<page_title> 1868 United States presidential election in Kentucky </page_title> <section_title> Results </section_title> <table> <cell> Horatio Seymour of New York <col_header> United States presidential election in Kentucky, 1868 </col_header> <col_header> Candidate </col_header> <col_header> Count </col_header> </cell> <cell> 74.55% <col_header> United States presidential election in Kentucky, 1868 </col_header> <col_header> Popular vote </col_header> </cell> </table>
In the 1868 United States presidential election in Kentucky, Seymour won with 74.55% of the popular vote.