I calculated that the probability of drawing a king and then a heart from a deck of cards is
$\frac{1(12) + 3(13)}{\text{Permutation}(52,2)}=\frac{1}{52}$
However, I also noticed that this is the same as the probability of drawing the king of hearts, which is also $\frac{1}{52}$
Is this just a coincidence or is there a reason why this happens?
One possible interpretation of your question is as follows. Let $A$ be the event the first draw results in a King, and $B$ be the event the second draw results in a heart. Why are the events $A$ and $B$ independent?
That they are independent can be verified by the cases computation in the post above. But let us see why the result is intuitively clear.
The probability of a heart on the second draw, is, like the probability of a heart of the first draw, or the seventeenth, equal to $\frac{1}{4}$.
Now suppose that we are told the first card drawn was a King. Should that change our estimate of the probability that the second is a heart? If so, being told the first draw was a Jack should change our estimate in exactly the same way, as should being told that the first draw was a $9$.
Since all ranks are equally likely to be drawn first, the conditional probability that the second is a heart given the first is of a specified rank is the same as the plain unconditional probability that the second card is a heart.
This says that $A$ and $B$ are independent, that is, $\Pr(A\cap B)=\frac{1}{13}\cdot\frac{1}{4}$.
Let's say you have a different deck. One where there are $v$ values and $s$ suits and therefore $sv$ cards. Let's ask what the probability of drawing a particular value (number of suits divided by total cards) followed by a particular suit (number of cards of that suit divided by left over cards).
This will be first the probability of getting a particular value (e.g. $\frac{4}{52}$)--the number of suits, $s$, divided by total number of cards, $sv$. Then there are two ways of getting a particular suit: 1) it's the same suit as the card you drew--there's a $\frac{1}{s}$ chance of this (e.g. $\frac{1}{4}$)--then there are $v - 1$ values (e.g. $12 = 13 - 1$) of the particular suit left from the $sv - 1$ cards left and 2) the chosen value is not the suit you chose, with probability of $\frac{s - 1}{s}$ (e.g. $\frac{3}{4}$) and now all $v$ values (e.g. all $13$) of the suit are left from the $sv - 1$, giving:
\begin{align*} \frac{s}{sv} \left(\frac{1}{s}\frac{v - 1}{sv - 1} + \frac{s - 1}{s}\frac{v}{sv - 1}\right) =&\ \frac{v - 1 + v(s - 1)}{sv(sv - 1)} \\ =&\ \frac{v - 1 + sv - v}{sv(sv - 1)}\\ =&\ \frac{sv - 1}{sv(sv - 1)} \\ =&\ \frac{1}{sv} \end{align*}
I think André Nicolas gave a much better answer, but here is a more general case which shows that this isn't just a coincidence.
