Alternate way of computing the probability of being dealt a 13 card hand with 3 kings given that you have been dealt 2 kings We are dealt 13 cards from a standard 52 card deck.  If $A$ is the event where we are dealt at least two kings and $B$ is the event where we are dealt at least 3 kings, we want to know $P(B|A)$.  I believe the correct way to do this is to calculate the probability of being dealt a hand with each number of kings separately as follows:
$\displaystyle \frac{{4 \choose 3}{48 \choose 10} + {4 \choose 4}{48 \choose 9}}{{4 \choose 2}{48 \choose 11} + {4 \choose 3}{48 \choose 10} + {4 \choose 4}{48 \choose 9}} \approx .17$.
However, it also makes sense to me that if we know we have been dealt 2 kings, it doesn't matter where in our hand they are, the $P(B|A)$ should be the same as the probability of getting dealt either 1 or 2 kings in an 11 card hand from a 50 card deck with two kings in it as follows:
$\displaystyle \frac{{2 \choose 1}{48 \choose 10} + {2 \choose 2}{48 \choose 9}}{{50 \choose 11}} \approx .4$
(Or compute $1-p$, where $p$ is the probability of getting no kings in an 11 card hand from a deck with 50 cards and only 2 kings.)
What is the issue with my logic here?

Your first answer is a correct application of Bayes Rule.  If, as you say, $A$ is the event "at least 2 kings" and $B$ is the event "at least 3 kings", then 
$$P(B|A)=\frac{P(A|B) \cdot P(B)}{P(A)}$$
$$P(A|B) = 1$$
$$P(A) = \frac{\begin{pmatrix} 4\\2 \end{pmatrix}\begin{pmatrix} 48\\11 \end{pmatrix}+\begin{pmatrix} 4\\3 \end{pmatrix}\begin{pmatrix} 48\\10 \end{pmatrix}+\begin{pmatrix} 4\\4 \end{pmatrix}\begin{pmatrix} 48\\9
\end{pmatrix} }{\begin{pmatrix} 52\\13 \end{pmatrix}}$$
$$P(B) = \frac{\begin{pmatrix} 4\\3 \end{pmatrix}\begin{pmatrix} 48\\10 \end{pmatrix}+\begin{pmatrix} 4\\4 \end{pmatrix}\begin{pmatrix} 48 \\9 \end{pmatrix} }{\begin{pmatrix} 52\\13 \end{pmatrix}}$$
$$P(B|A) = \frac{P(B)}{P(A)} = \frac{\begin{pmatrix} 4\\3 \end{pmatrix}\begin{pmatrix} 48\\10 \end{pmatrix}+\begin{pmatrix} 4\\4 \end{pmatrix}\begin{pmatrix} 48\\9 \end{pmatrix} }{\begin{pmatrix} 4\\2 \end{pmatrix}\begin{pmatrix} 48\\11 \end{pmatrix}+\begin{pmatrix} 4\\3 \end{pmatrix}\begin{pmatrix} 48\\10 \end{pmatrix}+\begin{pmatrix} 4\\4 \end{pmatrix}\begin{pmatrix} 48\\9 \end{pmatrix} } $$
Your second answer, if you were to try to "apply Bayes Rule" would show the falicy.  What are your events "$A$" and "$B$" in this case?
You can also answer the question with the $1-p$ type argument, but again, you need to consider what the events are, and compute Bayes' rule accordingly.
