CULVER CITY (CNS) - “Jeopardy!'s” three highest money winners will compete in a multiple consecutive night event in prime time on ABC beginning Jan. 7.
Ken Jennings, Brad Rutter and James Holzhauer will compete on “Jeopardy! The Greatest of All Time,” ABC announced Monday. The winner will receive $1 million and the title of “Jeopardy! 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 will also air on 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.
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 `Jeopardy!'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 “Jeopardy!'s” top single-day winnings records and won the 2019 Tournament of Champions.
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