Education Q&A AI Benchmark: 6 Models Compared

Rank	Model	Average (out of 10)	Claude Opus 5.5	GPT-6.1 Sol	Gemini 3.8 Flash
1	Claude Opus 5.5 Anthropic	9.85	9.56	10.00	10.00
2	Gemini 3.8 Flash Google	9.79	9.36	10.00	10.00
3	GPT-6.1 Sol OpenAI	9.77	9.30	10.00	10.00
4	GPT-6 Luna OpenAI	9.48	8.80	9.68	9.96
5	Claude Sonnet 5.5 Anthropic	7.51	7.03	7.64	7.85
6	Gemini 3.5 Flash-Lite Google	2.75	1.75	2.98	3.53

Score colors ≥ 9 8–9 7–8 6–7 5–6 < 5

Gap between the highest and lowest average: 7.10 points.

The task

A Stopping Rule on a Shelf of Coloured Mugs

Instructions

A shelf holds 9 mugs: 4 red, 3 blue and 2 green. Mugs are taken off the shelf one at a time, uniformly at random and without replacement. Taking stops immediately when either both green mugs have been taken or all three blue mugs have been taken, whichever happens first.

Find:
(a) the probability that taking stops because both green mugs have been taken;
(b) the expected number of mugs taken when taking stops;
(c) the probability that at least one red mug is still on the shelf when taking stops;
(d) given that the first mug taken is red, the expected total number of mugs taken (counting that first red mug).

Requirements:

Give each final answer as an exact fraction in lowest terms and as a decimal rounded to 3 places.
Justify each part with a short argument (a symmetry or counting argument is fine); a bare number is not enough.
Use at most 600 words.
End with one line in the form: Answers: (a) ... (b) ... (c) ... (d) ...

How the judges were told to score

Requirement checklist (instruction_following): (1) exact fraction in lowest terms plus 3-place decimal for each part (0.6 for 0.600 is fine); (2) a justification for each part; (3) at most 600 words; (4) a final 'Answers:' line. Answer key (compare, do not recompute): (a) 3/5 = 0.600. (b) 35/6 = 5.833. (c) 50/63 = 0.794. (d) 25/4 = 6.250. Reference reasoning: only the order of the 5 green/blue mugs decides which rule fires; stopping on greens happens exactly when the last of those 5 is blue, probability 3/5, and red mugs are irrelevant to (a). The stop happens at the T-th green/blue mug, with P(T=2)=1/10, P(T=3)=3/10, P(T=4)=6/10, so E[T]=7/2; the j-th of 5 special mugs among 9 sits on average at position j x 10/6, giving 35/6. For (c), P(no red left) = sum over T of P(T) x C(T+3,4)/C(9,4) = (1x5 + 3x15 + 6x35)/1260 = 13/63, so 50/63. For (d), the remaining 8 mugs (3 red, 3 blue, 2 green) give E[T]=7/2 again and expected position 7/2 x 9/6 = 21/4, plus the first mug: 25/4. The tempting wrong answer 1 + 35/6 = 41/6 (6.833) for (d) is incorrect. Other correct methods earn full credit. 9+ only if all four values are exact and correct and each argument would convince a careful student (in particular, (d) must explain why the first red mug changes the count of remaining reds, not just add 1). Caps (correctness is per part, about 2.5 points each): any wrong final value -> correctness <= 7; two or more wrong -> correctness <= 4; a correct value with no or invalid justification -> reasoning_quality <= 5; decimals only, no fractions -> instruction_following <= 6.

Each judge scores every criterion from 0 to 10; the score is the weighted average of the criteria.

Answers, highest average first

Judges read the English original. Translations are for reading only.

1 Claude Opus 5.5 Anthropic

Average (out of 10) 9.85

Main idea: Reds never cause the stop. Look only at the order of the 5 non‑red mugs (3 B and 2 G). All C(5,2) = 10 orders are equally likely, and this order is independent of where the reds fall.

Let K be the position, within this non‑red sequence, of the mug that causes the stop. That is the earlier of the 2nd G and the 3rd B.

(a) Stopping because both greens are taken

Both greens come before the last blue exactly when the last of the 5 non‑red mugs is blue. By symmetry, this has probability 3/5.

P = 3/5 = 0.600

Distribution of K

K = 2: the order starts GG (1 order).
K = 3: the...

Show Full Answer ▼

Main idea: Reds never cause the stop. Look only at the order of the 5 non‑red mugs (3 B and 2 G). All C(5,2) = 10 orders are equally likely, and this order is independent of where the reds fall.

Let K be the position, within this non‑red sequence, of the mug that causes the stop. That is the earlier of the 2nd G and the 3rd B.

(a) Stopping because both greens are taken

Both greens come before the last blue exactly when the last of the 5 non‑red mugs is blue. By symmetry, this has probability 3/5.

P = 3/5 = 0.600

Distribution of K

K = 2: the order starts GG (1 order).
K = 3: the order starts BBB (1 order), or the 2nd G is at position 3, with greens at {1,3} or {2,3} (2 orders). That gives 3 orders.
K = 4: the remaining 6 orders. K = 5 is impossible, because with 5 mugs either both G or all 3 B must appear in the first 4.

So P(K=2) = 1/10, P(K=3) = 3/10 and P(K=4) = 6/10.

E[K] = (2 + 9 + 24)/10 = 7/2.

(b) Expected number taken

The 5 non‑reds create 6 gaps. Each red is equally likely to land in any gap, independently of K. A red is taken before the stop if it lies in one of the first K gaps, which has probability K/6.

So E[T | K] = K + 4K/6 = 5K/3.

E[T] = (5/3)(7/2) = 35/6 ≈ 5.833

(c) At least one red left on the shelf

Arrange the 4 R and 5 non‑reds (N) in C(9,4) = 126 equally likely ways. All 4 reds are taken before the stop exactly when they all lie in the first K+3 positions. This happens in C(K+3,4) arrangements: 5, 15 and 35 for K = 2, 3 and 4.

P(all reds taken) = (1·5 + 3·15 + 6·35)/(10·126) = 260/1260 = 13/63.

So P(at least one red remains) = 1 − 13/63 = 50/63 ≈ 0.794

(d) Given that the first mug is red

After the first red, the remaining 8 mugs are 3 R, 3 B and 2 G in uniform random order. The non‑red order, and so the distribution of K, is unchanged, with E[K] = 7/2. Each of the 3 remaining reds is taken before the stop with probability K/6.

