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Use individual stratified randomization within each prison, with concealed centrally generated permuted-block randomization and varying block sizes. Prison can affect both baseline risk and intervention delivery, so balancing assignments within prison protects against imbalance in that important covariate. Varying block sizes and allocation concealment prevent staff from predicting the next allocation and selecting participants accordingly. Prespecify how incomplete final blocks are treated. To enforce the overall 4000:4000 allocation exactly, coordinate the stratum totals centrally, allocating odd final places through a concealed random rule; exact half allocation within every prison is impossible if some prison totals are odd.

Individual randomization stratified by prison is preferable to randomizing only twenty whole prisons for this design. A cluster-randomized trial would have few independent units and could lose substantial precision through intraclass correlation. Sharing or selling the intervention can cause treatment contamination, so record it and retain an intention-to-treat analysis; substantial spillover might justify a separately powered cluster design rather than being ignored.

Two useful questions separate intervention retention from actual opportunities to use it.

Responses can explain implementation, but cannot replace the randomized intention-to-treat analysis. Reincarcerated survivors are a selected subgroup, and recall errors and missing follow-up may affect their reports.

Use equal allocation and a conventional two-sided 5% comparison of independent proportions. The assumed risks are $p_C=0.005$ and $p_I=0.005(1-0.30)=0.0035$, giving target difference $\delta =0.0015$. Put $\overline{p}=(p_C+p_I)/2=0.00425$ and let $m$ be the number per arm.

The null critical difference is approximately $z_{0.975}\sqrt{2\overline{p}(1-\overline{p})/m}$. Under the alternative, the difference has standard deviation $\sqrt{[p_C(1-p_C)+p_I(1-p_I)]/m}$. For 80% statistical power, the target difference must exceed the critical difference by $z_{0.8}$ alternative standard deviations. Solving gives the sample size for comparing two proportionsThis is about $29,500$ per arm, or $59,000$ altogether. Rounding for planning:The rare-event Poisson approximation gives a similar answer. This assumes the stated baseline risk applies, no endpoint loss, and no attenuation by treatment contamination. The two-sided convention is explicit: a one-sided 5% test would require fewer participants.

Define the difference as control minus intervention, positive for fewer deaths under intervention assignment. The observed difference per thousand isFor independent binomial distributions, its unpooled standard error per thousand isThe approximate 95% confidence interval is thereforeThe rare-event Poisson standard error $\sqrt{148+123}/30$ gives almost the same interval. A reduction of 1.5 per thousand is consistent with the data, since it lies inside the interval. Zero also lies inside, so a nonzero benefit is not established at the two-sided 5% significance level. This does not establish equivalence or prove the proposed effect size, and cannot by itself establish treatment contamination.

Measure receipt and transfers directly rather than inferring them only from the outcome difference. Use identifiable pack codes and confidential early follow-up to record whether packs were retained, lost, taken, given away or administered, and to whom. Where feasible, corroborate administration reports with care records and establish whether recipients belonged to the opposite randomized group.

A process-evaluation sample across both arms can estimate the frequency and direction of treatment contamination. Separate receipt tracking from blinded outcome ascertainment where possible. Keep randomized assignments for the primary intention-to-treat analysis: excluding transfers or reclassifying participants by receipt can introduce confounding, since receipt behavior is not randomized. Receipt data can inform a supplementary analysis whose additional assumptions are explicit.

Assume treatment receipt gives event risk $p_1$, nonreceipt gives risk $p_0$, and a fraction $c$ in each equally sized randomized arm crosses to the opposite treatment. The two assignment risks are $(1-c)p_1+c{p}_0$ and $c{p}_1+(1-c)p_0$. Subtracting yields the displayed attenuation, with ${\Delta }_{\mathrm{receipt}}=p_0-p_1$. It depends on the stated equal crossover fractions and unchanged receipt-specific risks; it is not a universal formula for every form of treatment contamination.