Does removing random items effect probability? Let us say you have a finite set of things (maybe playing cards?) called $S$. It has $n$ things. No let us say we want a thing from an arbitrary set $E$. First we remove $k$ number random things from S, were $0 \leq k&lt;n$. Now I pick a random thing from $S$. Does the probability that the thing is in $E$ in terms of $E$, $k$, $n$ and $S$ actually depend upon $k$? If so, how?
Denote with $R$ the set that remains after removing $k$ elements from $S$. What the question boils down to is if the following equality holds:

$$\Pr(x \in E) = \Pr(x \in E \mid x \in R)$$

where the right-hand side is a conditional probability. This conditional probability can be calculated by the identity:

$$\Pr(x \in E \text{ and } x \in R) = \Pr(x \in E \mid x \in R) \Pr(x \in R)$$

Now because $R$ picks (uniformly) randomly from $S$, $x \in R$ is independent from $x \in E$, that is, $\Pr(x \in E \text{ and } x \in R) = \Pr(x \in E)\Pr(x \in R)$.

It finally follows that (using $\Pr(x \in R) &gt; 0$):

$$\begin{align}
\Pr(x \in E \mid x \in R) &amp;= \frac{Pr(x \in E \text{ and } x \in R)}{\Pr(x \in R)}\\
&amp;= \frac{\Pr(x \in E)\Pr(x \in R)}{\Pr(x \in R)}\\
&amp;= \Pr(x \in E)
\end{align}$$

which means that the probability that an $x \in R$ is in $E$ actually does not depend on $R$ whatsoever (as long as $R$ is not empty, i.e. $k &lt; n$).
If you know what the things drawn from $E$ are, then yes, it affects the probability.  If you don't know what the things are, then it does not affect the probability.