Why test everyone if the group is tainted? A binary search (subdividing into two groups) would seem more efficient. Perhaps they had to call everyone back in for a second sample and thought that three times would seem silly.
The largest possible n is the total number of population, and if the expected number of infected soldiers is == 0 the largest n is also the optimal one. Since we know the percentage of infections is a lot more than 0, the largest n isn't optimal. Otherwise if 1 person is infected, you'd be back to square one and your initial test of a cocktail of your entire populations' samples was fruitless.
The second largest n one could choose is half the population by dividing into two groups. It's the most optimal n if the expected number of infected soldiers is around 1. Since we know the percentage of infections is a lot more than 1, the largest n isn't optimal. Otherwise, if 1 person in each group is infected, you'd be back to square one and your initial test of two cocktails of each half of the populations' samples was fruitless.
...etc...
At p = 0.01, the optimal n is, according to this article, around 20-25.
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[ 0.26 ms ] story [ 17.4 ms ] threadIt only works well if a small percentage of the population is infected.
That was precisely the reason. The paper mentions that they only collected enough for the group-level and subsequent exhaustive search.
The second largest n one could choose is half the population by dividing into two groups. It's the most optimal n if the expected number of infected soldiers is around 1. Since we know the percentage of infections is a lot more than 1, the largest n isn't optimal. Otherwise, if 1 person in each group is infected, you'd be back to square one and your initial test of two cocktails of each half of the populations' samples was fruitless.
...etc...
At p = 0.01, the optimal n is, according to this article, around 20-25.