Estimate from eligible observations
Read the recorded counts, select the conditioning group and use its total as the denominator. Combine only compatible runs and explain what finite observations can establish
- Filter the denominator For last red given first red, RR and RB are eligible; only RR is a success. For an unconditional estimate, use every completed trial
- Keep zero distinct from undefined Zero successes among a positive number of eligible trials gives an estimate of zero. With no eligible observations, the conditional estimate is undefined
- Pool counts, not rates For separate runs with the same model, condition and success event, add their successes and eligible totals before dividing. Do not average rates without weighting their denominators
- Avoid counting observations twice A 2000-trial extension already contains its 200-trial prefix. They are not two separate runs to pool
- Interpret uncertainty Keep the exact fraction before rounding. Compare repeated runs with the exact model prediction, allowing for variation. Observations do not prove independence or impossibility
Worked example
Run 1 records 43 RR trials, 49 RB trials, 50 BR trials and 18 BB trials. A separate run under the same model records RR = 21 and RB = 27, alongside other first-blue trials. Estimate last red given first red for Run 1, then for the pooled records
- \(\displaystyle n(F)=43+49=92\)Only trials beginning red satisfy the condition
- \(\displaystyle n(L\cap F)=43\)Within that group, RR is a success
- \(\displaystyle \widehat{\Pr}(L\mid F)=43/92\)Use the eligible denominator. This is an estimate, not an exact model probability
- \(\displaystyle \begin{aligned}s&=43+21=64\\n&=92+21+27=140\end{aligned}\)Add successes and eligible counts from the two separate, compatible runs. First-blue trials do not enter either conditional denominator
- \(\displaystyle \begin{aligned}\hat p_{\mathrm{pooled}}&=\frac{64}{140}\\&=\frac{16}{35}\end{aligned}\)This estimate weights each eligible observation equally. It is not the unweighted average of the two run proportions
Common mistakes
Difficulty levels
Questions about this topic
Which trials belong in a conditional denominator?+
Every trial satisfying the condition, including eligible failures. For first red, both RR and RB qualify
Is zero the same as undefined?+
No. Zero successes with a positive eligible count gives zero. No eligible observations gives an undefined estimate
Can I average two observed rates?+
Equal eligible counts allow an unweighted average. With unequal counts it can differ from the pooled estimate, so add successes and eligible totals, then divide
When may I pool runs?+
Use separate, non-overlapping runs with the same trial model, replacement rule, condition and success event
Can I pool a run with its longer extension?+
No. The longer run already includes the shorter prefix, so adding them would count some observations twice
Does selecting a condition generate new results?+
No. It filters the same outcome table. A different run number changes the reproducible random sequence
Do unequal estimates prove dependence?+
No. Independent models can produce unequal finite-sample estimates. Establish exact independence from the model; use observations as evidence with uncertainty
What should I report?+
The model and condition, full and eligible counts, successes, exact and rounded estimates, comparisons across separate runs and a limitation of the evidence
Can I use a calculator?+
No. This topic is practised without a calculator
Which year level and chapter is this?+
Year 10 Mathematics, chapter 4: Probability