The three highest money winners in “Jeopardy” history will join forces for an ABC event early next year.
Ken Jennings, Brad Rutter and James Holzhauer will compete on “Jeopardy! The Greatest of All Time,” which debuts at 8 p.m. Jan. 7.
Alex Trebek will host the event, which ABC said will air over multiple consecutive nights.
The first player to win three matches earns $1 million and the title as the game show’s greatest of all time. The other two players each receive $250,000.
Holzhauer just won the “Jeopardy!” Tournament of Champions that concluded Friday and collected the $250,000 first prize.
“Based on their previous performances, these three are already the ‘greatest,’ but you can’t help wondering: Who is the best of the best?” Trebek said in a statement.
The event continues at 8 p.m. Jan. 8 and 9. If necessary, the special series will run at 8 p.m. Jan 10 as well as 8 p.m. Jan. 14-16.
“We’re excited to bring ‘Jeopardy!’ to prime time! It’s been a long time in the making —we wanted to create a unique experience sure to wow not just our fans, but all audiences,” said Mike Hopkins, chairman of Sony Pictures Television, which produces the game show. “We are thrilled to have James, Brad and Ken, three powerhouse players each worthy of the title ‘The Greatest of All Time.’ With Alex hosting this is truly going to be something special. I can’t wait to see who comes out on top!”