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Reasoning and Reinforcement Learning · Lesson 7

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Lesson Outline 8 subsections
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Reasoning and Reinforcement Learning / Learning Plan ↗

LESSON 07 / 8Suggested study 135 minutes

Report Training Gains, Test Regressions and Sampling Budgets Separately

When training answers improve but every held-out task fails, what can additional sampling and parameter learning each establish?

3 Questions for This Lesson
  • Failed to read the same training upgrade and combination that freezes running
  • IRD pass@k and distinguish between budget increases and parameters learning
  • Check the conclusions with precise enumeration, independent sampling, formatting pressure and zero-satisfactory upper limit
Course Project ZIP ↓

Fix the Experimental Setup for This Lesson

All values in this chapter are derived from the same model configuration, the same data contract, and the same 40-step SFT plus 100-step RL run. Before is the post-supervising and pre-RL weight, after is the final weight of the RL and is always referred to as before. Training 20 questions and testing five questions are reported separately. Never mix the correct nineteenth question with the successful test zero into another average and only good-looking part. Each frozen model has a 100-track list and 64 separate entries; greed is the third evaluation, not confused with random single chances of success.

A precise list gives the successful quality of the strategy distribution under a fixed input, so there is no Monte Carlo error; however, network probability is measured with limited accuracy and results are still rounded out, and this does not amount to an unbiased statistical estimate of the overall task distribution. Twenty-five questions are small manual domains that we define, and tests are intended intermediates and not samples of natural mathematics problems. The lessons are separated from accurate probability calculations: network weights are measured by training, and the limited path expectations after a given weight are calculated using model definitions.

Example 1: Real Training-Domain Gains and Real Extrapolation Failure

The average single probability of success for 20 training topics rose from 0.503452473 to 0 7706372887; greed from 16/20 to 19/20; and a total of 1280 samples per sixty-four issues, with a view to observing success from 662 to 987. The three indicators are consistent, supporting the RL to increase rigorous incentives in the field. The average complete track KL to be changed from about 0.30669642268 NT, indicating that the strategy did have a relative change in reference rather than merely reporting or increasing the number of evaluation samples.

The average precise probability of success for the five test tracks is reduced from 0.001999408 to 0. 0000325802, with greed always at 0/5; each three hundred and twenty random evaluations have been zero, and the final average trajectory KL is about 0. 3060306642. Test degradation is a precise probability, not based on sampled speculations. The conclusion is that “this training improvement field awards, while worsening this combination of extrapolations”, cannot be scaled up into “RLFull effectiveness” or expanded to “RLIt can't be generalized.”.

Why Held-out Sums Are Especially Diagnostic

The training has never been intermediate and fourth, so the correct fourth has not been monitored as a first step; ultimately the remaining eight are only for the domain and four, absent in the right second step of the training. The network can learn many of the matches that are entered into the other and not routinely requested to export [4,8] in this group. Enhanced learning continues to reward only 20 subjects of training, which may further reduce the quality of unrewarded responses. This mechanism is compatible with observations, but to prove specific internal causes and consequences it requires additional intervention by fixed models, which cannot be interpreted solely as confirmed neurological mechanisms.

For example, the final greed response to test two plus two is [6,2,EOS]. The second step inside is indeed twice as good as six, but the first step is wrong. If an evaluation only checks the middle to the final self-control, it will be misrecorded as successful; chapter one is strictly validated to preserve such failures. Different test questions will also export [0,0,EOS] or [3,6,EOS], showing that the visible trajectory structure can be clearly separated from the completion of input tasks.

Accounting for the Search Budget

Staple an input and model, and if the probability of success is p for each independent sample, the probability of failure is k (1−p)k(1-p)^k, at least once the probability of success is a supplement. This is the exact formula for the IID sampling and accurate oracle validation conditions. No parameter update is required. If the sample is relevant, deweighted, beam search or filter is wrong, the formula is no longer applied directly. The project report pass@8 is the ideal for oracle to know which one to pass. “At least one is correct.”, is not the actual selection rate of an answer rankingr to learn.

