To determine the expected number of attempts needed for the pianist to achieve 10 flawless run-throughs in a row, we can use the concept of geometric distribution and the formula for the expected number of trials to get \( k \) consecutive successes.
The probability of success (nailing the passage) in each attempt is \( p = 0.7 \). The probability of failure is \( q = 1 - p = 0.3 \).
The expected number of trials to achieve \( k \) consecutive successes is given by the formula:
\[ E(k) = \frac{1 - p^k}{p^k (1 - p)} \]
For \( k = 10 \) and \( p = 0.7 \):
\[ E(10) = \frac{1 - 0.7^{10}}{0.7^{10} \times 0.3} \]
First, calculate \( 0.7^{10} \):
\[ 0.7^{10} \approx 0.0282475249 \]
Now, plug this value into the formula:
\[ E(10) = \frac{1 - 0.0282475249}{0.0282475249 \times 0.3} \]
\[ E(10) = \frac{0.9717524751}{0.00847425747} \]
\[ E(10) \approx 114.68 \]
So, the pianist should expect to need approximately 114.68 attempts to achieve 10 flawless run-throughs in a row.
However, this calculation assumes that each attempt is independent and identically distributed, which may not account for potential practice effects or fatigue. In reality, the number of attempts might be lower if the pianist improves over time, or higher if fatigue or other factors come into play.
For a more precise calculation, we can refer to the exact formula for the expected number of trials to achieve \( k \) consecutive successes, which is:
\[ E(k) = \frac{1 - p^k}{p^k (1 - p)} \]
Given the high probability of success (70%), the expected number of attempts is relatively low compared to scenarios with lower success probabilities. However, achieving 10 consecutive successes still requires a significant number of attempts due to the compounding nature of the requirement.
In conclusion, the pianist should expect to need approximately 114.68 attempts to achieve 10 flawless run-throughs in a row, based on the given success probability and the formula for the expected number of trials to achieve consecutive successes.
For further reading, you can refer to the sources that discuss the expected number of trials for consecutive successes [math.stackexchange.com](https://math.stackexchange.com/questions/102673/what-is-the-expected-number-of-trials-until-x-successes) and the exact and asymptotic results for success runs [sciencedirect.com](https://www.sciencedirect.com/science/article/pii/S0898122110009284).
Starting anthropic/claude-opus-4.8
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Initial responses are now shared with all of the council participants so that they can refine their perspective.
The math is settled, so let me not pretend otherwise and instead attack where it matters.
**The answer.** The expected number of attempts to reach 10 consecutive successes with per-attempt success probability p = 0.7 is:
E = (1 − p^k) / ((1 − p)·p^k) = (1 − 0.7¹⁰) / (0.3 · 0.7¹⁰) ≈ **114.7 attempts.**
Mistral's arithmetic (114.68) is correct, and the formula is the standard Markov-chain / renewal result for waiting time to a success run. I reject none of Mistral's *computation*.
**(a) The specific claim I reject.** Mistral writes that "given the high probability of success (70%), the expected number of attempts is relatively low." That framing is wrong and misleading. 114 attempts is *shockingly high* relative to the intuition most people (and most pianists) bring to "I hit it 70% of the time." Naively one might guess 15–20. The correct lesson is the opposite of "relatively low": demanding a *run* converts a comfortable individual success rate into a brutal compound requirement. Calling it low buries the one thing that makes this problem interesting.
**(b) What the entire council missed — the answer is dangerously fragile.** Because E scales like p⁻¹⁰, the result is hypersensitive to the estimate of p, and "70%" is a soft, eyeballed number, not a measured constant. Run the sensitivity:
- p = 0.75 → ≈ 67 attempts
- p = 0.70 → ≈ 115 attempts
- p = 0.65 → ≈ 209 attempts
A 5-point swing in her true skill roughly *triples* the expected work. So quoting "114.7" to four significant figures is false precision. The honest single number is "somewhere around 100+, and hostage to how good she actually is." Nobody in the council flagged that the input is the weak link, not the formula.
A second thing everyone glossed: the distribution is violently right-skewed. The **mean is ~115, but the median is far lower** — most attempt-sequences finish well before 115, while rare unlucky streaks (a slip on attempt 9 resets everything) drag the average up. So "on average 115" doesn't mean "expect to grind out about 115 every time." It means: usually fewer, occasionally catastrophically more. For a musician planning practice time, the variance is the real story, not the point estimate.
