Probability and Ambiguity

A. Hayrapetyan

REVIEW 3 major objections 4 minor 6 references

Probability and Ambiguity

T0 review · 3 major / 4 minor · reviewed 2026-08-10 · deepseek-v4-flash

Challenge this review Re-run · record.json Download PDF Read on arXiv ↗ Review JSON

Pith's one-line read This paper claims the obtuse random triangle probability is exactly 3/4 under the angle-sum model.

desk verdict 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. read the letter →

arxiv 2412.20728 v1 pith:SBSUY435 submitted 2024-12-30 math.PR

A. Hayrapetyan This is my paper · ORCID

classification math.PR MSC 60D0560A05

keywords obtusetriangleproblembrokenstickBertrand'sparadoxtwoboysthreeprisonerssamplespaceambiguitygeometricprobabilityconditional

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

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.

What carries the argument

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.

What would settle it

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.

Watch

Extended reading notes

Core claim

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.

Load-bearing premise

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.

Editorial extensions

If this is right

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 $3/4$.

Reading between the lines

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

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.

Share X Bluesky LinkedIn Reddit HN

Signed reviews

No signed human review yet.

Editorial analysis

A structured set of objections, weighed in public.

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

Desk Editor's Note

read the 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.

The 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.

There 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.

If 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.

Referee Report

3 major / 4 minor

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.

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 (3)

[Obtuse random triangle problem, Simple model and Simpler model] 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.
[Conclusions, final paragraph] 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.
[Appendix, 'Generated' method] 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.

minor comments (4)

[Broken stick problem, Eq. [4]] 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.
[Simpler model] 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.
[Throughout] 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.
[Broken stick problem, sequential method] 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.

Circularity Check

1 steps flagged · score 6.0 of 10

The Simpler-model route to 3/4 is self-definitional: the uniform-largest-angle sample space is assumed, and the probability is read off as an interval ratio, so the later claim that 3/4 is 'the right probability' is partly forced by the prior rather than independently derived.

self definitional [Obtuse random triangle problem, 'Simpler model' paragraph and Conclusions]
"A triangle can have only one big angle and only this angle can be obtuse. The range of uniform distribution of the big angle value is [π/3, π]. This is the sample space. Favorable outcomes are in the [π/2, π] range. Hence, the probability of randomly getting an obtuse triangle is 1/2 ⋆ π / 2/3 ⋆ π."

The 3/4 value is the ratio of the favorable interval [π/2, π] to the assumed sample-space interval [π/3, π] under a postulated uniform distribution of the largest angle. That uniform-largest-angle postulate is not derived from 'three points at random on an infinite plane'; the paper itself quotes Hamming and Portnoy to the effect that no uniform planar distribution exists and that different limiting shapes give different answers. Thus the Simpler model encodes the answer in its prior. The Conclusions then use the 3/4 obtained from 'sum of angles' as evidence that it is 'the right probability', so the model's assumption is doing the work that a derivation should do.

full rationale

The broken-stick calculations (1/4 simultaneous, 0.193 sequential) are genuine conditional-probability computations and are not circular. The L and M method area ratios are also independently computed, and the paper honestly reports that they give about 0.64 and 0.82. The only circular step is the Simpler model, which assumes a uniform distribution of the largest angle and then obtains 3/4 by interval division; this is a self-defined model, not a first-principles result. The paper's self-citation [Hay2024] on Bertrand's paradox is not load-bearing for the obtuse-triangle conclusion. The deeper problem is that the original obtuse-triangle question is ill-posed, and the paper's assertion that 3/4 is 'the right probability' overreaches; however, overreach due to an arbitrary prior is only partially a circularity issue, so the score is moderate rather than extreme.

Assumptions & free parameters 0 free parameters · 6 assumptions · 0 invented entities

No fitted constants are used in the paper's central derivations. The main fragile input is the modeling assumption that random triangles are generated by uniform cuts on an angle-sum segment and that the largest angle is uniform on [pi/3, pi]; this is ad hoc for the original infinite-plane formulation.

assumptions (6)

domain assumption Uniform distribution on a line segment: probability proportional to length, applied to cuts on the stick and on the angle-sum segment.
Used throughout the broken-stick and obtuse-triangle models; not derived, standard for these puzzles.
domain assumption In the sequential broken-stick procedure, the second cut point is uniformly distributed along the chosen piece.
Explicitly stated in the Broken stick section as the conditional range for c2; this is what makes the x/(1-x) probability function valid.
ad hoc to paper The largest angle of a randomly chosen triangle is uniformly distributed on [pi/3, pi].
Introduced in the 'Simpler model' subsection with no derivation and no connection to the original infinite-plane sample space; it is the main unsupported premise for the 3/4 claim.
ad hoc to paper Two independent uniform cuts on the angle-sum segment define 'random triangle' in the obtuse-triangle problem.
Used to equate the obtuse triangle problem to the broken-stick model; this is one possible sampling convention, not entailed by the original problem statement.
domain assumption Laplace's definition of equally possible cases applies to continuous sample spaces via measure ratios.
Stated in the Randomness section as the working definition; it justifies the ratio-of-areas calculations.
standard math In an equilateral triangle, the sum of the perpendicular distances from any interior point to the three sides equals the altitude.
Used in the broken-stick diagram to map point positions to piece lengths.

how reviews work

0 comments

Cite this review

Pith. "Pith review of Probability and Ambiguity." pith.science (2026). https://pith.science/paper/SBSUY435

@misc{pith2026241220728,
  author       = {Pith},
  title        = {Pith review of: Probability and Ambiguity},
  year         = {2026},
  howpublished = {\url{https://pith.science/paper/SBSUY435}},
  note         = {Machine review of arXiv:2412.20728}
}

read the original abstract

The title of the article is identical to the title of Chapter 21 in Gardner (2001): because we are going to analyze the probability calculations and the ambiguity of the problem statements. We will analyze 3 out of 4 problems from Gardner (2001): broken stick, two boys, three prisoners; and the obtuse random triangle in Hamming (1991) problems. Keywords: Broken stick problem, Obtuse triangle problem, Bertrand's problem, Two sons problem, Three prisoners problem, Monty Hall problem; probability function, probability density, randomness, ambiguity

Discussion (0). Continue with ORCID to comment.

