REVIEW 5 minor 9 references
A Probability Puzzle
T0 review · 0 major / 5 minor · reviewed 2026-08-11 · deepseek-v4-flash
Pith's one-line read Given a uniform prior on the number of red balls in a 100-ball urn, a red first draw makes the second draw red with probability 2/3.
desk verdict A correct, clearly written expository solution to a known puzzle; nothing new mathematically, but useful for teaching. read the letter →
The pith
A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.
The reading
What carries the argument
The load-bearing machinery is the identity $P(r_2 \mid r_1) = P(r_1 \cap r_2)/P(r_1)$ together with the law of total probability, used to express both probabilities as sums over the hidden composition variable $R$. The uniform prior $P(R = r) = 1/101$ makes each term concrete: drawing red twice from an urn with $r$ red balls has probability $(r/100)((r-1)/99)$, and drawing red once has probability $r/100$. The ratio of the two summed expressions cancels to $2/3$. The mechanism behind the surprising value is that conditioning on the first red draw assigns more posterior weight to urns with many red balls.
What would settle it
Simulate the process many times: pick $R$ uniformly from 0 to 100, draw two balls without replacement, and condition on the first being red. The proportion of those trials with a red second ball should approach 2/3; a result that clearly does not would show the derivation is mistaken.
Extended reading notes
Core claim
The paper's central claim is that $P(r_2 \mid r_1) = 2/3$, where $r_1$ and $r_2$ are the events that the first and second draws are red. Letting $R$ be the number of red balls, assumed uniform on $\{0,\dots,100\}$, the derivation expands both numerator and denominator by total probability. The numerator is the sum over $r \ge 2$ of $(r/100)((r-1)/99)(1/101)$, and the denominator is the sum over $r \ge 1$ of $(r/100)(1/101)$. After applying formulas for sums of integers and their squares, the ratio simplifies to $2/3$. Thus, conditional on a red first draw, the second draw is red twice as often as it is green.
Load-bearing premise
The entire calculation assumes the number of red balls was picked uniformly from 0 to 100; if the prior over $R$ is anything else, the 2/3 answer no longer follows.
Editorial extensions
If this is right
- If the first ball is red, the second ball is red with probability 2/3, making red twice as likely as green.
- The same total-probability computation yields 2/3 for any urn size $n \ge 2$, so the puzzle's choice of 100 balls is not what drives the answer.
- The posterior distribution of $R$ shifts toward larger values after a red first draw, and this shift is the reason the second draw favors red.
- The puzzle can be taught using only conditional probability and total probability, making it a compact classroom illustration of those concepts.
Reading between the lines
- Changing the prior from uniform to, say, a binomial distribution on $R$ would produce a different second-draw probability; computing it would show how much of the 2/3 answer depends on the uniform assumption.
- The same conditioning technique can be applied to variants where two or more balls are drawn before the prediction, or where draws are made with replacement; the ratio-of-sums structure would still carry the calculation.
- A simulation experiment, generating $R$ uniformly, drawing two balls without replacement, and conditioning on a red first draw, would give empirical frequencies approaching 2/3 and could serve as a classroom check.
Signed reviews
Editorial analysis
A structured set of objections, weighed in public.
Referee Report
Summary. The paper analyzes a probability puzzle originally posted by Daniel Litt: an urn contains 100 balls, some red and some green, with the number of red balls chosen uniformly from 0 to 100; after drawing a red ball first, is a second draw more likely to be red or green? The author formulates the desired quantity as P(r2|r1), applies the law of total probability with a uniform prior on the number of red balls, and evaluates the resulting sums using formulas for sums of integers and squares. The conclusion is that P(r2|r1)=2/3, so a second red ball is more likely than a green one. The paper frames the solution as an instructive exercise in conditional probability and total probability, and it connects the puzzle to Bertrand's box paradox and the Monty Hall problem.
Significance. If the derivation is correct, this is a clean and useful pedagogical example: a puzzle that is notoriously counterintuitive is resolved by a short, explicit total-probability computation using only material from a typical undergraduate probability course. The calculation is transparent, has no fitted parameters, and is self-contained. The main interpretive assumption is the uniform prior on the number of red balls, which is stated in the setup and is a natural reading of the hat-drawing mechanism. The note also gives proper credit to earlier work, including the independent derivation in Paterson and Stinson's preprint. Its value lies mainly in the clarity and accessibility of the exposition rather than in new mathematical content, which is appropriate for the stated goal.
minor comments (5)
- [Section 2] The computation uses draws without replacement via the factor (r-1)/99, but this assumption is never stated explicitly; please add a sentence saying that the first ball is not returned before the second draw.