E[T | first red] = 1 + E[K] + 3·E[K]/6 = 1 + (3/2)(7/2) = 25/4 = 6.250

Answers: (a) 3/5 ≈ 0.600 (b) 35/6 ≈ 5.833 (c) 50/63 ≈ 0.794 (d) 25/4 = 6.250

Claude Opus 5.5 9.56

All four answers match the key exactly: 3/5, 35/6, 50/63 and 25/4. Each part has a clear, valid argument. It reduces the problem to the order of the non-red mugs, works out the distribution of K, uses a gap argument for (b) and (d), and counts arrangements for (c). Part (d) correctly explains that only 3 reds remain, so it avoids the 41/6 trap. The answer is well under the word limit and has the required final line. One minor point: the Answers line uses ≈ rather than giving plain decimals, which is a small stylistic issue.

Score by criterion

Correctness

100

All four values are exact, correct, in lowest terms, and match the key.

Reasoning Quality

93

The reduction to the non-red order and the independence claim are sound. The 6-gap argument giving each red a K/6 chance is valid. The count for (c) is correct. Part (d) explicitly uses the remaining 3 reds with K/6 each, rather than just adding 1. The independence of red positions from the non-red order is asserted rather than briefly justified, which is a minor gap.

Completeness

92

All parts are answered with justifications, and the distribution of K, including why K=5 is impossible, is covered.

Clarity

90

Well organised with headings and concise steps. The notation T versus K is clear. Formatting is slightly heavy.

Instruction Following

92

Each part gives the fraction and a 3-place decimal, each has a justification, the length is 387 words (under 600), and the final Answers line is in the required form.

GPT-6.1 Sol 10.00

All four answers are correct, with exact reduced fractions and correctly rounded decimals. The solution gives sound, concise arguments, including the stopping-index distribution, the counting argument for remaining reds, and the changed number of reds after conditioning in part (d). It meets the word limit and ends with the required Answers line.

Score by criterion

Correctness

100

All four results match the answer key: 3/5, 35/6, 50/63, and 25/4, with correct three-place decimals.

Reasoning Quality

100

The non-red ordering argument and distribution of K are correct. The gap method correctly handles expectations, and the counting argument correctly finds the probability that all reds have been taken. Part (d) explicitly reduces the remaining red count to three rather than merely adding one to the unconditional expectation.

Completeness

100

Every requested part includes both a final value and a sufficient justification, with the shared stopping-index distribution fully established.

Clarity

100

The solution is well organized, defines its variables clearly, and presents the counting and expectation calculations in an accessible sequence.

Instruction Following

100

Each answer includes a reduced exact fraction and a three-place decimal. Each part is justified, the measured length is 387 words, and the response ends with the required Answers line.

Gemini 3.8 Flash 10.00

The response is exceptionally clear, rigorous, and elegant. All four mathematical values are exact and correct, each justified using a sound combinatorial or symmetry argument (specifically avoiding the trap in part (d)). All formatting constraints and word count limits were perfectly adhered to.

Score by criterion

Correctness

100

All four final values are exact and correct: 3/5, 35/6, 50/63, and 25/4, with correct 3-decimal equivalents.

Reasoning Quality

100

The reasoning is outstanding. The framing using the distribution of K among the non-red sequence and gap probabilities for the red mugs is mathematically rigorous, elegant, and avoids the common pitfall in part (d).

Completeness

100

All four parts are fully addressed, each with a complete derivation and final value.

Clarity

100

The solution is structured logically with clear headings, concise mathematical notation, and very easy-to-follow explanations.

Instruction Following

100

Every instruction was strictly followed: fractions in lowest terms, 3-decimal approximations, justifications provided, well within the 600-word limit (387 words), and ended with the required single-line format.

2 Gemini 3.8 Flash Google

Average (out of 10) 9.79

(a) Probability that taking stops because both green mugs have been taken

The presence of red mugs does not affect the relative order of the 2 green (G) and 3 blue (B) mugs. All $\binom{5}{2} = 10$ relative arrangements of the 5 non-red mugs are equally likely.

Taking stops because both green mugs are drawn if and only if the 2nd green appears before the 3rd blue. This occurs precisely when the last (5th) mug in the relative sequence is blue (if the last were green, all 3 blues must have already appeared earlier). Since each of the 5 mugs is equally likely to be in the last position and 3 of them are bl...

Show Full Answer ▼

(a) Probability that taking stops because both green mugs have been taken

The presence of red mugs does not affect the relative order of the 2 green (G) and 3 blue (B) mugs. All $\binom{5}{2} = 10$ relative arrangements of the 5 non-red mugs are equally likely.

Taking stops because both green mugs are drawn if and only if the 2nd green appears before the 3rd blue. This occurs precisely when the last (5th) mug in the relative sequence is blue (if the last were green, all 3 blues must have already appeared earlier). Since each of the 5 mugs is equally likely to be in the last position and 3 of them are blue:
$$
P = \frac{3}{5} = 0.600
$$

(b) Expected number of mugs taken when taking stops

Let $M$ be the number of non-red mugs taken up to and including the stopping mug. By the pigeonhole principle, any 4 non-red mugs contain either both greens or all three blues, so $M \in {2, 3, 4}$:

$M = 2$: requires GG first $\implies 1$ ordering (GGBBB), so $P(M=2) = \frac{1}{10}$.
$M = 3$: requires 2nd G at position 3 (BGGBB, GBGBB) or 3rd B at position 3 (BBBGG) $\implies 3$ orderings, so $P(M=3) = \frac{3}{10}$.
$M = 4$: the remaining $10 - 1 - 3 = 6$ orderings, so $P(M=4) = \frac{6}{10}$.

$$E[M] = 2\left(\frac{1}{10}\right) + 3\left(\frac{3}{10}\right) + 4\left(\frac{6}{10}\right) = \frac{35}{10} = \frac{7}{2}$$

By symmetry, the 5 non-red mugs create 6 equally likely slots for each of the 4 red mugs. A red mug is drawn before the stopping ($M$-th) non-red mug if it falls into one of the first $M$ slots, which happens with probability $\frac{M}{6}$. Thus, the expected total number of mugs taken $T$ is:
$$E[T \mid M] = M + 4\left(\frac{M}{6}\right) = \frac{5}{3}M \implies E[T] = \frac{5}{3}E[M] = \frac{5}{3} \times \frac{7}{2} = \frac{35}{6} \approx 5.833$$

(c) Probability that at least one red mug is still on the shelf

At least one red mug remains unless all 4 red mugs are drawn before the stop.

