Does it matter how you pull marbles out of this vat?

Question

There's this large, well-mixed vat containing some unknown but finite varieties of marbles: $$\{v_{1},v_{2},v_{3},...,v_{k}\}$$ Some varieties are more common than others.

Marbles are going to be pulled out of this vat by some machine. Each time you run the machine, it produces a set containing $q$ different kinds of marbles: $$\{v_{q_{1}},v_{q_{2}},v_{q_3},...,v_{q}\}$$ We're going to look at a bunch of these sets with the idea of estimating how likely it is that any $v_{n}$ shows up in such a set.

So now we have to go and collect our data. The thing is, we don't know precisely how the machine picks marbles; it could pick up $q$ simultaneously, then discard and pick again if they aren't all different. Alternatively, it could keep picking up marbles until there were $q$ varieties. It could also pick a marble, then ignore the rest of it's kind. There's plenty of ways it could be picking the marbles.

(Assume the vat is large enough that any finite number of marbles we pull out doesn't affect the population to a measurable degree.)

The only thing we know about how the machine picks is that it doesn't discriminate against the marbles it picks as long as they are relatively unique. In other words, the likelihood that a given variety shows up in the $q$ set is determined by it's commonness in the vat, not the machine.

This brings me to my question, will the results of our statistical analysis be affected at all by the manner in which the machine collects marbles?

Answer 1

A simple way to verify that the method matters is to choose particular probabilities for types of marbles, and calculate the chance of each subset according to some methods. This can't prove that the method doesn't matter, though.

Suppose there are $3$ types and the chances of each type are $1/2$, $1/4$, and $1/4$, respectively. Suppose you are choosing $2$ types of marbles.

Suppose after choosing a marble, you ignore the rest of the kind. The chance you get $\lbrace v_2,v_3\rbrace$ is $2*1/4*1/3 = 1/6$.

Suppose you reject pairs with repeated types. The chance of $\lbrace v_2,v_3\rbrace$ is $$\frac{2*1/4*1/4}{2*1/4*1/4 + 2*1/2*1/4 + 2*1/2*1/4} = \frac{1/8}{1/8 + 1/4 + 1/4} = 1/5.$$

Since these are different, the method the machine uses matters. Rejecting pairs with repeated types tends to weight the pairs with common types less.

Two of the methods you mention are equivalent. Ignoring the rest of its kind after picking a marble is the same as picking until you have $q$ different types.