Reference graph

Works this paper leans on

6 extracted references · 6 canonical work pages

[1]

Solving the Hard Problem of Bertrand's Paradox

[Aer2014] Aerts, D., de Bianchi M.S., (2014), Solving the hard problem of Bertrand’s paradox. arXiv:1403.4139v2 [physics.hist-ph] 27 Jun 2014 [Ban2009] N. Banerjee. (2009), Random Obtuse Triangles and Convex Quadrilaterals. Master of Science in Computation for Design and Optimization thesis. Massachusetts Institute of Technology

work page Pith review arXiv 2014
[1994]

(2011) Resolving Bertrand's Probability Paradox

[Wan2011] Wang, J., Jackson, R.. (2011) Resolving Bertrand's Probability Paradox. Int. J. Open Problems Compt. Math., Vol. 4, No. 3, September

work page 2011
[2006]

Bertrand's paradox: a physical solution

[Che2023] R.A. Chechile. (2023), Bertrand’s Paradox Resolution and Its Implications for the Bing–Fisher Problem. Mathematics 2023, 11, 3282 [DiP2010] Di Porto, P., Crosignani, B., Ciattoni, A., Liu, H. C. (2010), Bertrand’s paradox: a physical solution. arXiv:1008.1878v1 [physics.data-an] 11 Aug

work page Pith review arXiv 2023
[2009]

(1889), Calcul des probabilités

[Ber1889] Bertrand, J. (1889), Calcul des probabilités. Gauthier-Villars, Paris. [Cha2006] G. Chaitin. (2006), Meta Math!, Vintage Books, NY,

work page 2006
[2010]

Random Triangle Theory with Geometry and Applications

[Dod1893] C.L. Dodgson (Lewis Carroll), (1893) Curiosa Mathematica. Part II : Pillow-Problems, thought out during Sleepless Nights, Macmillan, L. [Ede2015] A. Edelman, G. Strang, (2015) Random Triangle Theory with Geometry and Applications, arXiv:1501.03053v1 [math.HO] 9 Jan

work page Pith review arXiv 2015
[2015]

at random

[Gar2001] Gardner, M. (2001), The Colossal Book of Mathematics. W.W. Norton & Company, NY,. [Ham1991] R.W. Hamming, (1991) The Art of Probability. For scientists and engineers. CRC Press 2018 [Hay2024] Hayrapetyan, A. (2024), A Note on Bertrand’s paradox. academia.edu/118451522 [Jay1973] Jaynes, E. T. (1973), The Well-Posed Problem, Foundations of Physics...

work page arXiv 2001

Pith tools

Reviewed August 10, 2026 · model on record in the stance chip above.

[1] [1]

Solving the Hard Problem of Bertrand's Paradox

[Aer2014] Aerts, D., de Bianchi M.S., (2014), Solving the hard problem of Bertrand’s paradox. arXiv:1403.4139v2 [physics.hist-ph] 27 Jun 2014 [Ban2009] N. Banerjee. (2009), Random Obtuse Triangles and Convex Quadrilaterals. Master of Science in Computation for Design and Optimization thesis. Massachusetts Institute of Technology

work page Pith review arXiv 2014

[2] [1994]

(2011) Resolving Bertrand's Probability Paradox

[Wan2011] Wang, J., Jackson, R.. (2011) Resolving Bertrand's Probability Paradox. Int. J. Open Problems Compt. Math., Vol. 4, No. 3, September

work page 2011

[3] [2006]

Bertrand's paradox: a physical solution

[Che2023] R.A. Chechile. (2023), Bertrand’s Paradox Resolution and Its Implications for the Bing–Fisher Problem. Mathematics 2023, 11, 3282 [DiP2010] Di Porto, P., Crosignani, B., Ciattoni, A., Liu, H. C. (2010), Bertrand’s paradox: a physical solution. arXiv:1008.1878v1 [physics.data-an] 11 Aug

work page Pith review arXiv 2023

[4] [2009]

(1889), Calcul des probabilités

[Ber1889] Bertrand, J. (1889), Calcul des probabilités. Gauthier-Villars, Paris. [Cha2006] G. Chaitin. (2006), Meta Math!, Vintage Books, NY,

work page 2006

[5] [2010]

Random Triangle Theory with Geometry and Applications

[Dod1893] C.L. Dodgson (Lewis Carroll), (1893) Curiosa Mathematica. Part II : Pillow-Problems, thought out during Sleepless Nights, Macmillan, L. [Ede2015] A. Edelman, G. Strang, (2015) Random Triangle Theory with Geometry and Applications, arXiv:1501.03053v1 [math.HO] 9 Jan

work page Pith review arXiv 2015

[6] [2015]

at random

[Gar2001] Gardner, M. (2001), The Colossal Book of Mathematics. W.W. Norton & Company, NY,. [Ham1991] R.W. Hamming, (1991) The Art of Probability. For scientists and engineers. CRC Press 2018 [Hay2024] Hayrapetyan, A. (2024), A Note on Bertrand’s paradox. academia.edu/118451522 [Jay1973] Jaynes, E. T. (1973), The Well-Posed Problem, Foundations of Physics...

work page arXiv 2001