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REVIEW 3 major objections 3 minor 30 references

Quantum-Optical set-up for the Monty Hall problem

T0 review · 3 major / 3 minor · reviewed 2026-08-14 · deepseek-v4-flash

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Pith's one-line read A quantum-optical set-up can realize the Monty Hall game, reproducing the classical 1/3 vs 2/3 odds in its semiclassical limit.

desk verdict The central claim that this optical setup verifies Monty Hall doesn't survive: the host's reveal is independent of the player's choice, and the 1/3 vs 2/3 numbers evaporate once you discard the impossible events. read the letter →

arxiv 1908.05774 v2 pith:EVJJY6JX submitted 2019-08-15 quant-ph

L. F. Quezada , A. Martín-Ruiz , A. Frank , E. Nahmad-Achar This is my paper · ORCID
classification quant-ph PACS 03.67.-a42.50.-p03.65.-w
keywords MontyHallproblemquantumgametheoryopticspolarizationentanglementspontaneousparametricdown-conversionnoisychannelcoincidencecountingplayerswitchingstrategy
verification ladder T0 review T1 audit T2 compute T3 formal

The pith

A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.

The reading

This paper proposes an experimental quantum-optical realization of the Monty Hall problem using a pair of polarization-entangled photons produced by spontaneous parametric down-conversion. The authors encode the player's initial choice and the prize's location as three-mode states of the two photons, and model the host's door-opening by a polarization-dependent operator. In the semi-classical limit the model reproduces the classical probabilities $P_{\mathrm{ns}}=1/3$ and $P_{\mathrm{s}}=2/3$, and averaged over random parameter choices the non-entangled game gives the player a switching advantage of about 1.73 to 1. Entanglement between player and host reverses the advantage, and noise on the channel acts in the player's favor in the entangled case. If the set-up works as described, it offers a table-top demonstration of the paradox with pedagogical value.

What carries the argument

The load-bearing device is the door-opening operator $\hat{D}_o=\cos(\phi_1)|1_b\rangle\langle 1_b|+\sin(\phi_2)|2_b\rangle\langle 2_b|+\sin(\phi_3)|3_b\rangle\langle 3_b|$, together with the two-doors-remained-closed condition $\cos^2\phi_1+\sin^2\phi_2+\sin^2\phi_3=2$. It acts on the prize subsystem alone, de-normalizes the state, and is renormalized before computing winning probabilities. In the optical set-up this operator is implemented by three polarizers in front of Bob's detectors, while polarization rotators and beam splitters set the amplitudes $a_i$ and $b_i$; coincidences between Alice and Bob detectors are normalized to obtain the probabilities of winning by switching and by not switching.

What would settle it

Run the proposed set-up with uniform amplitudes $a_i=b_i=1/\sqrt{3}$ and door-opening angles that select two doors, then compare the coincidence-normalized counts to $P_{\mathrm{ns}}=1/3$ and $P_{\mathrm{s}}=2/3$; a statistically significant deviation would refute the semi-classical claim.

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Extended reading notes

Core claim

The central claim is that the Monty Hall paradox can be reproduced and extended in a concrete photon experiment. With Bob's prize state uniform, Alice's choice state uniform, and the door-opening operator projecting onto two of the three doors, equations (12) and (13) give exactly $P_{\mathrm{ns}}=1/3$ and $P_{\mathrm{s}}=2/3$. When the parameters are treated as random variables, the non-entangled protocol yields $\langle P_{\mathrm{ns}}\rangle_{\mathrm{ran}}\approx 0.3664$ and $\langle P_{\mathrm{s}}\rangle_{\mathrm{ran}}\approx 0.6336$, i.e. switching is about 1.73 times better. The entangled version inverts this, giving $\langle P_{\mathrm{e,ns}}\rangle_{\mathrm{ran}}\approx 0.5189$ and $\langle P_{\mathrm{e,s}}\rangle_{\mathrm{ran}}\approx 0.4811$. The paper further claims that in the strategy-based game the host can tune the door-opening to make switching three times better in the non-entangled case, or, with entanglement, can make the two choices equally good; and that Pauli noise on Alice's photon does not alter the non-entangled results but improves the switching odds in the entangled game.

