The expected number of runs is 27.
Let $X_n$ be the color of the n'th card. For n<52, the n'th card is the end of a run if and only if $X_n\not=X_{n+1}$ and the last card in the pack is always the end of a run. So, the total number of runs is $$ N=\sum_{n=1}^{51}1_{\{X_n\not=X_{n+1}\}}+1. $$ The expected number of runs is $$ \mathbb{E}[N]=\sum_{n=1}^{51}\mathbb{P}(X_n\not=X_{n+1})+1. $$ Whatever colour the n'th card is, there are 51 remaining cards in the deck of which 26 of them are a different colour from $X_n$. So, $\mathbb{P}(X_n\not=X_{n+1})=26/51$, giving $$ \mathbb{E}[N]=51 (26/51) + 1=27. $$