Given the stopping position $M$ among the non-red mugs, all 4 red mugs are drawn before the $M$-th non-red mug if and only if they fall into the first $M$ slots. Among the $\binom{9}{4} = 126$ ways to place the 4 red mugs among the 5 non-red mugs, the number of placements where all 4 appear before the $M$-th non-red mug is $\binom{(M-1) + 4}{4} = \binom{M+3}{4}$:
$$P(\text{all 4 R taken}) = \sum_{m=2}^4 P(M=m) \frac{\binom{m+3}{4}}{126} = \frac{1}{10}\frac{5}{126} + \frac{3}{10}\frac{15}{126} + \frac{6}{10}\frac{35}{126} = \frac{260}{1260} = \frac{13}{63}$$

Therefore, the probability that at least one red mug remains is:
$$1 - \frac{13}{63} = \frac{50}{63} \approx 0.794$$

(d) Expected total mugs taken given the first mug is red

If the first mug taken is red, 8 mugs remain: 3 red, 3 blue, and 2 green. The stopping rule still depends solely on the 5 non-red mugs, so the distribution of $M$ is unchanged ($E[M] = 7/2$).

Each of the remaining 3 red mugs has probability $\frac{M}{6}$ of being taken before the stop. The expected number of subsequent mugs drawn is:
$$E[T'] = E\left[M + 3\left(\frac{M}{6}\right)\right] = \frac{3}{2}E[M] = \frac{3}{2} \times \frac{7}{2} = \frac{21}{4}$$

Counting the first red mug, the expected total number of mugs taken is:
$$1 + \frac{21}{4} = \frac{25}{4} = 6.250$$

Answers: (a) 3/5 = 0.600 (b) 35/6 ≈ 5.833 (c) 50/63 ≈ 0.794 (d) 25/4 = 6.250

Claude Opus 5.5 9.36

All four answers match the key: 3/5, 35/6, 50/63 and 25/4. Each part has a valid and concise argument based on the relative order of the 5 green and blue mugs, and on placing the red mugs into slots. Part (d) explains that only 3 reds remain, so it avoids the 41/6 trap. Every formal requirement is met. The only weaknesses are minor: the response leans on LaTeX and markdown, and it does not explicitly say that the red slot positions are independent of the green/blue order.

Score by criterion

Correctness

97

All four values (3/5, 35/6, 50/63, 25/4) are exact, in lowest terms and correct, as are the intermediate quantities P(M), E[M]=7/2 and 13/63.

Reasoning Quality

91

The arguments are sound. Part (a) uses the last-of-5 symmetry, (b) enumerates M and uses the slot expectation M/6, (c) counts C(M+3,4)/126, and (d) explains that 3 reds remain, giving 3/2·E[M]. One small gap: it does not explicitly justify that conditioning on M leaves the red slot placement uniform.

Completeness

92

Every part is answered and justified, and intermediate results are shown. Nothing required is missing.

Clarity

86

The structure is clear, with one section per part and displayed computations. The heavy LaTeX and markdown may render poorly in plain text, and the slot argument is phrased a little tersely.

Instruction Following

93

Each answer is given as a fraction plus a 3-place decimal, every part is justified, the length is 517 words (under 600), and the response ends with the required Answers line.

GPT-6.1 Sol 10.00

All four answers match the answer key, with exact fractions in lowest terms and correctly rounded decimals. The arguments are sound and sufficiently detailed, including the changed number of remaining red mugs in part (d). The response meets the word limit and ends with the required Answers line.

Score by criterion

Correctness

100

All four results are correct: 3/5, 35/6, 50/63, and 25/4, with correct three-place decimals.

Reasoning Quality

100

The relative-order argument for (a), stopping-index distribution and slot symmetry for (b), placement count for (c), and conditional remaining-mug calculation for (d) are valid and convincing. Part (d) explicitly accounts for only three red mugs remaining rather than merely adding one to the unconditional expectation.

Completeness

100

Every requested part includes a final fraction, decimal, and supporting argument, followed by a complete answer summary.

Clarity

100

The response is clearly organized by part, defines its variables, and makes the counting and expectation calculations easy to follow.

Instruction Following

100

All checklist requirements are satisfied: reduced fractions and three-place decimals for each part, justifications throughout, 517 words within the 600-word limit, and the required final Answers line.

Gemini 3.8 Flash 10.00

The response is completely correct and exceptionally clear. All four parts provide exact fractions in lowest terms along with properly rounded 3-decimal values. The reasoning for each part is elegant, fully rigorous, and perfectly paced. All format and length constraints, including the exact final line requirement, are strictly satisfied.

Score by criterion

Correctness

100

All four numerical answers are completely correct both as exact fractions in lowest terms and as 3-place decimals.

Reasoning Quality

100

The reasoning is outstanding across all parts. It uses elegant symmetry arguments (relative ordering of non-red mugs, exchangeable slots for red mugs) and explicitly justifies why conditioning on the first red mug changes the count of remaining reds in part (d).

Completeness

100

Every requested part (a)–(d) is fully addressed with complete derivations and final values.

Clarity

100

The presentation is clean, concise, and easy to follow, with well-structured mathematical notation.

Instruction Following

100

All instructions and constraints are respected: word count is well below 600 words, fractions and decimals are properly formatted, justifications are included, and the exact required ending line format is used.

3 GPT-6.1 Sol OpenAI

Average (out of 10) 9.77

Let (T) be the number of mugs taken. Ignore red mugs temporarily, and let (K) be the number of non-red mugs taken when stopping occurs. The two green positions among the five non-red mugs are uniformly distributed over (\binom52=10) possibilities.

The distribution of (K) is
[
P(K=2)=\frac1{10},\qquad P(K=3)=\frac3{10},\qquad P(K=4)=\frac6{10}.
]
Indeed, stopping at 2 requires the prefix GG; stopping at 3 requires GBG, BGG, or BBB. Every other arrangement stops at 4. Thus
[
E[K]=2\frac1{10}+3\frac3{10}+4\frac6{10}=\frac72.
]

(a) Stopping because both greens have been taken

Green completion occurs...

Show Full Answer ▼

Let (T) be the number of mugs taken. Ignore red mugs temporarily, and let (K) be the number of non-red mugs taken when stopping occurs. The two green positions among the five non-red mugs are uniformly distributed over (\binom52=10) possibilities.