Pr⁡(至少一次成功∣x,k)=1−(1−px)k,pass@k‾=1M∑x[1−(1−px)k]\Pr(\text{至少一次成功}\mid x,k)=1-(1-p_x)^k,\quad \overline{\mathrm{pass@}k}=\frac1M\sum_x[1-(1-p_x)^k]
Freezing strategy, cross-topic IID sampling, rigorous verification; problem-by-issue and average.

Example 2: More Sampling Is Not Evidence of Learning

The uniform strategy p. 0.01 is one per cent success of one single success, eight at least one success rate of about 77:73, and sixty-four at approximately 47:44. All upgrades can take place in the context of completely unchanged parameters, with only extra reasoning budgets. If a 63-th meeting of the updated model, with a baseline taken only once, a higher level of ultimate success cannot be recognized.

The average pass@8 in this training area ranged from 0.980224252 to 0. 9902835309, which is much less than the absolute increase in single probability, because most of the before questions were already more likely, and several more were close to saturation. The average pass@8 in the test field fell from 0.0126487218 to 0.0002606006. Here it has to be counted on a case-by-case basis. 1−(1−px)81-(1-p_x)^8 On average, the formula of average progeny p cannot be used first; non-linearity makes the two generally different.

Zero Successes Still Require Statistical Interpretation

If the freeze model is independently picked at the same probability of success, the zero probability of success is (1−p)n(1-p)^n. Set it to 0. 05, which solves the normal 95% one side of the border p≤1−0.051/np\le1-0.05^{1/n} This is the frequency-based upper hand of two fixed p models, not the Bayesian version of “pYes.95%Probability in this range.”. Each issue of this project has different probabilities, and 320 times cannot be treated as the same p without any explanation; the upper limit of 64 per question is more conservative, about 4.57 %. The precise list gives a more direct idea of the real strategy of the domain.

The overall average success rate is also estimated to be subject-specific weights. For the same sixty-four questions, the sample averages the issue-by-topic average weight; if more problems are encountered, all success figures are directly removed from all samples and then weighted to budget. The report should fix the overall and weight of the topic before deciding on the distribution of sampling, which does not allow evaluation to facilitate a change of objective.

Format Stress Tests on One Frozen Model

Probe_formats. py runs four numerical numbers for all twenty-five questions, each responding with the same limited greed. The normative order has been successful 19/25, two operations have been exchanged still 19/25; the equivalent number 4/25 has been removed and the equivalent has been moved to the previous 3/25. The test question itself has not been successful. These aggregate numbers include a combination of training and cannot be claimed to be an independent general achievement. They indicate that a slight change in the point of reference could undermine many of the training topics that have already been learned, and that the network is not yet well formatted.

The same number of exchange operations has been successful in being compatible with adding-up exchange laws, but it is not sufficient to demonstrate that the network has a general knowledge of exchange; or that the limited domain may be memory symmetrical input. Declassify the equivalent, move the hint beyond the training serialization contract, and failure does not mean that the authentication device is broken, but that the model is vulnerable to this distribution change. Six other searchers explicitly reject the absence of EOS, multiple EOS, boolean, error end value and error middle and middle; and divide model format migration into two records that the certificationer rejects as illegitimate, avoiding mixing it into one test.

Draw Conclusions Only After Independent Verification

Tarts.py loads the actual checkpoint, recalculates the test ' s exact probability of success and compares it to the results.json ' s item by item; freezes the reference digest must be consistent, all path probabilities and close to one. Training is preceded and followed by a comparison of the same subject, the same successful contract and the same sampling budget. Additional sampling for models until it is finally successful, drops the failure issue, extends the certification rules and raises beautiful figures, but answers different questions, and the list of evidence for this course remains original and failed without being replaced.

