Need a pointer on conditional probability of a fair coin toss 
A fair coin is thrown $n$ times. Show that the conditional probability of a head on any specified trial, given a total of $k$ heads over the $n$ trials, is $\frac{k}{n}$ ($k > 0$).

I immediately think of Bernoulli trials, which would result in the following:
$$\binom{n}{k}\cdot\left(\frac{1}{2}\right)^4$$
But I don't see how to fit that in to conditional probability, something like:
$$\mathbb{P}\left[H|K\right]=\frac{\mathbb{P}\left[H\cap K\right]}{\mathbb{P}\left[K\right]}$$
And that's where I'm stuck. K is just the number of heads, so I'm not quite sure how to move forward from here...
Would $\mathbb{P}\left[K\right]=\binom{n}{k}\cdot\left(\frac{1}{2}\right)^4$?
 A: By symmetry, the conditional probability is the same for all trials. By linearity of expectation, the conditional expectation value for the number of heads is the sum over the conditional probabilities of heads for all trials, which is just $n$ times one of these conditional probabilities, since they're all the same. Since the conditional expectation value for the number of heads given $k$ heads is $k$, it follows that the conditional probabilities are $k/n$.
A: One could argue that the probability claim is obvious. We are sampling from a smaller universe, a universe in which the sequence of tosses has a total of $k$ heads and $n-k$ tails, and we draw without replacement.  Each draw is just as likely to be a head as any other, since all positions of the $k$ heads are equally likely.  But it is clear that the probability of a head on the first draw is $\dfrac{k}{n}$.  
However,  to keep in practice, let us calculate, using the conditional probability formula that you quoted.
The probability of $P(K)$ of $k$ heads is given by 
$$P(K)=\binom{n}{k}\left(\frac{1}{2}\right)^n.$$ 
Now for a particular position $m$, what is the probability $P(H\cap K)$ that we get a head on the $m$-th toss, and a total of $k$ heads?  We need to get a head in the $m$-th position, and a total of  $k-1$ heads in the remaining $n-1$ positions. Thus 
$$P(H\cap K)=\left(\frac{1}{2}\right)\binom{n-1}{k-1}\left(\frac{1}{2}\right)^{n-1}.$$
Divide. We get
$$P(H|K)=\frac{\binom{n-1}{k-1}}{\binom{n}{k}}.$$
This expression can be simplified. Note that
$$\binom{n}{k}=\frac{n!}{k!(n-k)!}=\frac{n}{k}\frac{(n-1)!}{(k-1)!(n-k)!}=\frac{n}{k}\binom{n-1}{k-1}.$$ The terms $\binom{n-1}{k-1}$ cancel, and the expression for $P(H|K)$ simplifies to $\dfrac{k}{n}$. 
