{"id":"0fe63c79-16ed-4b2e-b9d0-1918b9d74778","arxiv_id":"2412.20728","paper_version":1,"verdict":"CONDITIONAL","confidence":"MODERATE","novelty_score":2.0,"correctness_risk":"medium","formal_verification":"none","parameter_count":0,"one_line_summary":"The paper re-derives known answers to classic probability puzzles and claims the obtuse-triangle probability is 3/4, but that claim relies on an unstated uniform-largest-angle assumption.","lead":"This paper revisits four classic probability puzzles and argues that their paradoxes come from changing the sample space, not from randomness being undefined. It recalculates the broken-stick and obtuse-triangle answers, and claims a unique 'right' probability of 3/4 for a random obtuse triangle.","discovery_kind":"review","skeptic_critique":{"model":"deepseek-v4-flash","headline":"The claim that 3/4 is the right obtuse-triangle probability is not established: it relies on an arbitrary simplex prior, and the paper's own cited results show other natural regularizations disagree.","rationale":"The paper is a readable expository note with several sound sections: the broken-stick sequential calculation, the two-boys sample-space distinction, and the three-prisoners analysis all match standard sources. The L and M method computations, apart from a noted typographical issue, also reproduce known values. My concern is confined to the conclusion. The reader's weakest_assumption identifies the same underlying issue: the Simpler model's uniform-largest-angle assumption is not derived from the infinite-plane wording. I partially agree, and I sharpen the point in two ways. First, the Simpler model is internally incompatible with the Simple model it is meant to simplify: under two independent uniform cuts on a segment of length pi, the largest angle is not uniform on [pi/3, pi] but has density 6(pi - M)/pi^2. The 3/4 answer survives under both priors, but the derivation as written is a non sequitur. Second, the central 'right probability' claim requires uniqueness across natural regularizations, and the paper itself supplies counterexamples: its own L and M methods give 0.64 and 0.82, and Portnoy's transformation-group model gives 0.8426. The agreement of two simplex-based priors does not override these. Thus the conclusion should be weakened to: under the angle-simplex convention, the probability is 3/4. The reader's CONDITIONAL verdict already captures this, so I recommend no change to the verdict rather than a move to reject, because the expository content is sound and the overclaim is fixable by revision.","tokens_in":12412,"tokens_out":7679,"duration_ms":81376,"concrete_test":"Independently re-derive Portnoy's transformation-group result: implement a sampler for triangle shapes under a translation-, rotation-, and scale-invariant prior on triples of points in R^6, compute the fraction of obtuse triangles, and compare with 3/4. If the fraction is not 3/4, then the conclusion that 3/4 is the right probability for the original infinite-plane problem is unsupported. This is a single decisive check of the uniqueness claim, and the paper already cites the expected value, 0.8426.","verdict_should_be":"UNCHANGED","load_bearing_attack":"To sustain the conclusion that 3/4 is the right probability for the obtuse triangle, the paper needs the angle-sum and unit-stick models to be a canonical regularization of 'three points at random on an infinite plane.' That condition is not met. The paper itself quotes Hamming and Portnoy to the effect that no uniform distribution on the infinite plane exists and that the answer depends on the limiting shape; Portnoy reports 0.8426 for a transformation-group model, and the paper's own L and M methods give about 0.64 and 0.82. The fact that two simplex-based models (two uniform cuts on a segment of length pi, and a separately specified uniform largest angle) both give 3/4 is not evidence of uniqueness: each model is an arbitrary prior on triangle shapes, and the agreement is specific to those priors. There is also an internal defect in the derivation: under the Simple model, two independent uniform cuts on [0, pi] induce a largest angle M with density 6(pi - M)/pi^2 on [pi/3, pi], not the uniform density assumed by the Simpler model. The uniform-largest-angle model is a different distribution; it happens to give 3/4 as well, but the stated derivation does not follow from the Simple model. Therefore the central claim overreaches: 3/4 is the answer to a well-posed but conventional problem, not the unique answer to the original ill-posed one.","agreement_with_reader":"partial"},"referee_report":{"model":"deepseek-v4-flash","summary":"The