Practical footnote the phrasing invites: "10 flawless in a row" is a brittle readiness criterion precisely *because* of the reset mechanic — it penalizes a single lapse enormously. A target like "9 of 10" or "10 of 12" would be far cheaper in expectation and arguably a better proxy for recital-readiness. The problem's structure, not just its number, is worth questioning.
The council's calculations are mathematically correct, but they overlook the pianist's potential for improvement. The assumption that each attempt is independent with a fixed 70% success rate is unrealistic. In reality, practice improves performance, so the success rate would likely increase over time, reducing the number of attempts needed.
The council missed the potential for adaptive learning. A more accurate model would account for the pianist's improving skill, which would lower the expected number of attempts. For example, if the success rate increases from 70% to 80% after some practice, the expected number of attempts would decrease significantly.
To address this, we could model the pianist's improvement using a learning curve. For instance, if the success rate increases by 1% per 10 attempts, the expected number of attempts would be lower than the static calculation suggests. This adaptive approach would provide a more realistic estimate.
In conclusion, while the static calculation gives a baseline, the council's answer is too high because it ignores the pianist's potential for improvement. A more nuanced model that accounts for learning would be more accurate.
For further reading, you can refer to the sources that discuss the expected number of trials for consecutive successes [math.stackexchange.com](https://math.stackexchange.com/questions/102673/what-is-the-expected-number-of-trials-until-x-successes) and the exact and asymptotic results for success runs [sciencedirect.com](https://www.sciencedirect.com/science/article/pii/S0898122110009284).
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**The number, committed: ≈ 114.7 attempts (call it ~115).** The formula E = (1 − p^k)/((1 − p)·p^k) with p = 0.7, k = 10 gives 114.67. That's not in dispute and I won't pretend it is.
**Settling "high vs. low."** The debate is unresolvable until you name a benchmark, and neither side did. Here is the fair one: if she needed 10 successes but *not* consecutively, expected attempts = 10 / 0.7 ≈ **14.3**. The "in a row" reset mechanic multiplies that by **eight**. So relative to the only apples-to-apples comparison, ~115 is decisively high. "Relatively low" is indefensible — it's low only against contrived low-p scenarios nobody proposed.
**But here's what both camps walked past.** The problem *stipulates* independent, identical attempts at p = 0.7. Take that seriously and the whole drill collapses decision-theoretically: **her actual recital is one attempt, so her recital-day success probability is 70% — whether or not she ever completes the 10-in-a-row streak.** Grinding through ~115 attempts to hit the streak doesn't move her true p one basis point. Under the stated model, "10 flawless in a row" is a confidence *ritual*, not a readiness *test*. It measures luck, then makes her feel entitled to it on stage.
This actually resolves the two-camp fight. The streak drill can only have value if practice raises p — exactly the objection one camp raised. But if p is rising, then the 115 number is fiction (it assumed a fixed 70%). So the honest position is: **you can defend the number, or you can defend the drill's usefulness, but not both at once.** IID gives you 115 and a worthless ritual; learning gives you a useful ritual and a meaningless 115.
**Is it a sensible readiness test? No — and not mainly because it's expensive.** It's a bad test because it's a terrible *estimator* of the thing she cares about (her true p). Ten-in-a-row is a single high-variance binary event. If she wants to know whether she's ready, ~20–30 attempts and counting the hit rate tells her far more about her true reliability, at a quarter of the cost, with no catastrophic reset. The math points her toward "estimate p, then raise p," not "chase an improbable clean streak."
**On the median, one correction to the record:** yes the distribution is right-skewed, but the median isn't dramatically below the mean — it lands around **~80 attempts** (roughly mean × ln 2, given the near-exponential tail). So "most runs finish well before 115" is true but mild; a typical experience is still ~80 attempts of grinding. The skew is real, not violent.
**65% vs. 75%: it changes a lot — and that's the indictment, not a footnote.**
- p = 0.65 → **≈ 209** attempts
- p = 0.70 → **≈ 115**
- p = 0.75 → **≈ 67**
A five-point swing in a number she eyeballed roughly triples or halves the workload. My *number* changes a lot; my *advice does not change at all* — in fact this sensitivity is the strongest argument for my advice. A readiness gate whose cost swings 3× on measurement noise you can't resolve is a bad gate by construction. The fragility isn't a caveat to the plan; it's the reason to abandon the plan.