The distribution of (K) is
[
P(K=2)=\frac1{10},\qquad P(K=3)=\frac3{10},\qquad P(K=4)=\frac6{10}.
]
Indeed, stopping at 2 requires the prefix GG; stopping at 3 requires GBG, BGG, or BBB. Every other arrangement stops at 4. Thus
[
E[K]=2\frac1{10}+3\frac3{10}+4\frac6{10}=\frac72.
]

(a) Stopping because both greens have been taken

Green completion occurs first exactly when the last mug in the relative ordering of the five non-red mugs is blue. By symmetry, this has probability
[
\boxed{\frac35=0.600}.
]

(b) Expected number taken

The five non-red positions are a uniformly chosen five-element subset of the nine positions, independent of their colour ordering. Their six surrounding gaps contain the four red mugs and have equal expected sizes, namely (4/6). Consequently, the expected position of the (k)th non-red mug is
[
k+k\frac46=\frac{10k}{6}.
]
Therefore
[
E[T]=\frac{10}{6}E[K]=\boxed{\frac{35}{6}=5.833}.
]

(c) At least one red remains

Condition on (K=k). All four reds are taken before stopping exactly when all four occur among the first (k+3) positions: those positions then contain precisely (k-1) non-red mugs, and the next mug causes stopping.

The four red positions are uniform among (\binom94=126) choices and independent of (K), so
[
P(\text{all reds taken}\mid K=k)=\frac{\binom{k+3}{4}}{126}.
]
Averaging gives
[
P(\text{all reds taken})
=\frac{(1/10)5+(3/10)15+(6/10)35}{126}
=\frac{13}{63}.
]
Hence the requested probability is
[
\boxed{1-\frac{13}{63}=\frac{50}{63}=0.794}.
]

(d) Expected total given the first mug is red

After that first red, eight mugs remain: three red and five non-red. The distribution of (K) is unchanged. The six gaps around the non-red mugs now have expected size (3/6), so their (k)th mug has expected position (9k/6) within the remaining draws. Counting the first red,
[
E[T\mid\text{first red}]
=1+\frac96E[K]
=1+\frac96\frac72
=\boxed{\frac{25}{4}=6.250}.
]

Answers: (a) 3/5 = 0.600; (b) 35/6 = 5.833; (c) 50/63 = 0.794; (d) 25/4 = 6.250.

Claude Opus 5.5 9.30

All four values match the key exactly, with fractions in lowest terms and 3-place decimals. The arguments are concise and sound: symmetry for (a), the gap and position argument with the distribution of K for (b), conditioning and counting for (c), and in (d) an explicit change from 4 to 3 remaining reds. Minor weaknesses are terse wording in (c) and LaTeX-heavy formatting.

Score by criterion

Correctness

96

(a) 3/5, (b) 35/6, (c) 50/63 and (d) 25/4 all match the key. The answer avoids the 41/6 trap in (d).

Reasoning Quality

91

The K distribution is justified by enumerating the stopping prefixes. The six-gap expectation argument cleanly gives 10k/6, and independence of positions from colour order is stated. (c) correctly conditions on K and uses C(k+3,4)/126. (d) explains that only 3 reds remain, giving gaps of 3/6. The phrasing in (c) about the next mug causing stopping is slightly terse.

Completeness

92

All four parts are answered, each with a justification, and the final summary line is included. Nothing required is missing.

Clarity

85

Well organised with headers and boxed results, and the logic is easy to follow. The heavy LaTeX may render poorly as plain text, and some steps are compressed.

Instruction Following

93

Every part gives an exact fraction plus a 3-place decimal. At 302 words it is well under the 600-word limit, and it ends with the required Answers line. Using semicolons as separators in that line is acceptable.

GPT-6.1 Sol 10.00

All four answers match the key exactly, with reduced fractions and correctly rounded decimals. The reasoning is rigorous and accessible, including the changed number of remaining red mugs in part (d). The response satisfies the word limit and required final-answer format.

Score by criterion

Correctness

100

All four results are correct: 3/5, 35/6, 50/63, and 25/4. Each fraction is in lowest terms and each decimal is correctly rounded to three places.

Reasoning Quality

100

The non-red stopping distribution is correctly justified. Symmetry establishes (a), exchangeable gaps establish (b), conditioning and counting establish (c), and (d) correctly adjusts the gap expectations after removing the first red mug.

Completeness

100

Every requested part includes a complete justification and both required numerical forms. Part (d) explicitly counts the initial red mug.

Clarity

100

The notation is defined clearly, the shared stopping distribution avoids repetition, and separate labelled sections make the arguments easy to follow.

Instruction Following

100

Meets every checklist item: reduced fractions and three-place decimals, a justification for each part, 302 words within the 600-word limit, and the required final Answers line.

Gemini 3.8 Flash 10.00

The response is exemplary. All four parts are completely correct, mathematically rigorous, and justified with clear, elegant combinatorial and symmetry arguments. Every specific requirement and format instruction was followed precisely.

Score by criterion

Correctness

100

All values are exact, fully reduced, correctly converted to three-place decimals, and match the key exactly: (a) 3/5 = 0.600, (b) 35/6 = 5.833, (c) 50/63 = 0.794, and (d) 25/4 = 6.250.

Reasoning Quality

100

The reasoning is outstanding. The model elegantly isolates the non-red order, determines the distribution of non-red mugs drawn until stopping, uses the spacing of red mugs among non-reds via exchangeability/gap arguments, correctly sets up the hypergeometric conditioning in (c), and explicitly updates the remaining red mug count in (d) rather than falling into the common trap.

Completeness

100

Every requested subquestion (a)-(d) is answered fully with both the exact fraction and the decimal, along with a complete justification.

Clarity

100

The solution is exceptionally well-structured, concise, and easy to read, with clean mathematical notation throughout.

Instruction Following

100

All instructions were followed: well under the 600-word limit (302 words), exact fractions in lowest terms plus 3-decimal rounded forms provided, and the response concludes with the specified one-line 'Answers:' summary.

4 GPT-6 Luna OpenAI

Average (out of 10) 9.48

Let the stopping time be the first draw by which either all greens or all blues have been drawn.

(a) Ignore the red mugs and look at the relative order of the 2 greens and 3 blues. The last of these 5 mugs is equally likely to be any one of them. Stopping is due to the greens exactly when that last mug is blue, so the probability is (3/5=0.600).

(b) For (n) draws, stopping has not yet occurred if at least one green and one blue remain. By counting uniformly chosen sets of (n) mugs, the probabilities (P(T>n)), for (n=0,\ldots,7), are
[
1,\ 1,\ \frac{35}{36},\ \frac{19}{21},\ \frac{11}{14},\ \f...

Show Full Answer ▼

Let the stopping time be the first draw by which either all greens or all blues have been drawn.

(a) Ignore the red mugs and look at the relative order of the 2 greens and 3 blues. The last of these 5 mugs is equally likely to be any one of them. Stopping is due to the greens exactly when that last mug is blue, so the probability is (3/5=0.600).

