No there is not. First element is defacto winner, but you still have to loop through the rest with 1/n chance of being selected to fully give each element a chance of winner selection
pics[Math.random() * len(pics)]
The picture selection algorithm's kind of single-pass iterator usage might have been more performant back in the XP days, as it avoids possibly expensive operations.
Modern CPU/other optimizations might make a multi-pass approach more performant due to better memory locality or other factors.