Readiness
Lessons read measures coverage, not readiness. This estimates the thing you actually want to know: if you sat a fresh exam today, would you pass? The pool is public and finite, the exam structure is published, and your answer history is in this browser — so the number is computed, not guessed.
Nothing to estimate yet
Readiness is computed from answered questions, so it needs some answers. A single full-proportion exam is enough to produce a first estimate, and it will be a wide one.
Probability of passing
Simulated score distribution
Where the marks are going
Ordered by expected questions lost, which is what matters — a weak subelement that contributes six exam questions costs more than a weaker one contributing two.
| Subelement | On exam | Pool seen | Accuracy | Est. correct | Expected loss |
|---|
Coverage and history
- Pool seen
- Questions answered
- Full exams taken
- Best full exam
Full exams over time
How this number is computed
Each pool question gets a probability of a correct answer, estimated as the mean of a Beta posterior with recency-weighted evidence:
pi = (a0 + Σ wkxk) / (a0 + b0 + Σ wk), wk = exp(−Δtk/τ)
with τ = 45 days, so a question answered correctly three months ago counts for about a seventh of one answered today. The prior is Beta(0.5, 1.5) — mean 0.25, a blind guess among four choices — and it is weak enough that a handful of answers moves it.
A question you have never seen inherits the rate for its subelement, shrunk toward that prior. Being strong on the rest of E5 is evidence about an unseen E5 question; it is not proof.
The exam is then simulated the way a real one is drawn: each subelement contributes its published number of questions, taken from distinct groups. Twenty thousand runs, each scored by sampling Bernoulli(pi). The reported probability is the fraction that reach the passing mark, and the interval is ±2 standard errors of that fraction.
Monte Carlo rather than a closed form on purpose: the exact distribution is a mixture of Poisson-binomials over draws without replacement, and simulating it is both correct and legible.
The caveat that matters. Accuracy on a question you have already seen partly measures memory of the answer rather than knowledge of the material, so the all-attempts estimate runs optimistic. First attempts are the unbiased estimator for a fresh draw. Where the two disagree, believe the lower one — and note that onlylastAt is timestamped, so repeat attempts are aged as if they were all as recent as the last, which also tilts optimistic.