Drawing balls from a box

Given $n \in \mathbb{N}$ , let $E_j$ (for $0 \le j \le N$ ) be the number of additional draws that need to be taken to obtain a consecutive run of $N$ black balls, given that the current "streak" is $j$ black balls long. Then $E_N=0$ and $\begin{aligned} E_j & = \; \tfrac12(1 + E_{j+1}) + \tfrac12(1+E_0) \; = \; 1 + \tfrac12E_0 + \tfrac12E_{j+1} \\ \frac{E_j}{2^j} & = \; \frac{1 + \frac12E_0}{2^j} + \frac{E_{j+1}}{2^{j+1}} \\ \frac{E_j}{2^j} - \frac{E_{j+1}}{2^{j+1}} & = \; \frac{1 + \frac12E_0}{2^j} \end{aligned}$ for $0 \le j \le N-1$ , so that $E_0 \; = \; E_0 - \frac{E_N}{2^N} \; = \; \sum_{j=0}^{N-1} \frac{1 + \frac12E_0}{2^j} \; = \; (2 + E_0)(1 - 2^{-N})$ and hence the desired expected value is $E_0 \; = \; 2(2^N-1)$ When $N=10$ the answer is therefore $\boxed{2046}$ .