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pass@k Estimator

Problem Statement

Estimate pass@k, the probability that at least one of k sampled generations passes, given n total samples of which c passed. This is the standard unbiased HumanEval estimator.

Background

The unbiased estimator is

$\text{pass@}k = 1 - \frac{\binom{n-c}{k}}{\binom{n}{k}}$

If n - c < k (fewer failures than k), pass@k is exactly 1.0 (every draw of k must include a pass). Compute it stably.

Your Task

def pass_at_k(n, c, k):

Return the pass@k estimate as a float rounded to 6 decimals.

Input Format

n (int total samples), c (int correct), k (int), with 0 <= c <= n, 1 <= k <= n.

Output Format

A float rounded to 6 decimals.

Sample

print(pass_at_k(5, 1, 1))

Output:

0.2

Example:

Input:

print(pass_at_k(5, 1, 1))

Output:

0.2

Reasoning:

Check for guaranteed pass: Compare the number of failures ( $n - c$ ) against $k$ . With $n=5$ and $c=1$ , there are $5 - 1 = 4$ failures. Since $4 \ge 1$ ( $k$ ), it is not guaranteed that a pass is included in the sample, so we proceed to the probabilistic calculation.
Calculate the failure probability: The term $\frac{\binom{n-c}{k}}{\binom{n}{k}}$ represents the probability that all $k$ samples are failures. For $k=1$ , this simplifies to the ratio of failures to total samples: $\frac{5-1}{5} = \frac{4}{5} = 0.8$ .
Compute the pass probability: Subtract the failure probability from 1 to find the probability of at least one pass: $1 - 0.8 = 0.2$ .
Format the result: Round the result to 6 decimal places, yielding $0.200000$ , which is represented as 0.2.
The final output is 0.2

Constraints:

If n - c < k, return 1.0.
Otherwise 1 - C(n-c,k)/C(n,k); compute with a stable product.
Round to 6 decimals.

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1. Background Knowledge

pass@k is a standard metric in code generation evaluation (introduced in the HumanEval paper) that estimates the probability that at least one of $k$ independently sampled solutions is correct. Given $n$ total samples where $c$ are correct, the unbiased estimator is:

$\text{pass@}k = 1 - \frac{\binom{n-c}{k}}{\binom{n}{k}}$

The term $\frac{\binom{n-c}{k}}{\binom{n}{k}}$ represents the probability that all $k$ draws are failures (i.e., none pass). Subtracting from 1 gives the probability that at least one passes. This is a hypergeometric probability: drawing $k$ items without replacement from a population of $n$ containing $n-c$ failures.

A critical edge case exists: if $n - c < k$ , it is impossible to draw $k$ failures because there aren't enough of them. In this scenario, every subset of size $k$ must contain at least one correct sample, so pass@k is exactly $1.0$ .

2. Algorithm Approach

The core challenge is numerical stability. Directly computing binomial coefficients $\binom{n}{k}$ via factorials will overflow for moderately large $n$ (e.g., $n > 170$ in double-precision floating point). Instead, compute the ratio $\frac{\binom{n-c}{k}}{\binom{n}{k}}$ using a product of ratios, which avoids large intermediate values.

The ratio can be expanded as:

$\frac{\binom{n-c}{k}}{\binom{n}{k}} = \prod_{i=0}^{k-1} \frac{n-c-i}{n-i}$

This product form is numerically stable because each factor is a ratio of two numbers of similar magnitude, keeping intermediate results bounded between 0 and 1.

3. Step-by-Step Strategy

Handle the edge case: If $n - c < k$ , return 1.0 immediately.
Initialize a variable fail_prob = 1.0 to accumulate the product.
Loop $i$ from $0$ to $k-1$ :

Multiply fail_prob by $\frac{n - c - i}{n - i}$ .

Compute the result as $1 - \text{fail\_prob}$ .
Round to 6 decimal places and return.

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solution.py

Python 3.13

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