paper revisits four classical probability puzzles—the broken-stick problem, the two-boys problem, the three-prisoners/Monty-Hall problem, and the obtuse random triangle problem. It argues that Bertrand's three chord constructions correspond to different well-defined sample spaces rather than to an ambiguity in the underlying notion of randomness, and it gives a sequential-method calculation for the broken-stick problem leading to probability 0.193147. For the obtuse triangle problem, the paper presents the 'Simple model' in which two uniform cuts on an angle-sum segment of length π yield probability 3/4 for an obtuse triangle, and the 'Simpler model' in which the largest angle is assumed uniform on [π/3, π], again yielding 3/4. The Conclusions assert that because the unit-length-stick and angle-sum approaches agree on 3/4, this is 'the right probability' for the obtuse triangle problem, with 1/4 for the broken-stick problem. Problem statements and calculations for the two-boys and three-prisoners problems are also discussed and resolved through sample-space specification.","tokens_in":12685,"tokens_out":4491,"duration_ms":47517,"significance":"If the paper's central claim were established, it would resolve the obtuse random triangle problem by identifying a unique probability of 3/4 and proving equivalence with the broken-stick problem. The supporting calculations for the broken-stick sequential method, the two-boys conditional/unconditional distinction, and the three-prisoners analysis are sound and clearly presented. The paper also usefully emphasizes that the three Bertrand answers are associated with different events or sample spaces. However, the claimed uniqueness of 3/4 for the obtuse triangle problem is not established: the paper introduces an unproved uniform-largest-angle assumption and, by its own citations, other natural regularizations give different values. The manuscript is an expository treatment with a strong concluding claim that currently overreaches its evidence; nevertheless, the core conditional-probability derivations are correct and the paper could be revised to present a conditional, convention-dependent result rather than a unique answer.","major_comments":[{"comment":"The derivation of the 'Simpler model' is internally inconsistent with the 'Simple model'. Under two independent uniform cuts on [0, π], the ordered angles have joint density 2/π^2 and the largest angle M has density f_M(m) = 6(π - m)/π^2 on [π/3, π], not the uniform density asserted in the sentence 'The range of uniform distribution of the big angle value is [π/3, π]'. The fact that both models give probability 3/4 for the event M > π/2 is a coincidence for this particular event, not a validation of the uniform-largest-angle assumption. Since the paper's Conclusion relies on the 'Simpler model' as one of the two approaches that agree on 3/4, this internal inconsistency directly undermines the central claim.","section":"Obtuse random triangle problem, Simple model and Simpler model"},{"comment":"The statement 'Since different approaches using unit length stick or sum of angles to limit variable ranges give 3/4 then it is the right probability for the Obtuse triangle' does not follow from the preceding analysis. The paper itself quotes Hamming (p. 205) and Portnoy (1994) to the effect that 'at random on an infinite plane' is not well-defined, and Portnoy reports a different value, 0.8426, under a transformation-group model. The paper's own L and M methods yield approximately 0.639 and 0.821. Agreement between the broken-stick and angle-sum conventions shows only that two simplex-based priors give the same answer; it does not show that this answer is the unique or 'right' probability for the original ill-posed problem. The conclusion should be weakened to state that 3/4 is the answer under the stated angle-sum convention.","section":"Conclusions, final paragraph"},{"comment":"The simulation labeled 'Generated' in the Appendix does not provide independent evidence for the uniform-largest-angle model. Its description says that triangles are generated by incrementing the big angle A uniformly over [π/3, π] and then randomly splitting the remaining sum; this directly implements the assumption whose validity is at issue. The resulting row P = 0.75 therefore only confirms the arithmetic of the 'Simpler model', not the appropriateness of that model for the original problem. The paper should either identify this as a model assumption or provide a separate test that does not build in the assumption.","section":"Appendix, 