Neighbours of the same rank

Idea. Write the count as a sum of 51 indicators, one per pair of neighbouring positions. Each has the same probability, and expectation is linear.

Step 1. For neighbouring positions $i$ and $i+1$: whatever card is in position $i$, the card in position $i+1$ is equally likely to be any of the other 51 cards, and 3 of them have the same rank. So the pair matches with probability $\frac{3}{51}$.

Step 2. $$\mathbb{E}[\text{matches}]=51\cdot \frac{3}{51}=3.$$

In general. With $k$ suits the answer is $k-1$, whatever the number of ranks. The probability that there is at least one matching pair is much harder to compute; it is about 95.5%.