Jeopardy! superstar James Holzhauer, just coming off of his win of the Tournament of Champions, returns to TV on Jan. 7, but this time, he’ll be in primetime.
ABC is hosting a multi-night event in which Holzhauer will face off against the two leading money winners in Jeopardy!’s history: Ken Jennings and Brad Rutter. The first night of Jeopardy! The Greatest of All Time will premiere Tuesday, Jan. 7, at 8 p.m. and will be hosted by Alex Trebek. Harry Friedman will executive produce.
“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.
Whichever of the three contestants manages to win three games will take home $1 million and henceforth be known by the title of “Jeopardy! The Greatest of All Time.” The tournament is currently scheduled to air Tuesday, Jan. 7; Wednesday, Jan. 8; and Thursday, Jan. 9; with Friday, Jan. 10; Tuesday, Jan. 14; Wednesday, Jan. 15; and Thursday, Jan. 16, slated for games “if necessary,” according to ABC.
“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 Karey Burke, president, ABC Entertainment, in a statement.
“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.”
Each player brings his own brand of credentials to the tourney. Holzhauer is the new kid on the scene, having won 32 games in a row last spring before he was eventually defeated by Chicago librarian Emma Boettcher. Holzhauer faced Boettcher again at the Tournament of Champions and this time came out on top, increasing his total winnings to $2,712,216.
Jennings has won the most consecutive games of any player, having won 74 in a row in 2004. His total winnings amount to $3,370,700.
Rutter is the game’s highest money winner, with a total of $4,688,436. Rutter first appeared on Jeopardy! in October 2000 when the rules only allowed players to win as many as five games in a row. The rules were changed in 2003, which allowed Jennings to rack up 74 straight wins. Most of Rutter’s winnings have been earned in tournament situations.