- ${48 \choose 5}/{52 \choose 5}$ is probability of no Aces
- ${48 \choose 5}/{52 \choose 5}$ is probability of no Kings
- ${44 \choose 5}/{52 \choose 5}$ is probability of no Aces or Kings
so by inclusion-exclusion, the probability of at least one Ace and at least one King is
$$\dfrac{ \displaystyle{52 \choose 5} - 2{48 \choose 5} + {44 \choose 5}}{ \displaystyle{52 \choose 5} }$$
which is close to $10\%$.
Edit:
In my original answer, I ignored the condition that the ace and king are in the same suit. That is partly due to the awkward language, which technically can support my reading. But there is no reason for the question to mention suits at all if the problem was about just having an ace and a king,
so I'm taking the suggestion by commenter AnkitSeth that the correct reading that we are looking for the probability that the hand has, for at least one suit, the ace and king of that suit.
Let $A$ be the set of all deals.
Let $A_i$ be the deals that contain the ace and king of suit $i,$ for $i=1,2,3,4.$
Then the inclusion-exclusion gives the probability:
$$\frac{\binom{4}{1}|A_1|-\binom{4}{2}|A_1\cap A_2|+\binom{4}{3}|A_1\cap A_2\cap A_3|-\binom{4}{4}|A_1\cap A_2\cap A_3\cap A_4|}{|A|}\\=\frac{\binom{4}{1}\binom{50}{11}-\binom{4}{2}\binom{48}{9}+\binom{4}{3}\binom{46}{7}-\binom{4}{4}\binom{44}{5}}{\binom{52}{13}}$$
Original Answer
This answer finds the probability that the hand has one ace and one king.
If $p$ is the probability that you get at least one ace and one king, consider $$1-p = p_{\lnot A}+p_{\lnot K}-p_{\lnot AK}$$ where $p_{\lnot A}$ is that probability that you didn't get any ace, $p_{\lnot K}$ is the probability that you got no king, and $p_{\lnot AK}$ is the probability that you got no ace and no king.
So $$p_{\lnot A} = p_{\lnot K}=\frac{\binom{48}{13}}{\binom{52}{13}}$$ and $$p_{\lnot AK}=\frac{\binom{44}{13}}{\binom{52}{13}}$$
So $$p = \frac{\binom{52}{13} - 2\binom{48}{13} + \binom{44}{13}}{\binom{52}{13}}$$
Best Answer
Answer:
Exactly 1 Ace could be picked in ${4\choose1}$ ways, similarly exactly 1 King could be picked in ${4\choose1}$. Now you have 8 cards removed from the deck and the remainder 3 cards should come from 44 cards which are non - ace and non-king. That you can choose in ${44\choose3}$. Lastly, 5 cards could be chosen in ${52\choose5}$
Thus the probability is $$\approx \dfrac{{4\choose1}.{4\choose1}.{44\choose3}}{{52\choose5}}$$
EDIT:
Other way is :
The Ace could be chosen with a probability $= \frac{4}{52}$
The King could be chosen with a probability $= \frac{4}{51}$
Rest of the three cards could be chosen with a probability $= \frac{44}{50}.\frac{43}{49} .\frac{42}{48}$
These five cards could be picked in $=\frac{5!}{1!.1!.3!} $
Now multiply $$=\frac{4}{52}.\frac{4}{51}.\frac{44}{50}.\frac{43}{49} .\frac{42}{48}.\frac{5!}{1!.1!.3!}$$
The first method is an approximate where as the second one is the most accurate.