TITLE: Alternate way of computing the probability of being dealt a 13 card hand with 3 kings given that you have been dealt 2 kings
QUESTION [5 upvotes]: 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?

REPLY [2 votes]: You're solving 2 different problems. In the above Venn Diagram, 
Your First Probability is $\frac{A}{A+B+C}$ 
Your Second Probability is $\frac{A}{A+B}$
To understand better, I'll recommend you to solve the following version of your problem :

You're given 12 different cards out of which 11 are colored Red and 1 is colored Blue. You have to pick 11 cards from the total available 12 cards. 


Find the probability that there are at least 11 Red cards (or in other words all cards are red) given atleast 10 cards are known to be Red.
Find the probability that there are at least 11 Red cards (or in other words all cards are red) if you pick 10 Red cards before starting the experiment.