'Generated' method"}],"minor_comments":[{"comment":"The integrand in Eq. [4] is written as x/(x−1), while the probability function f_p in Eq. [3] is x/(1−x); the displayed antiderivative corresponds to the latter, so the sign in the integrand should be corrected and the variable of integration should be labeled.","section":"Broken stick problem, Eq. [4]"},{"comment":"The expression '1/2 ⋆ π / 2/3 ⋆ π' is ambiguous; it should be written as (π/2)/(2π/3) = 3/4 to make the ratio of favorable to total length transparent.","section":"Simpler model"},{"comment":"The text contains several typographical errors and artifacts, including 'O btuse' in the Abstract, 'Betrand #2' in the Appendix table, 'M methode' in the subsection heading, and the garbled sequence 'Jr\\b' in the Bertrand section; these should be corrected in a revision.","section":"Throughout"},{"comment":"The transition from geometric lengths ab/DE in Diagram 1 to the probability function f_p = x/(1−x) in Eq. [3] is compressed; adding one sentence explaining that the conditional probability is the ratio of the favorable subinterval length to the length of the available segment would improve readability.","section":"Broken stick problem, sequential method"}],"recommendation":"major_revision","confidential_remarks":"The manuscript is closer to a pedagogical survey than a research article, and its one novel claim—that 3/4 is the unique 'right' probability for the obtuse triangle problem—is not supported by the arguments presented. The author should be asked to either supply a rigorous invariance or symmetry principle that canonically selects the angle-sum model, or to explicitly retract the uniqueness claim and present 3/4 as the answer under a specified convention. The broken-stick, two-boys, and three-prisoners analyses are solid and could form the basis of a useful expository paper."},"author_rebuttal":null,"desk_editor":{"model":"deepseek-v4-flash","letter":"This is a clear expository treatment of classic ambiguous probability puzzles, and it gets a lot right. The broken-stick analysis is careful and correct, including the sequential 0.193 calculation and the explanation of why sequential vs. simultaneous cutting differ because the cuts are dependent versus independent. The two-boys and three-prisoners discussions are standard but well presented. The framing of Bertrand's three methods as different sample spaces, not merely different randomizing procedures, is useful. The paper also includes simulation results for many sampling schemes, which is commendable.\n\nThe genuinely new bit is the claim that the broken-stick and obtuse-triangle problems are equivalent under the angle-sum model. That is a nice observation, but it does not settle the original ill-posed problem. The conclusion overreaches: saying that because the unit-stick and angle-sum models both give 3/4, this is \"the right probability\" for the obtuse triangle ignores that both models are arbitrary conventions for regularizing a problem with no uniform distribution on the infinite plane. The paper itself quotes Hamming and Portnoy acknowledging the problem is not well-posed, and the L and M methods give 0.64 and 0.82, while Portnoy cites a transformation-group model giving 0.8426. The agreement of two simplex-based priors is not evidence of uniqueness.\n\nThere is also a concrete internal defect. The \"Simpler model\" assumes the largest angle is uniform on [π/3, π], but under the \"Simple model\" of two independent uniform cuts on [0, π], the largest angle is not uniform; it has a piecewise density, approximately 12(π − 2M)/π² for M in [π/3, π/2] and 6(π − M)/π² for M in [π/2, π]. So the stated derivation does not follow from the Simple model. This is a technical flaw, not just a philosophical quibble.\n\nIf submitted as a research paper, I would recommend heavy revision or rejection. The expository parts are fine and could become a useful teaching note if the \"right probability\" language is replaced with \"under a chosen convention.