Arbitrary sieving: Scenario 1. This is the first of two related, but distinct, questions that are puzzling me at present. 
The sieve of Eratosthanes operates by casting out from the list of natural numbers first every other number, then every third number, then every fifth number, etc. for all prime numbers. In each step, the starting point is the prime number itself, and that starting number does not get cast out. Each step discards $\frac{1}{p}$ of the numbers in the list, retaining $1-\frac{1}{p}$ of the numbers. One might naively intuit that if the starting list contained the numbers $1$ to $k$, and one sieved with every prime less than or equal to $k$, the sieved list would contain $k\prod(1-\frac{1}{p})$ numbers, but this is not the case. 
Without concerning ourselves here how to determine the correct count of numbers remaining on the list, let us call the count of numbers on the sieved list $C(k)$.
Now we introduce an arbitrary change. We perform the same sieve according to essentially the same rules, but instead of starting each sieving step from a corresponding prime, we start from some arbitrary, randomly selected number within the list. For example, when sieving with $2$, we might start with $68$, and discard all even numbers other than $68$. Or we might start with $37$ and discard all odd numbers other than $37$. When sieving with $3$, we might start at $47$ and discard all numbers congruent to $-1 \pmod 3$ other than $47$. And so forth. The point is, each sieving step discards exactly the same fraction of numbers as happens in the sieve of Eratosthanes, but not the same numbers.
My intuition tells me that the count of numbers on this arbitrarily sieved list should be exactly $C(k)$, but I have no idea how to prove or disprove that sense.
Question: Is the count of numbers on sieved lists (as described above) independent of where the sieving startpoint of each step is set? Can a proof be given?
 A: Here is a fairly simple example to show the count may change. Consider $k = 10$, so the sieve of Eratosthenes on the list of natural numbers $\gt 1$ would leave $\{2,3,5,7\}$, i.e., $C(k) = 4$. Next, using your suggestion, have the starting point for $2$ be $10$, so the first sweep will leave $\{3,5,7,9,10\}$. For $3$, have it work normally, so will next get $\{3,5,7,10\}$. Next, for $5$, have the starting point be $10$, so will have $\{3,7,10\}$ remaining. Finally, for $7$, just have it work normally, so the set doesn't change. Here, the final set has only $3$ member, i.e., it's $1$ smaller than the set of primes.
The basic issue is that, for each step in the sieve of Eratosthanes, a certain fraction of values are "removed", but sometimes they have already been removed previously, so the actual new ones removed may be less. To get the actual count at any given time, you can use the Inclusion-exclusion principle to determine the number of integers which don't have a factor of any prime up to some value, such as shown with the simple example at the Counting integers section for $2,3,5$. Your suggested change would leave the fraction being removed the same. However, it could affect how many of those values to remove have already been removed, and thus how many new ones are removed at the stage. At the end, this may cause the count to be different.