- [Section 2] The puzzle text says “some red and some green,” which a literal reader could take to exclude R=0 and R=100; the solution instead uses a uniform prior on the full set {0,1,...,100}. If the support were {1,...,99}, the exact answer would become 323400/490050 ≈ 0.66 rather than 2/3. Please add a sentence acknowledging this ambiguity and explaining the intended interpretation, since the exact value 2/3 depends on it.
- [Section 2] In the final display, the simplification from (1/99)(201/3 - 1) to 2/3 is somewhat abrupt; expanding 201/3 - 1 = 66 would make the algebra easier for the intended undergraduate audience.
- [Section 1] The claim that “only about 22% chose correctly” is introduced in the paragraph after the sentence citing [1]; citing [1] directly at the point of that statistic would improve traceability.
- [Abstract and Introduction] Several line-break artifacts appear in the abstract (e.g., “Toron to,” “probabilit y,” “undergrad u-ate”); these should be cleaned up in the final typeset version.
Circularity Check
No circularity: the result is a direct conditional-probability computation from the puzzle's stated uniform prior, with no fitted parameters and no load-bearing self-citation.
full rationale
The paper's only substantive derivation is the computation of P(r2|r1) in Section 2. It defines R as a uniform discrete random variable on {0,...,100}, conditioned on the puzzle's description of the number of red balls being picked from a hat, and then applies the law of total probability. The numerator and denominator are evaluated directly from the hypergeometric draw mechanism: P(r1∩r2|R=r) = (r/100)((r-1)/99) and P(r1|R=r) = r/100, with P(R=r) = 1/101. The algebra then reduces to 2/3. No parameter is fitted to the target quantity, no result is imported from the author's own prior work, and no assumption is made that already contains P(r2|r1). The only judgment call is the uniform prior itself, which is stated explicitly and supported by the puzzle wording; changing the prior would change the number, as the paper's own formula shows, but that is a modeling assumption, not circularity. The references to the author's GitHub repository and to prior solutions are illustrative, not load-bearing. The derivation is self-contained and therefore receives a circularity score of 0.
Assumptions & free parameters
assumptions (4)
- domain assumption R, the number of red balls, is uniformly distributed on {0,...,100}.
- domain assumption The two balls are drawn without replacement.
- standard math Conditional probability formula P(r2|r1)=P(r1∩r2)/P(r1).
- standard math Sum formulas for the first 100 integers and squares.
Cite this review
Pith. "Pith review of A Probability Puzzle." pith.science (2026). https://pith.science/paper/CYYSLGYR
@misc{pith2026250103238,
author = {Pith},
title = {Pith review of: A Probability Puzzle},
year = {2026},
howpublished = {\url{https://pith.science/paper/CYYSLGYR}},
note = {Machine review of arXiv:2501.03238}
}
read the original abstract
In this short article, we present a solution to one of the probability puzzles that Daniel Litt, a mathematician at the University of Toronto, posted on his X account earlier this year. The main goal of this note is to show how some of the typical concepts taught in an undergraduate probability course can be used to solve these types of probability problems, which sound simple, but can be very difficult to solve.
Reference graph
Works this paper leans on
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[2]
Daniel Litt's Probability Puzzle
Maura B. Paterson and Douglas R. Stinson , Daniel Litt’s Prob- ability Puzzle , arXiv preprint arXiv:2409.08094 (2024)
work page Pith review arXiv 2024
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[1]
Klarreich Perplexing the Web, One Probability Puzzle at a Time , in Quanta Magazine (2024)
E. Klarreich Perplexing the Web, One Probability Puzzle at a Time , in Quanta Magazine (2024)
work page 2024
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[3]
V azsonyi, Andrew, Which door has the Cadillac , in Decision Line, 30.1 (1999)
work page 1999
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[4]
https://en.wikipedia.org/wiki/Square_pyramidal_number
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[5]
https://en.wikipedia.org/wiki/Triangular_number
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[6]
https://en.wikipedia.org/wiki/Monty_Hall_problem
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[7]
Russell, Stuart, and Peter Norvig , AI a modern approach , Learn- ing 2.3 (2005)
work page 2005
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[8]
https://github.com/danotero/MA2014
Show all 9 references
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[9]
Ekhad and Doron Zeilberger , How to Answer Ques- tions of the Type: If you toss a coin n times, how likely is HH to s how up more than HT? , https://arxiv.org/abs/2405.13561
Shalosh B. Ekhad and Doron Zeilberger , How to Answer Ques- tions of the Type: If you toss a coin n times, how likely is HH to s how up more than HT? , https://arxiv.org/abs/2405.13561
Reviewed August 11, 2026 · model on record in the stance chip above.
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