Load-bearing premise

The argument rests on treating the polarizer-based operation as a faithful stand-in for the host's informed reveal, even though in the model it acts on the prize alone and is not conditioned on which door the player chose.

Editorial extensions

If this is right

  • If the set-up is built as described, a university teaching lab can display the $1/3$ vs $2/3$ split by counting photon coincidences.
  • The random-game average predicts a robust switching advantage near $1.73{:}1$ even without fine-tuning the angles.
  • Entangling the player's and host's photons removes the switching advantage, so the same hardware can illustrate how correlations change a game.
  • With a noisy channel, the entangled game's advantage tilts toward switching as noise increases, making the effect of decoherence visible.
  • The host can tune the door-opening polarizers to either amplify the advantage to a $3{:}1$ ratio or, in the entangled case, erase it entirely.

Reading between the lines

Editorial extensions of the paper, not claims the author makes directly.

  • The classical Monty Hall reveal depends on which door the player picked; building a feed-forward from Alice's detected choice into the door-opening polarizer settings would turn the present set-up into a fully conditional version.
  • A natural next step is to add a classical feed-forward that sets the polarizer angles based on Alice's detected choice, which would implement the conditional host rule and could restore exact $1/3$/$2/3$ probabilities outside the symmetric limit.
  • The same coincidence-counting geometry could test whether a generalized $n$-door version retains $P_{\mathrm{ns}}=1/n$ and $P_{\mathrm{s}}=(n-1)/(n-m-1)\cdot 1/n$ once more spatial modes are added.
  • In a remote-play configuration, the measured switching-versus-not-switching odds could serve as a diagnostic of the channel's Pauli noise parameter $p$.
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Editorial analysis

A structured set of objections, weighed in public.

Desk editor's note, referee report, and a circularity audit.

Referee Report

3 major / 3 minor

Summary. The paper proposes a quantum-optical implementation of the Monty Hall problem. Two photonic qudits (or their polarization/spatial modes) represent Alice's initially chosen door and Bob's prize location; a 'door-opening operator' acting on Bob's subsystem is introduced, followed by renormalization, and the winning probabilities for switching and not switching are taken as sums of squared amplitudes over diagonal and off-diagonal coincidence events. The authors analyze a symmetric unentangled ('semi-classical') case, which yields the classical probabilities 1/3 and 2/3, and then study random and strategy-based settings, with and without entanglement, including the effect of a Pauli channel acting on Alice's photon.

Significance. The concrete optical construction is a strength: the paper gives an explicit map from beam-splitter, polarizer, and rotator settings to the amplitudes in Eqs. (26)-(31), and the algebraic steps in Sections III and IV are internally consistent. The inclusion of a separable/entangled comparison and a noise analysis via a Pauli channel is useful for a table-top pedagogical demonstration. If the mapping to the actual Monty Hall problem were faithful, the proposal would be a nice way to 'verify' the classical counter-intuitive result. That mapping, however, is the paper's central weakness: the host's operation does not depend on the player's initial choice, so the presented 1/3 versus 2/3 result is an unconditional average rather than the conditional probability that defines the Monty Hall problem. The paper therefore needs substantial reworking of its central claim before the contribution can be accepted.

