Compute default n=5 coverage and selection.
Coverage is 1-0.6^5=0.92224. Selected success is 0.92224*0.9=0.830016, because the selector finds a correct candidate with probability 0.9 when one exists.
Why prefer n1 under shared error?
With a shared error, coverage stays 0.4 whatever the count. A bank with a correct candidate then holds only correct candidates, so for two or more candidates selected success is 0.4*1+0.6*0=0.4 whatever the selector accuracy. Success ties at 0.4, but the extra samples and the selector add cost, so one candidate (cost 1) is preferred.
What if no allocation is feasible?
selected_allocation is unavailable; revise the allowed contract or abstain.