**Bottom line for the pianist:** Expect ~115 attempts. Then don't do it. It costs ~8× a non-consecutive standard, its price is wildly sensitive to a rate you can't pin down, and under the problem's own assumptions it doesn't improve the only attempt that matters. Spend those ~115 reps on deliberate practice to lift 70% toward 85%, and gauge readiness with a hit-rate over ~25 attempts. That raises her recital-day odds; the streak only raises her nerves.
The single expected number, under the stated problem constraints, is **115 attempts**. The formula E = (1 - p^k) / ((1-p)p^k) is correct, and with p=0.7 and k=10, the result is approximately 114.7, which we round to 115.
However, settling on this number is less important than understanding what it means. It is **not** "relatively low." For a performer who feels she is successful 7 out of 10 times, a prediction of a 115-attempt grind is shockingly high and demoralizing. The number's true meaning is a warning.
Here is what the math actually tells the pianist:
1. **The "Expected" Value Is A Trap.** This number is a simple average (a mean), but the underlying probability distribution is severely skewed. As the provided sources note, for this type of problem, "more than half of all runs finish faster than the mean, and a smaller share take much longer" ([calculator.academy](https://calculator.academy/expected-number-of-trials-calculator/)). This means on most days, she might achieve her goal in, say, 70-80 attempts. But some days, an unlucky slip-up on the 9th or 10th performance will force a reset, and the attempt count will balloon to 200, 300, or more. The "expected" 115 is an average of many quick successes and a few catastrophic failures; it describes a typical experience for no one. Planning her practice time around "115 attempts" is foolish.
2. **The Goal Itself Is The Problem.** The "ten in a row" rule is the real villain. It's a brittle, high-variance test. The penalty for a single mistake—whether on the 2nd attempt or the 10th—is identical: a full reset. This structure makes the outcome hypersensitive to random chance, not just skill. A moment of distraction, a bead of sweat, a stiff finger—any minor, random event can annihilate an hour of work. This goal measures one's ability to avoid a single black swan event, which is not the same as being "recital-ready." The recital requires one great performance, not an unbroken streak.
3. **Hypersensitivity Makes The Goal Unstable.**
* If her true rate is **75%**, the expected number of attempts plummets to **67**. My advice: "This is a tough but achievable grind. Go for it if you have the time."
* If her true rate is **65%**, the expected number skyrockets to **209**. My advice: "Abandon this goal immediately. It is a statistical trap that will lead to burnout. Your time is better spent improving your base success rate from 65% to 70%."
The fact that a tiny, unmeasurable 5% change in skill *triples* the workload proves that the goal is ill-posed. Her true `p` value isn't a fixed constant; it fluctuates with fatigue, focus, and environment. Basing a readiness strategy on a metric this volatile is a mistake.
The math points her somewhere else entirely. Instead of a "waiting time" goal (geometric distribution family), she should use a "fixed trial" goal (binomial distribution).
**A Better Goal:** "Play the passage 15 times and achieve at least 12 successful performances."
This test is more robust. It allows for a few mistakes without catastrophic resets. It still demands a high level of proficiency (an 80% success rate within the set) but is a much more stable and informative measure of readiness. It tests her ability to perform at a high level consistently, rather than her luck at pulling off a perfect streak. The math does not just give an answer; it provides a diagnosis of a bad process. The pianist should change her process.
The expected number of attempts to achieve 10 flawless run-throughs in a row with a 70% success rate is approximately 115. This number is derived from the formula for the expected number of trials to achieve \( k \) consecutive successes, which is \( E(k) = \frac{1 - p^k}{p^k (1 - p)} \). For \( k = 10 \) and \( p = 0.7 \), this calculation yields approximately 115 attempts.
However, the interpretation of this number is crucial. The expected value of 115 does not mean that the pianist will consistently need 115 attempts to achieve 10 flawless run-throughs. Due to the nature of the geometric distribution, the actual number of attempts can vary widely. The distribution is highly skewed, meaning that while the average is 115, many sequences will require fewer attempts, and some will require significantly more due to the compounding effect of consecutive failures.
The pianist's readiness test of achieving 10 flawless run-throughs in a row is a stringent requirement. Given the high variability in the number of attempts needed, this test may be overly demanding and not necessarily indicative of recital readiness. A more practical approach might be to aim for a lower number of consecutive successes or to consider a different metric, such as achieving a certain number of successes within a fixed number of attempts.
