Nov. 18 (UPI) -- Jeopardy! winners Ken Jennings, Brad Rutter and James Holzhauer will return to compete on the show in January.
ABC announced in a press release Monday that Alex Trebek will host Jeopardy! The Greatest of All Time, a multiple consecutive night TV event premiering Jan. 7 at 8 p.m. ET.
The Greatest of All Time brings together Jennings, Rutter and Holzhauer, the three highest money winners in Jeopardy! history, for a new series of matches. The first to win three will receive $1 million and the title of "Jeopardy! The Greatest of All Time." The two runners up will each receive $250,000.
"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.
Rutter is the highest-earning contestant on Jeopardy! at $4.7 million, followed by Jennings at $3.5 million and Holzhauer at $2.9 million.
"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,'" ABC Entertainment president Karey Burke said.
"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," she added.
Trebek was diagnosed with stage 4 pancreatic cancer in the spring. He promoted pancreatic cancer awareness in a PSA in October ahead of World Pancreatic Cancer Day, which falls Thursday.
Jeopardy! The Greatest of All Time will air as follows:
Jan. 7 at 8 p.m. ET
Jan. 8 at 8 p.m. ET
Jan 9 at 8 p.m. ET
Jan. 10 at 8 p.m. ET (if needed)
Jan. 14 at 8 p.m. ET (if needed)
Jan. 15 at 8 p.m. ET (if needed)
Jan. 16 at 8 p.m. ET (if needed)