	\(P_{\text{H}}(\mu)\)	\(P_{\text{H}}(1)\)	\(P_{\text{H}}(0)\)
3	\(\mu/6 + 1/3\)	\(1/2\)	\(1/3\)
4	\(\mu/4 + 1/4\)	\(1/2\)	\(1/4\)
5	\(3\mu/10 + 1/5\)	\(1/2\)	\(1/5\)
6	\(\mu/3 + 1/6\)	\(1/2\)	\(1/6\)

At \(n=3\) the two ends of the dial are exactly the two famous answers, \(\tfrac12\) and \(\tfrac13\), and every extra slot on the list pushes the blended answer further down, to \(1/n\). Hidden knowledge is worth something concrete here. The coin is what the halfer knows and the thirder does not; knowing it sets \(\mu=1\) and returns \(\tfrac12\) however many event-types the chart carries, and not knowing it costs exactly \(\tfrac12-\tfrac1n\). So \(\mu\) is the fraction of the event stream the chart resolves, and \(1-\mu\) is the part it does not. Both readings recur in §7, where the same \(\mu\) prices the same thing for a Bell test.

The pairing itself is clean. For any contrast \(\overrightarrow{c}\) with \(\overrightarrow{c}^{\top}\mathbf{1}_{n}=0\) and \(\overrightarrow{c}^{\top}\overrightarrow{d}=1\), such as \((1,1,-2,0,\ldots,0)^{\top}\),

\[ S_{\text{SB}}(\mu) = \overrightarrow{c}^{\top}M_{n}(\mu)\overrightarrow{d} = \mu . \]

One index, one factor of \(\mu\). Hold on to that. It is the rule that decides everything from §6 onward, and the only work left will be to say honestly how many indices each object has.

One detail is easy to get wrong, so it is worth pinning down. The dial is legal — the result is still a probability distribution — exactly for

\[ \mu\in\Bigl[-\tfrac{2}{n-2},\;1\Bigr] \quad\text{for }\overrightarrow{d}, \qquad \mu\in\Bigl[-\tfrac{1}{n-1},\;1\Bigr] \quad\text{for an arbitrary nonnegative vector.} \]

Only the top end is \(1\). Turning the dial past \(0\) into negative numbers is still fine for a while: at \(n=3,\ \mu=-1\) the result is \(\bigl(\tfrac16,\tfrac16,\tfrac23\bigr)\), a perfectly good distribution. So the range is not \([0,1]\), as one might guess.

It worth mentioning that each \(\mu\) that gives us a nonnegative vector represents a transformation from a halfer observer to a valid observer.

The dial is a group

Before going any further it pays to look at the family \(\{M_{n}(\mu)\}\) as a whole, because almost everything later is a consequence of its structure rather than of any particular calculation.

Write \(J_{n}=\mathbf{1}_{n}\mathbf{1}_{n}^{\top}\) for the all-ones matrix. The one fact you need is \(J_{n}^{2}=n\,J_{n}\). Multiply two dials and collect terms:

\[ M_{n}(\mu)M_{n}(\nu) = \mu\nu\,I_{n} + \frac{\mu(1-\nu)+\nu(1-\mu)+(1-\mu)(1-\nu)}{n}\,J_{n} = \mu\nu\,I_{n}+\frac{1-\mu\nu}{n}\,J_{n}, \]

because the bracket collapses to \(1-\mu\nu\). So

\[ \boxed{\;M_{n}(\mu)\,M_{n}(\nu) = M_{n}(\mu\nu)\;} \]

The consequences are immediate and they are worth naming.

It is a homomorphism. The map \(\mu\mapsto M_{n}(\mu)\) sends multiplication of numbers to multiplication of matrices. Chart changes compose by multiplying their dials, and they commute: \(M(\mu)M(\nu)=M(\nu)M(\mu)\). Two successive losses of resolution are one bigger loss, and the order in which they happened cannot be recovered from the result.

The identity is \(\mu=1\) and the inverse is \(1/\mu\). \(M_{n}(1)=I_{n}\), and for \(\mu\neq0\)

\[ M_{n}(\mu)^{-1} = M_{n}(1/\mu), \]

which you can check either from the boxed law with \(\nu=1/\mu\), or directly. So \(\{M_{n}(\mu):\mu\neq0\}\) is an abelian group isomorphic to \((\mathbb{R}^{\times},\times)\), and adding \(\mu=0\) makes it a commutative monoid isomorphic to \((\mathbb{R},\times)\). The element \(\mu=0\), \(M_{n}(0)=J_{n}/n\), is the absorbing one: it has rank one, it is not invertible, and once you are there no further chart change takes you back.

But the inverse is usually not itself a chart change. This is the point that does real work later. If \(\mu\in(0,1)\) then \(1/\mu>1\), and \(M_{n}(1/\mu)\) has negative off-diagonal entries — it is a perfectly good linear map on the affine hull of the simplex, but it is not a table of conditional probabilities. The dials that are legal chart changes form only a sub-semigroup, \(\mu\in\bigl[-\tfrac1{n-1},1\bigr]\), which is closed under multiplication but has no inverses except at \(\mu=\pm1\). Un-mixing is a computation you may perform on paper; it is not an observation anybody can make. Recovering the finer chart requires the finer data, not the algebra.

The dial has a rapidity. Let \(\Pi_{n}=I_{n}-J_{n}/n\), the orthogonal projector onto the vectors summing to zero, so \(\Pi_{n}^{2}=\Pi_{n}\). Then

\[ M_{n}(\mu) = I_{n}-(1-\mu)\,\Pi_{n}, \qquad\text{and}\qquad M_{n}(e^{-\lambda}) = e^{-\lambda\Pi_{n}} . \]

In the variable \(\lambda=-\ln\mu\) the dials form a one-parameter semigroup whose parameters add, with generator \(-\Pi_{n}\). This is the same bookkeeping that makes rapidity, rather than velocity, the additive parameter of a boost, and it is the sharpest sense in which the word "relativistic" is being used here. \(\lambda=0\) is the observer's own chart; \(\lambda\to\infty\) is total loss of resolution.