\" As is, the central claim is not supported. Still, I would send it to peer review because the topic is classic and the writing is clear; a referee should focus on the obtuse-triangle conclusion and the unjustified uniformity assumption.","headline":"A readable expository note that correctly re-derives several classic puzzle answers but overreaches when it names 3/4 as 'the right' obtuse-triangle probability.","tokens_in":13232,"tokens_out":4480,"would_cite":false,"duration_ms":42541,"reading_group":"maybe","serious_thinker":"yes","would_accept_peer_review":true},"rs_alignment":null,"lean_confirmation":null,"pith_extraction":{"msc":["60D05","60A05"],"pacs":[],"model":"deepseek-v4-flash","headline":"This paper claims the obtuse random triangle probability is exactly 3/4 under the angle-sum model.","keywords":["obtuse triangle problem","broken stick problem","Bertrand's paradox","two boys problem","three prisoners problem","sample space ambiguity","geometric probability","conditional probability"],"falsifier":"A decisive check is to sample three points uniformly from a circle of radius $R$, from a square of side $2R$, and from a long thin rectangle, increasing $R$ in each case, and record the fraction of obtuse triangles. If the fractions converge to one common value and that value is $3/4$, the claim survives; the paper's own appendix shows they do not converge (different shapes give roughly $0.72$ to $0.87$), so the literal plane problem has no unique empirical limit and the $3/4$ answer lives in the angle-sum model.","tokens_in":12161,"feed_emoji":"📐","tokens_out":9791,"duration_ms":90449,"temperature":0.7,"pith_summary":"Many classic probability puzzles seem to give different answers depending on how 'random' is interpreted. This paper claims that the apparent ambiguity is usually a difference in sample space, not a failure of randomness, and it works through the broken-stick problem, the obtuse random triangle, Bertrand's chords, the two-boys puzzle, and the three-prisoners puzzle. The central result is that the obtuse-triangle problem, once its angles are represented by two uniform cuts on a segment of length $\\pi$, is the same problem as the broken stick: the probability that a random triangle is obtuse is $3/4$, and the probability that three broken-stick pieces form a triangle is $1/4$ for simultaneous cuts. A sequential breaking of the stick gives the different value $0.193147\\ldots$, because the second cut's conditional distribution depends on the first cut. What matters, the paper argues, is that the modeling convention fixes the sample space; the original words 'three points at random on an infinite plane' do not specify such a convention and cannot be tested directly.","feed_headline":"Three random points make an obtuse triangle 75% of the time","feed_subtitle":"A one-line angle-sum model maps the classic puzzle to broken sticks and gives one unambiguous answer.","key_machinery":"The carrying device is the 'Simpler model': a straight line segment, of length $1$ for the broken-stick problem and length $\\pi$ for the angle problem, with two independent uniformly distributed points marking the cuts. In both settings the three resulting segments are the three sides (or the three angles) of a triangle, and the condition for failure is that one segment is longer than half the total, which for angles is exactly an obtuse angle larger than $\\pi/2$. The same one-dimensional diagram yields the area calculation for the event, and for the sequential stick-breaking it produces the conditional density $f_p = x/(1-x)$ for the second cut after a first cut of length $x$, whose integral over $[0,1/2]$ gives $0.193147\\ldots$. The equivalence of perimeter and angle sum is what transfers the probability from one puzzle to the other.","core_discovery":"On the author's own terms, the discovery is that the obtuse random triangle problem has a definite answer once the triangle is described by its three angles as a triplet of positive numbers summing to $\\pi$. Two independent uniform points on a segment of length $\\pi$ split the angle sum into three angles, and the triangle is obtuse exactly when one of those angles exceeds $\\pi/2$. That event has probability $3/4$, matching the event that one of three pieces of a unit stick broken at two independent uniform points exceeds $1/2$, whose complement—the triangle formed from the pieces—has probability $1/4$. The paper also shows that the two classical side-fixing calculations, which give about $0.639$ and $0.821$, differ because fixing the longest or the middle side defines different sample spaces for the other sides, and it identifies the sequential broken-stick value $0.193147\\ldots$ as the conditional-probability version of the same problem.","pith_inferences":["If the angle-sum convention is accepted, a natural testable extension is that any sampling rule whose induced distribution on the three angle gaps is uniform and exchangeable should reproduce $3/4$; rules with other induced distributions should not, which would give a diagnostic for 'random triangle' in other models.","The paper leaves implicit that the 3/4 answer is a property of the chosen model rather than of the original phrase 'at random on an infinite plane'; the author's