major comments (3)
  1. [Section II, Eqs. (8)-(13)] The central claim that the semi-classical case reproduces the Monty Hall probabilities is not supported. Equation (8) acts only on Bob's prize subsystem, so the host's operation is independent of Alice's chosen door. In the classical game, as correctly described in Table I, the host opens a door k with k ≠ i,j; in particular, the host can never open the door the player chose. For the semi-classical choice D_o = |2_b><2_b| + |3_b><3_b| (door 1 opened), Eqs. (12)-(13) count the events A1-B2 and A1-B3 as winning by switching, even though those are runs where the host has opened the door that Alice selected. If those events are discarded and the remaining four events are renormalized, the four surviving events are equally weighted and P_ns = P_s = 1/2. Thus the claimed 1/3 versus 2/3 arises only because one averages over all three possible opened doors while keeping Alice's choice independent; it is not the conditional posterior produced by a knowledgeable host. This is the load-bearing point of the paper and must be fixed by either making the door-opening operator depend on Alice's choice or by explicitly reframing the model as a non-adaptive variant rather than a realization of the Monty Hall problem.
  2. [Section II, Eq. (9), and Section V, Eqs. (59)-(61)] The 'two-doors-remained-closed condition' does not encode the intended host action. For generic φ1, φ2, φ3 satisfying Eq. (9), all three β_i are nonzero, so no door is actually opened; the condition merely fixes the sum of the squared transmission factors. The special choices listed in the semi-classical case are projections, but the random-game region defined by Eqs. (59)-(61) includes non-projective attenuations. Consequently, the random-game expectation values in Eqs. (64)-(65) are averages over a process in which the host never opens a door, and the switching advantage ⟨P_s⟩ ≈ 0.6336 versus ⟨P_ns⟩ ≈ 0.3664 is largely a combinatorial effect of there being six off-diagonal versus three diagonal coincidence channels. This should be stated explicitly, or the model should be revised so that the host's operation genuinely removes one door.
  3. [Section II, 'semi-classical case' paragraph] The statement that the symmetric state a_i = b_i = 1/√3, together with the three special choices of φ, constitutes a case that 'actually resembles the classical problem' is not justified. In the classical Monty Hall game the host's choice of door carries information about the prize location conditioned on the player's initial choice; the likelihood ratio is 1:2 (two empty doors when the player's first choice hides the prize versus only one when it does not). The operator in Eq. (8) has no access to Alice's state and therefore cannot produce this likelihood ratio. The numbers 1/3 and 2/3 in Eqs. (12)-(13) follow from the uniform and independent distributions of Alice's choice and the door that is projected out, not from a conditional host strategy. The paper should either introduce a conditional operation or explicitly classify its scheme as a different game, such as an 'ignorant host' or a host who may reveal the player's door.
minor comments (3)
  1. [Section III, Fig. 1 caption] The caption says the BBO crystal produces photons entangled 'in both position and polarization', while the text and Eq. (17) describe only polarization entanglement; please align the two descriptions.
  2. [Section V, Eq. (62)] Defining the probability density ρ through its reciprocal is unconventional; it would be clearer to state the actual density and its support explicitly.
  3. [Section III, after Eq. (35)] The sentence explaining that renormalization is 'experimentally justified' by Eq. (35) would benefit from an explicit note that Eq. (35) is a conditional normalization on detected coincidences, not a proper quantum-state renormalization after a unitary evolution.

Circularity Check

0 steps flagged · score 0.0 of 10

No circularity: the semiclassical 1/3 vs 2/3 result is a transparent evaluation of the model's own definitions, and no fitted parameter, self-citation, or hidden reduction carries the argument.

full rationale

The derivation is self-contained. Equations (12) and (13) define the winning probabilities in the proposed quantum-optical model by direct analogy with the classical sample space of Table I; no parameter is fitted to external data and no prior result of the same authors is load-bearing. The semiclassical result is obtained by an explicit choice a_i = b_i = 1/sqrt(3) and a door-opening projection that annihilates one door; substituting these values into (12) and (13) gives exactly 1/3 and 2/3. This is a transparent modeling choice, not a hidden input masquerading as a prediction. The random-game and strategy-based results are computed from the model's probability distributions over the angles, and the noise section evaluates the model under a Pauli channel; none of these reduce by construction to the claimed classical output. The paper's substantive weakness—that the door-opening operator (8) acts on Bob's prize subsystem alone and does not implement the host's conditional dependence on Alice's initial choice—is a fidelity or correctness concern about whether the setup really implements the classical Monty Hall game, not a circularity in the derivation. There are no self-citations, no fitted inputs called predictions, and no imported uniqueness theorems. The central claim is therefore independent and non-circular.