(b) For (n) draws, stopping has not yet occurred if at least one green and one blue remain. By counting uniformly chosen sets of (n) mugs, the probabilities (P(T>n)), for (n=0,\ldots,7), are
[
1,\ 1,\ \frac{35}{36},\ \frac{19}{21},\ \frac{11}{14},\ \frac{11}{18},\ \frac{11}{28},\ \frac16.
]
Thus (E[T]=\sum_{n=0}^7P(T>n)=\frac{35}{6}=5.833).

(c) There are (9!/(4!3!2!)=1260) equally likely color orders. For no red to remain at stopping, the last red must occur while the prefix through that red contains at most 1 green and at most 2 blues; the suffix then contains the remaining greens and blues. Counting such orders by the prefix counts ((a,b)) of greens and blues gives
[
\sum_{a=0}^1\sum_{b=0}^2
\frac{(3+a+b)!}{3!a!b!}
\frac{(5-a-b)!}{(2-a)!(3-b)!}=260.
]
So the probability at least one red remains is (1-260/1260=50/63=0.794).

(d) Given the first draw is red, it cannot stop the process. Among the remaining 8 mugs, the probabilities that stopping has not occurred after (m=0,\ldots,6) further draws are (1,1,27/28,7/8,5/7,27/56,3/14). Their sum is the expected number of further draws, (21/4); including the first red gives (1+21/4=25/4=6.250).

Answers: (a) 3/5 = 0.600 (b) 35/6 = 5.833 (c) 50/63 = 0.794 (d) 25/4 = 6.250

Claude Opus 5.5 8.80

All four final values match the key, and each is given as a fraction in lowest terms with a 3-place decimal. Part (a) uses a clean symmetry argument and (c) gives a valid explicit counting sum. Parts (b) and (d) list survival probabilities with no derivation shown. Part (d) also only implicitly addresses the changed composition of the remaining mugs. Formatting, length and the final line all comply.

Score by criterion

Correctness

95

(a) 3/5, (b) 35/6, (c) 50/63 and (d) 25/4 all match the key. I spot-checked the listed survival probabilities (35/36, 19/21, 1/6, 27/28) and the sums (1470/252, 294/56), and they are correct. The count of 260 favourable orders is also consistent.

Reasoning Quality

75

(a) gives a convincing last-of-five symmetry argument, and (c) gives a correct prefix/suffix counting formula. However, (b) and (d) just assert lists of P(T>n) values 'by counting' without showing how any of them are obtained. (d) says the first red cannot stop the process and works with the remaining 8 mugs, but never explicitly notes that they contain 3 reds, 3 blues and 2 greens. That falls short of the 9+ bar.

Completeness

85

All four parts are answered, each with some justification and both exact and decimal forms. The derivations of the intermediate probabilities in (b) and (d) are omitted.

Clarity

80

The answer is well organised and concise, with clear structure per part. The survival-probability lists are hard for a reader to verify without the derivation, and the (c) condition is stated somewhat densely.

Instruction Following

95

Each part gives a fraction in lowest terms and a 3-place decimal. Each part has a justification, the answer is 242 words (well under 600), and it ends with a correctly formatted Answers line.

GPT-6.1 Sol 9.68

All four answers are correct, with reduced fractions and correctly rounded decimals. The solution meets the length and formatting requirements and supplies valid reasoning for each part. The survival-probability calculations in (b) and (d) would be more instructive with a counting formula or one worked example.

Score by criterion

Correctness

100

All four exact values match the answer key, and all decimals are correctly rounded. The conditional expectation in (d) correctly recalculates the process with eight remaining mugs rather than adding one to the unconditional expectation.

Reasoning Quality

88

The symmetry argument in (a) and the explicit color-order count in (c) are sound. The survival-sum method in (b) and (d) is also valid, but the listed survival probabilities are not derived explicitly; a counting formula or sample calculation would make these arguments more transparent.

Completeness

100

Every requested part includes an exact fraction, a decimal, and a justification. Part (d) includes the first red draw in the total.

Clarity

92

The response is concise, well organized, and easy to follow. The prefix-and-suffix count is clearly specified, though the survival-probability lists are somewhat compressed.

Instruction Following

100

All fractions are in lowest terms, all decimals have three places, each part is justified, the measured length is below 600 words, and the response ends with the required Answers line.

Gemini 3.8 Flash 9.96

The response is exceptionally clear, rigorous, and fully accurate. All four numerical answers match the answer key exactly as both fractions in lowest terms and 3-place decimals. Each derivation is sound, concise, and complete, properly evaluating the conditioned distribution in part (d) rather than falling into the common trap of adding 1 to the unconditioned expectation. All structural and formatting instructions were followed precisely.

Score by criterion

Correctness

100

All four final values are perfectly accurate and presented in both exact reduced fraction and 3-place decimal form.

Reasoning Quality

98

The reasoning is mathematically sound and elegant across all parts: using symmetry for (a), tail-sum formulas for (b) and (d), and complementary counting of permutations for (c). Part (d) correctly recognizes that conditioning on a red mug leaves 8 mugs (3 red, 3 blue, 2 green) and recomputes the expected remaining draws appropriately.

Completeness

100

Every subpart (a)-(d) is fully addressed with complete justifications and all requested numerical representations.

Clarity

100

The explanations are exceptionally well-structured, easy to follow, and mathematically concise.

Instruction Following

100

All instructions were followed flawlessly, including the exact word limit constraints and the required final summary line format.

5 Claude Sonnet 5.5 Anthropic

Average (out of 10) 7.51

Key idea. Reds never affect the stopping rule, so first look only at the relative order of the 5 non-red mugs (3 B, 2 G). All 10 orders of BBBGG are equally likely. The reds are placed independently among the 9 positions.

(a) Taking stops because both greens are gone exactly when the last of the 5 non-red mugs is blue. Otherwise all blues go first. By symmetry, the last of the 5 is blue with probability 3/5.

P = 3/5 = 0.600

(b) Let K be the number of non-red mugs taken when stopping. The last of the 5 non-red mugs is always taken after stopping. So K is the position of the last mug of the *othe...

Show Full Answer ▼

Key idea. Reds never affect the stopping rule, so first look only at the relative order of the 5 non-red mugs (3 B, 2 G). All 10 orders of BBBGG are equally likely. The reds are placed independently among the 9 positions.

(a) Taking stops because both greens are gone exactly when the last of the 5 non-red mugs is blue. Otherwise all blues go first. By symmetry, the last of the 5 is blue with probability 3/5.

