First card outranking the second card drawn

Two cards are taken without replacement from a shuffled $52$-card deck, where only rank matters (two through ace) and suits do not break ties. What is the probability that the first card has strictly higher rank than the second?

$8/17$. The second card has the same rank as the first with probability $\tfrac{3}{51}=\tfrac1{17}$. Otherwise one is higher, and by symmetry each card is equally likely to be the higher one: $\tfrac12\cdot\tfrac{16}{17}=\tfrac8{17}$. (If suits broke ties, the answer would be exactly $\tfrac12$.)