This chapter does not estimate the variance of more training seeds, no natural language mathematical problems, no scale expansion, remote API or GPU measurements; these are all not running. The failure of a fixed CPU domain is still valuable because readers can accurately state that the code gradient is correct, the parameters are genuinely updated, the training incentive upgrades are not developed, but the designated test capability is not formed. Chapter VIII will include these “Through” and “Failed” in the same way as a full recovery report, rather than simply using a total PASS to hide the results.

Standalone ProgramDirectly Run Check
import math
p=.01
for k in [1,8,64]: print(k,1-(1-p)**k)
upper=1-.05**(1/64)
assert .045<upper<.047
ps=[.01,.9];k=8
exact=sum(1-(1-p)**k for p in ps)/2
wrong=1-(1-sum(ps)/2)**k
assert abs(exact-wrong)>.1
print("per-task",exact,"using mean p",wrong)
Terminal CommandRead-only freeze checkpoint evaluation; new format reports are saved separately
python run.py evaluate --output evidence --split test
python probe_formats.py --checkpoint evidence --output my-format-probes.json

TRY IT YOURSELF

Practice I: Budget effects

How many pass@1 and pass@8 at p=0.1? Does it prove modeling?

Hint 1
Pass@k indicates that the same strategy has a k-second chance of independence, and the probability of all failure is calculated first.
Hint 2
Model parameters can also increase success rates due to an increase in k; when comparing learning, they need to be fixed k with a different reasoning budget.
Compare with the Solution
pass@ 1=0.1 ,pass@ 8=1−0.9^8 ≈ 0.56953279. The parameters can be completely static, and the upgrades come from eight independent opportunities and cannot be considered learning gains.

TRY IT YOURSELF

Practice II: Average Order

The success rates for each of the two questions are 0 and 1, k2, and how much is the average pass and average p after the question?

Hint 1
Applies non-linear 1−(1−p)^k to each issue and then averages it.
Hint 2
Select the course of the average p first, comparing two results; non-linear functions cannot be implied to average the exchange.
Compare with the Solution
The pass is 0 and 1, an average of 0 5. The average pp is 0 5, the proxy formula 1−0.5 2 = 0.75, is wrongly overestimated. Non-linear functions cannot be exchanged with expectations at any time; this item saves complete p per question and returns the respective pass.

TRY IT YOURSELF

Practice III: Training and extrapolation

How can an in-evidence conclusion be written when training is on average p up, testing is down and sampling is not successful?

Hint 1
Separate the results of the 20 training fields from those of the 5 not found in the intermediate and domain.
Hint 2
The writing of fixed seeds and which targets under the budget are met and which are rejected in reverse, does not render all RL invalid for an extra failure.
Compare with the Solution
This fixed seed and budget raises strict incentives in the 20 training areas, while the 5 and 4 combinations have lost their precise quality, greed, and 0/5. It proves that this operation has reached the extrapolation target; it cannot be said that all RLs are ineffective, nor can they be called general reasoning enhancement based on training achievement alone.

CHECK YOUR UNDERSTANDING

Check Your Understanding

Answer first, then open the explanation to check your reasoning. These questions help reveal gaps in understanding; independent implementation and transfer require separate evaluation.

1. How's the change going on here?

Show ExplanationFrom approximately 0.0015994 to 0.00003258, greed still exists at 0/5.
What does 2. pass@8 mean?

Show ExplanationAssuming independent sampling and perfect validation does not represent actual strategy quality.
3. 320 successful zeros can be an unconditional test for p-2?

Show ExplanationThe heterogeneous task structure should be described or analysed on a subject-by-question basis; it may also be listed directly here.
4. How well did the specifications and de-equivalencies work overall?

Show ExplanationThe same freezing model, the full 25 issues of training overlap, is not an independent natural mission test.

Manual completion records your progress; it does not certify your project.