I was looking at Wolves and Sheep and had an interesting thought.
The original problem says that out of $100$ sheep, it is known that there are $5$ wolves and they want to minimize the number of tests used. Let's say that instead of giving a discrete number such as $5$ wolves, they said that each sheep has a $10 \%$ chance of secretly being a wolf.
In that case, how would we do the testing? All of the solutions on Wolves and Sheep rely on the fact that there are exactly $5$ wolves.
I wish I had some sort of insight or steps that I had taken thus far to approach this problem, but I simply haven't been able to do anything. I don't get how to solve this whatsoever.
You would need a multistage testing and scoring mechanism and only have a probabilistic answer at the end so you only know (say with probability 0.99) you have found all the defectives.
Assuming noiseless group tests, if $T(A)$ for a subset $A$ returns “defect present” you use the conditional probability distribution that the number of defectives in $A$ is at least 1, and score each element equally in terms of likelihood of being defective.
Then apply $T(A’)$ to another subset.
If both $T(A)$ and $T(A’)$ say “defective present” any element belonging to $A \cap A’$ gets a higher score.
You can determine based on $p=0.1$ that the most informative test is to test an untested subset of size $k$ where the probability of “defect absent” is roughly $1/2$. This means $(1-p)^k=1/2$ and you take the nearest integer to determine $k.$ This should in general minimize the total number of tests.
See wikipedia for more, the above quote is from there in the “non adaptive and probabilistic testing” section.