The spectrum says what is preserved. \(M_{n}(\mu)\mathbf{1}_{n}=\mathbf{1}_{n}\), so total probability is fixed; on the complement \(\mathbf{1}_{n}^{\perp}\) the dial is multiplication by \(\mu\). Hence \(\det M_{n}(\mu)=\mu^{\,n-1}\), and every contrast — every question of the form "how much more likely is this than that" — is scaled by exactly \(\mu\). That is the structural reason behind \(S_{\text{SB}}(\mu)=\mu\) above, and it is what §6 will carry over to the Bell case.

Now the two-index case.

2. Two kinds of description

Sleeping Beauty has two observers and one event stream, and the two charts differ only in how finely that stream is cut. Bell is not like that. There are two descriptions of the same apparatus in play, a classical one and a quantum one, and they are not symmetric. Setting the asymmetry out now explains why the calculation below has the shape it has.

Classical physics describes the inside of the box. Maxwell's equations hand you the field at every point in space and time — amplitude, phase, polarisation, the lot. The description is complete, and it is about the interior. Nothing is hidden except by the errors of sloppy instruments, and we know how to handle the errors of sloppy instruments in practice, so it's not a gap in the theory. Three things follow. The list of possible outcomes is fixed by the state of the field, not chosen by whoever is watching, so there is no freedom to add to it. Correlations are computed from field intensities, not any probability distribution; probability shows up only as bookkeeping across many runs, never as part of the description itself. And a reading of \(+1\) or \(-1\) has nothing else it could have been — there is no third thing for the field to do.

So on the classical side the list cannot grow, and the limit of \(2\) that Bell's inequality places on \(\lvert S\rvert\) is simply a fixed number. It is fixed not because we have decided to hold it fixed, but because there is nothing available to add to it.

Quantum mechanics, read the standard Copenhagen way, describes the surface. The system is a black box by design. Heisenberg took that phrase literally. Keep only what goes in and what comes out, he proposed, and collect the whole content of the theory in one table of amplitudes from incoming states to outgoing ones — the scattering matrix, or S-matrix.⁶ This mindset was around in the Quantum mechanics' community long before that though! What the theory hands you is the probability of each outcome at the place where your apparatus touches it, and those probabilities are taken as fundamental — not as a summary of some finer story going on inside. Einstein, Podolsky and Rosen pressed exactly this point⁷: if the description only covers the boundary of the experiment, perhaps it is incomplete. Three things follow here too. The list of outcomes is whatever your apparatus can tell apart. That is a fact about the apparatus, not something the theory dictates. And so an experimenter can, at least in principle, learn to distinguish an outcome nobody had been counting — a new measurement on the surface of the box to uncover some hidden knowledge of the system.

That is the imbalance. The classical limit is frozen because that description is complete and about the interior. The quantum number depends on the list because that description covers only the surface and is made of probabilities fundamentally.

Everything below follows from that. The number \(2\) in the classical view stays put since the inside of the experiment has been described, while \(|S|\) moves with the list in the quantum picture since there's always room to have more detector on the boundary of the experiment. That is what it would mean for Bell's inequality to suffer from a lack of knowledge of the system: not a hidden variable inside the box, but an unrecorded event-type on its surface. Call that the hidden knowledge of the system. It is the natural thing for a counting space to measure, since a list of outcomes is exactly a statement of what its owner can tell apart, and §7 to §9 give it a size.

3. The chart, and the vector a Bell experiment records

Fix a Bell experiment. Alice chooses a setting \(x\in\{a,a'\}\), Bob chooses \(y\in\{b,b'\}\).

A chart is just the list of outcomes being counted, written down. Alice's list has \(n_{A}\) entries; Bob's has \(n_{B}\). The usual list has \(n_{A}=n_{B}=2\) entries a side: "the detector fired \(+\)" and "the detector fired \(-\)". Nothing in the mathematics forces \(2\), and the whole question of this post is what changes when it is bigger.

The counting space of the experiment is not Alice's list, and it is not Bob's. It is the descartian package of the two: the list of pairs

\[ \{1,\dots,n_{A}\}\times\{1,\dots,n_{B}\}, \qquad N = n_{A}n_{B} \]

slots in all. This is the point where it is easy to take a wrong turn, so let us be explicit about it, because the rest of the post is decided here.

What the experiment records is one probability vector on that product list. For each pair of settings the data is

\[ \pi^{xy}_{(\alpha\beta)} = P(\alpha,\beta \mid x,y), \qquad \pi^{xy}_{(\alpha\beta)}\ge0, \qquad \sum_{(\alpha\beta)}\pi^{xy}_{(\alpha\beta)}=1 . \]

It is convenient to print those \(N\) numbers in a rectangle, with Alice's outcome choosing the row and Bob's the column, and we shall do that constantly. But printing a list in a rectangle does not give it a second index. By the rule of Geometrical Probability³, the rank of a probability object is the number of things it is conditioned on. A joint probability \(P(\alpha\cap\beta)\) is conditioned on nothing; it is a \((1,0)\) tensor — a vector — exactly like Sleeping Beauty's \(\overrightarrow{d}\), only on a longer list. The object with two indices is the conditional one, \(P(\beta\mid\alpha)\), which is a \((1,1)\) tensor, and it is not what the experiment hands you directly.

