Single coin toss

Let $X_n$ be the expected number of additional tosses until you get a "streak" of $N$ Heads, given that you have a current "streak" of $n$ Heads. Conditioning on the next toss, we have $X_n \; = \; 1 + \tfrac12X_0 + \tfrac12X_{n+1} \hspace{2cm} 0 \le n \le N-1$ and we know that $X_N = 0$ . Solving these equations, we obtain that $X_n \; = \; 2^{N+1} - 2^{n+1} \hspace{2cm} 0 \le n \le N$ and so the expected number of tosses to get a "streak" of $N$ heads is $2^{N+1}-2$ . For this problem, $N=5$ , so the answer is $\boxed{62}$ .

With this problem it is important to distinguish between 'tosses' and 'trials'. For example if you throw HHT, it is basically one trial, but 3 tosses. You need on average 2^5 = 32 trials, but it is not how many actual tosses you need. If you tossed 5 coins at once, you would, on average, need 32 tosses to throw 5 'heads'. But here we have one coin only.

I had no idea how to approach this problem in the beginning. I only knew I have to figure out how many tosses is the average trial. I finally decided to work backwards. I ran a computer simulation and it turned out I need on average 62 throws. So, the average trial requires 1.9375 tosses. After getting some help from a friend, we figured out the solution is 2^5 * (1 + 1/2 + 1/4 + 1/8 + 1/16) Likewise for 7 'heads' in a row, you would need 2^7 * (1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64) tosses