Probability of getting first head on even coin flip A summary of the given question: 
A biased coin has a probability ($p$) of getting heads. Whats the probability of getting the 1st heads on an even numbered flip. $H_1$ is head of first flip. $E$ is first head on even numbered flip. $X^c$ means event $X$ didn't occur. For example: $H_1^c$ means tails on first flip


I don't understand the part
$$P(E | H_1^c) = P(E^c)$$
I seem to think in fact, $P(E | H_1^c) = P(E)$ since the only way $P(E)$ can occur is when $H_1^c$ has occurred. 
Moving forward, how do I get 
$$P(E) = (1-p)(1-P(E)) = \frac{1-p}{2-p}$$ 
 A: Before addressing why $P[E|H_1^c] = P[E^c]$, I'll mention why it is not true that $P[E|H_1^c] = P[E]$. In general whenever we condition on some information, we either increase or decrease probabilities (so long as the information is not independent). In this case, $E$ is certainly not independent of $H_1$ (or $H_1^c$), and so we do not expect $P[E|H_1^c] = P[E]$, else the first coin toss would have no effect on the probabilities.
As an example, consider the easier problem, that the first head falls on the second throw, $H_2$. One sees that $P[H_2] = 1/4$, whilst $P[H_2|H_1^c] = 1/2$, and so $P[H_2|H_1^c] \neq P[H_2]$. This example extends to the situation above.
The way to interpret the statement $P[E|H_1^c] = P[E^c]$ is by considering the sequence of coin tosses after the first toss. If we know that the first coin was a tail, then lets now forget about that coin, and consider all the subsequent ones. We now would want that the first head lands on an odd throw, the event $E^c$.
In brief the idea is: condition on the first throw being tail. Then, consider only the throws after that point, and calculate the probability that the first head falls on an odd throw.
Alternatively, here's a direct method to obtain the result,though it doesn't follow the route above.
I rely on the fact that the occurrence of the first head is given by a geometric distribution:
$$P[\text{k'th throw is first head}] = (1-p)^{k-1}p$$
So, using this fact we see,
$$P[\text{first head is on even throw}] = \sum_{k=1}^\infty (1-p)^{2k-1}p $$
Rearranging, this is equivalent to
$$p(1-p)\sum_{k=0}^\infty (1-p)^{2k} = \frac{p(1-p)}{1-(1-p)^2} =\frac{1-p}{2-p} $$
