James writes down fifteen 1's in a row and randomly writes $+$ or $-$ between each pair of consecutive 1's. One such example is \[1+1+1-1-1+1-1+1-1+1-1-1-1+1+1.\] What is the probability that the value of the expression James wrote down is $7$?

$\begin{array}{@{\hspace{-1em}}l@{\hspace{14em}}l@{\hspace{14em}}l}

\textbf{(A) }0&\textbf{(B) }\dfrac{6435}{2^{14}}&\textbf{(C) }\dfrac{6435}{2^{13}}\\\\ 

\textbf{(D) }\dfrac{429}{2^{12}}&\textbf{(E) }\dfrac{429}{2^{11}}&\textbf{(F) }\dfrac{429}{2^{10}}\\\\

\textbf{(G) }\dfrac1{15}&\textbf{(H) } \dfrac1{31}&\textbf{(I) }\dfrac1{30}\\\\

\textbf{(J) }\dfrac1{29}&\textbf{(K) }\dfrac{1001}{2^{15}}&\textbf{(L) }\dfrac{1001}{2^{14}}\\\\

\textbf{(M) }\dfrac{1001}{2^{13}}&\textbf{(N) }\dfrac1{2^7}&\textbf{(O) }\dfrac1{2^{14}}\\\\

\textbf{(P) }\dfrac1{2^{15}}&\textbf{(Q) }\dfrac{2007}{2^{14}}&\textbf{(R) }\dfrac{2007}{2^{15}}\\\\

\textbf{(S) }\dfrac{2007}{2^{2007}}&\textbf{(T) }\dfrac1{2007}&\textbf{(U) }\dfrac{-2007}{2^{14}}\end{array}$
1. Let's denote the fifteen 1's as \(1, 1, 1, \ldots, 1\). James writes either \(+\) or \(-\) between each pair of consecutive 1's. This means there are 14 operations between the 15 ones.
2. We need to find the probability that the value of the expression is 7. This can be represented as finding the number of ways to arrange the \(+\) and \(-\) signs such that the sum of the expression equals 7.
3. Let \(x_i\) be the \(i\)-th 1 in the sequence. The expression can be written as:
   \[
   x_1 + x_2 + x_3 + \cdots + x_{15}
   \]
   where each \(x_i\) is either \(+1\) or \(-1\).
4. We need the sum of these \(x_i\) to be 7. Let \(k\) be the number of \(+1\)'s and \(m\) be the number of \(-1\)'s. We have:
   \[
   k + m = 15
   \]
   and
   \[
   k - m = 7
   \]
5. Solving these two equations, we get:
   \[
   k + m = 15 \quad \text{(1)}
   \]
   \[
   k - m = 7 \quad \text{(2)}
   \]
   Adding (1) and (2), we get:
   \[
   2k = 22 \implies k = 11
   \]
   Subtracting (2) from (1), we get:
   \[
   2m = 8 \implies m = 4
   \]
6. Therefore, we need 11 \(+1\)'s and 4 \(-1\)'s. The number of ways to choose 11 positions out of 14 for the \(+1\)'s is given by the binomial coefficient:
   \[
   \binom{14}{11} = \binom{14}{3} = \frac{14 \times 13 \times 12}{3 \times 2 \times 1} = 364
   \]
7. The total number of ways to place \(+\) and \(-\) between the 15 ones is \(2^{14}\) because each of the 14 positions can be either \(+\) or \(-\).
8. Therefore, the probability that the value of the expression is 7 is:
   \[
   \frac{\binom{14}{11}}{2^{14}} = \frac{364}{2^{14}}
   \]
9. Simplifying the fraction:
   \[
   \frac{364}{2^{14}} = \frac{364}{16384} = \frac{91}{4096}
   \]

The final answer is \(\boxed{\frac{91}{4096}}\).