That the correlation is built from the joint vector, and not from anything of higher rank, is visible in the definition itself:

\[ E(\hat a,\hat b) = \sum_{A,B=\pm1}AB\,P(A,B) = \sum_{A,B=\pm1}AB\,P(A\cap B) = \sum_{A,B=\pm1}AB\,P(A\mid B)\,P(B). \]

Reading that from left to right: \(E\) is a single sum over the product list, i.e. the pairing of a vector of values with the vector \(P(A\cap B)\). Reading the last equality from right to left: the rank-two object \(P(A\mid B)\) enters only through the chain rule, contracted against the marginal \(P(B)\) to rebuild that same vector. Both readings give one \((1,0)\) object, and the transformation rule that applies to it is therefore the one-index rule of §1.

Having said that, the chain-rule split is still the useful way to read the data, so let us write it down. Split the joint the way the chain rule splits any joint probability, into "the chance Alice saw \(\alpha\)" times "the chance Bob saw \(\beta\) given that she did":

\[ P(\alpha,\beta\mid x,y) = P(\beta\mid\alpha,x,y)\,P(\alpha\mid x,y), \qquad\text{that is}\qquad P^{xy} = \operatorname{diag}(m^{xy})\,K^{xy}, \]

where \(m^{xy}_{\alpha}=P(\alpha\mid x,y)\) is the marginal, \(P^{xy}\) is the rectangle holding \(\pi^{xy}\), and \(K^{xy}_{\alpha\beta}=P(\beta\mid\alpha,x,y)\) is the matrix of conditional probabilities.

Two features of \(K\) are worth naming.

It has two indices, one for each side of the experiment. This is what it means to say a conditional probability is not really a number. It is a table, with a row for each thing Alice might see and a column for each thing Bob might see.

Each of its rows adds up to one. That is the only structure a conditional probability has. It is not a rotation, not symmetric, nothing else — just a table whose rows are probability distributions.

And one feature of the relation between them, which §6 will use: \(K\) is a quotient. It is what is left of \(\pi\) after the marginal has been divided out, and the marginal is itself extracted from \(\pi\) by summing. Whatever transforms linearly, it is \(\pi\), not \(K\).

4. The readout, and \(E(x,y)\)

A list of outcomes on its own gives you no correlation number. You also have to say what each outcome is worth. That assignment is the readout:

\[ v_{A}\colon \{1,\dots,n_{A}\}\to[-1,1], \qquad v_{B}\colon \{1,\dots,n_{B}\}\to[-1,1]. \]

On the usual two-entry list the readout is \(v_{A}=v_{B}=(1,-1)^{\top}\): a \(+\) is worth \(+1\), a \(-\) is worth \(-1\). Restricting the values to \([-1,1]\) is not decoration. It is what makes the resulting number the kind of thing Bell's inequality is about.

A readout on the product list is the product of the two:

\[ \omega_{(\alpha\beta)} = v_{A}(\alpha)\,v_{B}(\beta), \qquad\text{i.e.}\qquad \omega = v_{A}\otimes v_{B}. \]

Keeping \(\omega\) of this product form is not a convenience either. It is precisely the assumption that Alice's verdict is a function of Alice's outcome alone and Bob's of Bob's, and it is what keeps the classical bound of \(2\) meaningful. Abandon it and the number you compute is no longer one that any local model was ever supposed to bound.

The correlation number \(E(x,y)\) is then the pairing of the readout with the data:

\[ E(x,y) = \bigl\langle \omega,\,\pi^{xy}\bigr\rangle = \sum_{\alpha\beta} v_{A}(\alpha)\,v_{B}(\beta)\,\pi^{xy}_{(\alpha\beta)} = \sum_{\alpha} m^{xy}_{\alpha}\,v_{A}(\alpha)\,\bigl(K^{xy}v_{B}\bigr)_{\alpha}. \]

The first form is the one that transforms; the last is the one that explains. It says: for each thing Alice might see, take how often she sees it, times what it is worth to her, times what Bob's outcome is worth on average given that she saw it. The conditional table \(K\) is doing real work there, with both of its indices — which is what the second post³ meant by calling conditional probability a two-index object — but it is doing that work inside a single contraction that returns the one-index vector \(\pi\).

For a standard two-outcome experiment with no bias toward \(+\) or \(-\) on either side, the rectangle is

\[ P = \frac14 \begin{pmatrix} 1+E & 1-E \\ 1-E & 1+E \end{pmatrix}, \qquad v^{\top}Pv = E, \]

and the rows of the corresponding \(K\) sum to one, as they must. Both marginals are \(\bigl(\tfrac12,\tfrac12\bigr)\), so \(\sum_{\alpha\beta}v_{A}(\alpha)\pi_{(\alpha\beta)}=0\) and likewise for Bob: the experiment is unbiased on each wing separately. Keep that fact; §9 needs it and nothing before §9 does.

5. \(|S|\), the quantity Bell bounds

Now go up a level. Each of the four setting pairs gives one correlation number, so collect them into their own little \(2\times2\) table — this time the rows and columns are Alice's and Bob's settings, not their outcomes:

\[ \mathcal{E} = \begin{pmatrix} E(a,b) & E(a,b') \\ E(a',b) & E(a',b') \end{pmatrix}. \]

The CHSH combination⁸ adds three of them and subtracts the fourth. Writing that as a table of weights \(W\) alongside \(\mathcal{E}\):

\[ W = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}, \qquad S = \langle W,\mathcal{E}\rangle = \operatorname{tr}\bigl(W^{\top}\mathcal{E}\bigr) = E(a,b) + E(a,b') + E(a',b) - E(a',b') . \]

There are now two lists stacked on top of each other, and telling them apart is the single most important piece of bookkeeping in this post. The outcome list is where \(\alpha\) and \(\beta\) live, and it is the one that will grow later. The setting list is where \(x\) and \(y\) live, and it will not grow: adding rows there would mean adding measurement settings, which is a different experiment entirely.