own quotations (that no uniform distribution on the infinite plane exists) suggest the claim is best read as solving a well-posed reformulation.","One could build a physical sequential-breaking apparatus and test whether the empirical triangle-formation rate converges to $0.193147\\ldots$; this would separate the conditional-probability claim from the simultaneous-cut claim in a way the paper does not attempt."],"forward_implications":["The obtuse-triangle and broken-stick problems are probabilistically the same, so results proven for either transfer to the other under the angle-sum or unit-perimeter convention.","Simultaneous random cuts yield a $1/4$ chance that the pieces form a triangle; a sequential breaking procedure yields about $0.193$, so reports of a single 'correct' broken-stick answer are incomplete without specifying the sampling protocol.","The three Bertrand answers correspond to three different objects (random chord, chord perpendicular to a random point on a radius, chord perpendicular to a random point in the disk), so no additional information is needed to resolve Bertrand's paradox.","The two-boys answer is $1/3$ when the sample space is families with at least one boy and $1/2$ when the question is the conditional probability in all two-child families; the three-prisoners answer keeps prisoner A at $1/3$ and raises prisoner C to $2/3$.","The literal infinite-plane formulation of the obtuse triangle cannot be realized physically or computationally; the paper's equivalent angle-sum or unit-line versions can be simulated and give $3/4$."],"supporting_citations":[{"why":"Supplies the classic formulations and the broken-stick ambiguity statement from which the paper starts.","marker":"[Gar2001]"},{"why":"Provides the two obtuse-triangle methods, the warning about uniform distributions on the infinite plane, and the broken-stick context.","marker":"[Ham1991]"},{"why":"Gives the analytic derivation of 3/4 via six-point sampling and spherical symmetry, plus the reference for one of the side-fixing methods.","marker":"[Por1994]"},{"why":"Offers the area-geometry derivation of 3/4 that the paper uses as a bridge between the broken-stick and obtuse-triangle problems.","marker":"[Ede2015]"},{"why":"States the original pillow-problem of three points on an infinite plane and the earlier 0.639 result that the paper traces.","marker":"[Dod1893]"},{"why":"Defines Bertrand's three chord constructions, which ground the paper's claim that the three answers are different sample spaces.","marker":"[Ber1889]"}],"fun_headline_variants":["Obtuse triangle probability: 75% via angle triplet method","One unambiguous answer for obtuse random triangle: 3/4","Broken stick and obtuse triangle: same 75% probability","Angle-sum perspective gives 75% chance of obtuse triangle"],"cache_read_input_tokens":3200,"weakest_assumption_plain":"The load-bearing assumption is that 'random triangle' means two independent uniform cuts on a segment whose length is the sum of the angles, and in particular that the largest angle is uniformly distributed on $[\\pi/3,\\pi]$; the original wording about three points on an infinite plane does not force this convention.","fun_headline_variants_meta":{"raw":{"variants":["Obtuse triangle probability: 75% via angle triplet method","One unambiguous answer for obtuse random triangle: 3/4","Broken stick and obtuse triangle: same 75% probability","Angle-sum perspective gives 75% chance of obtuse triangle"]},"model":"deepseek-v4-flash","effort":"low","cost_usd":0.000195,"raw_usage":{"total_tokens":1296,"prompt_tokens":821,"completion_tokens":475,"prompt_tokens_details":{"cached_tokens":384},"prompt_cache_hit_tokens":384,"prompt_cache_miss_tokens":437,"completion_tokens_details":{"reasoning_tokens":401}},"tokens_in":437,"tokens_out":475,"duration_ms":5293,"temperature":1.0,"reasoning_tokens":401,"cache_read_input_tokens":384,"cache_creation_input_tokens":0},"cache_creation_input_tokens":0},"created_at":"2026-08-10T23:13:38.776080+00:00","model_set":{"reader":"deepseek-v4-flash"},"falsifier":"A decisive check is to sample three points uniformly from a circle of radius $R$, from a square of side $2R$, and from a long thin rectangle, increasing $R$ in each case, and record the fraction of obtuse triangles. If the fractions converge to one common value and that value is $3/4$, the claim survives; the paper's own appendix shows they do not converge (different shapes give roughly $0.72$ to $0.87$), so the literal plane problem has no unique empirical limit and the $3/4$ answer lives in the angle-sum model.","supporting_citations":[],"review_version":1}