4.3.12 problem 52

Let $I_j$ be the indicator variable for the $j$-toss landing on an outcome different from the previous toss for $2\le j\le n$. Then, the total number of such tosses is $X={\sum }_{j=2}^n{I}_j$. The total number of runs is $Y=X+1$. Since $\text{E}\left(X\right)={\sum }_{j=2}^{n}P(I_j=1)={\sum }_{j=2}^n\frac{1}{2}=\frac{n-1}{2}$, $\text{E}\left(Y\right)=\frac{n-1}{2}+1=\frac{n+1}{2}$.