Two numbers are worth writing down now, to compare against later.

The classical bound is \(\lvert S\rvert\le2\). It holds for any model in which Alice's verdict depends on her setting and some shared variable but not on Bob's setting, with values in \([-1,1]\).

The quantum maximum is \(2\sqrt2\). At the best choice of measurement angles the four correlation numbers all come out to \(\pm1/\sqrt2\), arranged so that

\[ \mathcal{E} = \frac{1}{\sqrt2} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} = \frac{W}{\sqrt2}, \qquad W^{2}=2I_{2}, \]

In other words the best possible table of correlation numbers is just the weight table \(W\) divided by \(\sqrt2\): quantum mechanics does as well as it can by lining the data up with the question being asked. That maximum, \(S=2\sqrt2\), is Tsirelson's bound⁹.

There are eight ways to pick which of the four terms gets the minus sign (four choices, two overall signs). At this best setting, two of them hit \(\pm2\sqrt2\) and the other six come out to exactly zero.

Everything from here on takes \(2\sqrt2\) as an input and asks what a change of chart does to it.

6. Changing the chart

Now the central question. Suppose the observer's list is not the whole story — suppose the apparatus produces outcomes she is not counting, or that what she calls one outcome is really several counted together as one. How does her data change?

Whatever the rule is, it has to turn probabilities into probabilities. So it is itself a table of conditional probabilities: "given the old event-type, what is the chance of each new one." On the product list of §3 that means a column-stochastic \(N\times N\) matrix \(T\), and

\[ \pi \;\longmapsto\; \pi' = T\,\pi . \]

That is the whole constraint. Whatever rectangle we print it in, a joint distribution transforms the way \(\overrightarrow{d}\) transformed in §1, because it is the same kind of object.

Among all such \(T\), §1 already handed us the canonical one-parameter family: the Sleeping Beauty dial on the product list,

\[ M_{N}(\mu) = \mu\, I_{N} + \frac{1-\mu}{N}\,\mathbf{1}_{N}\mathbf{1}_{N}^{\top}, \qquad N = n_{A}n_{B}. \]

It is symmetric, its rows and columns both add to one for \(\mu\in[0,1]\), and by §1 the family is a group under multiplication of the dial. It is square, which looks at first like a problem: it spreads an \(N\)-entry list across an \(N\)-entry list rather than making the list longer. But §1 already solved that. Pad the list with unused slots first, exactly as we padded \(\overrightarrow{d}\), and the square dial works at any \(N\).

The master formula

Put a readout \(\omega\) on the longer list and pair it with the transformed data.

Proposition 1 (master formula). For any \(\omega\) and any probability vector \(\pi\),

\[ E' = \bigl\langle\omega,\,M_{N}(\mu)\pi\bigr\rangle = \bigl\langle M_{N}(\mu)^{\top}\omega,\,\pi\bigr\rangle = \mu\,E \;+\; (1-\mu)\,\bar\omega, \qquad \bar\omega=\frac1N\sum_{(\alpha\beta)}\omega_{(\alpha\beta)} . \]

Proof. Move the brackets, then use \(M_{N}(\mu)^{\top}=M_{N}(\mu)\) and \(M_{N}(\mu)\omega=\mu\omega+(1-\mu)\bar\omega\mathbf{1}_{N}\), and finally \(\mathbf{1}_{N}^{\top}\pi=1\). \(\square\)

That is a three-line proof and it carries the entire post. It says a change of list never touches the physics — the vector \(\pi\) is untouched. What changes is the readout: instead of \(\omega\) you are effectively using \(\mu\omega+(1-\mu)\bar\omega\mathbf 1\), every entry of which is a weighted average of the old values.

One index, one factor of \(\mu\). The joint distribution has one index — on the product list — so it picks up one factor of \(\mu\), exactly as Sleeping Beauty's did. This is the same statement as \(S_{\text{SB}}(\mu)=\mu\) in §1, and it comes from the same place: the dial is the identity on \(\mathbf 1\) and multiplication by \(\mu\) on everything orthogonal to it.

Write \(\lVert\omega\rVert_{\infty}=\max\lvert\omega\rvert\) for the biggest value the readout gives to anything — read it as "the strongest verdict this readout ever delivers."

Proposition 2 (contraction lemma). For \(\mu\in[0,1]\), \(\lVert M_{N}(\mu)^{\top}\omega\rVert_{\infty} \le \lVert\omega\rVert_{\infty}\), with equality at a component only if \(\omega\) is constant \(\pm\lVert\omega\rVert_{\infty}\) across everything that component mixes into.

Proof. An average of numbers lying between \(-c\) and \(c\) also lies between \(-c\) and \(c\), and it can only reach the edge if there was nothing to average in the first place. \(\square\)

In plain terms: spreading the same events over more event-types weakens the readout, and so weakens the correlation number. That is the engine of §7 to §9. Recall that probability is relativistic, and the dial is only transforming between what different observers observe of these events. They just see the events in different arrangements that we called event-types.

A factorised readout has a factorised average

One small computation is used repeatedly below. If \(\omega=v_{A}\otimes v_{B}\) on the product list, then

\[ \bar\omega = \frac{1}{n_{A}n_{B}}\sum_{\alpha\beta}v_{A}(\alpha)v_{B}(\beta) = \Bigl(\frac{1}{n_{A}}\sum_{\alpha}v_{A}(\alpha)\Bigr) \Bigl(\frac{1}{n_{B}}\sum_{\beta}v_{B}(\beta)\Bigr) = \bar v_{A}\,\bar v_{B}. \]

So the master formula reads, for a product readout,

\[ \boxed{\;E'(x,y) = \mu\,E(x,y)\;+\;(1-\mu)\,\bar v_{A}\bar v_{B}\;} \]

with the same \(\mu\) and the same offset for all four setting pairs, since the chart is a property of the observer and not of what she chose to measure.

