Splitting a card deck of 100 cards into two decks of 50 cards

Perhaps it helps to consider the $4$ a priori cases. We will label the cases $(i,j)$ for $i,j\in \{1,2\}$, where, say, $(1,2)$ means that $J_1\in \text{Pile}_1$ and $J_2\in \text{Pile}_2$ while $(1,1)$ means that both jokers are in $\text{Pile}_1$.

The probability of each case is:

$P_{1,1}=P_{2,2}= \binom {98}{48}\big / \binom {100}{50}=.2\overline {47}$

$P_{1,2}=P_{2,1}=\binom {98}{49}\big/ \binom {100}{50}=.\overline {25}$

Sanity checks: Note that these add to $1$ as they should, and note that we expected each case to be around $\frac 14$.

Now, your first question asks: What is the probability that you are in case $(1,1)$ given that you are not in $(2,2)$?

The answer is, of course, $$\frac {P_{1,1}}{1-P_{2,2}}=\frac {P_{1,1}}{P_{1,1}+P_{1,2}+P_{2,1}}=.328859$$

The second question asks: What is the probability that you are in $(1,1)$ given that you are either in $(1,1)$ or $(1,2)$?

The answer, of course, is: $$\frac {P_{1,1}}{P_{1,1}+P_{1,2}}=.\overline {49}$$

Note that these answers are easy to justify intuitively: if we ignore the slight dependence on the placement of the jokers, each case would have probability exactly $\frac 14$, so the answer to the first question would be $\frac 13$ while the answer to the second would be $\frac 12$. As expected, we see that the (very minor) dependence changes the intuitive results only slightly.