Beating the dealer with a strictly higher card

You and a dealer are each dealt one card from the same shuffled $52$-card deck. Card values run from two up to ace, suits are ignored, and ties go to the dealer. What is the chance that your card is strictly higher?

$8/17$. The two cards share a rank with probability $\tfrac3{51}=\tfrac1{17}$. In the remaining $\tfrac{16}{17}$ of cases symmetry makes each player equally likely to hold the higher card, so you win with probability $\tfrac8{17}$.