Why \(\pi\) rather than \(K\)

\(K\) is the two-index object, so why not put the dial on it directly?

Because \(K\) is not what the chart change acts on, and it does not even determine \(E\) on its own — you need the marginal too, and the marginal changes as well when the list changes. Writing \(m'\) for the new marginal, the new conditional table is

\[ K' = \operatorname{diag}(m')^{-1}\,(\text{the transformed rectangle}), \]

a ratio: a messy, nonlinear expression that depends on the very marginal it is supposed to be conditioned on. The joint vector \(\pi\) is the thing that transforms cleanly, because it is the thing with the right rank; \(K\) is what you get afterwards by dividing the marginal back out. So the calculation runs on \(\pi\).

What the CHSH combination does with it

Since the transformation is affine in each \(E(x,y)\) with a common offset, and since \(\sum_{xy}W_{xy}=2\),

\[ S' = \sum_{xy}W_{xy}E'(x,y) = \mu\,S \;+\; 2(1-\mu)\,\bar v_{A}\bar v_{B}. \]

Two consequences, both immediate.

The local bound does not move. If \(\lvert S\rvert\le2\) then, for \(\mu\in[0,1]\) and \(\lvert\bar v_{A}\bar v_{B}\rvert\le1\),

\[ \lvert S'\rvert \le \mu\lvert S\rvert + 2(1-\mu)\lvert\bar v_{A}\bar v_{B}\rvert \le 2\mu + 2(1-\mu) = 2 . \]

This is not a coincidence of the algebra. \(S'\) is a convex combination of \(S\) and the value \(2\bar v_{A}\bar v_{B}\) produced by the deterministic local strategy "Alice always answers \(\bar v_{A}\), Bob always answers \(\bar v_{B}\)", and the set of local correlations is convex and contains that strategy. A chart change maps local models to local models. So \(2\) really is the right number to compare against on the new chart, and the comparison in §7 to §9 is honest.

The quantum value does move. Starting from Tsirelson,

\[ \lvert S'\rvert \le 2\sqrt2\,\mu + 2(1-\mu)\,\bar v_{A}\bar v_{B}, \]

with equality attainable, since \(2\sqrt2\) itself is attainable. Everything left to do is to read that inequality.

7. Adding unused outcomes, then spreading

Here is the natural question, and it is the right one. In §1 one observer's list was short an entry, and we handled it by padding with zeros so that the square dial could act. Do exactly the same to \(\pi\).

Pad the \(2\times2\) rectangle out to \(n\times n\) with zero rows and columns — those event-types simply have not been counted — so the product list has \(N=n^{2}\) slots, and turn the dial:

\[ \pi \;\longmapsto\; M_{N}(\mu)\,\pi = \mu\,\pi + \frac{1-\mu}{N}\,\mathbf{1}_{N}. \]

The total still comes to one, nothing is negative, and the result is a proper distribution on the longer list — except that the padded slots now carry real probability, \((1-\mu)/N\) each, spread there by the dial. That is the same thing that dragged \(P_{\text{H}}\) from \(\tfrac12\) down to \(\tfrac1n\) back in §1.

Her readout on her own list is \(v_{A}=v_{B}=(1,-1,0,\dots,0)^{\top}\): the two outcomes she resolves are worth \(+1\) and \(-1\), and the rest are things her two-way verdict simply does not cover, so they get \(0\). Those padded slots are the hidden knowledge of §2, made concrete: event-types the apparatus really produces that her list has no slot for. And \(1-\mu\) is how much of the event stream lands in them.

Since \(\bar v_{A}=\bar v_{B}=0\), the master formula gives

\[ E'(x,y) = \mu\,E(x,y), \qquad \lvert S'(\mu)\rvert \;\le\; 2\sqrt2\,\mu . \]

Notice what did not happen: no square. The joint distribution is one \((1,0)\) object on the product list, so it takes one factor of \(\mu\), and §1's rule survives intact. Notice also that the answer does not depend on \(n\) at all. Padding with unvalued slots and mixing gives the same \(\mu E\) whether the chart has four slots or four thousand: what matters is how much of the stream the chart resolves, not how many boxes are drawn around the rest.

Now the comparison, done on her list rather than someone else's. By §6 her local bound is still exactly \(2\). So

\[ \lvert S'(\mu)\rvert > 2 \quad\Longleftrightarrow\quad \mu > \frac{1}{\sqrt2} \approx 0.7071 . \]

\(\mu\)	\(1\)	\(0.9\)	\(1/\sqrt2\approx0.7071\)	\(0.6\)	\(0.5\)
\(\lvert S'\rvert=2\sqrt2\mu\)	\(2.8284\)	\(2.5456\)	\(2.0000\)	\(1.6971\)	\(1.4142\)
\(>2\)?	yes	yes	boundary	no	no

That number is not a coincidence

Write out what \(M_{N}(\mu)\) did to the recorded statistics, for the plain \(n=2\) chart with \(N=4\):

\[ \pi' = \mu\,\pi + (1-\mu)\,\tfrac14\,\mathbf{1}_{4} . \]

That is, verbatim, the outcome distribution of the state

\[ \rho_{\mu} = \mu\,\lvert\Psi^{-}\rangle\langle\Psi^{-}\rvert + (1-\mu)\,\tfrac{I_{4}}{4}, \]

the Werner state¹⁰ at visibility \(\mu\): a singlet mixed with white noise. Its CHSH maximum is \(2\sqrt2\mu\), and by the Horodecki criterion¹¹ it violates CHSH exactly for \(\mu>1/\sqrt2\).