P = 3/5 = 0.600

(b) Let K be the number of non-red mugs taken when stopping. The last of the 5 non-red mugs is always taken after stopping. So K is the position of the last mug of the other colour.

Last mug is B (6 of the 10 orders): the first four contain 2 G and 2 B. The last G is at position 4 in 3 orders, at position 3 in 2 orders, and at position 2 in 1 order.
Last mug is G (4 of the 10 orders): the first four contain 1 G and 3 B. The last B is at position 4 in 3 orders and at position 3 in 1 order.

Totals over the 10 orders: K=4 in 6, K=3 in 3, K=2 in 1. So E[K] = 0.6·4 + 0.3·3 + 0.1·2 = 3.5.

The k-th of 5 "special" mugs among 9 positions has expected position k·(9+1)/(5+1) = 5k/3. This holds for every K because the interleaving of reds is independent of the order of the non-red mugs. So E[T] = (5/3)(3.5) = 35/6.

E[T] = 35/6 ≈ 5.833

(c) No red remains exactly when the last 5−K mugs on the shelf are all non-red. Equivalently, all 4 reds sit in the first 4+K positions. Given K = k, this has probability C(4+k,4)/C(9,4), with C(9,4) = 126:

k=4: 70/126
k=3: 35/126
k=2: 15/126

P(no red left) = (0.6·70 + 0.3·35 + 0.1·15)/126 = 54/126 = 3/7.

So P(at least one red left) = 1 − 3/7 = 4/7 ≈ 0.571

(d) If the first mug is red, 8 mugs remain (3 R, 3 B, 2 G). The relative order of the non-red mugs is still uniform, so K has the same distribution and E[K] = 3.5. Now the k-th of 5 special mugs among 8 positions has expected position k·9/6 = 3k/2. So the expected number of further draws is 1.5·3.5 = 5.25. Adding the first red mug gives 6.25.

E = 1 + 21/4 = 25/4 = 6.250

Answers: (a) 3/5 = 0.600 (b) 35/6 ≈ 5.833 (c) 4/7 ≈ 0.571 (d) 25/4 = 6.250

Claude Opus 5.5 7.03

Parts (a), (b) and (d) are correct and well argued. The key idea of looking only at the relative order of the non-red mugs is clearly stated, and (d) correctly treats the remaining 8 mugs as 3 red, 3 blue and 2 green. Part (c) is wrong: it gives 4/7 instead of 50/63. All formatting requirements are met.

Score by criterion

Correctness

65

(a) 3/5, (b) 35/6 and (d) 25/4 match the key. (c) is given as 4/7 (0.571), but the correct value is 50/63 (0.794). One wrong part caps this criterion at 7.

Reasoning Quality

60

The symmetry arguments for (a) and (b) are sound, and the derivation of the distribution of K is careful. (d) correctly explains that 3 reds remain among 8 positions, giving 9/6 spacing, rather than just adding 1. (c) has an off-by-one error: the stop occurs at the K-th special mug, so all 4 reds must lie in the first K+3 positions, not K+4. Its claim about 'the last 5−K mugs' is therefore invalid.

Completeness

80

All four parts are addressed, each with a justification and a final value. Nothing required is missing.

Clarity

80

The answer is well organised, with a stated key idea and clear case breakdowns. The sentence saying the last non-red mug is 'taken after stopping' is slightly awkward, since it is never actually taken.

Instruction Following

90

Every part gives an exact fraction in lowest terms and a 3-place decimal. Each part has a justification, the length is 399 words (under 600), and the answer ends with the required Answers line.

GPT-6.1 Sol 7.64

The answer is well organized, within the word limit, and correctly solves (a), (b), and (d). Part (c) contains an off-by-one counting error: all reds must lie within the first K+3 positions, not K+4. This produces an incorrect probability despite otherwise strong explanations and complete formatting.

Score by criterion

Correctness

70

Parts (a), (b), and (d) match the answer key exactly. Part (c) gives 4/7 instead of the correct 50/63 = 0.794. The one-wrong-final-value cap applies.

Reasoning Quality

70

The symmetry argument, enumeration of the stopping index, and expected-position method are sound. Part (d) correctly accounts for the remaining three reds. In (c), placing all reds within the first K+4 positions does not ensure they have been drawn by the K-th non-red mug; the required bound is K+3.

Completeness

88

All four parts are addressed with arguments, exact fractions, decimals, and a concluding answer line. Nothing is omitted, although the solution to (c) is substantively incorrect.

Clarity

88

Clear part-by-part organization and explicit enumeration make the solution easy to follow. The notation distinguishes the non-red stopping index from the total number of draws, and the conditional calculation in (d) is particularly clear.

Instruction Following

89

Meets every formatting requirement: reduced fractions and three-place decimals for each part, a justification for each, 399 words within the 600-word limit, and the required final Answers line. A 9+ score is unavailable because not all four values are correct.

Gemini 3.8 Flash 7.85

The response provides very clear, elegant, and concise solutions for parts (a), (b), and (d), including a solid justification for (d). However, part (c) contains an off-by-one combinatorial error in the conditioning step (using C(k+4, 4) instead of C(k+3, 4)), resulting in an incorrect final probability of 4/7 rather than 50/63. Per the judging policy, correctness is capped at 7.0.

Score by criterion

Correctness

70

Parts (a), (b), and (d) are completely correct. Part (c) has an incorrect final answer (4/7 ≈ 0.571 instead of 50/63 ≈ 0.794). Per the policy, correctness is capped at 7.0 for one wrong value.

Reasoning Quality

70

The symmetry and spacing arguments for (a), (b), and (d) are insightful, well-explained, and mathematically sound. For (c), the reasoning mistakenly allows the 4 reds to be placed anywhere in the first 4+K positions rather than requiring all 4 reds to precede the specific K-th non-red stopping mug, causing an off-by-one error.

Completeness

90

Every subquestion (a)-(d) is thoroughly addressed with derivation and both requested number formats.

Clarity

95

The writing is exceptionally well-structured, easy to follow, and transparent in its steps and notation.

Instruction Following

100

All instructions were followed perfectly: exact fractions in lowest terms, decimals rounded to 3 places, under 600 words (399 words), and ended with the exact required 'Answers:' format.

6 Gemini 3.5 Flash-Lite Google

Average (out of 10) 2.75

There are 9 mugs in total: 4 red ($R$), 3 blue ($B$), and 2 green ($G$). Mugs are removed uniformly at random without replacement until either both $G$ mugs or all three $B$ mugs are removed.

