Reviewing Movies I

Let $p = 1/5, q = 4/5.$ The expected length of an up run is $q(1+2p+3p^2+\cdots) = \frac{q}{(1-p)^2} = 1/q = 5/4.$ Similarly, the expected length of a down run is $1/p = 5.$ So $\mu_1 + \mu_2 = 5 + 5/4 = 25/4,$ because the first two runs consist of an up run and a down run in some order.

The probability that the seventh run is an up run is the probability that the first run is an up run, which is $p,$ so $\mu_7 = p(1/q) + q(1/p) = 1/4+4 = 17/4.$ The sum is $42/4 = 21/2,$ so the answer is $\fbox{23}.$

(More generally, the same arguments show that $\mu_n = 17/4$ if $n$ is odd and $2$ if $n$ is even.)

I did basically the same thing as Patrick; we have $\mu_1=\sum_{k=1}^{\infty }\left(\left(\dfrac45 \right)^k\times \dfrac15 \times k\right)+\sum_{k=1}^{\infty }\left(\left(\dfrac15 \right)^k\times \dfrac45 \times k\right)=\dfrac{17}{4}.$ Then $\mu_2=\dfrac45\times \left(\dfrac14+1\right)+\dfrac15\times (4+1)=2,$ and moreover, $\mu_{2j}=\dfrac45\times \left(\dfrac14+1\right)+\dfrac15\times (4+1)=2\;\; \text{ and }\;\; \mu_{2j-1}=\dfrac45\times (4+1)+\dfrac15\times \left(\dfrac14+1\right)=\dfrac{17}{4},\;\; j\in \mathbb{Z}^+.$ Then $\mu_1+\mu_2+\mu_7=\dfrac{17}{4}+2+\dfrac{17}{4}=\dfrac{21}{2},$ making the answer $21+2=\boxed{23}$ .