Mary has a deck of $x$ cards of only two types. There are $y$ cards of the first type and $z$ cards of the second type.
If Mary draws $n$ cards before Sarah draws $m$ cards from the same deck (without replacement of cards), what are the chances of Mary having at least a quantity $q$ of the first type of cards if she draws another $t$ cards from the deck?
I am thinking that this question could be solved with the following function:
$$ \sum_{k=q_2}^t \sum_{j=0}^m \sum_{i=q_1}^n \left ( \frac{\binom{y}{i} \cdot \binom{z}{n-i}}{\binom{x}{n}} \right ) \left ( \frac{\binom{y-i}{j} \cdot \binom{x-y+i-n}{m-j}}{\binom{x-n}{m}} \right ) \left ( \frac{\binom{y-i-j}{k} \cdot \binom{x-y+i-n+j-m}{t-k}}{\binom{x-n-m}{t}} \right ) $$
in all cases in which $q_1 + q_2\geq q$.
As you can see, the problem is that multiple summation functions cannot calculate conditional probability because they do not multiply each new case by the previous cases from which it is conditional upon.
Any way you can define a new function for the benefit of all humanity?
Hint:
If I understand well (Mary takes $n$ cards without replacement. Then Sarah takes $m$ cards from the remaining cards without replacement. Then Mary takes $t$ cards again from the remaining cards without replacement) then you can rephrase the question:
"What are the chances of Mary having at least a quantity $q$ of the first type of cards if she draws in total $n+t$ cards from the deck?"