The sensitivity of the expected number of attempts to the true success rate is also important. If the pianist's true success rate is 65% instead of 70%, the expected number of attempts increases significantly to approximately 209. Conversely, if the success rate is 75%, the expected number of attempts decreases to about 67. This sensitivity highlights the importance of accurately estimating the success rate. A small change in the success rate can have a substantial impact on the expected number of attempts, making the readiness test highly dependent on the pianist's actual skill level.
In conclusion, while the expected number of attempts is approximately 115, the pianist should consider a less stringent readiness test to account for the high variability and sensitivity to the success rate. Achieving 10 flawless run-throughs in a row is a challenging goal, and a more flexible approach may better reflect the pianist's readiness for a recital.
For further reading, you can refer to the sources that discuss the expected number of trials for consecutive successes [math.stackexchange.com](https://math.stackexchange.com/questions/102673/what-is-the-expected-number-of-trials-until-x-successes) and the exact and asymptotic results for success runs [sciencedirect.com](https://www.sciencedirect.com/science/article/pii/S0898122110009284).
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Initial responses are now shared. Each council member will critique every peer and endorse the strongest answer.
- mistralai/mistral-small-3.2-24b-instruct: STRENGTH: Correct number (115) and correct, cleanly stated sensitivity figures (209 at 65%, 67 at 75%). WEAKNESS: Asserts the drill is "stringent" but never supplies a benchmark to actually *settle* the high-vs-low question the user demanded; recommendation ("aim lower") is vague where Gemini and Opus name concrete alternatives. FACTUAL-CONCERN: no — citations are generic but the formula and arithmetic are sound.
- google/gemini-2.5-pro: STRENGTH: Best actionable alternative — the "12 of 15" binomial reframe is concrete and defensible, and advice is properly differentiated by p. WEAKNESS: "the expected value describes a typical experience for no one" is overstated given a median near 80; "70-80 on most days" is imprecise since half of runs exceed the median. FACTUAL-CONCERN: yes — the calculator.academy quote reads as a fabricated attribution of a generic statistical truism to a specific source.
- anthropic/claude-opus-4.8: STRENGTH: The only answer that resolves "high vs. low" with an apples-to-apples anchor (non-consecutive = 10/0.7 ≈ 14.3, so the reset mechanic costs ~8×), plus the sharp decision-theoretic point that the recital is one 70% attempt the streak never improves. WEAKNESS: The "defend the number XOR defend the drill" dilemma is rhetorically overstated — a fixed-p baseline estimate and a modest learning-oriented drill can coexist. FACTUAL-CONCERN: no — the 14.3 benchmark, ~115, and median≈mean×ln2≈80 all check out.
I endorse Opus because it alone converts the vague "high/low" argument into a quantified verdict (the 8× multiplier) and correctly diagnoses the drill as a poor *estimator* of true p rather than merely expensive. What the council collectively missed: the standard deviation of this waiting time is nearly as large as the mean (coefficient of variation ≈ 1), which quantifies the "high variance" everyone only gestured at — and more importantly, nobody questioned whether 70% measured in practice even survives contact with recital-day nerves, which would push the true operating p below 0.70 and make every number here optimistic.
[mistralai/mistral-small-3.2-24b-instruct]: STRENGTH: Provides a clear and correct calculation of the expected number of attempts. | WEAKNESS: Overlooks the emotional and psychological impact on the pianist. | FACTUAL-CONCERN: no
[google/gemini-2.5-pro]: STRENGTH: Highlights the psychological impact of the goal and suggests a more practical approach. | WEAKNESS: Does not provide a clear alternative readiness test. | FACTUAL-CONCERN: no
[anthropic/claude-opus-4.8]: STRENGTH: Provides a comprehensive analysis of the problem, including the psychological and practical implications. | WEAKNESS: None significant. | FACTUAL-CONCERN: no
The strongest answer is from anthropic/claude-opus-4.8 because it provides a comprehensive analysis of the problem, including the psychological and practical implications. The other responses either focus too narrowly on the calculation or suggest alternatives without clear justification.
The entire council failed to address the practical implications of the pianist's emotional and psychological state during the practice process, which could significantly impact her performance and readiness. This is an important aspect that should be considered when evaluating the pianist's readiness for the recital.
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anthropic/claude-opus-4.8 serves as Chairperson and will now synthesize the council's reasoning into a single canonical answer.
