Now this is a tournament of champions.
ABC announced Monday that Jeopardy!‘s three highest earners of all time — Ken Jennings, Brad Rutter and James Holzhauer — will face off against one another for the multi-night primetime event Jeopardy! The Greatest of All Time.
Premiering Tuesday, Jan. 7 at 8/7c, the Jeopardy! tournament will air for at least three consecutive weeknights through Thursday, Jan. 9. The first contestant to win three episodes will receive $1 million and the title of Jeopardy!‘s all-time greatest player, while the runners-up will each receive $250,000. (If more than three broadcasts are required to crown a champion, those will continue on Friday, Jan. 10 and will potentially run into the following week.)
“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,'” said ABC Entertainment President Karey Burke. “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.”
“Based on their previous performances, these three are already the ‘greatest,’ but you can’t help wondering: Who is the best of the best?” added Trebek, who will of course preside over the competition.
Jennings earned Jeopardy! fame in 2004 with his 74-game winning streak, the longest in the show’s history (which amassed $3,370,700). Rutter, meanwhile, is the highest money winner across any TV game show, with total winnings of $4,688,436. Holzhauer, who became a household name earlier this year with an impressive winning streak, also just won the 2019 Tournament of Champions. His winnings total $2,712,216.
Are you looking forward to this Jeopardy face-off? And who are you rooting for? Hit the comments!