Let $X_1$ be the number of coin tosses until the first head, $X_2$ be the number of coin tosses after that until the second head, and so on. We want $E[X]$ where $X = X_1 + X_2 + \dots + X_k$. By linearity of expectation, we have $E[X] = E[X_1] + E[X_2] + \dots + E[X_k]$.
For any $i$, the number $E[X_i]$, the expected number of coin tosses until a head appears, is $\dfrac{1}{p}$. You can see this by calculating the mean of the geometric distribution, or by noticing that $E[X_i] = 1 + (1-p)E[X_i]$, etc.
All this gives $E[X] = \dfrac1p + \dfrac1p + \dots + \dfrac1p = \dfrac{k}p$.
I ask you for additional explanation. Meanwhile I'll post here another approach.
Denote by $\tau_i^5$ the random variable that counts the time required to get five heads starting from $i$ heads, ok?
What we want is exactly $E[\tau_0^5]$, right?
Now, you can evaluate $E[\tau_0^5]$ conditioning at the first step.
$$
E[\tau_0^5] = \frac{E[\tau_0^5]}{2} + \frac{E[\tau_1^5]}{2} +1
$$
Explaining the equation above. With probability $1/2$ you have a tail, so the time you will take to get five heads is the same, because you have any heads. On the other hand, with the same probability you get a head, and now, the number of flips needed to get 5 heads is $E[\tau_1^5]$, because now you that you have one head. The +1 it is because you have to count the first step.
Repeating the argument above you get
$$
E[\tau_1^5] = \frac{E[\tau_0^5]}{2} + \frac{E[\tau_2^5]}{2} +1
$$
Proceeding this way, and remembering $E[\tau_5^5]=0$, you get
$$
E[\tau_0^5] = 62
$$
This may seems more complicated at the first sight, but the idea of "to conditionate at what happens at the first time (or movement)" solve a big variety of problems.
Best Answer
You are correct that the probability of 0 heads is equal to the probability of 4 heads -- there is one way to get each, and each coin flip has equal probability so the probability for each case is
$$\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{16}$$
The probability of getting 1, 2, or 3 heads is different. For example, there are four ways to get 1 heads:
HTTT
THTT
TTHT
TTTH
Each of these cases has probability $1/16$ as before, but there are 4 of them so the total probability of getting 1 heads is $4/16$.
It is a similar procedure to count the number of ways to get 2 or 3 heads. More generally, the number of ways to get $x$ heads out of $n$ coin tosses is given by the binomial coefficient:
$$n \choose x$$
To generalize even further, this coin tossing experiment follows the binomial distribution. If the probability of tossing a heads is $p$ then the PMF is given by
$$P(X = x) = {n \choose x} p^{x}(1-p)^{n-x}$$
In this case, for a fair coin $p = 1/2$ so the distribution simplifies a bit.