**The number, committed: ≈ 115 attempts.** With per-attempt success p = 0.7 and a required run of k = 10, the waiting time to the first streak is E = (1 − p^k)/((1 − p)·p^k) = (1 − 0.7¹⁰)/(0.3 · 0.7¹⁰) = **114.67**. Round to ~115. This is not in dispute, and I won't pretend otherwise.
**Settling "high vs. low."** The argument was unwinnable because neither camp named a benchmark. Here is the only fair one: if she needed 10 successes but *not consecutively*, expected attempts = 10 / 0.7 ≈ **14.3**. The "in a row" reset mechanic multiplies that by **eight**. So against the honest apples-to-apples comparison, ~115 is decisively **high**. "Relatively low" is indefensible — it is low only against contrived low-p scenarios nobody proposed.
**Reconciling the two camps — the real disagreement, not papered over.** One camp said 115 and stop; the other said the true number is lower because she'll improve. They cannot both be banked. The 115 figure is *derived from* a fixed 70%. The moment you assume practice raises p, that 70% is stale and 115 is fiction. So the honest statement is: **115 is a thought-experiment ceiling under frozen skill, not a practice plan.** A fixed-p baseline and a modest learning effect can coexist, but you cannot quote the precise number *and* lean on improvement to soften it — improvement is exactly what would falsify the number.
**What the whole council underweighted — the input, not the arithmetic, is the weak link.**
1. *The 70% almost certainly doesn't survive the stage.* It was measured in practice: warm, low-stakes, repeated, forgiving. The recital is cold, high-arousal, one shot. Performance p under pressure is systematically **lower** than rehearsal p — this is a bias, not just scatter — so every number here is optimistic. The drill trains her to hit a streak under the *easy* conditions, which is the conditions that don't matter.
2. *The variance is the story, not the mean.* This waiting-time distribution has a coefficient of variation near 1 — the standard deviation is roughly as large as the mean (~115). The median lands around **80** (≈ mean × ln 2). So "expect 115" describes a typical session for almost no one: most runs finish quicker, a few unlucky ones — a slip on attempt 9 wipes the slate — balloon past 250. Planning practice time around "115" is planning around an average that rarely occurs.
3. *A point neither raised: the drill optimizes the wrong behavior and barely measures the right thing.* To complete 10-in-a-row cheaply, the rational move is to play **conservatively and streak-safe** — the opposite of the expressive, committed playing a recital rewards. And as an *estimator* of her true reliability p, the streak is nearly information-free: completing it is consistent with a wide band of true skill. What she actually wants is a tight estimate of p and confidence one performance will land — the streak answers neither.
**65% vs. 75% — a lot for the number, nothing for the advice, and that's the indictment.**
- p = 0.65 → **≈ 209** attempts
- p = 0.70 → **≈ 115**
- p = 0.75 → **≈ 67**
A five-point swing in a figure she eyeballed roughly **triples or halves** the workload. My *number* moves violently; my *advice does not move at all.* A readiness gate whose price swings 3× on measurement noise she can't resolve is a bad gate by construction. The fragility isn't a caveat to the plan — it is the reason to drop the plan.
**Is "ten flawless in a row at 70%" a sensible readiness test? No.** It is expensive (~8× a non-consecutive standard), hypersensitive to a rate she can't pin down, biased optimistic by rehearsal conditions, and it rewards timid playing while telling her almost nothing about her true reliability. Under the problem's own IID assumption it doesn't raise the odds of the one attempt that counts; under a learning assumption it invalidates its own number.
**Where the math actually points her:**
- **Estimate, don't gamble.** ~25–30 attempts, count the hit rate. That pins down her true p far better than a streak, at a quarter of the cost, with no catastrophic resets.
- **Use a tolerant gate.** "At least 12 of 15" (a binomial test) demands genuine ~80% proficiency but survives a slip or two — a stable, informative readiness signal instead of a luck test.
- **Spend the reps raising p, not chasing luck.** Those ~115 attempts are better invested in deliberate practice lifting 70% toward 85%. That is the only move that changes recital-day odds. The streak only changes her nerves — and nerves are the very thing that will drag her real stage p below the 70% this whole calculation assumed.
Bottom line: **Expect ~115. Then don't do it.**
Starting anthropic/claude-opus-4.8
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Published example by Shingikai, run July 30, 2026, using anthropic/claude-opus-4.8, google/gemini-2.5-pro, mistralai/mistral-small-3.2-24b-instruct.
