I have come across this problem and would like to know how you would solve it:
$n$ coins are tossed one after the other. The probability of a coin getting head is $p$. If we got $k$ heads, what is the probability of the first coin being head?
Now, I believe the solution would be simply $k/n$ given that the event of getting $k$ heads is assured, but on the other hand that would be too easy and the probability p would not be used at all, which is somehow strange. I do not think a conditional probability approach is right, though.
The approach you mention seems to be correct and smart. A more pedantic way following the definitions can be the following.
Let $X_1,...,X_n$ be the $n$ iid random variables corresponding to each coin tossing. Assume tail is $0$ and head is $1$.
The quantity we are after is $$ \mathbb P(X_1=1|X_1+\dots+X_n=k)=\frac{\mathbb P(X_1=1\text{ and }X_1+\dots+X_n=k)}{\mathbb P(X_1+\dots+X_n=k)}. $$
Now $$ \mathbb P(X_1+\dots+X_n=k)={n\choose k}p^k(1-p)^{n-k} $$ and $$ \mathbb P(X_1=1\text{ and }X_1+\dots+X_n=k)= \mathbb P(X_1=1\text{ and }X_2+\dots+X_n=k-1)={{n-1}\choose {k-1}}p^k(1-p)^{n-k} $$ so the result is indeed $k/n$.