What's the probability of a 5 card hand is dealt 4 Kings, given that the hand has the King in Spades and Hearts? What's the probability of a 5 card hand is dealt 4 Kings, given that the hand has the King in Spades and Hearts?
Here's my attempt:
E= KKKK_
F= KK_ _ _ 
so $P(E) = 48/(\phantom{}_{52}C_5)$
and $P(F) = (\phantom{}_4C_2)(\phantom{}_{50}C_3)/(\phantom{}_{52}C_5)$.
The intersection is $P(E)$, so $P(E)/P(F)$.
I got the same answer as if the question didn't specify the Hearts and Spades.
 A: You're essentially dealing a three-card hand from a deck of 50 (missing the spade and heart kings), wanting to know the probability of getting the remaining two kings among those three. The number of ways of dealing three cards is $_{50}C_3=19600$; the number of hands with two kings is $48$ (one for each of the remaining cards in the hand); so the probability is $48/19600=3/1225$.
A: We can use the formal machinery of conditional probability. Let $A$ be the event the hand has $4$ Kings (and of course a fifth card). Let $B$ be the event the hand has the $\spadesuit$ K, and the $\heartsuit$ K, together with $3$ other cards, some of which may be Kings.
We want $\Pr(A|B)$, the probability of $A$ given $B$. By the definition of conditional probability, we have $\Pr(A|B)=\frac{\Pr(A\cap B)}{\Pr(B)}$. We compute the probabilities on the right.
We have $\Pr(B)=\frac{\binom{2}{2}\binom{50}{3}}{\binom{52}{5}}$ and $\Pr(A\cap B)=\frac{\binom{4}{4}\binom{48}{1}}{\binom{52}{5}}$. Divide. 
Remark: Note that the solution of msh210 is more intuitive, and therefore better. 