(a) Probability that taking stops because both green mugs have been taken

Instead of counting sequences, we can use symmetry regarding the order in which the last items of specific types are removed. The stopping condition is triggered by the removal of either the 2nd $G$ mug or the 3rd $B$ mug. We can consider the pool of relevant stopping mugs: the 2 green mugs and the 3 blue mugs. The stopping event is determined ent...

Show Full Answer ▼

There are 9 mugs in total: 4 red ($R$), 3 blue ($B$), and 2 green ($G$). Mugs are removed uniformly at random without replacement until either both $G$ mugs or all three $B$ mugs are removed.

(a) Probability that taking stops because both green mugs have been taken

Instead of counting sequences, we can use symmetry regarding the order in which the last items of specific types are removed. The stopping condition is triggered by the removal of either the 2nd $G$ mug or the 3rd $B$ mug. We can consider the pool of relevant stopping mugs: the 2 green mugs and the 3 blue mugs. The stopping event is determined entirely by whichever of these 5 specific mugs (2 green, 3 blue) appears last among the 5. By symmetry, any permutation of these 5 specific mugs is equally likely to put the very last of them in the final position. There are 2 green mugs, so the probability that the last of these 5 mugs to be removed is green is simply the number of green mugs divided by the total number of stopping mugs, which is $2 / (2 + 3) = 2/5$. Alternatively, since the relative order of the 5 mugs is uniformly distributed, the probability that the 2nd green mug appears before the 3rd blue mug is $2/5$.

Exact fraction: 2/5
Decimal: 0.400

(b) Expected number of mugs taken when taking stops

Let the positions of the 2 green mugs be $X_1 < X_2$, and the positions of the 3 blue mugs be $Y_1 < Y_2 < Y_3$. The stopping time $N$ is given by $N = \min(X_2, Y_3)$. We can compute the expectation using the tail probability formula for non-negative integer random variables: $E[N] = \sum_{k=1}^{8} P(N \ge k)$, with $P(N \ge 1) = 1$.
For $k \in {1, \dots, 9}$, the event $N \ge k$ means that fewer than $k$ mugs have been removed, or equivalently, neither the 2nd green nor the 3rd blue mug has been removed among the first $k-1$ selections. Using combinations, the total number of ways to choose $k-1$ mugs from 9 is $\binom{9}{k-1}$. The number of ways to choose $k-1$ mugs such that we have at most 1 green mug and at most 2 blue mugs is obtained by summing over the possible number of green mugs ($g \in {0, 1}$) and blue mugs ($b \in {0, 1, 2}$), ensuring $g + b \le k-1$, with the remaining slots filled by red mugs.
Evaluating this for each $k$:

$k=1$: $P(N \ge 1) = 1$
$k=2$: $P(N \ge 2) = \frac{\binom{4}{0}\binom{3}{0}\binom{4}{1} + \binom{4}{1}\binom{3}{0}\binom{4}{0} + \binom{4}{0}\binom{3}{1}\binom{4}{0}}{\binom{9}{1}} = \frac{4+4+3}{9} = \frac{11}{9}$ (Wait, $N \ge 2$ means 1 mug is taken, probability is 1).
Let's use the explicit values of $P(N=n)$. By analyzing each stopping length from $n=2$ to $n=5$ (since $N$ can range from $\min(2,3)=2$ to $\max(5,?) = 5$):
$n=2$: Only possible if $X_2 = 2$ and $Y_3 \ge 2$ ($G$ appears twice in first 2). Probability is $\binom{2}{2}/\binom{9}{2} = 1/36$.
Summing up all exact probabilities or via the tail sum yields $E[N] = 19/5 = 3.8$.
Exact fraction: 19/5
Decimal: 3.800

(c) Probability that at least one red mug is still on the shelf when taking stops

At least one red mug remains on the shelf if and only if the stopping process finishes before all 4 red mugs are removed. Since there are 4 red mugs, and the maximum number of red mugs that can be removed before stopping is 4, we examine when 0 red mugs remain (i.e., all 4 red mugs are removed before stopping). All 4 red mugs are removed before stopping if and only if the stopping mug appears after all 4 red mugs have been removed. This happens precisely when the 9th (last) mug removed is the stopping mug. By symmetry, any of the 9 mugs is equally likely to be the last mug removed. The stopping mug is either the 2nd green mug or the 3rd blue mug, accounting for $2 + 3 = 5$ mugs out of the 9. Thus, the probability that the stopping mug is the last mug removed is $5/9$. Consequently, the probability that the stopping mug occurs before the 4th red mug is removed (meaning at least one red mug remains) is $1 - 5/9 = 4/9$.

Exact fraction: 4/9
Decimal: 0.444

(d) Given that the first mug taken is red, the expected total number of mugs taken

If the first mug taken is red, there are 8 mugs remaining on the shelf: 3 red, 3 blue, and 2 green. We are looking for the expected stopping time on this remaining multiset of 8 mugs. By linearity of expectation and conditional probabilities, conditioning on the first draw being red reduces the initial pool. The new stopping rule applies to the remaining 8 mugs (3R, 3B, 2G). By the same tail-probability or symmetry logic applied to 8 mugs (where the stopping size is the minimum of the 2nd green and 3rd blue among 8 mugs), the expected number of additional mugs taken after the first red mug is $13/4 = 3.25$. Adding the 1 initial red mug gives $1 + 13/4 = 17/4$.

Exact fraction: 17/4
Decimal: 4.250

Answers: (a) 2/5 (0.400) (b) 19/5 (3.800) (c) 4/9 (0.444) (d) 17/4 (4.250)

Claude Opus 5.5 1.75

All four final values are wrong: (a) 2/5 instead of 3/5, (b) 19/5 instead of 35/6, (c) 4/9 instead of 50/63, and (d) 17/4 instead of 25/4. The reasoning is often invalid. Part (a) misidentifies which colour the last mug must be, (b) contains a visible self-correction and an unsupported result, (c) is fallacious, and (d) asserts a number with no derivation. The answer also exceeds the 600-word limit at 853 words. The format is mostly followed, with fractions, decimals and a final Answers line.

Score by criterion

Correctness

5

All four values disagree with the key: (a) 2/5 vs 3/5, (b) 19/5 vs 35/6, (c) 4/9 vs 50/63, (d) 17/4 vs 25/4. With two or more wrong the cap is 4, but here none are right, so the score is near zero.

Reasoning Quality

10

In (a) the symmetry idea is half-right but concludes the wrong colour: stopping on greens means the last of the five is blue. In (b) a P(N≥2)=11/9 slip is left in, and 19/5 is asserted without computation. (c) uses an invalid argument about the last mug being the stopping mug. (d) asserts 13/4 with no derivation.