So the chart change is not like depolarisation; on the recorded statistics it is depolarisation, and the wall the chart geometry produces is the depolarising visibility threshold on the nose, not approximately. The usual reading of that threshold is that the state was noisy. The reading here is that the observer's chart was coarse. The two are the same arithmetic, and no experiment distinguishes them, because what an experiment reports is \(\pi\) and \(\pi\) is all they share.

8. Letting the extra event-types count for something

Section 7 gave the padded event-types the value zero — the experimenter shrugs and throws them away. That was a choice, and not her only one. Nothing stops her deciding that an unresolved outcome counts as, say, a \(+1\).

Let \(s,t\in[-1,1]\) be what Alice and Bob decide the padded event-types are worth, so on an \(n\)-slot chart a side,

\[ v_{A} = (1,\,-1,\,s,\,\ldots,\,s)^{\top}, \qquad v_{B} = (1,\,-1,\,t,\,\ldots,\,t)^{\top}, \]

and their averages are

\[ \bar v_{A} = \frac{(n-2)s}{n} = c_{n}s, \qquad \bar v_{B} = c_{n}t, \qquad c_{n} \;\equiv\; \frac{n-2}{n} . \]

Note \(c_{n}<1\) strictly: the two resolved slots are pinned at \(+1\) and \(-1\) and they drag the average down, however many extra slots there are. Writing \(q=\bar v_{A}\bar v_{B}=c_{n}^{2}st\), so that \(q\in[-c_{n}^{2},\,c_{n}^{2}]\), the master formula gives

\[ \boxed{\; \lvert S'(\mu,q)\rvert \;\le\; 2\sqrt2\,\mu \;+\; 2\,(1-\mu)\,q \;} \]

with §7 as the special case \(q=0\). Her strongest verdict is still one, so her local limit stays at exactly \(2\).

Fix the event stream and fix \(\mu\), and let the observers range over all the readouts available on an \(n\)-slot chart. The value they report ranges over

\[ 2\sqrt2\mu - 2(1-\mu)c_{n}^{2} \;\le\; S' \;\le\; 2\sqrt2\mu + 2(1-\mu)c_{n}^{2}, \]

so there are two walls, not one.

Some observer reports a violation as soon as the top of that range clears \(2\):

\[ 2\sqrt2\mu + 2(1-\mu)c^{2} > 2 \quad\Longleftrightarrow\quad \mu > \mu_{-}(c) = \frac{1-c^{2}}{\sqrt2-c^{2}} . \]

Every observer reports a violation only once the bottom of the range clears it too:

\[ 2\sqrt2\mu - 2(1-\mu)c^{2} > 2 \quad\Longleftrightarrow\quad \mu > \mu_{+}(c) = \frac{1+c^{2}}{\sqrt2+c^{2}} . \]

Between them lies the interval that matters:

\[ \mu\in\bigl(\mu_{-}(c_{n}),\;\mu_{+}(c_{n})\bigr] \quad\Longrightarrow\quad \text{whether }\lvert S\rvert>2\text{ depends on the chart, not on the world.} \]

Both endpoints are monotone in \(c\): \(\mu_{-}'(c)<0\) and \(\mu_{+}'(c)>0\), so the window only widens as the chart grows.

\(n\)	2	3	4	5	6	10	50	\(\to\infty\)
\(c_{n}=(n-2)/n\)	\(0\)	\(1/3\)	\(1/2\)	\(3/5\)	\(2/3\)	\(4/5\)	\(24/25\)	\(1\)
\(\mu_{-}\)	\(0.7071\)	\(0.6821\)	\(0.6442\)	\(0.6071\)	\(0.5729\)	\(0.4650\)	\(0.1592\)	\(0\)
\(\mu_{+}\)	\(0.7071\)	\(0.7284\)	\(0.7511\)	\(0.7665\)	\(0.7771\)	\(0.7984\)	\(0.8227\)	\(0.8284\)

At \(n=2\) there is nothing to disagree about: no unresolved event-types, no freedom in the readout, and the window collapses to the single point \(1/\sqrt2\) of §7. As the chart grows the window opens out, and in the limit it becomes

\[ \bigl(0,\;2(\sqrt2-1)\bigr], \qquad 2(\sqrt2-1)\approx0.8284 . \]

That upper number has a name. \(2(\sqrt2-1)\) is the CHSH detection-efficiency threshold for a maximally entangled state¹² — the efficiency below which a Bell test cannot demonstrate a violation. It is normally derived from detector physics. Here it is the value of \(\mu\) above which the chart stops mattering: above it, every valid observer of the same event stream agrees there is a violation; below it, some do and some do not. (Claude found this relation and I am happy about it!)

So the two ends of the window are the two thresholds a Bell experimentalist already knows: \(1/\sqrt2\), where a violation first becomes possible for anybody, and \(2(\sqrt2-1)\), where it becomes compulsory for everybody. Both fall out of the geometry of the chart, with no hardware and no state in the derivation.

9. Re-charting the two wings separately

There is a second chart change worth doing, and it is not the same map.

Suppose Alice re-charts her own list and Bob re-charts his, independently and with the same dial setting. The map on the product list is then the tensor product of the two one-wing dials,

\[ T = M_{n_{A}}(\mu)\otimes M_{n_{B}}(\mu), \]

which is still column-stochastic, so it is still a legal chart change on the \((1,0)\) vector \(\pi\). It is simply a different one. It is not a member of the family \(\{M_{N}(\lambda)\}\): already at \(n_{A}=n_{B}=2\), \(\mu=\tfrac12\), the two matrices differ in every entry by \(\tfrac1{16}\). Two dials, one per wing, is two factors of \(\mu\); one dial on the product list is one.