Assumptions & free parameters 0 free parameters · 5 assumptions · 0 invented entities

The central claim rests on a custom mapping between quantum states and Monty Hall events, plus a host action that is not conditioned on the player's choice. There are no fitted numbers, no new particles, and no data. The model is self-contained but also self-referential: the semi-classical limit is constructed to reproduce 1/3 and 2/3.

assumptions (5)
  • ad hoc to paper Door-opening operator (Eq. 8) with the two-doors-remained-closed condition (Eq. 9) models the host's action.
    The operator is diagonal in the prize basis and does not depend on the player's chosen door; the constraint is introduced solely so that two doors remain effectively available. This does not match the classical host's conditional reveal.
  • ad hoc to paper Probabilities of winning are identified with sums of squared amplitudes for matching and non-matching detector pairs (Eqs. 12-13, 15-16).
    This defines the game rules; it maps the Monty Hall sample space onto coincidence detections.
  • ad hoc to paper The random-game analysis assumes all four player/host angles are independent and uniformly distributed over [0, pi/2] (Section V).
    This uniform distribution over strategies is chosen by the authors, not derived from a physical process or a game equilibrium.
  • domain assumption The noise is modeled as a Pauli channel acting only on Alice's photon (Eq. 48), with px=py=pz=p/3.
    This is a standard quantum information noise model, appropriate for a depolarizing channel, though the paper uses a simplified symmetric case.
  • domain assumption The SPDC source and the polarizing elements produce the states (17) and (18) and implement the operations in Section III.
    Assumes the physics of type-II SPDC, PBS and polarizers as described, which is standard quantum optics.

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Cite this review

Pith. "Pith review of Quantum-Optical set-up for the Monty Hall problem." pith.science (2026). https://pith.science/paper/EVJJY6JX

@misc{pith2026190805774,
  author       = {Pith},
  title        = {Pith review of: Quantum-Optical set-up for the Monty Hall problem},
  year         = {2026},
  howpublished = {\url{https://pith.science/paper/EVJJY6JX}},
  note         = {Machine review of arXiv:1908.05774}
}
read the original abstract

A quantum version of the Monty Hall problem is proposed inspired by an experimentally-feasible, quantum-optical set-up that resembles the classical game. The expected payoff of the player is studied by analyzing the classical expectation values of the obtained quantum probabilities. Results are examined by considering both entanglement and non-entanglement between player and host, and using two different approaches: random and strategy-based. We also discuss the influence of noise on the game outcome when the parties play through a noisy quantum channel. The experimental set-up can be used to quickly verify the counter-intuitive result of the Monty Hall problem, adding pedagogic value to the proposal.

Figures

Figures reproduced from arXiv: 1908.05774 by the authors.

Figure 1
Figure 1. Diagram of the experimental set-up proposed for the Monty Hall problem. The BBO type II crystal produces a pair [PITH_FULL_IMAGE:figures/full_fig_p006_1.png] view at source ↗
Figure 2
Figure 2. Probabilities of Alice winning the semi-classical Monty Hall game by switching (continuous line) and by not switching [PITH_FULL_IMAGE:figures/full_fig_p010_2.png] view at source ↗
Figure 3
Figure 3. Expectation values of the probabilities (random case) of Alice winning by switching (black) and by not switching [PITH_FULL_IMAGE:figures/full_fig_p010_3.png] view at source ↗
Figures from the paper (3 more)
Figure 4
Figure 4. Figure 4: Maximum (black) and minimum (gray) of the expectation value of the probability of Alice winning by switching [PITH_FULL_IMAGE:figures/full_fig_p012_4.png]
Figure 5
Figure 5. Figure 5: Expectation value of the probability of Alice winning by switching, as a function of the door-opening parameters [PITH_FULL_IMAGE:figures/full_fig_p012_5.png]
Figure 6
Figure 6. Figure 6: Absolute value of the difference between expectation values of the probabilities of Alice winning by not switching [PITH_FULL_IMAGE:figures/full_fig_p013_6.png]

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Reference graph

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