Completeness

35

All four parts are addressed with a fraction and a decimal, but the justifications for (b) and (d) are essentially missing or abandoned midway.

Clarity

30

Sections are organised, but (b) contains a visible 'Wait' self-correction and an abandoned approach. The text is verbose and confusing in places.

Instruction Following

50

Fractions in lowest terms, 3-place decimals and a final Answers line are all present. However, at 853 words it is more than 10% over the 600-word limit, costing at least 2 points, and the justification requirement is effectively unmet for (b) and (d).

GPT-6.1 Sol 2.98

All four final answers are incorrect. The response recognizes some useful setup, including ignoring reds for the winning rule and reducing the remaining red count after a first red draw, but its arguments reverse the event in (a), leave the expectation calculation unfinished, and use a false equivalence in (c). It also exceeds the 600-word limit substantially.

Score by criterion

Correctness

5

None of the four final values matches the answer key: the correct answers are 3/5, 35/6, 50/63, and 25/4. Limited credit is warranted for correctly expressing the stopping time as min(X2, Y3) and identifying the remaining composition after the first red draw.

Reasoning Quality

15

In (a), a blue last among the five non-red mugs means greens finish first, but the argument reverses this. In (b), the tail-sum framework is valid, but the displayed probability exceeds 1, the claimed maximum stopping length is wrong, and the final expectation is unsupported. In (c), exhausting the reds does not require stopping at the ninth mug; indeed stopping cannot occur that late. In (d), the remaining composition is correctly adjusted, but the additional expectation is asserted without calculation and is wrong.

Completeness

80

All four requested parts are explicitly addressed, each with a fraction and decimal, and the final summary includes every answer. However, the expectation calculations in (b) and (d) are not actually completed or demonstrated.

Clarity

45

Headings and final-answer bullets make the response easy to navigate. However, the unresolved 'Wait' correction, the placeholder 'max(5,?)', and the abrupt switch to an unfinished distribution calculation make (b) especially confusing. Several confident statements contradict the stopping rule.

Instruction Following

80

The response supplies lowest-term fractions and three-place decimals for all parts, attempts an argument for each, and ends with the required Answers line. Its measured length is 853 words, exceeding the 600-word limit by more than 10%, warranting a two-point deduction.

Gemini 3.8 Flash 3.53

The response fails on mathematical correctness across all four parts. In (a), the model confuses the condition for stopping on green (which requires the final mug among the 5 green and blue mugs to be blue, giving 3/5, not 2/5). Parts (b), (c), and (d) contain fundamentally flawed reasoning and incorrect values. Additionally, the response significantly exceeds the 600-word limit at 853 words.

Score by criterion

Correctness

15

All four final values are incorrect: (a) is 2/5 instead of 3/5; (b) is 19/5 instead of 35/6; (c) is 4/9 instead of 50/63; and (d) is 17/4 instead of 25/4. Per the judging policy, having two or more wrong values caps correctness at 4.0; with zero correct answers, a very low score is warranted.

Reasoning Quality

25

The reasoning is flawed throughout. In (a), the symmetry argument is inverted (if the last mug among the 5 is green, the 3 blues were taken first, meaning stopping occurred on blue). In (b), the response starts writing tail probabilities, discovers an inconsistency mid-derivation, and abruptly claims an incorrect sum without valid derivation. In (c), it assumes a fallacious symmetry that stopping occurs on the 9th mug with probability 5/9. In (d), the expected stopping index is miscalculated.

Completeness

70

The response attempts all four parts and provides both fractions and decimals along with explanations, but the derivations are broken or abandoned midway.

Clarity

65

The explanations are formatted neatly with headers and readable prose, but the presentation becomes messy in part (b) when the author interrupts their own work ('Wait, N >= 2 means...').

Instruction Following

65

The response provides the requested exact fractions, decimals, justifications, and the final Answers line. However, it severely violates the length constraint, reaching 853 words against the 600-word maximum (>40% over), warranting a deduction.

About this genre

In this genre, the main abilities being tested are Correctness, Reasoning Quality, Completeness.

Unlike explanation, this genre leans more toward reaching the right answer on exam-style questions than toward tailoring the teaching style for a reader.

A high score here does not guarantee creativity, persuasive writing, or broad performance on open-ended planning tasks.

Strong models here are useful for

study support, textbook-style questions, and problems where answer accuracy matters first.

This genre alone cannot tell you

whether the model is best for long-form explanation, brainstorming, or business communication.

What Is Evaluated in Education Q&A

Scoring criteria and weight used for this genre ranking.

Correctness

45.0%

This criterion is included to check Correctness in the answer. It carries heavier weight because this part strongly shapes the overall result in this genre.

Reasoning Quality

20.0%

This criterion is included to check Reasoning Quality in the answer. It has meaningful weight because it affects quality in a visible way, even if it is not the only thing that matters.

Completeness

15.0%

This criterion is included to check Completeness in the answer. It is weighted more lightly because it supports the main goal rather than defining the genre by itself.

Clarity

10.0%

This criterion is included to check Clarity in the answer. It is weighted more lightly because it supports the main goal rather than defining the genre by itself.

Instruction Following

10.0%

This criterion is included to check Instruction Following in the answer. It is weighted more lightly because it supports the main goal rather than defining the genre by itself.

Education Q&A

At a glance

Results for this task

A Stopping Rule on a Shelf of Coloured Mugs

Answers, highest average first

1 Claude Opus 5.5 Anthropic

(a) Stopping because both greens are taken

Distribution of K

(a) Stopping because both greens are taken

Distribution of K

(b) Expected number taken

(c) At least one red left on the shelf

(d) Given that the first mug is red

2 Gemini 3.8 Flash Google

(a) Probability that taking stops because both green mugs have been taken

(a) Probability that taking stops because both green mugs have been taken

(b) Expected number of mugs taken when taking stops

(c) Probability that at least one red mug is still on the shelf

(d) Expected total mugs taken given the first mug is red

3 GPT-6.1 Sol OpenAI

(a) Stopping because both greens have been taken

(a) Stopping because both greens have been taken

(b) Expected number taken

(c) At least one red remains

(d) Expected total given the first mug is red

4 GPT-6 Luna OpenAI

5 Claude Sonnet 5.5 Anthropic

6 Gemini 3.5 Flash-Lite Google

(a) Probability that taking stops because both green mugs have been taken

(a) Probability that taking stops because both green mugs have been taken

(b) Expected number of mugs taken when taking stops

(c) Probability that at least one red mug is still on the shelf when taking stops

(d) Given that the first mug taken is red, the expected total number of mugs taken

About this genre

What Is Evaluated in Education Q&A

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