Working out the pairing the same way — move the two dials onto the readout, where each wing's readout becomes \(\mu v+(1-\mu)\bar v\mathbf{1}\) — gives

\[ E'(x,y) = \mu^{2}E(x,y) +\mu(1-\mu)\bigl[\bar v_{B}\langle v_{A}\rangle+\bar v_{A}\langle v_{B}\rangle\bigr] +(1-\mu)^{2}\bar v_{A}\bar v_{B}, \]

where \(\langle v_{A}\rangle=\sum_{\alpha\beta}v_{A}(\alpha)\pi_{(\alpha\beta)}\) is Alice's marginal mean. For the unbiased experiment of §4 both marginal means vanish, and

\[ E'(x,y) = \mu^{2}E(x,y) + (1-\mu)^{2}\,\bar v_{A}\bar v_{B}, \qquad \lvert S'\rvert \le 2\sqrt2\,\mu^{2} + 2(1-\mu)^{2}\,q . \]

Unvalued padding, \(q=0\). \(\lvert S'\rvert\le2\sqrt2\mu^{2}\), and the wall is at

\[ \mu^{2}>\frac1{\sqrt2} \quad\Longleftrightarrow\quad \mu > 2^{-1/4}\approx0.8409 . \]

Fully valued padding, \(q\to1\). \(2\sqrt2\mu^{2}+2(1-\mu)^{2}>2\) factors as \(2\mu\bigl[(\sqrt2+1)\mu-2\bigr]>0\), so for \(\mu>0\)

\[ \mu > \frac{2}{\sqrt2+1} = 2(\sqrt2-1)\approx0.8284 . \]

In between, with \(q=c_{n}^{2}\), the condition is the quadratic \((\sqrt2+q)\mu^{2}-2q\mu+(q-1)>0\), whose relevant root is

\[ \mu^{\star}(c) = \frac{c^{2}+\sqrt{\,\sqrt2-(\sqrt2-1)c^{2}\,}}{\sqrt2+c^{2}} . \]

\(n\)	2	3	4	5	6	10	50	\(\to\infty\)
\(c_{n}=(n-2)/n\)	\(0\)	\(1/3\)	\(1/2\)	\(3/5\)	\(2/3\)	\(4/5\)	\(24/25\)	\(1\)
\(\mu^{\star}\)	\(0.8409\)	\(0.8397\)	\(0.8381\)	\(0.8369\)	\(0.8358\)	\(0.8334\)	\(0.8296\)	\(0.8284\)

This family too has a physical twin, and again an exact one. Independent loss on each wing at efficiency \(\eta\), with undetected events assigned a fixed verdict, gives correlations \(\eta^{2}E+(1-\eta)^{2}\) when the marginals vanish — the same expression with \(\mu\) renamed \(\eta\). That is precisely the model in which Garg and Mermin¹² derived \(2(\sqrt2-1)\) as the necessary and sufficient detection efficiency for two settings a side. So the per-wing dial reproduces the detection-efficiency calculation exactly, just as the joint dial reproduced the depolarising one.

Which map is the right one to use is not a matter of taste; it is a matter of what the observer's chart actually did. If her coarseness is a property of the joint event-type — a whole coincidence she cannot classify — the joint dial applies and the walls are those of §8. If it is a property of each detector separately, the per-wing dial applies and the walls are those of §9. The point for what follows is that both are legal, both are one-parameter, and they disagree.

10. What is invariant, and what is not

Proposition 3 (no absolute statement from \(S\)). Two observers of the same event stream, differing only in the chart they keep and the values they assign to what that chart does not resolve, compute

\[ S' = \mu\,S + 2(1-\mu)\,q, \qquad S'' = \mu'\,S + 2(1-\mu')\,q' , \]

and these are different numbers. So \(S\) is not an invariant of the system. Nor is \(\max\lvert S\rvert\), which §7 to §9 have just computed to be \(2\sqrt2\mu+2(1-\mu)q\), a function of the chart. And by §8 the sign of \(S-2\) — not merely its size — is chart-dependent throughout a whole interval of \(\mu\).

The conclusion is a piece of ordinary logic, not of physics. An argument whose conclusion is an absolute statement about the world — locality, non-locality, realism — cannot take a chart-dependent quantity as its premise, any more than a claim about the invariant mass of a system can be based on somebody's measured energy alone. You may build invariants out of relativistic variables and then argue from those. You may not skip that step. \(\lvert S\rvert>2\) is a statement about a chart, and until it is supplemented by a statement about which chart, it does not reach the world.

Note carefully what this does not say. It does not say the experiments are wrong; the recorded numbers are the recorded numbers. It does not say quantum mechanics predicts anything other than what it predicts. It says that the step from "this observer's \(S\) exceeds \(2\)" to "nature is not locally realistic" is missing a premise, and that the missing premise is about the completeness of the observer's list of event-types — which is exactly the thing §2 argued the quantum formalism, describing only the boundary, does not fix.

The missing piece is not a hidden variable inside the box. It is hidden knowledge on its surface — an unrecorded event-type.

Even Fine's theorem¹³ carries the same caveat. Its "if and only if" is stated on the chart whose atomic events are the sixteen tuples \((\alpha,\alpha',\beta,\beta')\), so if the real chart is finer with mass outside that set, the equivalence is about a different object than the data. It's just out of its assumptions.

