Jeopardy is staging a tournament of champions.
ABC confirmed Monday that the series' three highest earners of all time -- Ken Jennings, Brad Rutter, and James Holzhauer -- will compete on Jeopardy! The Greatest of All Time.
The special series will air in primetime beginning Tuesday, January 7 at 8/7c, and is slated to air for three consecutive weeknights through Thursday, January 9.
However, that could change when the series actually gets underway, depending on what plays out.
The player with the most combined winnings from the two games wins the match, and the game continues until one of them has won three matches.
Once that happens, that particular player will take home $1 million, while the runners up will get $250,000 each, so there's a lot of money at stake here.
“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.
Jennings amassed $3,370,700 during his turn on the show in 2004 with a 74-game winning streak. Rutter managed $4,688,436, becoming the highest winner across any TV game show.
Holzhauer is one of the more recent entries to the Jeopardy! hall of fame, scoring $2,712,216.
What are your thoughts on this special set of episodes coming to primetime?
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Paul Dailly is the Associate Editor for TV Fanatic. Follow him on Twitter.