"Jeopardy! The Greatest of All Time" will bring together the three highest money winners in the long-running game show's history: Ken Jennings, Brad Rutter and James Holzhauer. The first of the three to win three matches walks away with a $1 million prize and the title of "Jeopardy! The Greatest of All Time." Both runners-up will receive $250,000.
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- Ken Jennings became a household name during his record 74-game winning streak, the longest in "Jeopardy!" history. His winnings total $3,370,700.
- Brad Rutter is the highest money winner of all time across any television game show, with total "Jeopardy!" winnings of $4,688,436. He has never lost "Jeopardy!" to a human opponent.
- James Holzhauer holds the record for all 15 of the top single-day winnings records on "Jeopardy!," and just won the 2019 Tournament of Champions. His winnings total $2,712,216.
The series is scheduled to air Tuesday, Jan. 7, Wednesday, Jan. 8, and Thursday, Jan. 9, at 8 p.m. ET/PT | 7 p.m. CT. If additional games are needed for one player to win three matches, those games will air on Friday, Jan. 10, Tuesday, Jan. 14, Wednesday, Jan. 15, and Thursday, Jan. 16.
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"Based on their previous performances, these three are already the 'greatest,' but you can't help wondering: who is the best of the best?" longtime host Alex Trebek said in a news release.
"I am always so blown away by the incredibly talented and legendary Alex Trebek who has entertained, rallied and championed the masses for generations-and the world-class 'Jeopardy!' team who truly are 'the greatest of all time,'" added Karey Burke, the president of ABC Entertainment. "This timeless and extraordinary format is the gift that keeps on giving and winning the hearts of America every week. We can't wait to deliver this epic and fiercely competitive showdown-with these unprecedented contestants-to ABC viewers and loyal fans everywhere."
The special is produced by Sony Pictures Television with executive producer Harry Friedman.