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Another popular question of mine. Well here's to self-help, starting with a sketch illustrating the 3 paths in the $N = 4, k =2$ situation: The number of ways to arrange the 2 test sets to occur in 4 time periods is ${4 \choose 2} = 6$, and $\frac{k}{N} = .5$ is the fraction of the combinations that will start with a test set. Since a "path" is a continuous ...