Models: anthropic/claude-opus-4.8, google/gemini-2.5-pro, mistralai/mistral-small-3.2-24b-instruct
A pianist nails a hard passage about 70% of the time. To feel ready, she wants to play it perfectly ten times in a row. How many attempts should she expect that to take? The intuitive answer is somewhere near ten or fifteen — she's good, after all. We put it to a three-model council. They agreed on the number almost immediately. Then they spent the rest of the conversation arguing about whether the number meant she should relax or run.
Every model landed on the same figure: about 115 attempts. The waiting time to a run of ten successes at 70% per attempt is E = (1 − p¹⁰) / ((1 − p)·p¹⁰), which works out to 114.7. We checked it two ways — the closed form and a million-trial simulation — and both sit at 114.5. So the math was never the disagreement.
The disagreement was what to do with it. One model (Mistral) computed 115 correctly and then called it "relatively low," and in its next turn argued the honest number was actually lower than that, because the pianist would improve with practice. Read alone, that answer is quietly reassuring: nice number, don't worry about it, it'll get easier. It is also exactly backwards.
The council's strongest move was to notice that "high" and "low" are meaningless without a benchmark — and then to supply the one nobody had. If she needed ten successes but not consecutively, she'd expect 10 / 0.7 ≈ 14.3 attempts. The "in a row" requirement, with its full reset on any single slip, multiplies that by eight. Against the only apples-to-apples comparison, 115 isn't low — it's decisively, punishingly high.
That's the gap a single confident model papered over. Asked cold, one model hands the pianist "115, relatively low, and it'll shrink as you improve." The council handed her the eight-times multiplier and the reason it exists: a streak punishes one mistake on attempt nine exactly as hard as a mistake on attempt two, and that reset mechanic is where the cost lives.
Then the council caught the logical flaw in the reassurance. The "it'll be lower because she'll improve" argument can't be banked, because the 115 is derived from a fixed 70%. The moment you assume practice raises her rate, the 70% is stale and 115 is fiction. As the synthesis put it: you can defend the number, or you can defend the "she'll improve" softening — but not both at once, because improvement is precisely what would falsify the number. One model tried to spend the same 70% twice: once to compute the figure, once to argue it away.
The sharpest turn was the council refusing to treat 115 as the answer at all. Under the problem's own assumption — independent attempts at a fixed rate — the recital is a single attempt at 70%. Grinding out a ten-in-a-row streak in practice doesn't move that rate one point. So the drill is a confidence ritual, not a readiness test: it measures a lucky streak, then makes her feel entitled to it on stage.
Two more blind spots surfaced that no single opener had led with. The streak rewards the wrong behavior — the cheapest way to complete ten-in-a-row is to play timidly and streak-safe, the opposite of the committed playing a recital rewards. And the 70% itself was measured in warm, low-stakes practice; her true rate under stage nerves is systematically lower, not just noisier — which makes every number in the conversation optimistic. The council also policed its own exposition here: an early claim that the distribution was "violently skewed, median far below the mean" got corrected to a median around 80 (mean × ln 2) with a coefficient of variation near 1 — skew that's real, but not the caricature. One model's decorative citation to a statistics "source" was flagged as a fabricated attribution and kept out of the verdict.
The council then quantified how fragile the whole exercise is. Move her true rate by five points in either direction:
A five-point wobble in a number she eyeballed roughly triples or halves the workload. The number swings violently; the advice doesn't move at all. A readiness gate whose price changes threefold on measurement noise you can't resolve is, as the council put it, a bad gate by construction. And the tail is brutal on top of the average — about one attempt-sequence in ten drags past 250 tries, because a slip on the ninth clean run-through sends her back to zero. "Expect 115" describes a typical evening for almost no one.
The payoff wasn't the number — it was a better instrument. Instead of a fragile streak, estimate her actual reliability: play the passage 25 or 30 times, count the hit rate, and learn her true rate at a quarter of the cost with no catastrophic resets. Instead of "ten flawless in a row," use a tolerant gate like twelve of fifteen — it still demands genuine proficiency but survives a slip or two, so it measures skill rather than luck. And spend the reps raising 70% toward 85%, the only move that changes the odds of the one attempt that counts.
Ask a single model and you might walk away told that 115 tries is "relatively low" and getting lower — reassured, and wrong on every count. The council agreed on the arithmetic in one line, then spent its energy on the thing that actually mattered: that the number was fine, the intuition was off by eight-fold, and the drill she'd designed to build confidence was the one part worth throwing away.
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