'Jeopardy!' Champs Will Face Off In 'Greatest Of All Time' Series

CULVER CITY, CA — The three highest money winners on "Jeopardy!" will compete in a multiple night event in prime time on ABC beginning Jan. 7, the network announced Monday.

Show champs Ken Jennings, Brad Rutter and James Holzhauer will compete on "Jeopardy! The Greatest of All Time." The winner will receive $1 million and the bragging rights that come with being declared "The Greatest of All Time." The two runners-up will each receive $250,000.

Each hourlong episode will consist of two games to make one match. The first player to win three matches will be declared "The Greatest of All Time."

Episodes also will air Jan. 8 and Jan. 9. Additional matches will air Jan. 10, 14, 15 and 16, if necessary, to determine the champion. All episodes will air at 8 p.m. EST.

The episodes will be taped in December at the Sony Pictures Studios in Culver City. Like all non-sports prime-time programming, "Jeopardy! The Greatest of All Time" will air three hours later on the West Coast, so fans will have to stay off social media to avoid learning the outcome.

Jennings won a record 74 consecutive games in 2004. Rutter is the show's money-winning record holder with $4,688,436, the most on any game show in American television. Holzhauer holds the record for all 15 of the show's top single-day winnings records and won the 2019 Tournament of Champions.

City News Service contributed to this post