One day Jason finishes his math homework early, and decides to take a jog through his neighborhood.  While jogging, Jason trips over a leprechaun.  After dusting himself off and apologizing to the odd little magical creature, Jason, thinking there is nothing unusual about the situation, starts jogging again.  Immediately the leprechaun calls out, "hey, stupid, this is your only chance to win gold from a leprechaun!"

Jason, while not particularly greedy, recognizes the value of gold.  Thinking about his limited college savings, Jason approaches the leprechaun and asks about the opportunity.  The leprechaun hands Jason a fair coin and tells him to flip it as many times as it takes to flip a head.  For each tail Jason flips, the leprechaun promises one gold coin.

If Jason flips a head right away, he wins nothing.  If he first flips a tail, then a head, he wins one gold coin.  If he's lucky and flips ten tails before the first head, he wins $\textit{ten gold coins.}$  What is the expected number of gold coins Jason wins at this game?

$\textbf{(A) }0\hspace{14em}\textbf{(B) }\dfrac1{10}\hspace{13.5em}\textbf{(C) }\dfrac18$

$\textbf{(D) }\dfrac15\hspace{13.8em}\textbf{(E) }\dfrac14\hspace{14em}\textbf{(F) }\dfrac13$

$\textbf{(G) }\dfrac25\hspace{13.7em}\textbf{(H) }\dfrac12\hspace{14em}\textbf{(I) }\dfrac35$

$\textbf{(J) }\dfrac23\hspace{14em}\textbf{(K) }\dfrac45\hspace{14em}\textbf{(L) }1$

$\textbf{(M) }\dfrac54\hspace{13.5em}\textbf{(N) }\dfrac43\hspace{14em}\textbf{(O) }\dfrac32$

$\textbf{(P) }2\hspace{14.1em}\textbf{(Q) }3\hspace{14.2em}\textbf{(R) }4$

$\textbf{(S) }2007$
1. Let's denote the expected number of gold coins Jason wins as \( E \).

2. Jason flips a fair coin, so the probability of getting a head (H) is \( \frac{1}{2} \) and the probability of getting a tail (T) is \( \frac{1}{2} \).

3. If Jason flips a head on the first try, he wins 0 gold coins. This happens with probability \( \frac{1}{2} \).

4. If Jason flips a tail on the first try, he wins 1 gold coin plus the expected number of gold coins from the remaining flips. This happens with probability \( \frac{1}{2} \).

5. Therefore, we can set up the following equation for \( E \):
   \[
   E = \left(\frac{1}{2} \times 0\right) + \left(\frac{1}{2} \times (1 + E)\right)
   \]

6. Simplifying the equation:
   \[
   E = \frac{1}{2} \times 0 + \frac{1}{2} \times (1 + E)
   \]
   \[
   E = \frac{1}{2} \times 1 + \frac{1}{2} \times E
   \]
   \[
   E = \frac{1}{2} + \frac{1}{2}E
   \]

7. To isolate \( E \), subtract \( \frac{1}{2}E \) from both sides:
   \[
   E - \frac{1}{2}E = \frac{1}{2}
   \]
   \[
   \frac{1}{2}E = \frac{1}{2}
   \]

8. Multiply both sides by 2 to solve for \( E \):
   \[
   E = 1
   \]

Thus, the expected number of gold coins Jason wins is \( \boxed{1} \).