11. Comparison with other approaches

The textbook reading is that \(\lvert S\rvert>2\) rules out local realism^{4 8}, and one then chooses a victim. De Broglie–Bohm keeps hidden variables and pays with nonlocality¹⁴. Everett keeps unitary evolution and drops collapse¹⁵. Superdeterminism and retrocausality drop independence of settings^{16 17}. QBism and relational quantum mechanics deny that a probability belongs to the system^{18 19}.

The move here is different: keep everything and account for the chart. Every entry in that list is an answer to "what is the world like." The counting space answers a different question — "what did the observer's chart resolve, and what did it leave out" — and the hidden knowledge it measures lives on the boundary of the apparatus rather than inside the system. So it is not a rival to any of them.

The closest neighbours are not on that list at all. They are the two experimental thresholds recovered above. The detection loophole and the visibility requirement are usually presented as engineering facts about detectors and states. What §7 to §9 show is that the same two numbers are already there in the geometry of the chart, before any hardware is specified — which suggests the loopholes are less an imperfection of apparatus than a visible symptom of the chart-dependence of \(S\).

It should be said plainly how far that goes, and how far it does not. Everything above is the two-setting CHSH case with a maximally entangled state, and the detection thresholds of the literature are known to move once either of those is relaxed: Eberhard²⁰ gets the requirement down to \(2/3\) by using a non-maximally entangled state, the chained inequalities of Braunstein and Caves²¹ have thresholds of their own²², and for Mermin's \(n\)-party inequalities²³ the optimal figure is \(n/(2n-2)\)²⁴. The chart calculation here reproduces exactly the two numbers belonging to the case it was set up for, and says nothing yet about the others. Whether the same geometry delivers \(2/3\) and \(n/(2n-2)\) is a concrete question, and a good test of the whole idea; I have not done it. Note also that the loophole-free experiments²⁵ close the detection and locality loopholes as they are usually posed, and nothing above disputes their results — the point of §10 is about what may be concluded from \(\lvert S\rvert>2\), not about whether \(\lvert S\rvert>2\) was measured.

12. Conclusion

Probability is relativistic because observers do not share the same event-types. Sleeping Beauty is the one-index demonstration: one event stream, two charts, and a mixer that carries \(P_{\text{H}}\) from \(\tfrac12\) at \(\mu=1\) to \(\tfrac1n\) at \(\mu=0\), with \(S_{\text{SB}}(\mu)=\mu\). Those mixers form a group, \(M_{n}(\mu)M_{n}(\nu)=M_{n}(\mu\nu)\) with \(M_{n}(\mu)^{-1}=M_{n}(1/\mu)\), of which only a sub-semigroup consists of legal chart changes — so resolution can be lost by observation but not regained by algebra.

The two descriptions of a Bell apparatus are not symmetric. Classically the field is given everywhere, so the chart is fixed and the bound of \(2\) has no extra measure to absorb. Quantum mechanically only the boundary is described, and the list of distinguishable event-types is a fact about the apparatus rather than about the theory — so it can be lengthened.

What a Bell experiment records is one probability vector on the product chart,

\[ \pi^{xy}_{(\alpha\beta)} = P(\alpha,\beta\mid x,y), \qquad E(x,y) = \langle v_{A}\otimes v_{B},\,\pi^{xy}\rangle, \qquad S = \langle W,\mathcal{E}\rangle . \]

It is a \((1,0)\) tensor, exactly like \(\overrightarrow{d}\), so it takes exactly one factor of \(\mu\) under a chart change, and

\[ E' = \mu E + (1-\mu)\,\bar v_{A}\bar v_{B}, \qquad S' = \mu S + 2(1-\mu)\,\bar v_{A}\bar v_{B} . \]

The local bound is carried to itself by that map, so \(2\) is still the right thing to compare against; the quantum value is not, and slides. With the extra event-types discarded the wall is the depolarising visibility threshold \(1/\sqrt2\)^{10 11}, reproduced exactly rather than by analogy, since on the recorded statistics the dial is the depolarising channel. With the extra event-types valued, the walls open into an interval \(\bigl(\mu_{-},\mu_{+}\bigr]\) with

\[ \mu_{-} = \frac{1-c^{2}}{\sqrt2-c^{2}}, \qquad \mu_{+} = \frac{1+c^{2}}{\sqrt2+c^{2}}, \qquad c=\frac{n-2}{n}, \]

which grows from the single point \(1/\sqrt2\) to \(\bigl(0,2(\sqrt2-1)\bigr]\) — the CHSH detection-efficiency threshold¹² — as the chart grows. Re-charting each wing separately instead is a different map, with two factors of \(\mu\), and it reproduces the detection-efficiency model itself, walls at \(2^{-1/4}\) and \(2(\sqrt2-1)\). Two known experimental numbers, out of the geometry of the chart rather than out of the hardware or the state. In my opinion, this is even better than rediscovering them!

The counting space in the coordinate of each observer is the knowledge the observer has of the system. By adding extra event-types, we claim there are observers who could have more knowledge about the system. Especially we showed \(|S|\) is not an invariant for all observers. The most important conclusion here is that if it's not an invariant, it could not be used as an input of an argument whose output is an absolute statement. The Bell's theorem cannot judge for absolute statements like locality, non-locality, etc. The missing piece is not a hidden variable inside the box. It is hidden knowledge on its surface — an unrecorded event-type.

---

References

Cite

If you found this work useful, please consider citing:

@misc{hadilq2026GeoProbBell,
    author = {{Hadi Lashkari Ghouchani}},
    note = {Published electronically at \url{https://hadilq.com/posts/geometrical-probability-of-bell/}},
    gitlab = {Gitlab source at \href{https://gitlab.com/hadilq/hadilq.gitlab.io/-/blob/main/content/posts/2026-09-08-geometrical-probability-of-bell/index.md}},
    title = {Geometrical Probability of Bell's Theorem},
    year={2026},
}